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6.003: Signals and Systems Feedback, Poles, and Fundamental Modes September 15, 2011 1

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Page 1: Feedback, Poles, and Fundamental Modes (PDF)

6003 Signals and Systems

Feedback Poles and Fundamental Modes

September 15 2011 1

Homework

Doing the homework is essential to understanding the content

Weekly Homework Assigments

bull tutor (exam-type) problems

answers are automatically checked to provide quick feedback

bull engineering design (real-world) problems

graded by a human

Learning doesnrsquot end when you have submitted your work

bull solutions will be posted on Wednesdays at 5pm

bull read solutions to find errors and to see alternative approaches

bull mark the errors in your previously submitted work

bull submit the markup by Friday at 5pm

bull identify ALL errors and get back half of the points you lost

2

Last Time Multiple Representations of DT Systems

Verbal descriptions preserve the rationale

ldquoTo reduce the number of bits needed to store a sequence of

large numbers that are nearly equal record the first number

and then record successive differencesrdquo

Difference equations mathematically compact

y[n] = x[n] minus x[n minus 1]

Block diagrams illustrate signal flow paths

minus1 Delay

+x[n] y[n]

Operator representations analyze systems as polynomials

Y = (1 minus R) X

3

Last Time Feedback Cyclic Signal Paths and Modes

Systems with signals that depend on previous values of the same

signal are said to have feedback

Example The accumulator system has feedback

Delay

+X Y

minus1 Delay

+X Y

By contrast the difference machine does not have feedback

4

Last Time Feedback Cyclic Signal Paths and Modes

The effect of feedback can be visualized by tracing each cycle

through the cyclic signal paths

Delay

+

p0

X Y

minus1 0 1 2 3 4n

x[n] = δ[n]

minus1 0 1 2 3 4n

y[n]

Each cycle creates another sample in the output

5

Last Time Feedback Cyclic Signal Paths and Modes

The effect of feedback can be visualized by tracing each cycle

through the cyclic signal paths

Delay

+

p0

X Y

minus1 0 1 2 3 4n

x[n] = δ[n]

minus1 0 1 2 3 4n

y[n]

Each cycle creates another sample in the output

6

Last Time Feedback Cyclic Signal Paths and Modes

The effect of feedback can be visualized by tracing each cycle

through the cyclic signal paths

Delay

+

p0

X Y

minus1 0 1 2 3 4n

x[n] = δ[n]

minus1 0 1 2 3 4n

y[n]

Each cycle creates another sample in the output

7

Last Time Feedback Cyclic Signal Paths and Modes

The effect of feedback can be visualized by tracing each cycle

through the cyclic signal paths

Delay

+

p0

X Y

minus1 0 1 2 3 4n

x[n] = δ[n]

minus1 0 1 2 3 4n

y[n]

Each cycle creates another sample in the output

The response will persist even though the input is transient

8

minus1 0 1 2 3 4n

y[n]

Geometric Growth Poles

These unit-sample responses can be characterized by a single number

mdash the pole mdash which is the base of the geometric sequence

Delay

+

p0

X Y

y[n] =

pn0 if n gt= 0

0 otherwise

minus1 0 1 2 3 4n

y[n]

minus1 0 1 2 3 4n

y[n]

p0 = 05 p0 = 1 p0 = 12 9

Check Yourself

How many of the following unit-sample responses can be

represented by a single pole

n n

n n

n

10

Check Yourself

How many of the following unit-sample responses can be

represented by a single pole 3

n n

n n

n

11

Geometric Growth

The value of p0 determines the rate of growth

y[n] y[n] y[n] y[n]

minus1 0 1z

p0 lt minus1 magnitude diverges alternating sign

minus1 lt p0 lt 0 magnitude converges alternating sign

0 lt p0 lt 1 magnitude converges monotonically

p0 gt 1 magnitude diverges monotonically 12

Second-Order Systems

The unit-sample responses of more complicated cyclic systems are

more complicated

R

R

16

minus063

+X Y

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Not geometric This response grows then decays 13

Factoring Second-Order Systems

Factor the operator expression to break the system into two simpler

systems (divide and conquer)

R

R

16

minus063

+X Y

Y = X + 16RY minus 063R2Y

(1 minus 16R + 063R2) Y = X

(1 minus 07R)(1 minus 09R) Y = X

14

Factoring Second-Order Systems

The factored form corresponds to a cascade of simpler systems

(1 minus 07R)(1 minus 09R) Y = X

+

07 R

+

09 R

X YY2

(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2

+

09 R

+

07 R

X YY1

(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1

The order doesnrsquot matter (if systems are initially at rest) 15

Factoring Second-Order Systems

The unit-sample response of the cascaded system can be found by

multiplying the polynomial representations of the subsystems

Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _

_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)

Multiply then collect terms of equal order

Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X

+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot

16

Multiplying Polynomial

Graphical representation of polynomial multiplication

Y

Y X

= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1

a R

a2 R2

a3 R3

+

1

b R

b2 R2

b3 R3

+

X Y

Collect terms of equal order

= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

17

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Multiplying Polynomials

Tabular representation of polynomial multiplication

(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1 bR b2R2 b3R3 middot middot middot

1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot

a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot

middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot

Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

18

Partial Fractions

Use partial fractions to rewrite as a sum of simpler parts

R

R

16

minus063

+X Y

Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R

19

Second-Order Systems Equivalent Forms

The sum of simpler parts suggests a parallel implementation

Y X

= 45

1 minus 09R minus

35 1 minus 07R

+

09 R

45 +

minus35

R07

+

X YY1

Y2

If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0

Thus y[n] = 45(09)n minus 35(07)n for n ge 0

20

Partial Fractions

Graphical representation of the sum of geometric sequences

minus1 0 1 2 3 4 5 6 7 8n

y1[n] = 09n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y2[n] = 07n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y[n] = 45(09)n minus 35(07)nfor n ge 0

21

Partial Fractions

Partial fractions provides a remarkable equivalence

R

R

16

minus063

+X Y

+

09 R

45 +

minus35

R07

+

X YY1

Y2

rarr follows from thinking about system as polynomial (factoring) 22

Poles

The key to simplifying a higher-order system is identifying its poles

Poles are the roots of the denominator of the system functional

when R rarr 1 z

Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)

p0=07 p1=09

1 Substitute R rarr and find roots of denominator

z 2 2Y 1 z z= = =

X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2

z0=07 z1=09

The poles are at 07 and 09

23

︸ ︷︷ ︸ ︸ ︷︷ ︸

︸ ︷︷ ︸ ︸ ︷︷ ︸

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

24

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z == = 14

1 18

1 2 + 14

18

121 + z +minus z minus z minus2 zz z

1 The unit sample response converges to zero 12 and z = 1

42 There are poles at z =123 There is a pole at z =

4 There are two poles

5 None of the above

25

14

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z= = = 14

1 18

1 2 + 14

18

12

141 + z +minus z minus z minus2 zz z

radic 1 The unit sample response converges to zero

12 and z = 1

4 X2 There are poles at z =12 Xradic3 There is a pole at z =

4 There are two poles

5 None of the above X

26

( ) ( )

( )( )

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true 2

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

27

Population Growth

28

Population Growth

29

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 2: Feedback, Poles, and Fundamental Modes (PDF)

Homework

Doing the homework is essential to understanding the content

Weekly Homework Assigments

bull tutor (exam-type) problems

answers are automatically checked to provide quick feedback

bull engineering design (real-world) problems

graded by a human

Learning doesnrsquot end when you have submitted your work

bull solutions will be posted on Wednesdays at 5pm

bull read solutions to find errors and to see alternative approaches

bull mark the errors in your previously submitted work

bull submit the markup by Friday at 5pm

bull identify ALL errors and get back half of the points you lost

2

Last Time Multiple Representations of DT Systems

Verbal descriptions preserve the rationale

ldquoTo reduce the number of bits needed to store a sequence of

large numbers that are nearly equal record the first number

and then record successive differencesrdquo

Difference equations mathematically compact

y[n] = x[n] minus x[n minus 1]

Block diagrams illustrate signal flow paths

minus1 Delay

+x[n] y[n]

Operator representations analyze systems as polynomials

Y = (1 minus R) X

3

Last Time Feedback Cyclic Signal Paths and Modes

Systems with signals that depend on previous values of the same

signal are said to have feedback

Example The accumulator system has feedback

Delay

+X Y

minus1 Delay

+X Y

By contrast the difference machine does not have feedback

4

Last Time Feedback Cyclic Signal Paths and Modes

The effect of feedback can be visualized by tracing each cycle

through the cyclic signal paths

Delay

+

p0

X Y

minus1 0 1 2 3 4n

x[n] = δ[n]

minus1 0 1 2 3 4n

y[n]

Each cycle creates another sample in the output

5

Last Time Feedback Cyclic Signal Paths and Modes

The effect of feedback can be visualized by tracing each cycle

through the cyclic signal paths

Delay

+

p0

X Y

minus1 0 1 2 3 4n

x[n] = δ[n]

minus1 0 1 2 3 4n

y[n]

Each cycle creates another sample in the output

6

Last Time Feedback Cyclic Signal Paths and Modes

The effect of feedback can be visualized by tracing each cycle

through the cyclic signal paths

Delay

+

p0

X Y

minus1 0 1 2 3 4n

x[n] = δ[n]

minus1 0 1 2 3 4n

y[n]

Each cycle creates another sample in the output

7

Last Time Feedback Cyclic Signal Paths and Modes

The effect of feedback can be visualized by tracing each cycle

through the cyclic signal paths

Delay

+

p0

X Y

minus1 0 1 2 3 4n

x[n] = δ[n]

minus1 0 1 2 3 4n

y[n]

Each cycle creates another sample in the output

The response will persist even though the input is transient

8

minus1 0 1 2 3 4n

y[n]

Geometric Growth Poles

These unit-sample responses can be characterized by a single number

mdash the pole mdash which is the base of the geometric sequence

Delay

+

p0

X Y

y[n] =

pn0 if n gt= 0

0 otherwise

minus1 0 1 2 3 4n

y[n]

minus1 0 1 2 3 4n

y[n]

p0 = 05 p0 = 1 p0 = 12 9

Check Yourself

How many of the following unit-sample responses can be

represented by a single pole

n n

n n

n

10

Check Yourself

How many of the following unit-sample responses can be

represented by a single pole 3

n n

n n

n

11

Geometric Growth

The value of p0 determines the rate of growth

y[n] y[n] y[n] y[n]

minus1 0 1z

p0 lt minus1 magnitude diverges alternating sign

minus1 lt p0 lt 0 magnitude converges alternating sign

0 lt p0 lt 1 magnitude converges monotonically

p0 gt 1 magnitude diverges monotonically 12

Second-Order Systems

The unit-sample responses of more complicated cyclic systems are

more complicated

R

R

16

minus063

+X Y

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Not geometric This response grows then decays 13

Factoring Second-Order Systems

Factor the operator expression to break the system into two simpler

systems (divide and conquer)

R

R

16

minus063

+X Y

Y = X + 16RY minus 063R2Y

(1 minus 16R + 063R2) Y = X

(1 minus 07R)(1 minus 09R) Y = X

14

Factoring Second-Order Systems

The factored form corresponds to a cascade of simpler systems

(1 minus 07R)(1 minus 09R) Y = X

+

07 R

+

09 R

X YY2

(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2

+

09 R

+

07 R

X YY1

(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1

The order doesnrsquot matter (if systems are initially at rest) 15

Factoring Second-Order Systems

The unit-sample response of the cascaded system can be found by

multiplying the polynomial representations of the subsystems

Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _

_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)

Multiply then collect terms of equal order

Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X

+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot

16

Multiplying Polynomial

Graphical representation of polynomial multiplication

Y

Y X

= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1

a R

a2 R2

a3 R3

+

1

b R

b2 R2

b3 R3

+

X Y

Collect terms of equal order

= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

17

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Multiplying Polynomials

Tabular representation of polynomial multiplication

(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1 bR b2R2 b3R3 middot middot middot

1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot

a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot

middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot

Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

18

Partial Fractions

Use partial fractions to rewrite as a sum of simpler parts

R

R

16

minus063

+X Y

Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R

19

Second-Order Systems Equivalent Forms

The sum of simpler parts suggests a parallel implementation

Y X

= 45

1 minus 09R minus

35 1 minus 07R

+

09 R

45 +

minus35

R07

+

X YY1

Y2

If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0

Thus y[n] = 45(09)n minus 35(07)n for n ge 0

20

Partial Fractions

Graphical representation of the sum of geometric sequences

minus1 0 1 2 3 4 5 6 7 8n

y1[n] = 09n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y2[n] = 07n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y[n] = 45(09)n minus 35(07)nfor n ge 0

21

Partial Fractions

Partial fractions provides a remarkable equivalence

R

R

16

minus063

+X Y

+

09 R

45 +

minus35

R07

+

X YY1

Y2

rarr follows from thinking about system as polynomial (factoring) 22

Poles

The key to simplifying a higher-order system is identifying its poles

Poles are the roots of the denominator of the system functional

when R rarr 1 z

Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)

p0=07 p1=09

1 Substitute R rarr and find roots of denominator

z 2 2Y 1 z z= = =

X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2

z0=07 z1=09

The poles are at 07 and 09

23

︸ ︷︷ ︸ ︸ ︷︷ ︸

︸ ︷︷ ︸ ︸ ︷︷ ︸

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

24

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z == = 14

1 18

1 2 + 14

18

121 + z +minus z minus z minus2 zz z

1 The unit sample response converges to zero 12 and z = 1

42 There are poles at z =123 There is a pole at z =

4 There are two poles

5 None of the above

25

14

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z= = = 14

1 18

1 2 + 14

18

12

141 + z +minus z minus z minus2 zz z

radic 1 The unit sample response converges to zero

12 and z = 1

4 X2 There are poles at z =12 Xradic3 There is a pole at z =

4 There are two poles

5 None of the above X

26

( ) ( )

( )( )

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true 2

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

27

Population Growth

28

Population Growth

29

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 3: Feedback, Poles, and Fundamental Modes (PDF)

Last Time Multiple Representations of DT Systems

Verbal descriptions preserve the rationale

ldquoTo reduce the number of bits needed to store a sequence of

large numbers that are nearly equal record the first number

and then record successive differencesrdquo

Difference equations mathematically compact

y[n] = x[n] minus x[n minus 1]

Block diagrams illustrate signal flow paths

minus1 Delay

+x[n] y[n]

Operator representations analyze systems as polynomials

Y = (1 minus R) X

3

Last Time Feedback Cyclic Signal Paths and Modes

Systems with signals that depend on previous values of the same

signal are said to have feedback

Example The accumulator system has feedback

Delay

+X Y

minus1 Delay

+X Y

By contrast the difference machine does not have feedback

4

Last Time Feedback Cyclic Signal Paths and Modes

The effect of feedback can be visualized by tracing each cycle

through the cyclic signal paths

Delay

+

p0

X Y

minus1 0 1 2 3 4n

x[n] = δ[n]

minus1 0 1 2 3 4n

y[n]

Each cycle creates another sample in the output

5

Last Time Feedback Cyclic Signal Paths and Modes

The effect of feedback can be visualized by tracing each cycle

through the cyclic signal paths

Delay

+

p0

X Y

minus1 0 1 2 3 4n

x[n] = δ[n]

minus1 0 1 2 3 4n

y[n]

Each cycle creates another sample in the output

6

Last Time Feedback Cyclic Signal Paths and Modes

The effect of feedback can be visualized by tracing each cycle

through the cyclic signal paths

Delay

+

p0

X Y

minus1 0 1 2 3 4n

x[n] = δ[n]

minus1 0 1 2 3 4n

y[n]

Each cycle creates another sample in the output

7

Last Time Feedback Cyclic Signal Paths and Modes

The effect of feedback can be visualized by tracing each cycle

through the cyclic signal paths

Delay

+

p0

X Y

minus1 0 1 2 3 4n

x[n] = δ[n]

minus1 0 1 2 3 4n

y[n]

Each cycle creates another sample in the output

The response will persist even though the input is transient

8

minus1 0 1 2 3 4n

y[n]

Geometric Growth Poles

These unit-sample responses can be characterized by a single number

mdash the pole mdash which is the base of the geometric sequence

Delay

+

p0

X Y

y[n] =

pn0 if n gt= 0

0 otherwise

minus1 0 1 2 3 4n

y[n]

minus1 0 1 2 3 4n

y[n]

p0 = 05 p0 = 1 p0 = 12 9

Check Yourself

How many of the following unit-sample responses can be

represented by a single pole

n n

n n

n

10

Check Yourself

How many of the following unit-sample responses can be

represented by a single pole 3

n n

n n

n

11

Geometric Growth

The value of p0 determines the rate of growth

y[n] y[n] y[n] y[n]

minus1 0 1z

p0 lt minus1 magnitude diverges alternating sign

minus1 lt p0 lt 0 magnitude converges alternating sign

0 lt p0 lt 1 magnitude converges monotonically

p0 gt 1 magnitude diverges monotonically 12

Second-Order Systems

The unit-sample responses of more complicated cyclic systems are

more complicated

R

R

16

minus063

+X Y

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Not geometric This response grows then decays 13

Factoring Second-Order Systems

Factor the operator expression to break the system into two simpler

systems (divide and conquer)

R

R

16

minus063

+X Y

Y = X + 16RY minus 063R2Y

(1 minus 16R + 063R2) Y = X

(1 minus 07R)(1 minus 09R) Y = X

14

Factoring Second-Order Systems

The factored form corresponds to a cascade of simpler systems

(1 minus 07R)(1 minus 09R) Y = X

+

07 R

+

09 R

X YY2

(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2

+

09 R

+

07 R

X YY1

(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1

The order doesnrsquot matter (if systems are initially at rest) 15

Factoring Second-Order Systems

The unit-sample response of the cascaded system can be found by

multiplying the polynomial representations of the subsystems

Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _

_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)

Multiply then collect terms of equal order

Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X

+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot

16

Multiplying Polynomial

Graphical representation of polynomial multiplication

Y

Y X

= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1

a R

a2 R2

a3 R3

+

1

b R

b2 R2

b3 R3

+

X Y

Collect terms of equal order

= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

17

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Multiplying Polynomials

Tabular representation of polynomial multiplication

(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1 bR b2R2 b3R3 middot middot middot

1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot

a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot

middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot

Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

18

Partial Fractions

Use partial fractions to rewrite as a sum of simpler parts

R

R

16

minus063

+X Y

Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R

19

Second-Order Systems Equivalent Forms

The sum of simpler parts suggests a parallel implementation

Y X

= 45

1 minus 09R minus

35 1 minus 07R

+

09 R

45 +

minus35

R07

+

X YY1

Y2

If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0

Thus y[n] = 45(09)n minus 35(07)n for n ge 0

20

Partial Fractions

Graphical representation of the sum of geometric sequences

minus1 0 1 2 3 4 5 6 7 8n

y1[n] = 09n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y2[n] = 07n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y[n] = 45(09)n minus 35(07)nfor n ge 0

21

Partial Fractions

Partial fractions provides a remarkable equivalence

R

R

16

minus063

+X Y

+

09 R

45 +

minus35

R07

+

X YY1

Y2

rarr follows from thinking about system as polynomial (factoring) 22

Poles

The key to simplifying a higher-order system is identifying its poles

Poles are the roots of the denominator of the system functional

when R rarr 1 z

Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)

p0=07 p1=09

1 Substitute R rarr and find roots of denominator

z 2 2Y 1 z z= = =

X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2

z0=07 z1=09

The poles are at 07 and 09

23

︸ ︷︷ ︸ ︸ ︷︷ ︸

︸ ︷︷ ︸ ︸ ︷︷ ︸

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

24

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z == = 14

1 18

1 2 + 14

18

121 + z +minus z minus z minus2 zz z

1 The unit sample response converges to zero 12 and z = 1

42 There are poles at z =123 There is a pole at z =

4 There are two poles

5 None of the above

25

14

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z= = = 14

1 18

1 2 + 14

18

12

141 + z +minus z minus z minus2 zz z

radic 1 The unit sample response converges to zero

12 and z = 1

4 X2 There are poles at z =12 Xradic3 There is a pole at z =

4 There are two poles

5 None of the above X

26

( ) ( )

( )( )

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true 2

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

27

Population Growth

28

Population Growth

29

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 4: Feedback, Poles, and Fundamental Modes (PDF)

Last Time Feedback Cyclic Signal Paths and Modes

Systems with signals that depend on previous values of the same

signal are said to have feedback

Example The accumulator system has feedback

Delay

+X Y

minus1 Delay

+X Y

By contrast the difference machine does not have feedback

4

Last Time Feedback Cyclic Signal Paths and Modes

The effect of feedback can be visualized by tracing each cycle

through the cyclic signal paths

Delay

+

p0

X Y

minus1 0 1 2 3 4n

x[n] = δ[n]

minus1 0 1 2 3 4n

y[n]

Each cycle creates another sample in the output

5

Last Time Feedback Cyclic Signal Paths and Modes

The effect of feedback can be visualized by tracing each cycle

through the cyclic signal paths

Delay

+

p0

X Y

minus1 0 1 2 3 4n

x[n] = δ[n]

minus1 0 1 2 3 4n

y[n]

Each cycle creates another sample in the output

6

Last Time Feedback Cyclic Signal Paths and Modes

The effect of feedback can be visualized by tracing each cycle

through the cyclic signal paths

Delay

+

p0

X Y

minus1 0 1 2 3 4n

x[n] = δ[n]

minus1 0 1 2 3 4n

y[n]

Each cycle creates another sample in the output

7

Last Time Feedback Cyclic Signal Paths and Modes

The effect of feedback can be visualized by tracing each cycle

through the cyclic signal paths

Delay

+

p0

X Y

minus1 0 1 2 3 4n

x[n] = δ[n]

minus1 0 1 2 3 4n

y[n]

Each cycle creates another sample in the output

The response will persist even though the input is transient

8

minus1 0 1 2 3 4n

y[n]

Geometric Growth Poles

These unit-sample responses can be characterized by a single number

mdash the pole mdash which is the base of the geometric sequence

Delay

+

p0

X Y

y[n] =

pn0 if n gt= 0

0 otherwise

minus1 0 1 2 3 4n

y[n]

minus1 0 1 2 3 4n

y[n]

p0 = 05 p0 = 1 p0 = 12 9

Check Yourself

How many of the following unit-sample responses can be

represented by a single pole

n n

n n

n

10

Check Yourself

How many of the following unit-sample responses can be

represented by a single pole 3

n n

n n

n

11

Geometric Growth

The value of p0 determines the rate of growth

y[n] y[n] y[n] y[n]

minus1 0 1z

p0 lt minus1 magnitude diverges alternating sign

minus1 lt p0 lt 0 magnitude converges alternating sign

0 lt p0 lt 1 magnitude converges monotonically

p0 gt 1 magnitude diverges monotonically 12

Second-Order Systems

The unit-sample responses of more complicated cyclic systems are

more complicated

R

R

16

minus063

+X Y

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Not geometric This response grows then decays 13

Factoring Second-Order Systems

Factor the operator expression to break the system into two simpler

systems (divide and conquer)

R

R

16

minus063

+X Y

Y = X + 16RY minus 063R2Y

(1 minus 16R + 063R2) Y = X

(1 minus 07R)(1 minus 09R) Y = X

14

Factoring Second-Order Systems

The factored form corresponds to a cascade of simpler systems

(1 minus 07R)(1 minus 09R) Y = X

+

07 R

+

09 R

X YY2

(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2

+

09 R

+

07 R

X YY1

(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1

The order doesnrsquot matter (if systems are initially at rest) 15

Factoring Second-Order Systems

The unit-sample response of the cascaded system can be found by

multiplying the polynomial representations of the subsystems

Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _

_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)

Multiply then collect terms of equal order

Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X

+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot

16

Multiplying Polynomial

Graphical representation of polynomial multiplication

Y

Y X

= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1

a R

a2 R2

a3 R3

+

1

b R

b2 R2

b3 R3

+

X Y

Collect terms of equal order

= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

17

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Multiplying Polynomials

Tabular representation of polynomial multiplication

(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1 bR b2R2 b3R3 middot middot middot

1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot

a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot

middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot

Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

18

Partial Fractions

Use partial fractions to rewrite as a sum of simpler parts

R

R

16

minus063

+X Y

Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R

19

Second-Order Systems Equivalent Forms

The sum of simpler parts suggests a parallel implementation

Y X

= 45

1 minus 09R minus

35 1 minus 07R

+

09 R

45 +

minus35

R07

+

X YY1

Y2

If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0

Thus y[n] = 45(09)n minus 35(07)n for n ge 0

20

Partial Fractions

Graphical representation of the sum of geometric sequences

minus1 0 1 2 3 4 5 6 7 8n

y1[n] = 09n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y2[n] = 07n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y[n] = 45(09)n minus 35(07)nfor n ge 0

21

Partial Fractions

Partial fractions provides a remarkable equivalence

R

R

16

minus063

+X Y

+

09 R

45 +

minus35

R07

+

X YY1

Y2

rarr follows from thinking about system as polynomial (factoring) 22

Poles

The key to simplifying a higher-order system is identifying its poles

Poles are the roots of the denominator of the system functional

when R rarr 1 z

Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)

p0=07 p1=09

1 Substitute R rarr and find roots of denominator

z 2 2Y 1 z z= = =

X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2

z0=07 z1=09

The poles are at 07 and 09

23

︸ ︷︷ ︸ ︸ ︷︷ ︸

︸ ︷︷ ︸ ︸ ︷︷ ︸

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

24

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z == = 14

1 18

1 2 + 14

18

121 + z +minus z minus z minus2 zz z

1 The unit sample response converges to zero 12 and z = 1

42 There are poles at z =123 There is a pole at z =

4 There are two poles

5 None of the above

25

14

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z= = = 14

1 18

1 2 + 14

18

12

141 + z +minus z minus z minus2 zz z

radic 1 The unit sample response converges to zero

12 and z = 1

4 X2 There are poles at z =12 Xradic3 There is a pole at z =

4 There are two poles

5 None of the above X

26

( ) ( )

( )( )

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true 2

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

27

Population Growth

28

Population Growth

29

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 5: Feedback, Poles, and Fundamental Modes (PDF)

Last Time Feedback Cyclic Signal Paths and Modes

The effect of feedback can be visualized by tracing each cycle

through the cyclic signal paths

Delay

+

p0

X Y

minus1 0 1 2 3 4n

x[n] = δ[n]

minus1 0 1 2 3 4n

y[n]

Each cycle creates another sample in the output

5

Last Time Feedback Cyclic Signal Paths and Modes

The effect of feedback can be visualized by tracing each cycle

through the cyclic signal paths

Delay

+

p0

X Y

minus1 0 1 2 3 4n

x[n] = δ[n]

minus1 0 1 2 3 4n

y[n]

Each cycle creates another sample in the output

6

Last Time Feedback Cyclic Signal Paths and Modes

The effect of feedback can be visualized by tracing each cycle

through the cyclic signal paths

Delay

+

p0

X Y

minus1 0 1 2 3 4n

x[n] = δ[n]

minus1 0 1 2 3 4n

y[n]

Each cycle creates another sample in the output

7

Last Time Feedback Cyclic Signal Paths and Modes

The effect of feedback can be visualized by tracing each cycle

through the cyclic signal paths

Delay

+

p0

X Y

minus1 0 1 2 3 4n

x[n] = δ[n]

minus1 0 1 2 3 4n

y[n]

Each cycle creates another sample in the output

The response will persist even though the input is transient

8

minus1 0 1 2 3 4n

y[n]

Geometric Growth Poles

These unit-sample responses can be characterized by a single number

mdash the pole mdash which is the base of the geometric sequence

Delay

+

p0

X Y

y[n] =

pn0 if n gt= 0

0 otherwise

minus1 0 1 2 3 4n

y[n]

minus1 0 1 2 3 4n

y[n]

p0 = 05 p0 = 1 p0 = 12 9

Check Yourself

How many of the following unit-sample responses can be

represented by a single pole

n n

n n

n

10

Check Yourself

How many of the following unit-sample responses can be

represented by a single pole 3

n n

n n

n

11

Geometric Growth

The value of p0 determines the rate of growth

y[n] y[n] y[n] y[n]

minus1 0 1z

p0 lt minus1 magnitude diverges alternating sign

minus1 lt p0 lt 0 magnitude converges alternating sign

0 lt p0 lt 1 magnitude converges monotonically

p0 gt 1 magnitude diverges monotonically 12

Second-Order Systems

The unit-sample responses of more complicated cyclic systems are

more complicated

R

R

16

minus063

+X Y

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Not geometric This response grows then decays 13

Factoring Second-Order Systems

Factor the operator expression to break the system into two simpler

systems (divide and conquer)

R

R

16

minus063

+X Y

Y = X + 16RY minus 063R2Y

(1 minus 16R + 063R2) Y = X

(1 minus 07R)(1 minus 09R) Y = X

14

Factoring Second-Order Systems

The factored form corresponds to a cascade of simpler systems

(1 minus 07R)(1 minus 09R) Y = X

+

07 R

+

09 R

X YY2

(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2

+

09 R

+

07 R

X YY1

(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1

The order doesnrsquot matter (if systems are initially at rest) 15

Factoring Second-Order Systems

The unit-sample response of the cascaded system can be found by

multiplying the polynomial representations of the subsystems

Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _

_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)

Multiply then collect terms of equal order

Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X

+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot

16

Multiplying Polynomial

Graphical representation of polynomial multiplication

Y

Y X

= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1

a R

a2 R2

a3 R3

+

1

b R

b2 R2

b3 R3

+

X Y

Collect terms of equal order

= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

17

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Multiplying Polynomials

Tabular representation of polynomial multiplication

(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1 bR b2R2 b3R3 middot middot middot

1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot

a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot

middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot

Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

18

Partial Fractions

Use partial fractions to rewrite as a sum of simpler parts

R

R

16

minus063

+X Y

Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R

19

Second-Order Systems Equivalent Forms

The sum of simpler parts suggests a parallel implementation

Y X

= 45

1 minus 09R minus

35 1 minus 07R

+

09 R

45 +

minus35

R07

+

X YY1

Y2

If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0

Thus y[n] = 45(09)n minus 35(07)n for n ge 0

20

Partial Fractions

Graphical representation of the sum of geometric sequences

minus1 0 1 2 3 4 5 6 7 8n

y1[n] = 09n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y2[n] = 07n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y[n] = 45(09)n minus 35(07)nfor n ge 0

21

Partial Fractions

Partial fractions provides a remarkable equivalence

R

R

16

minus063

+X Y

+

09 R

45 +

minus35

R07

+

X YY1

Y2

rarr follows from thinking about system as polynomial (factoring) 22

Poles

The key to simplifying a higher-order system is identifying its poles

Poles are the roots of the denominator of the system functional

when R rarr 1 z

Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)

p0=07 p1=09

1 Substitute R rarr and find roots of denominator

z 2 2Y 1 z z= = =

X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2

z0=07 z1=09

The poles are at 07 and 09

23

︸ ︷︷ ︸ ︸ ︷︷ ︸

︸ ︷︷ ︸ ︸ ︷︷ ︸

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

24

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z == = 14

1 18

1 2 + 14

18

121 + z +minus z minus z minus2 zz z

1 The unit sample response converges to zero 12 and z = 1

42 There are poles at z =123 There is a pole at z =

4 There are two poles

5 None of the above

25

14

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z= = = 14

1 18

1 2 + 14

18

12

141 + z +minus z minus z minus2 zz z

radic 1 The unit sample response converges to zero

12 and z = 1

4 X2 There are poles at z =12 Xradic3 There is a pole at z =

4 There are two poles

5 None of the above X

26

( ) ( )

( )( )

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true 2

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

27

Population Growth

28

Population Growth

29

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 6: Feedback, Poles, and Fundamental Modes (PDF)

Last Time Feedback Cyclic Signal Paths and Modes

The effect of feedback can be visualized by tracing each cycle

through the cyclic signal paths

Delay

+

p0

X Y

minus1 0 1 2 3 4n

x[n] = δ[n]

minus1 0 1 2 3 4n

y[n]

Each cycle creates another sample in the output

6

Last Time Feedback Cyclic Signal Paths and Modes

The effect of feedback can be visualized by tracing each cycle

through the cyclic signal paths

Delay

+

p0

X Y

minus1 0 1 2 3 4n

x[n] = δ[n]

minus1 0 1 2 3 4n

y[n]

Each cycle creates another sample in the output

7

Last Time Feedback Cyclic Signal Paths and Modes

The effect of feedback can be visualized by tracing each cycle

through the cyclic signal paths

Delay

+

p0

X Y

minus1 0 1 2 3 4n

x[n] = δ[n]

minus1 0 1 2 3 4n

y[n]

Each cycle creates another sample in the output

The response will persist even though the input is transient

8

minus1 0 1 2 3 4n

y[n]

Geometric Growth Poles

These unit-sample responses can be characterized by a single number

mdash the pole mdash which is the base of the geometric sequence

Delay

+

p0

X Y

y[n] =

pn0 if n gt= 0

0 otherwise

minus1 0 1 2 3 4n

y[n]

minus1 0 1 2 3 4n

y[n]

p0 = 05 p0 = 1 p0 = 12 9

Check Yourself

How many of the following unit-sample responses can be

represented by a single pole

n n

n n

n

10

Check Yourself

How many of the following unit-sample responses can be

represented by a single pole 3

n n

n n

n

11

Geometric Growth

The value of p0 determines the rate of growth

y[n] y[n] y[n] y[n]

minus1 0 1z

p0 lt minus1 magnitude diverges alternating sign

minus1 lt p0 lt 0 magnitude converges alternating sign

0 lt p0 lt 1 magnitude converges monotonically

p0 gt 1 magnitude diverges monotonically 12

Second-Order Systems

The unit-sample responses of more complicated cyclic systems are

more complicated

R

R

16

minus063

+X Y

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Not geometric This response grows then decays 13

Factoring Second-Order Systems

Factor the operator expression to break the system into two simpler

systems (divide and conquer)

R

R

16

minus063

+X Y

Y = X + 16RY minus 063R2Y

(1 minus 16R + 063R2) Y = X

(1 minus 07R)(1 minus 09R) Y = X

14

Factoring Second-Order Systems

The factored form corresponds to a cascade of simpler systems

(1 minus 07R)(1 minus 09R) Y = X

+

07 R

+

09 R

X YY2

(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2

+

09 R

+

07 R

X YY1

(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1

The order doesnrsquot matter (if systems are initially at rest) 15

Factoring Second-Order Systems

The unit-sample response of the cascaded system can be found by

multiplying the polynomial representations of the subsystems

Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _

_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)

Multiply then collect terms of equal order

Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X

+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot

16

Multiplying Polynomial

Graphical representation of polynomial multiplication

Y

Y X

= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1

a R

a2 R2

a3 R3

+

1

b R

b2 R2

b3 R3

+

X Y

Collect terms of equal order

= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

17

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Multiplying Polynomials

Tabular representation of polynomial multiplication

(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1 bR b2R2 b3R3 middot middot middot

1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot

a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot

middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot

Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

18

Partial Fractions

Use partial fractions to rewrite as a sum of simpler parts

R

R

16

minus063

+X Y

Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R

19

Second-Order Systems Equivalent Forms

The sum of simpler parts suggests a parallel implementation

Y X

= 45

1 minus 09R minus

35 1 minus 07R

+

09 R

45 +

minus35

R07

+

X YY1

Y2

If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0

Thus y[n] = 45(09)n minus 35(07)n for n ge 0

20

Partial Fractions

Graphical representation of the sum of geometric sequences

minus1 0 1 2 3 4 5 6 7 8n

y1[n] = 09n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y2[n] = 07n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y[n] = 45(09)n minus 35(07)nfor n ge 0

21

Partial Fractions

Partial fractions provides a remarkable equivalence

R

R

16

minus063

+X Y

+

09 R

45 +

minus35

R07

+

X YY1

Y2

rarr follows from thinking about system as polynomial (factoring) 22

Poles

The key to simplifying a higher-order system is identifying its poles

Poles are the roots of the denominator of the system functional

when R rarr 1 z

Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)

p0=07 p1=09

1 Substitute R rarr and find roots of denominator

z 2 2Y 1 z z= = =

X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2

z0=07 z1=09

The poles are at 07 and 09

23

︸ ︷︷ ︸ ︸ ︷︷ ︸

︸ ︷︷ ︸ ︸ ︷︷ ︸

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

24

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z == = 14

1 18

1 2 + 14

18

121 + z +minus z minus z minus2 zz z

1 The unit sample response converges to zero 12 and z = 1

42 There are poles at z =123 There is a pole at z =

4 There are two poles

5 None of the above

25

14

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z= = = 14

1 18

1 2 + 14

18

12

141 + z +minus z minus z minus2 zz z

radic 1 The unit sample response converges to zero

12 and z = 1

4 X2 There are poles at z =12 Xradic3 There is a pole at z =

4 There are two poles

5 None of the above X

26

( ) ( )

( )( )

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true 2

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

27

Population Growth

28

Population Growth

29

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 7: Feedback, Poles, and Fundamental Modes (PDF)

Last Time Feedback Cyclic Signal Paths and Modes

The effect of feedback can be visualized by tracing each cycle

through the cyclic signal paths

Delay

+

p0

X Y

minus1 0 1 2 3 4n

x[n] = δ[n]

minus1 0 1 2 3 4n

y[n]

Each cycle creates another sample in the output

7

Last Time Feedback Cyclic Signal Paths and Modes

The effect of feedback can be visualized by tracing each cycle

through the cyclic signal paths

Delay

+

p0

X Y

minus1 0 1 2 3 4n

x[n] = δ[n]

minus1 0 1 2 3 4n

y[n]

Each cycle creates another sample in the output

The response will persist even though the input is transient

8

minus1 0 1 2 3 4n

y[n]

Geometric Growth Poles

These unit-sample responses can be characterized by a single number

mdash the pole mdash which is the base of the geometric sequence

Delay

+

p0

X Y

y[n] =

pn0 if n gt= 0

0 otherwise

minus1 0 1 2 3 4n

y[n]

minus1 0 1 2 3 4n

y[n]

p0 = 05 p0 = 1 p0 = 12 9

Check Yourself

How many of the following unit-sample responses can be

represented by a single pole

n n

n n

n

10

Check Yourself

How many of the following unit-sample responses can be

represented by a single pole 3

n n

n n

n

11

Geometric Growth

The value of p0 determines the rate of growth

y[n] y[n] y[n] y[n]

minus1 0 1z

p0 lt minus1 magnitude diverges alternating sign

minus1 lt p0 lt 0 magnitude converges alternating sign

0 lt p0 lt 1 magnitude converges monotonically

p0 gt 1 magnitude diverges monotonically 12

Second-Order Systems

The unit-sample responses of more complicated cyclic systems are

more complicated

R

R

16

minus063

+X Y

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Not geometric This response grows then decays 13

Factoring Second-Order Systems

Factor the operator expression to break the system into two simpler

systems (divide and conquer)

R

R

16

minus063

+X Y

Y = X + 16RY minus 063R2Y

(1 minus 16R + 063R2) Y = X

(1 minus 07R)(1 minus 09R) Y = X

14

Factoring Second-Order Systems

The factored form corresponds to a cascade of simpler systems

(1 minus 07R)(1 minus 09R) Y = X

+

07 R

+

09 R

X YY2

(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2

+

09 R

+

07 R

X YY1

(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1

The order doesnrsquot matter (if systems are initially at rest) 15

Factoring Second-Order Systems

The unit-sample response of the cascaded system can be found by

multiplying the polynomial representations of the subsystems

Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _

_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)

Multiply then collect terms of equal order

Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X

+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot

16

Multiplying Polynomial

Graphical representation of polynomial multiplication

Y

Y X

= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1

a R

a2 R2

a3 R3

+

1

b R

b2 R2

b3 R3

+

X Y

Collect terms of equal order

= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

17

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Multiplying Polynomials

Tabular representation of polynomial multiplication

(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1 bR b2R2 b3R3 middot middot middot

1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot

a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot

middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot

Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

18

Partial Fractions

Use partial fractions to rewrite as a sum of simpler parts

R

R

16

minus063

+X Y

Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R

19

Second-Order Systems Equivalent Forms

The sum of simpler parts suggests a parallel implementation

Y X

= 45

1 minus 09R minus

35 1 minus 07R

+

09 R

45 +

minus35

R07

+

X YY1

Y2

If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0

Thus y[n] = 45(09)n minus 35(07)n for n ge 0

20

Partial Fractions

Graphical representation of the sum of geometric sequences

minus1 0 1 2 3 4 5 6 7 8n

y1[n] = 09n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y2[n] = 07n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y[n] = 45(09)n minus 35(07)nfor n ge 0

21

Partial Fractions

Partial fractions provides a remarkable equivalence

R

R

16

minus063

+X Y

+

09 R

45 +

minus35

R07

+

X YY1

Y2

rarr follows from thinking about system as polynomial (factoring) 22

Poles

The key to simplifying a higher-order system is identifying its poles

Poles are the roots of the denominator of the system functional

when R rarr 1 z

Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)

p0=07 p1=09

1 Substitute R rarr and find roots of denominator

z 2 2Y 1 z z= = =

X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2

z0=07 z1=09

The poles are at 07 and 09

23

︸ ︷︷ ︸ ︸ ︷︷ ︸

︸ ︷︷ ︸ ︸ ︷︷ ︸

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

24

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z == = 14

1 18

1 2 + 14

18

121 + z +minus z minus z minus2 zz z

1 The unit sample response converges to zero 12 and z = 1

42 There are poles at z =123 There is a pole at z =

4 There are two poles

5 None of the above

25

14

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z= = = 14

1 18

1 2 + 14

18

12

141 + z +minus z minus z minus2 zz z

radic 1 The unit sample response converges to zero

12 and z = 1

4 X2 There are poles at z =12 Xradic3 There is a pole at z =

4 There are two poles

5 None of the above X

26

( ) ( )

( )( )

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true 2

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

27

Population Growth

28

Population Growth

29

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 8: Feedback, Poles, and Fundamental Modes (PDF)

Last Time Feedback Cyclic Signal Paths and Modes

The effect of feedback can be visualized by tracing each cycle

through the cyclic signal paths

Delay

+

p0

X Y

minus1 0 1 2 3 4n

x[n] = δ[n]

minus1 0 1 2 3 4n

y[n]

Each cycle creates another sample in the output

The response will persist even though the input is transient

8

minus1 0 1 2 3 4n

y[n]

Geometric Growth Poles

These unit-sample responses can be characterized by a single number

mdash the pole mdash which is the base of the geometric sequence

Delay

+

p0

X Y

y[n] =

pn0 if n gt= 0

0 otherwise

minus1 0 1 2 3 4n

y[n]

minus1 0 1 2 3 4n

y[n]

p0 = 05 p0 = 1 p0 = 12 9

Check Yourself

How many of the following unit-sample responses can be

represented by a single pole

n n

n n

n

10

Check Yourself

How many of the following unit-sample responses can be

represented by a single pole 3

n n

n n

n

11

Geometric Growth

The value of p0 determines the rate of growth

y[n] y[n] y[n] y[n]

minus1 0 1z

p0 lt minus1 magnitude diverges alternating sign

minus1 lt p0 lt 0 magnitude converges alternating sign

0 lt p0 lt 1 magnitude converges monotonically

p0 gt 1 magnitude diverges monotonically 12

Second-Order Systems

The unit-sample responses of more complicated cyclic systems are

more complicated

R

R

16

minus063

+X Y

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Not geometric This response grows then decays 13

Factoring Second-Order Systems

Factor the operator expression to break the system into two simpler

systems (divide and conquer)

R

R

16

minus063

+X Y

Y = X + 16RY minus 063R2Y

(1 minus 16R + 063R2) Y = X

(1 minus 07R)(1 minus 09R) Y = X

14

Factoring Second-Order Systems

The factored form corresponds to a cascade of simpler systems

(1 minus 07R)(1 minus 09R) Y = X

+

07 R

+

09 R

X YY2

(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2

+

09 R

+

07 R

X YY1

(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1

The order doesnrsquot matter (if systems are initially at rest) 15

Factoring Second-Order Systems

The unit-sample response of the cascaded system can be found by

multiplying the polynomial representations of the subsystems

Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _

_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)

Multiply then collect terms of equal order

Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X

+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot

16

Multiplying Polynomial

Graphical representation of polynomial multiplication

Y

Y X

= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1

a R

a2 R2

a3 R3

+

1

b R

b2 R2

b3 R3

+

X Y

Collect terms of equal order

= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

17

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Multiplying Polynomials

Tabular representation of polynomial multiplication

(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1 bR b2R2 b3R3 middot middot middot

1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot

a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot

middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot

Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

18

Partial Fractions

Use partial fractions to rewrite as a sum of simpler parts

R

R

16

minus063

+X Y

Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R

19

Second-Order Systems Equivalent Forms

The sum of simpler parts suggests a parallel implementation

Y X

= 45

1 minus 09R minus

35 1 minus 07R

+

09 R

45 +

minus35

R07

+

X YY1

Y2

If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0

Thus y[n] = 45(09)n minus 35(07)n for n ge 0

20

Partial Fractions

Graphical representation of the sum of geometric sequences

minus1 0 1 2 3 4 5 6 7 8n

y1[n] = 09n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y2[n] = 07n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y[n] = 45(09)n minus 35(07)nfor n ge 0

21

Partial Fractions

Partial fractions provides a remarkable equivalence

R

R

16

minus063

+X Y

+

09 R

45 +

minus35

R07

+

X YY1

Y2

rarr follows from thinking about system as polynomial (factoring) 22

Poles

The key to simplifying a higher-order system is identifying its poles

Poles are the roots of the denominator of the system functional

when R rarr 1 z

Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)

p0=07 p1=09

1 Substitute R rarr and find roots of denominator

z 2 2Y 1 z z= = =

X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2

z0=07 z1=09

The poles are at 07 and 09

23

︸ ︷︷ ︸ ︸ ︷︷ ︸

︸ ︷︷ ︸ ︸ ︷︷ ︸

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

24

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z == = 14

1 18

1 2 + 14

18

121 + z +minus z minus z minus2 zz z

1 The unit sample response converges to zero 12 and z = 1

42 There are poles at z =123 There is a pole at z =

4 There are two poles

5 None of the above

25

14

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z= = = 14

1 18

1 2 + 14

18

12

141 + z +minus z minus z minus2 zz z

radic 1 The unit sample response converges to zero

12 and z = 1

4 X2 There are poles at z =12 Xradic3 There is a pole at z =

4 There are two poles

5 None of the above X

26

( ) ( )

( )( )

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true 2

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

27

Population Growth

28

Population Growth

29

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 9: Feedback, Poles, and Fundamental Modes (PDF)

minus1 0 1 2 3 4n

y[n]

Geometric Growth Poles

These unit-sample responses can be characterized by a single number

mdash the pole mdash which is the base of the geometric sequence

Delay

+

p0

X Y

y[n] =

pn0 if n gt= 0

0 otherwise

minus1 0 1 2 3 4n

y[n]

minus1 0 1 2 3 4n

y[n]

p0 = 05 p0 = 1 p0 = 12 9

Check Yourself

How many of the following unit-sample responses can be

represented by a single pole

n n

n n

n

10

Check Yourself

How many of the following unit-sample responses can be

represented by a single pole 3

n n

n n

n

11

Geometric Growth

The value of p0 determines the rate of growth

y[n] y[n] y[n] y[n]

minus1 0 1z

p0 lt minus1 magnitude diverges alternating sign

minus1 lt p0 lt 0 magnitude converges alternating sign

0 lt p0 lt 1 magnitude converges monotonically

p0 gt 1 magnitude diverges monotonically 12

Second-Order Systems

The unit-sample responses of more complicated cyclic systems are

more complicated

R

R

16

minus063

+X Y

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Not geometric This response grows then decays 13

Factoring Second-Order Systems

Factor the operator expression to break the system into two simpler

systems (divide and conquer)

R

R

16

minus063

+X Y

Y = X + 16RY minus 063R2Y

(1 minus 16R + 063R2) Y = X

(1 minus 07R)(1 minus 09R) Y = X

14

Factoring Second-Order Systems

The factored form corresponds to a cascade of simpler systems

(1 minus 07R)(1 minus 09R) Y = X

+

07 R

+

09 R

X YY2

(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2

+

09 R

+

07 R

X YY1

(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1

The order doesnrsquot matter (if systems are initially at rest) 15

Factoring Second-Order Systems

The unit-sample response of the cascaded system can be found by

multiplying the polynomial representations of the subsystems

Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _

_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)

Multiply then collect terms of equal order

Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X

+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot

16

Multiplying Polynomial

Graphical representation of polynomial multiplication

Y

Y X

= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1

a R

a2 R2

a3 R3

+

1

b R

b2 R2

b3 R3

+

X Y

Collect terms of equal order

= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

17

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Multiplying Polynomials

Tabular representation of polynomial multiplication

(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1 bR b2R2 b3R3 middot middot middot

1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot

a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot

middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot

Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

18

Partial Fractions

Use partial fractions to rewrite as a sum of simpler parts

R

R

16

minus063

+X Y

Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R

19

Second-Order Systems Equivalent Forms

The sum of simpler parts suggests a parallel implementation

Y X

= 45

1 minus 09R minus

35 1 minus 07R

+

09 R

45 +

minus35

R07

+

X YY1

Y2

If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0

Thus y[n] = 45(09)n minus 35(07)n for n ge 0

20

Partial Fractions

Graphical representation of the sum of geometric sequences

minus1 0 1 2 3 4 5 6 7 8n

y1[n] = 09n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y2[n] = 07n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y[n] = 45(09)n minus 35(07)nfor n ge 0

21

Partial Fractions

Partial fractions provides a remarkable equivalence

R

R

16

minus063

+X Y

+

09 R

45 +

minus35

R07

+

X YY1

Y2

rarr follows from thinking about system as polynomial (factoring) 22

Poles

The key to simplifying a higher-order system is identifying its poles

Poles are the roots of the denominator of the system functional

when R rarr 1 z

Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)

p0=07 p1=09

1 Substitute R rarr and find roots of denominator

z 2 2Y 1 z z= = =

X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2

z0=07 z1=09

The poles are at 07 and 09

23

︸ ︷︷ ︸ ︸ ︷︷ ︸

︸ ︷︷ ︸ ︸ ︷︷ ︸

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

24

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z == = 14

1 18

1 2 + 14

18

121 + z +minus z minus z minus2 zz z

1 The unit sample response converges to zero 12 and z = 1

42 There are poles at z =123 There is a pole at z =

4 There are two poles

5 None of the above

25

14

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z= = = 14

1 18

1 2 + 14

18

12

141 + z +minus z minus z minus2 zz z

radic 1 The unit sample response converges to zero

12 and z = 1

4 X2 There are poles at z =12 Xradic3 There is a pole at z =

4 There are two poles

5 None of the above X

26

( ) ( )

( )( )

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true 2

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

27

Population Growth

28

Population Growth

29

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 10: Feedback, Poles, and Fundamental Modes (PDF)

Check Yourself

How many of the following unit-sample responses can be

represented by a single pole

n n

n n

n

10

Check Yourself

How many of the following unit-sample responses can be

represented by a single pole 3

n n

n n

n

11

Geometric Growth

The value of p0 determines the rate of growth

y[n] y[n] y[n] y[n]

minus1 0 1z

p0 lt minus1 magnitude diverges alternating sign

minus1 lt p0 lt 0 magnitude converges alternating sign

0 lt p0 lt 1 magnitude converges monotonically

p0 gt 1 magnitude diverges monotonically 12

Second-Order Systems

The unit-sample responses of more complicated cyclic systems are

more complicated

R

R

16

minus063

+X Y

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Not geometric This response grows then decays 13

Factoring Second-Order Systems

Factor the operator expression to break the system into two simpler

systems (divide and conquer)

R

R

16

minus063

+X Y

Y = X + 16RY minus 063R2Y

(1 minus 16R + 063R2) Y = X

(1 minus 07R)(1 minus 09R) Y = X

14

Factoring Second-Order Systems

The factored form corresponds to a cascade of simpler systems

(1 minus 07R)(1 minus 09R) Y = X

+

07 R

+

09 R

X YY2

(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2

+

09 R

+

07 R

X YY1

(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1

The order doesnrsquot matter (if systems are initially at rest) 15

Factoring Second-Order Systems

The unit-sample response of the cascaded system can be found by

multiplying the polynomial representations of the subsystems

Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _

_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)

Multiply then collect terms of equal order

Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X

+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot

16

Multiplying Polynomial

Graphical representation of polynomial multiplication

Y

Y X

= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1

a R

a2 R2

a3 R3

+

1

b R

b2 R2

b3 R3

+

X Y

Collect terms of equal order

= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

17

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Multiplying Polynomials

Tabular representation of polynomial multiplication

(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1 bR b2R2 b3R3 middot middot middot

1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot

a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot

middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot

Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

18

Partial Fractions

Use partial fractions to rewrite as a sum of simpler parts

R

R

16

minus063

+X Y

Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R

19

Second-Order Systems Equivalent Forms

The sum of simpler parts suggests a parallel implementation

Y X

= 45

1 minus 09R minus

35 1 minus 07R

+

09 R

45 +

minus35

R07

+

X YY1

Y2

If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0

Thus y[n] = 45(09)n minus 35(07)n for n ge 0

20

Partial Fractions

Graphical representation of the sum of geometric sequences

minus1 0 1 2 3 4 5 6 7 8n

y1[n] = 09n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y2[n] = 07n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y[n] = 45(09)n minus 35(07)nfor n ge 0

21

Partial Fractions

Partial fractions provides a remarkable equivalence

R

R

16

minus063

+X Y

+

09 R

45 +

minus35

R07

+

X YY1

Y2

rarr follows from thinking about system as polynomial (factoring) 22

Poles

The key to simplifying a higher-order system is identifying its poles

Poles are the roots of the denominator of the system functional

when R rarr 1 z

Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)

p0=07 p1=09

1 Substitute R rarr and find roots of denominator

z 2 2Y 1 z z= = =

X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2

z0=07 z1=09

The poles are at 07 and 09

23

︸ ︷︷ ︸ ︸ ︷︷ ︸

︸ ︷︷ ︸ ︸ ︷︷ ︸

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

24

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z == = 14

1 18

1 2 + 14

18

121 + z +minus z minus z minus2 zz z

1 The unit sample response converges to zero 12 and z = 1

42 There are poles at z =123 There is a pole at z =

4 There are two poles

5 None of the above

25

14

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z= = = 14

1 18

1 2 + 14

18

12

141 + z +minus z minus z minus2 zz z

radic 1 The unit sample response converges to zero

12 and z = 1

4 X2 There are poles at z =12 Xradic3 There is a pole at z =

4 There are two poles

5 None of the above X

26

( ) ( )

( )( )

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true 2

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

27

Population Growth

28

Population Growth

29

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 11: Feedback, Poles, and Fundamental Modes (PDF)

Check Yourself

How many of the following unit-sample responses can be

represented by a single pole 3

n n

n n

n

11

Geometric Growth

The value of p0 determines the rate of growth

y[n] y[n] y[n] y[n]

minus1 0 1z

p0 lt minus1 magnitude diverges alternating sign

minus1 lt p0 lt 0 magnitude converges alternating sign

0 lt p0 lt 1 magnitude converges monotonically

p0 gt 1 magnitude diverges monotonically 12

Second-Order Systems

The unit-sample responses of more complicated cyclic systems are

more complicated

R

R

16

minus063

+X Y

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Not geometric This response grows then decays 13

Factoring Second-Order Systems

Factor the operator expression to break the system into two simpler

systems (divide and conquer)

R

R

16

minus063

+X Y

Y = X + 16RY minus 063R2Y

(1 minus 16R + 063R2) Y = X

(1 minus 07R)(1 minus 09R) Y = X

14

Factoring Second-Order Systems

The factored form corresponds to a cascade of simpler systems

(1 minus 07R)(1 minus 09R) Y = X

+

07 R

+

09 R

X YY2

(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2

+

09 R

+

07 R

X YY1

(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1

The order doesnrsquot matter (if systems are initially at rest) 15

Factoring Second-Order Systems

The unit-sample response of the cascaded system can be found by

multiplying the polynomial representations of the subsystems

Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _

_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)

Multiply then collect terms of equal order

Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X

+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot

16

Multiplying Polynomial

Graphical representation of polynomial multiplication

Y

Y X

= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1

a R

a2 R2

a3 R3

+

1

b R

b2 R2

b3 R3

+

X Y

Collect terms of equal order

= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

17

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Multiplying Polynomials

Tabular representation of polynomial multiplication

(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1 bR b2R2 b3R3 middot middot middot

1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot

a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot

middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot

Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

18

Partial Fractions

Use partial fractions to rewrite as a sum of simpler parts

R

R

16

minus063

+X Y

Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R

19

Second-Order Systems Equivalent Forms

The sum of simpler parts suggests a parallel implementation

Y X

= 45

1 minus 09R minus

35 1 minus 07R

+

09 R

45 +

minus35

R07

+

X YY1

Y2

If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0

Thus y[n] = 45(09)n minus 35(07)n for n ge 0

20

Partial Fractions

Graphical representation of the sum of geometric sequences

minus1 0 1 2 3 4 5 6 7 8n

y1[n] = 09n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y2[n] = 07n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y[n] = 45(09)n minus 35(07)nfor n ge 0

21

Partial Fractions

Partial fractions provides a remarkable equivalence

R

R

16

minus063

+X Y

+

09 R

45 +

minus35

R07

+

X YY1

Y2

rarr follows from thinking about system as polynomial (factoring) 22

Poles

The key to simplifying a higher-order system is identifying its poles

Poles are the roots of the denominator of the system functional

when R rarr 1 z

Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)

p0=07 p1=09

1 Substitute R rarr and find roots of denominator

z 2 2Y 1 z z= = =

X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2

z0=07 z1=09

The poles are at 07 and 09

23

︸ ︷︷ ︸ ︸ ︷︷ ︸

︸ ︷︷ ︸ ︸ ︷︷ ︸

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

24

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z == = 14

1 18

1 2 + 14

18

121 + z +minus z minus z minus2 zz z

1 The unit sample response converges to zero 12 and z = 1

42 There are poles at z =123 There is a pole at z =

4 There are two poles

5 None of the above

25

14

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z= = = 14

1 18

1 2 + 14

18

12

141 + z +minus z minus z minus2 zz z

radic 1 The unit sample response converges to zero

12 and z = 1

4 X2 There are poles at z =12 Xradic3 There is a pole at z =

4 There are two poles

5 None of the above X

26

( ) ( )

( )( )

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true 2

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

27

Population Growth

28

Population Growth

29

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 12: Feedback, Poles, and Fundamental Modes (PDF)

Geometric Growth

The value of p0 determines the rate of growth

y[n] y[n] y[n] y[n]

minus1 0 1z

p0 lt minus1 magnitude diverges alternating sign

minus1 lt p0 lt 0 magnitude converges alternating sign

0 lt p0 lt 1 magnitude converges monotonically

p0 gt 1 magnitude diverges monotonically 12

Second-Order Systems

The unit-sample responses of more complicated cyclic systems are

more complicated

R

R

16

minus063

+X Y

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Not geometric This response grows then decays 13

Factoring Second-Order Systems

Factor the operator expression to break the system into two simpler

systems (divide and conquer)

R

R

16

minus063

+X Y

Y = X + 16RY minus 063R2Y

(1 minus 16R + 063R2) Y = X

(1 minus 07R)(1 minus 09R) Y = X

14

Factoring Second-Order Systems

The factored form corresponds to a cascade of simpler systems

(1 minus 07R)(1 minus 09R) Y = X

+

07 R

+

09 R

X YY2

(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2

+

09 R

+

07 R

X YY1

(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1

The order doesnrsquot matter (if systems are initially at rest) 15

Factoring Second-Order Systems

The unit-sample response of the cascaded system can be found by

multiplying the polynomial representations of the subsystems

Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _

_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)

Multiply then collect terms of equal order

Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X

+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot

16

Multiplying Polynomial

Graphical representation of polynomial multiplication

Y

Y X

= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1

a R

a2 R2

a3 R3

+

1

b R

b2 R2

b3 R3

+

X Y

Collect terms of equal order

= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

17

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Multiplying Polynomials

Tabular representation of polynomial multiplication

(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1 bR b2R2 b3R3 middot middot middot

1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot

a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot

middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot

Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

18

Partial Fractions

Use partial fractions to rewrite as a sum of simpler parts

R

R

16

minus063

+X Y

Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R

19

Second-Order Systems Equivalent Forms

The sum of simpler parts suggests a parallel implementation

Y X

= 45

1 minus 09R minus

35 1 minus 07R

+

09 R

45 +

minus35

R07

+

X YY1

Y2

If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0

Thus y[n] = 45(09)n minus 35(07)n for n ge 0

20

Partial Fractions

Graphical representation of the sum of geometric sequences

minus1 0 1 2 3 4 5 6 7 8n

y1[n] = 09n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y2[n] = 07n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y[n] = 45(09)n minus 35(07)nfor n ge 0

21

Partial Fractions

Partial fractions provides a remarkable equivalence

R

R

16

minus063

+X Y

+

09 R

45 +

minus35

R07

+

X YY1

Y2

rarr follows from thinking about system as polynomial (factoring) 22

Poles

The key to simplifying a higher-order system is identifying its poles

Poles are the roots of the denominator of the system functional

when R rarr 1 z

Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)

p0=07 p1=09

1 Substitute R rarr and find roots of denominator

z 2 2Y 1 z z= = =

X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2

z0=07 z1=09

The poles are at 07 and 09

23

︸ ︷︷ ︸ ︸ ︷︷ ︸

︸ ︷︷ ︸ ︸ ︷︷ ︸

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

24

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z == = 14

1 18

1 2 + 14

18

121 + z +minus z minus z minus2 zz z

1 The unit sample response converges to zero 12 and z = 1

42 There are poles at z =123 There is a pole at z =

4 There are two poles

5 None of the above

25

14

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z= = = 14

1 18

1 2 + 14

18

12

141 + z +minus z minus z minus2 zz z

radic 1 The unit sample response converges to zero

12 and z = 1

4 X2 There are poles at z =12 Xradic3 There is a pole at z =

4 There are two poles

5 None of the above X

26

( ) ( )

( )( )

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true 2

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

27

Population Growth

28

Population Growth

29

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 13: Feedback, Poles, and Fundamental Modes (PDF)

Second-Order Systems

The unit-sample responses of more complicated cyclic systems are

more complicated

R

R

16

minus063

+X Y

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Not geometric This response grows then decays 13

Factoring Second-Order Systems

Factor the operator expression to break the system into two simpler

systems (divide and conquer)

R

R

16

minus063

+X Y

Y = X + 16RY minus 063R2Y

(1 minus 16R + 063R2) Y = X

(1 minus 07R)(1 minus 09R) Y = X

14

Factoring Second-Order Systems

The factored form corresponds to a cascade of simpler systems

(1 minus 07R)(1 minus 09R) Y = X

+

07 R

+

09 R

X YY2

(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2

+

09 R

+

07 R

X YY1

(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1

The order doesnrsquot matter (if systems are initially at rest) 15

Factoring Second-Order Systems

The unit-sample response of the cascaded system can be found by

multiplying the polynomial representations of the subsystems

Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _

_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)

Multiply then collect terms of equal order

Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X

+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot

16

Multiplying Polynomial

Graphical representation of polynomial multiplication

Y

Y X

= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1

a R

a2 R2

a3 R3

+

1

b R

b2 R2

b3 R3

+

X Y

Collect terms of equal order

= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

17

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Multiplying Polynomials

Tabular representation of polynomial multiplication

(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1 bR b2R2 b3R3 middot middot middot

1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot

a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot

middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot

Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

18

Partial Fractions

Use partial fractions to rewrite as a sum of simpler parts

R

R

16

minus063

+X Y

Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R

19

Second-Order Systems Equivalent Forms

The sum of simpler parts suggests a parallel implementation

Y X

= 45

1 minus 09R minus

35 1 minus 07R

+

09 R

45 +

minus35

R07

+

X YY1

Y2

If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0

Thus y[n] = 45(09)n minus 35(07)n for n ge 0

20

Partial Fractions

Graphical representation of the sum of geometric sequences

minus1 0 1 2 3 4 5 6 7 8n

y1[n] = 09n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y2[n] = 07n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y[n] = 45(09)n minus 35(07)nfor n ge 0

21

Partial Fractions

Partial fractions provides a remarkable equivalence

R

R

16

minus063

+X Y

+

09 R

45 +

minus35

R07

+

X YY1

Y2

rarr follows from thinking about system as polynomial (factoring) 22

Poles

The key to simplifying a higher-order system is identifying its poles

Poles are the roots of the denominator of the system functional

when R rarr 1 z

Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)

p0=07 p1=09

1 Substitute R rarr and find roots of denominator

z 2 2Y 1 z z= = =

X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2

z0=07 z1=09

The poles are at 07 and 09

23

︸ ︷︷ ︸ ︸ ︷︷ ︸

︸ ︷︷ ︸ ︸ ︷︷ ︸

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

24

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z == = 14

1 18

1 2 + 14

18

121 + z +minus z minus z minus2 zz z

1 The unit sample response converges to zero 12 and z = 1

42 There are poles at z =123 There is a pole at z =

4 There are two poles

5 None of the above

25

14

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z= = = 14

1 18

1 2 + 14

18

12

141 + z +minus z minus z minus2 zz z

radic 1 The unit sample response converges to zero

12 and z = 1

4 X2 There are poles at z =12 Xradic3 There is a pole at z =

4 There are two poles

5 None of the above X

26

( ) ( )

( )( )

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true 2

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

27

Population Growth

28

Population Growth

29

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 14: Feedback, Poles, and Fundamental Modes (PDF)

Factoring Second-Order Systems

Factor the operator expression to break the system into two simpler

systems (divide and conquer)

R

R

16

minus063

+X Y

Y = X + 16RY minus 063R2Y

(1 minus 16R + 063R2) Y = X

(1 minus 07R)(1 minus 09R) Y = X

14

Factoring Second-Order Systems

The factored form corresponds to a cascade of simpler systems

(1 minus 07R)(1 minus 09R) Y = X

+

07 R

+

09 R

X YY2

(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2

+

09 R

+

07 R

X YY1

(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1

The order doesnrsquot matter (if systems are initially at rest) 15

Factoring Second-Order Systems

The unit-sample response of the cascaded system can be found by

multiplying the polynomial representations of the subsystems

Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _

_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)

Multiply then collect terms of equal order

Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X

+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot

16

Multiplying Polynomial

Graphical representation of polynomial multiplication

Y

Y X

= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1

a R

a2 R2

a3 R3

+

1

b R

b2 R2

b3 R3

+

X Y

Collect terms of equal order

= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

17

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Multiplying Polynomials

Tabular representation of polynomial multiplication

(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1 bR b2R2 b3R3 middot middot middot

1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot

a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot

middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot

Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

18

Partial Fractions

Use partial fractions to rewrite as a sum of simpler parts

R

R

16

minus063

+X Y

Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R

19

Second-Order Systems Equivalent Forms

The sum of simpler parts suggests a parallel implementation

Y X

= 45

1 minus 09R minus

35 1 minus 07R

+

09 R

45 +

minus35

R07

+

X YY1

Y2

If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0

Thus y[n] = 45(09)n minus 35(07)n for n ge 0

20

Partial Fractions

Graphical representation of the sum of geometric sequences

minus1 0 1 2 3 4 5 6 7 8n

y1[n] = 09n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y2[n] = 07n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y[n] = 45(09)n minus 35(07)nfor n ge 0

21

Partial Fractions

Partial fractions provides a remarkable equivalence

R

R

16

minus063

+X Y

+

09 R

45 +

minus35

R07

+

X YY1

Y2

rarr follows from thinking about system as polynomial (factoring) 22

Poles

The key to simplifying a higher-order system is identifying its poles

Poles are the roots of the denominator of the system functional

when R rarr 1 z

Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)

p0=07 p1=09

1 Substitute R rarr and find roots of denominator

z 2 2Y 1 z z= = =

X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2

z0=07 z1=09

The poles are at 07 and 09

23

︸ ︷︷ ︸ ︸ ︷︷ ︸

︸ ︷︷ ︸ ︸ ︷︷ ︸

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

24

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z == = 14

1 18

1 2 + 14

18

121 + z +minus z minus z minus2 zz z

1 The unit sample response converges to zero 12 and z = 1

42 There are poles at z =123 There is a pole at z =

4 There are two poles

5 None of the above

25

14

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z= = = 14

1 18

1 2 + 14

18

12

141 + z +minus z minus z minus2 zz z

radic 1 The unit sample response converges to zero

12 and z = 1

4 X2 There are poles at z =12 Xradic3 There is a pole at z =

4 There are two poles

5 None of the above X

26

( ) ( )

( )( )

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true 2

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

27

Population Growth

28

Population Growth

29

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 15: Feedback, Poles, and Fundamental Modes (PDF)

Factoring Second-Order Systems

The factored form corresponds to a cascade of simpler systems

(1 minus 07R)(1 minus 09R) Y = X

+

07 R

+

09 R

X YY2

(1 minus 07R) Y2 = X (1 minus 09R) Y = Y2

+

09 R

+

07 R

X YY1

(1 minus 09R) Y1 = X (1 minus 07R) Y = Y1

The order doesnrsquot matter (if systems are initially at rest) 15

Factoring Second-Order Systems

The unit-sample response of the cascaded system can be found by

multiplying the polynomial representations of the subsystems

Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _

_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)

Multiply then collect terms of equal order

Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X

+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot

16

Multiplying Polynomial

Graphical representation of polynomial multiplication

Y

Y X

= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1

a R

a2 R2

a3 R3

+

1

b R

b2 R2

b3 R3

+

X Y

Collect terms of equal order

= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

17

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Multiplying Polynomials

Tabular representation of polynomial multiplication

(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1 bR b2R2 b3R3 middot middot middot

1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot

a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot

middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot

Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

18

Partial Fractions

Use partial fractions to rewrite as a sum of simpler parts

R

R

16

minus063

+X Y

Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R

19

Second-Order Systems Equivalent Forms

The sum of simpler parts suggests a parallel implementation

Y X

= 45

1 minus 09R minus

35 1 minus 07R

+

09 R

45 +

minus35

R07

+

X YY1

Y2

If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0

Thus y[n] = 45(09)n minus 35(07)n for n ge 0

20

Partial Fractions

Graphical representation of the sum of geometric sequences

minus1 0 1 2 3 4 5 6 7 8n

y1[n] = 09n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y2[n] = 07n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y[n] = 45(09)n minus 35(07)nfor n ge 0

21

Partial Fractions

Partial fractions provides a remarkable equivalence

R

R

16

minus063

+X Y

+

09 R

45 +

minus35

R07

+

X YY1

Y2

rarr follows from thinking about system as polynomial (factoring) 22

Poles

The key to simplifying a higher-order system is identifying its poles

Poles are the roots of the denominator of the system functional

when R rarr 1 z

Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)

p0=07 p1=09

1 Substitute R rarr and find roots of denominator

z 2 2Y 1 z z= = =

X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2

z0=07 z1=09

The poles are at 07 and 09

23

︸ ︷︷ ︸ ︸ ︷︷ ︸

︸ ︷︷ ︸ ︸ ︷︷ ︸

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

24

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z == = 14

1 18

1 2 + 14

18

121 + z +minus z minus z minus2 zz z

1 The unit sample response converges to zero 12 and z = 1

42 There are poles at z =123 There is a pole at z =

4 There are two poles

5 None of the above

25

14

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z= = = 14

1 18

1 2 + 14

18

12

141 + z +minus z minus z minus2 zz z

radic 1 The unit sample response converges to zero

12 and z = 1

4 X2 There are poles at z =12 Xradic3 There is a pole at z =

4 There are two poles

5 None of the above X

26

( ) ( )

( )( )

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true 2

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

27

Population Growth

28

Population Growth

29

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 16: Feedback, Poles, and Fundamental Modes (PDF)

Factoring Second-Order Systems

The unit-sample response of the cascaded system can be found by

multiplying the polynomial representations of the subsystems

Y 1 1 1 = = times X (1 minus 07R)(1 minus 09R) (1 minus 07R) (1 minus 09R) _ _

_ _ = (1 + 07R + 072R2 + 073R3 + middot middot middot) times (1 + 09R + 092R2 + 093R3 + middot middot middot)

Multiply then collect terms of equal order

Y = 1 + (07 + 09)R + (072 + 07 times 09 + 092)R2 X

+ (073 + 072 times 09 + 07 times 092 + 093)R3 + middot middot middot

16

Multiplying Polynomial

Graphical representation of polynomial multiplication

Y

Y X

= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1

a R

a2 R2

a3 R3

+

1

b R

b2 R2

b3 R3

+

X Y

Collect terms of equal order

= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

17

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Multiplying Polynomials

Tabular representation of polynomial multiplication

(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1 bR b2R2 b3R3 middot middot middot

1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot

a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot

middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot

Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

18

Partial Fractions

Use partial fractions to rewrite as a sum of simpler parts

R

R

16

minus063

+X Y

Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R

19

Second-Order Systems Equivalent Forms

The sum of simpler parts suggests a parallel implementation

Y X

= 45

1 minus 09R minus

35 1 minus 07R

+

09 R

45 +

minus35

R07

+

X YY1

Y2

If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0

Thus y[n] = 45(09)n minus 35(07)n for n ge 0

20

Partial Fractions

Graphical representation of the sum of geometric sequences

minus1 0 1 2 3 4 5 6 7 8n

y1[n] = 09n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y2[n] = 07n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y[n] = 45(09)n minus 35(07)nfor n ge 0

21

Partial Fractions

Partial fractions provides a remarkable equivalence

R

R

16

minus063

+X Y

+

09 R

45 +

minus35

R07

+

X YY1

Y2

rarr follows from thinking about system as polynomial (factoring) 22

Poles

The key to simplifying a higher-order system is identifying its poles

Poles are the roots of the denominator of the system functional

when R rarr 1 z

Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)

p0=07 p1=09

1 Substitute R rarr and find roots of denominator

z 2 2Y 1 z z= = =

X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2

z0=07 z1=09

The poles are at 07 and 09

23

︸ ︷︷ ︸ ︸ ︷︷ ︸

︸ ︷︷ ︸ ︸ ︷︷ ︸

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

24

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z == = 14

1 18

1 2 + 14

18

121 + z +minus z minus z minus2 zz z

1 The unit sample response converges to zero 12 and z = 1

42 There are poles at z =123 There is a pole at z =

4 There are two poles

5 None of the above

25

14

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z= = = 14

1 18

1 2 + 14

18

12

141 + z +minus z minus z minus2 zz z

radic 1 The unit sample response converges to zero

12 and z = 1

4 X2 There are poles at z =12 Xradic3 There is a pole at z =

4 There are two poles

5 None of the above X

26

( ) ( )

( )( )

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true 2

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

27

Population Growth

28

Population Growth

29

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 17: Feedback, Poles, and Fundamental Modes (PDF)

Multiplying Polynomial

Graphical representation of polynomial multiplication

Y

Y X

= (1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1

a R

a2 R2

a3 R3

+

1

b R

b2 R2

b3 R3

+

X Y

Collect terms of equal order

= 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

17

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Multiplying Polynomials

Tabular representation of polynomial multiplication

(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1 bR b2R2 b3R3 middot middot middot

1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot

a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot

middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot

Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

18

Partial Fractions

Use partial fractions to rewrite as a sum of simpler parts

R

R

16

minus063

+X Y

Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R

19

Second-Order Systems Equivalent Forms

The sum of simpler parts suggests a parallel implementation

Y X

= 45

1 minus 09R minus

35 1 minus 07R

+

09 R

45 +

minus35

R07

+

X YY1

Y2

If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0

Thus y[n] = 45(09)n minus 35(07)n for n ge 0

20

Partial Fractions

Graphical representation of the sum of geometric sequences

minus1 0 1 2 3 4 5 6 7 8n

y1[n] = 09n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y2[n] = 07n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y[n] = 45(09)n minus 35(07)nfor n ge 0

21

Partial Fractions

Partial fractions provides a remarkable equivalence

R

R

16

minus063

+X Y

+

09 R

45 +

minus35

R07

+

X YY1

Y2

rarr follows from thinking about system as polynomial (factoring) 22

Poles

The key to simplifying a higher-order system is identifying its poles

Poles are the roots of the denominator of the system functional

when R rarr 1 z

Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)

p0=07 p1=09

1 Substitute R rarr and find roots of denominator

z 2 2Y 1 z z= = =

X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2

z0=07 z1=09

The poles are at 07 and 09

23

︸ ︷︷ ︸ ︸ ︷︷ ︸

︸ ︷︷ ︸ ︸ ︷︷ ︸

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

24

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z == = 14

1 18

1 2 + 14

18

121 + z +minus z minus z minus2 zz z

1 The unit sample response converges to zero 12 and z = 1

42 There are poles at z =123 There is a pole at z =

4 There are two poles

5 None of the above

25

14

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z= = = 14

1 18

1 2 + 14

18

12

141 + z +minus z minus z minus2 zz z

radic 1 The unit sample response converges to zero

12 and z = 1

4 X2 There are poles at z =12 Xradic3 There is a pole at z =

4 There are two poles

5 None of the above X

26

( ) ( )

( )( )

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true 2

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

27

Population Growth

28

Population Growth

29

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 18: Feedback, Poles, and Fundamental Modes (PDF)

minus1 0 1 2 3 4 5 6 7 8n

y[n]

Multiplying Polynomials

Tabular representation of polynomial multiplication

(1 + aR + a 2R2 + a 3R3 + middot middot middot) times (1 + bR + b2R2 + b3R3 + middot middot middot)

1 bR b2R2 b3R3 middot middot middot

1 1 bR b2R2 b3R3 middot middot middot aR aR abR2 ab2R3 ab3R4 middot middot middot

a2R2 a2R2 a2bR3 a2b2R4 a2b3R5 middot middot middot a3R3 a3R3 a3bR4 a3b2R5 a3b3R6 middot middot middot

middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot middot

Group same powers of R by following reverse diagonals Y = 1 + (a + b)R + (a 2 + ab + b2)R2 + (a 3 + a 2b + ab2 + b3)R3 + middot middot middot X

18

Partial Fractions

Use partial fractions to rewrite as a sum of simpler parts

R

R

16

minus063

+X Y

Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R

19

Second-Order Systems Equivalent Forms

The sum of simpler parts suggests a parallel implementation

Y X

= 45

1 minus 09R minus

35 1 minus 07R

+

09 R

45 +

minus35

R07

+

X YY1

Y2

If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0

Thus y[n] = 45(09)n minus 35(07)n for n ge 0

20

Partial Fractions

Graphical representation of the sum of geometric sequences

minus1 0 1 2 3 4 5 6 7 8n

y1[n] = 09n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y2[n] = 07n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y[n] = 45(09)n minus 35(07)nfor n ge 0

21

Partial Fractions

Partial fractions provides a remarkable equivalence

R

R

16

minus063

+X Y

+

09 R

45 +

minus35

R07

+

X YY1

Y2

rarr follows from thinking about system as polynomial (factoring) 22

Poles

The key to simplifying a higher-order system is identifying its poles

Poles are the roots of the denominator of the system functional

when R rarr 1 z

Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)

p0=07 p1=09

1 Substitute R rarr and find roots of denominator

z 2 2Y 1 z z= = =

X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2

z0=07 z1=09

The poles are at 07 and 09

23

︸ ︷︷ ︸ ︸ ︷︷ ︸

︸ ︷︷ ︸ ︸ ︷︷ ︸

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

24

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z == = 14

1 18

1 2 + 14

18

121 + z +minus z minus z minus2 zz z

1 The unit sample response converges to zero 12 and z = 1

42 There are poles at z =123 There is a pole at z =

4 There are two poles

5 None of the above

25

14

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z= = = 14

1 18

1 2 + 14

18

12

141 + z +minus z minus z minus2 zz z

radic 1 The unit sample response converges to zero

12 and z = 1

4 X2 There are poles at z =12 Xradic3 There is a pole at z =

4 There are two poles

5 None of the above X

26

( ) ( )

( )( )

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true 2

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

27

Population Growth

28

Population Growth

29

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 19: Feedback, Poles, and Fundamental Modes (PDF)

Partial Fractions

Use partial fractions to rewrite as a sum of simpler parts

R

R

16

minus063

+X Y

Y 1 1 45 35 = = = minus X 1 minus 16R + 063R2 (1 minus 09R)(1 minus 07R) 1 minus 09R 1 minus 07R

19

Second-Order Systems Equivalent Forms

The sum of simpler parts suggests a parallel implementation

Y X

= 45

1 minus 09R minus

35 1 minus 07R

+

09 R

45 +

minus35

R07

+

X YY1

Y2

If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0

Thus y[n] = 45(09)n minus 35(07)n for n ge 0

20

Partial Fractions

Graphical representation of the sum of geometric sequences

minus1 0 1 2 3 4 5 6 7 8n

y1[n] = 09n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y2[n] = 07n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y[n] = 45(09)n minus 35(07)nfor n ge 0

21

Partial Fractions

Partial fractions provides a remarkable equivalence

R

R

16

minus063

+X Y

+

09 R

45 +

minus35

R07

+

X YY1

Y2

rarr follows from thinking about system as polynomial (factoring) 22

Poles

The key to simplifying a higher-order system is identifying its poles

Poles are the roots of the denominator of the system functional

when R rarr 1 z

Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)

p0=07 p1=09

1 Substitute R rarr and find roots of denominator

z 2 2Y 1 z z= = =

X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2

z0=07 z1=09

The poles are at 07 and 09

23

︸ ︷︷ ︸ ︸ ︷︷ ︸

︸ ︷︷ ︸ ︸ ︷︷ ︸

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

24

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z == = 14

1 18

1 2 + 14

18

121 + z +minus z minus z minus2 zz z

1 The unit sample response converges to zero 12 and z = 1

42 There are poles at z =123 There is a pole at z =

4 There are two poles

5 None of the above

25

14

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z= = = 14

1 18

1 2 + 14

18

12

141 + z +minus z minus z minus2 zz z

radic 1 The unit sample response converges to zero

12 and z = 1

4 X2 There are poles at z =12 Xradic3 There is a pole at z =

4 There are two poles

5 None of the above X

26

( ) ( )

( )( )

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true 2

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

27

Population Growth

28

Population Growth

29

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 20: Feedback, Poles, and Fundamental Modes (PDF)

Second-Order Systems Equivalent Forms

The sum of simpler parts suggests a parallel implementation

Y X

= 45

1 minus 09R minus

35 1 minus 07R

+

09 R

45 +

minus35

R07

+

X YY1

Y2

If x[n] = δ[n] then y1[n] = 09n and y2[n] = 07n for n ge 0

Thus y[n] = 45(09)n minus 35(07)n for n ge 0

20

Partial Fractions

Graphical representation of the sum of geometric sequences

minus1 0 1 2 3 4 5 6 7 8n

y1[n] = 09n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y2[n] = 07n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y[n] = 45(09)n minus 35(07)nfor n ge 0

21

Partial Fractions

Partial fractions provides a remarkable equivalence

R

R

16

minus063

+X Y

+

09 R

45 +

minus35

R07

+

X YY1

Y2

rarr follows from thinking about system as polynomial (factoring) 22

Poles

The key to simplifying a higher-order system is identifying its poles

Poles are the roots of the denominator of the system functional

when R rarr 1 z

Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)

p0=07 p1=09

1 Substitute R rarr and find roots of denominator

z 2 2Y 1 z z= = =

X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2

z0=07 z1=09

The poles are at 07 and 09

23

︸ ︷︷ ︸ ︸ ︷︷ ︸

︸ ︷︷ ︸ ︸ ︷︷ ︸

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

24

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z == = 14

1 18

1 2 + 14

18

121 + z +minus z minus z minus2 zz z

1 The unit sample response converges to zero 12 and z = 1

42 There are poles at z =123 There is a pole at z =

4 There are two poles

5 None of the above

25

14

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z= = = 14

1 18

1 2 + 14

18

12

141 + z +minus z minus z minus2 zz z

radic 1 The unit sample response converges to zero

12 and z = 1

4 X2 There are poles at z =12 Xradic3 There is a pole at z =

4 There are two poles

5 None of the above X

26

( ) ( )

( )( )

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true 2

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

27

Population Growth

28

Population Growth

29

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 21: Feedback, Poles, and Fundamental Modes (PDF)

Partial Fractions

Graphical representation of the sum of geometric sequences

minus1 0 1 2 3 4 5 6 7 8n

y1[n] = 09n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y2[n] = 07n for n ge 0

minus1 0 1 2 3 4 5 6 7 8n

y[n] = 45(09)n minus 35(07)nfor n ge 0

21

Partial Fractions

Partial fractions provides a remarkable equivalence

R

R

16

minus063

+X Y

+

09 R

45 +

minus35

R07

+

X YY1

Y2

rarr follows from thinking about system as polynomial (factoring) 22

Poles

The key to simplifying a higher-order system is identifying its poles

Poles are the roots of the denominator of the system functional

when R rarr 1 z

Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)

p0=07 p1=09

1 Substitute R rarr and find roots of denominator

z 2 2Y 1 z z= = =

X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2

z0=07 z1=09

The poles are at 07 and 09

23

︸ ︷︷ ︸ ︸ ︷︷ ︸

︸ ︷︷ ︸ ︸ ︷︷ ︸

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

24

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z == = 14

1 18

1 2 + 14

18

121 + z +minus z minus z minus2 zz z

1 The unit sample response converges to zero 12 and z = 1

42 There are poles at z =123 There is a pole at z =

4 There are two poles

5 None of the above

25

14

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z= = = 14

1 18

1 2 + 14

18

12

141 + z +minus z minus z minus2 zz z

radic 1 The unit sample response converges to zero

12 and z = 1

4 X2 There are poles at z =12 Xradic3 There is a pole at z =

4 There are two poles

5 None of the above X

26

( ) ( )

( )( )

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true 2

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

27

Population Growth

28

Population Growth

29

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 22: Feedback, Poles, and Fundamental Modes (PDF)

Partial Fractions

Partial fractions provides a remarkable equivalence

R

R

16

minus063

+X Y

+

09 R

45 +

minus35

R07

+

X YY1

Y2

rarr follows from thinking about system as polynomial (factoring) 22

Poles

The key to simplifying a higher-order system is identifying its poles

Poles are the roots of the denominator of the system functional

when R rarr 1 z

Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)

p0=07 p1=09

1 Substitute R rarr and find roots of denominator

z 2 2Y 1 z z= = =

X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2

z0=07 z1=09

The poles are at 07 and 09

23

︸ ︷︷ ︸ ︸ ︷︷ ︸

︸ ︷︷ ︸ ︸ ︷︷ ︸

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

24

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z == = 14

1 18

1 2 + 14

18

121 + z +minus z minus z minus2 zz z

1 The unit sample response converges to zero 12 and z = 1

42 There are poles at z =123 There is a pole at z =

4 There are two poles

5 None of the above

25

14

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z= = = 14

1 18

1 2 + 14

18

12

141 + z +minus z minus z minus2 zz z

radic 1 The unit sample response converges to zero

12 and z = 1

4 X2 There are poles at z =12 Xradic3 There is a pole at z =

4 There are two poles

5 None of the above X

26

( ) ( )

( )( )

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true 2

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

27

Population Growth

28

Population Growth

29

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 23: Feedback, Poles, and Fundamental Modes (PDF)

Poles

The key to simplifying a higher-order system is identifying its poles

Poles are the roots of the denominator of the system functional

when R rarr 1 z

Start with system functional Y 1 1 1 = = = X 1 minus 16R+063R2 (1minusp0R)(1minusp1R) (1minus07R) (1minus09R)

p0=07 p1=09

1 Substitute R rarr and find roots of denominator

z 2 2Y 1 z z= = =

X 16 063 z2 minus16z+063 (zminus07) (zminus09)1 minus + z z2

z0=07 z1=09

The poles are at 07 and 09

23

︸ ︷︷ ︸ ︸ ︷︷ ︸

︸ ︷︷ ︸ ︸ ︷︷ ︸

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

24

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z == = 14

1 18

1 2 + 14

18

121 + z +minus z minus z minus2 zz z

1 The unit sample response converges to zero 12 and z = 1

42 There are poles at z =123 There is a pole at z =

4 There are two poles

5 None of the above

25

14

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z= = = 14

1 18

1 2 + 14

18

12

141 + z +minus z minus z minus2 zz z

radic 1 The unit sample response converges to zero

12 and z = 1

4 X2 There are poles at z =12 Xradic3 There is a pole at z =

4 There are two poles

5 None of the above X

26

( ) ( )

( )( )

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true 2

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

27

Population Growth

28

Population Growth

29

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 24: Feedback, Poles, and Fundamental Modes (PDF)

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

24

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z == = 14

1 18

1 2 + 14

18

121 + z +minus z minus z minus2 zz z

1 The unit sample response converges to zero 12 and z = 1

42 There are poles at z =123 There is a pole at z =

4 There are two poles

5 None of the above

25

14

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z= = = 14

1 18

1 2 + 14

18

12

141 + z +minus z minus z minus2 zz z

radic 1 The unit sample response converges to zero

12 and z = 1

4 X2 There are poles at z =12 Xradic3 There is a pole at z =

4 There are two poles

5 None of the above X

26

( ) ( )

( )( )

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true 2

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

27

Population Growth

28

Population Growth

29

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 25: Feedback, Poles, and Fundamental Modes (PDF)

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z == = 14

1 18

1 2 + 14

18

121 + z +minus z minus z minus2 zz z

1 The unit sample response converges to zero 12 and z = 1

42 There are poles at z =123 There is a pole at z =

4 There are two poles

5 None of the above

25

14

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z= = = 14

1 18

1 2 + 14

18

12

141 + z +minus z minus z minus2 zz z

radic 1 The unit sample response converges to zero

12 and z = 1

4 X2 There are poles at z =12 Xradic3 There is a pole at z =

4 There are two poles

5 None of the above X

26

( ) ( )

( )( )

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true 2

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

27

Population Growth

28

Population Growth

29

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 26: Feedback, Poles, and Fundamental Modes (PDF)

Check Yourself

1 1 1 y[n] = minus4y[n minus 1] + 8y[n minus 2] + x[n minus 1] minus 2x[n minus 2]

1 1 11 + 4R minus 8R2 Y = R minus 2R2 X

12R

2R minusH(R) =

Y = X 1

418R21 + R minus

1 12

1 12

12

minus z minus z minus2z z= = = 14

1 18

1 2 + 14

18

12

141 + z +minus z minus z minus2 zz z

radic 1 The unit sample response converges to zero

12 and z = 1

4 X2 There are poles at z =12 Xradic3 There is a pole at z =

4 There are two poles

5 None of the above X

26

( ) ( )

( )( )

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true 2

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

27

Population Growth

28

Population Growth

29

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 27: Feedback, Poles, and Fundamental Modes (PDF)

Check Yourself

Consider the system described by

y[n] = minus 1 4y[n minus 1] +

1 8y[n minus 2] + x[n minus 1] minus

1 2x[n minus 2]

How many of the following are true 2

1 The unit sample response converges to zero

2 There are poles at z = 12 and z = 1

4

3 There is a pole at z = 12

4 There are two poles

5 None of the above

27

Population Growth

28

Population Growth

29

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 28: Feedback, Poles, and Fundamental Modes (PDF)

Population Growth

28

Population Growth

29

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 29: Feedback, Poles, and Fundamental Modes (PDF)

Population Growth

29

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 30: Feedback, Poles, and Fundamental Modes (PDF)

Population Growth

30

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 31: Feedback, Poles, and Fundamental Modes (PDF)

Population Growth

31

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 32: Feedback, Poles, and Fundamental Modes (PDF)

Population Growth

32

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 33: Feedback, Poles, and Fundamental Modes (PDF)

Check Yourself

What are the pole(s) of the Fibonacci system

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

33

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 34: Feedback, Poles, and Fundamental Modes (PDF)

Check Yourself

What are the pole(s) of the Fibonacci system

Difference equation for Fibonacci system

y[n] = x[n] + y[n minus 1] + y[n minus 2]

System functional Y 1

H = = X 1 minus R minus R2

Denominator is second order rarr 2 poles

34

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 35: Feedback, Poles, and Fundamental Modes (PDF)

Check Yourself

Find the poles by substituting R rarr 1z in system functional 2Y 1 1 z

H = = = X 1 minus R minus R2 rarr

1 minus 1 minus 1 z2 minus z minus 1 z 2z

Poles are at radic1 plusmn 5 1

z = = φ or minus2 φ

where φ represents the ldquogolden ratiordquo radic1 + 5

φ = asymp 16182 The two poles are at

1 z0 = φ asymp 1618 and z1 = minus asymp minus0618

φ

35

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 36: Feedback, Poles, and Fundamental Modes (PDF)

Check Yourself

What are the pole(s) of the Fibonacci system 4

1 1

2 1 and minus1

3 minus1 and minus2

4 1618 and minus0618

5 none of the above

36

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 37: Feedback, Poles, and Fundamental Modes (PDF)

Example Fibonaccirsquos Bunnies

Each pole corresponds to a fundamental mode

1 φ asymp 1618 and minus asymp minus0618

φ

One mode diverges one mode oscillates

minus1 0 1 2 3 4n

φn

minus1 0 1 2 3 4n

(minus 1φ

)n

37

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 38: Feedback, Poles, and Fundamental Modes (PDF)

Example Fibonaccirsquos Bunnies

The unit-sample response of the Fibonacci system can be written

as a weighted sum of fundamental modes

φ 1 Y 1 radic

5 φ radic

5H = = = +

X 1 minus R minus R2 1 minus φR 1 + 1 Rφ

φ 1 h[n] = radic φn + radic (minusφ)minusn n ge 0

5 φ 5

But we already know that h[n] is the Fibonacci sequence f

f 1 1 2 3 5 8 13 21 34 55 89

Therefore we can calculate f [n] without knowing f [n minus 1] or f [n minus 2]

38

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 39: Feedback, Poles, and Fundamental Modes (PDF)

Complex Poles

What if a pole has a non-zero imaginary part

Example Y 1 = X 1 minus R + R2

1 z2 = =

1 minus 1 z + 1

z2 z2 minus z + 1

1 radic

3 plusmnjπ3Poles are z = 2 plusmn 2 j = e

What are the implications of complex poles

39

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 40: Feedback, Poles, and Fundamental Modes (PDF)

Complex Poles

Partial fractions work even when the poles are complex

jπ3 minusjπ3Y 1 1 1 e e= times = radic minus jπ3R minusjπ3R jπ3R minusjπ3RX 1 minus e 1 minus e j 3 1 minus e 1 minus e

There are two fundamental modes (both geometric sequences)

e jnπ3 = cos(nπ3) + j sin(nπ3) and e minusjnπ3 = cos(nπ3) minus j sin(nπ3)

n n

40

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 41: Feedback, Poles, and Fundamental Modes (PDF)

Complex Poles

Complex modes are easier to visualize in the complex plane

e jnπ3 = cos(nπ3) + j sin(nπ3)

Re

Im

e j0π3

e j1π3e j2π3

e j3π3

e j4π3 e j5π3

Re

Im

e j0π3

e j5π3e j4π3

e j3π3

e j2π3 e j1π3

n

eminusjnπ3 = cos(nπ3) minus j sin(nπ3)

n

41

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 42: Feedback, Poles, and Fundamental Modes (PDF)

Complex Poles

The output of a ldquorealrdquo system has real values

y[n] = x[n] + y[n minus 1] minus y[n minus 2] Y 1

H = = X 1 minus R + R2

1 1 = jπ3R

times minusjπ3R1 minus e 1 minus ejπ3 minusjπ31 e e=

j radic

3 1 minus e jπ3Rminus

1 minus eminusjπ3R 1 h[n] = radic

j 3e j(n+1)π3 minus e minusj(n+1)π3 = radic sin2 (n + 1)π

3 3

1

minus1

n

h[n]

42

( )

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 43: Feedback, Poles, and Fundamental Modes (PDF)

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

43

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 44: Feedback, Poles, and Fundamental Modes (PDF)

Check Yourself

Unit-sample response of a system with poles at z = re plusmnjΩ

n

Which of the following isare true 2

1 r lt 05 and Ω asymp 05

2 05 lt r lt 1 and Ω asymp 05

3 r lt 05 and Ω asymp 008

4 05 lt r lt 1 and Ω asymp 008

5 none of the above

44

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 45: Feedback, Poles, and Fundamental Modes (PDF)

Check Yourself

R R R+X Y

How many of the following statements are true

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

45

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 46: Feedback, Poles, and Fundamental Modes (PDF)

Check Yourself

R R R+X Y

How many of the following statements are true 4

1 This system has 3 fundamental modes

2 All of the fundamental modes can be written as geometrics

3 Unit-sample response is y[n] 0 0 0 1 0 0 1 0 0 1 0 0 1

4 Unit-sample response is y[n] 1 0 0 1 0 0 1 0 0 1 0 0 1

5 One of the fundamental modes of this system is the unit

step

46

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 47: Feedback, Poles, and Fundamental Modes (PDF)

Summary

Systems composed of adders gains and delays can be characterized

by their poles

The poles of a system determine its fundamental modes

The unit-sample response of a system can be expressed as a weighted

sum of fundamental modes

These properties follow from a polynomial interpretation of the sysshy

tem functional

47

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

Page 48: Feedback, Poles, and Fundamental Modes (PDF)

MIT OpenCourseWarehttpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms