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Sugata Munshi

Dept. of Electrical Engineering,

Jadavpur University, Kolkata.

of 1121

DISCRETE-TIME SIGNALS AND

SYSTEMS

A PRESENTATION BY

SUGATA MUNSHI

DEPARTMENT OF ELECTRICAL

ENGINEERING

JADAVPUR UNIVERSITY

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Discrete-Time Signals

Digital Signal is particular type of discrete-time signal.

A discrete time signal is one which is defined only at

discrete instants of time.

Samples

s(t)

t

Its values are known as

samples.

Hence it is also known

as :

sampled-data signal.

If the samples are uniformly spaced in time, we have a

uniformly sampled signal.

One of the distinctive feature of digital signals is that, they

are quantized both in timeand magnitude.That is, they are

discrete-time discrete-magnitude signals. That means not

only are they defined for discrete instants of time but they

can assume only discrete values.

Another distinctive feature of digital signals is that their

sample values are coded in binary numbers.

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s(t)

t

magnitude quantization

levels

0

q

2q

3q

4q

5q

Block diagram of a typical Digital Signal Processing System

ADC

Digital

Signal

Processor

DAC

Digital

Memory

Analog

signal

Digital

signal

Digital

signalAnalog

signal

The digitized analog signal is processed by a digital signal processor.

Processing can be of various types e.g. integrating or differentiating or

filtering or finding out the DFT (to determine the frequency spectra of the

signal). The processed signal is converted into an analog signal if necessary

or stored in memory for use in future.

BENEFITS OF PROCESSING SIGNALS

DIGITALLY:

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Guaranteed Accuracy determined only by

number of bits used.

Perfect Reproducibility: Identical

performance from unit to unit, since there is no

variations due to component tolerance.

No drift in performance with temperature or

age.

Greater Flexibility: Can be programmed

and reprogrammed to perform a variety of

functions, without modifying the hardware.

Superior Performance: DSP can be used to

perform functions not possible with analog

systems. Example: Linear phase response can

be achieved, complex adaptive fitering

algorithms can be implemented.

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Tremendous advancements in electronics

industry are fruitfully utilized: Compact

size, lower cost, low power nconsumption,

greater speed.

The Sampling Process (Uniform Sampling)

A sampling operation transforms a continuous time signal

into a discrete-time signal.

Ideal switch

(

closes periodically at intervals of ,

remains closed for a moment and opens immediately)

is the sampling period

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1sf

is the sampling frequency.

A more realistic case

The duration of closure of the switch should be finite (

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*px t x t p t

Pulse

Amplitude

Modulator

p(t) (carrier)

x(t)

modulating

signal

x*p(t) = x(t) p(t)

x(t)x*

p(t)

(

)

= Time period of

unit pulse train

= Sampling period

= Pulsewidth =

Sampling duration.

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K

p t u t K u t K

*pK

x t x t u t K u t K

If the sampling duration is very small compared to the

smallest time constant of the continuous-time system that

has generated x(t), then *

px t can be approximated by a

sequence of flat-topped pulses.

*px t

t

That is,

for 1 ; K=0,1,2,.......K t K ,

Then,

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*pK

x t x K u t K u t K

Taking the Laplace transform on both sides,

*1 s K s

pK

eX s x K e

s

Now,

2

1 1 1 ...........2!

s se s

If is very small compared to , 1 se s

K

sK

p

eKxsX )()(*

Taking inverse Laplace Transform of both sides,

)()()()( ** KtKxtxtxK

p

for small each pulse becomes a vertical line. R.H.S. represents a train of impulses, with the strength of

the impulse at t K equal to )( Kx

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Hence the finite pulsewidth sampler can be approximated

by an impulse sampler followed by an attenuator with

gain .

x(t)

( )

x*p(t)

= x*(t)

x(t)x*

i(t)

impulse sampler

x*(t)

(very

small)

)()()(* KtKxtxK

i

Here the role of is just scaling.Hence it is considered as unity.

K

KtKxtx )()()(*

Thus, for the sake of mathematical maneuvering a sampled-

data signal can be represented by a train of scaled impulses

and the strength of the impulse at any instant is equal to the

value of the sample at that instant. That is, actual sampling

process can be represented mathematically by impulse

sampling or impulse modulation.

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Impulse

Modulator

x(t) x*(t)

T(t)

t

Output of the impulse sampler is

K

Kttxttxtx )()()()()(*

Reconstruction of Continuous Time Signal from

its Samples:

4

- 4

- 3

- 2

-

0

2

3

In the absence of any additional conditions or

information, a continuous time signal can not beuniquely specified by a sequence of equally spaced

samples. In the above figure, three continuous time

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signals that can generate the given set of samples have

been shown.

In general, from a given sampled data signal an

infinite number of continuous time signals can be

reconstructed. Hence, reconstruction of a continuous-

time signal from its samples becomes a problem, unless

some other information is available.

Reconstruction

of Continuous-time

sinusoid from its samples:

A

x(t)

tto

x*(t)

t

Let a continuous-time signal

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2

sin sin 2 sinx t A t A ft A tT

be sampled with a sampling period .

Sampling rate1

sf

If the sampling starts at a time angle 2 oft from

the zero instant as shown, then the sampled-data signal is

where, n = 0,1,2,..

or,

2

sin

w

sin

re

2

he ,

s

s

nA

f T

f

f

x n A nf

is the time period in number of samples.

CaseI:

>> 1, i.e., fs>> f,

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x*(t)

t

Reconstructed analog

sinusoidal signal from

x*(t)

The analog sinusoid x(t) can be uniquely reconstructed from

x (n

).

CaseII:

< 1, i.e., fs< f, > T

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2 s

ff f

sin 2 2 2

1 sin 2 1

sin 2s

s

nx n A n n

A n

A n f

f f

2 3

s

f ff

sin 4 2 4

1 sin 2 2

si

2

n 2

s

s

nx n A n n

A n

A n f

f f

3 4s

f ff

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sin 6 2 6

1 sin 2 3

si3

n 2 s

s

nx n A n n

A n

A nf

f f

for ; 1,2,3, etc1 s

f f

f KK K

. sin 2 2 2

sin 2s

s

nx n A K n K n

A nf

f Kf

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Thus, since x n is also the sampled version

of a continuous-time sinusoid of frequency

sf f Kf f , the sinusoid that is actually

represented by x n , has a frequency

sf f Kf

2Also, sin 2

and sin 2 2

nx n A Mn

nx n A Mn

M = 1,2,3, etc.

i.e,

1sin 2x n A n M

and 1( ) [2 ( ) ]x n ASin n M M = 1,2,3, etc.

or,

sin 2 s

s

Mf fx n A n

f

and

sin 2 s

s

Mf fx n A n

f

M = 1,2,3, etc.

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Frequency sf Kf is the Alias of

frequency f .

Phenomenon is called aliasing. It is the

effect of undersampling.

CaseIII :

= 1, i.e., fs= f, = T

sin 2 sinx n A n A for all values of n.

x(t)

x*(t) A sin

Reconstructed

signal

Reconstruction gives a steady value.Samplings at zero

crossings yield no information.

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CASEIV:

Tf 2

T,2ff,i.e.,21

s

x(t)

x*(t)

t

t

1sin 2 sin 2 2 1

1 sin 2

sin 2

s

s

nx n A A n n

A n

f fA n

f

sin 2

s

sA nf

f f

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Also

2

sin 2 n

x n A Mn

M = 1,2,3,etc

2

sin 2 n

x n A Mn

M = 2,3,4 etc.

sin 2 s

s

Mf fx n A n

f

M=1,2,3,..

and

sin 2 s

s

Mf fx n A n

f

M=2,3,4..

x n is also sampled version of sinusoid with freq.

sf f f f .

Thus analog sinusoid reconstructed by interpolation, from

samples x n , will have a frequency sf f f .

So, frequency f of original signal is folded back into the

interval 02

sff .

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Frequency sf f is alias of f for

2sf f f

CaseV:

2, i.e., 2 ,2s

Tf f

sin 2 sin

2

nx n A A n

ASinnx )( for even n &

sinx n A for odd n .

x n can also be represented by

sin , i.e. sin 22

nB n B

such that sin sinB A

So, x n can also be sampled version of sinusoid

sinB t .

Infinite no. of combinations of B & yield

sin sinB A .

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Hence, if 0 , it is possible to have infinite number ofreconstructions of analog sinusoids of same frequencyf,but

differing in amplitude and phase.

x(t)

x*(t)

Reconstruction

with 2sf f , original sinusoid can not be uniquely

reconstructed from its samples, by interpolation.

CaseVI:

2, i.e. 2 ,2s

Tf f

sin 2

sin 2 2

sin 2 (M = 1,2,3......)s

s

nx n A

nA Mn

nA Mf f

f

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Also,

sin 2 ss

nx n A Mf f

f

(M = 1,2,3,..)

x(t)

x*(t)

Reconstruction

x n can be sampled version of x(t)as well as of analog

sinusoids with frequencies sMf f f & sMf f f .

Sinusoid ( ) sin 2x t A ft can be uniquely reconstructed

fromx*(t)by interpolation.

Observations :

sin 2x t A ft sampled at1

sf

starting

from t =0, such that

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10

2 s

f f .

sin 2 sy t A f Mf t (where M is a +ve integer)

sampled at same ratefs.

Then the sampled versions are,

sin 2x n A fn (1)

&

sin 2

sin 2 2

sy n A Mf f n

A nM fn

sin 2

or

y n A fn (2)

Sequences x n and y n are identical.

Not possible to distinguish between the

samples of sinusoids whose frequencies differ

by integral multiple of sampling frequency.

Frequency sf Mf (M=1,2,3,.) is folded back into

interval 02

sff .

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Frequency1

2 sf

folding frequency or Nyquist

frequency.

Frequency f is aliasof sMf f .

Again, if s i n 2 sz t A M f f are sampled at

, 02

ss

ff f

, then

sin 2

sin 2 2

sz n A Mf f n

A nM fn

sin 2

or

z n A fn (3)

Hence frequencies sMf f are folded back into

interval 0 2

sff .

Frequency Spectra of Discrete-Time

Signals

x(t) is sampled at 1sf to obtain a D.T. signal x*(t).

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Fourier Transform of x(t) is

( ) ( ) ( ) j t

X j F x t x t e dt

F.T. of x*(t) is ,

*

n

n

X j x n t n

x t t n

F

F

Consider the periodic signal

n

t t n

, with a period

.

tcan be expanded into Fourier series as

+

n=-

( ) (t-n )= sjm t

m

m

F e

0

0 0

0

1 1( ) ( )

1 1 = for all m,

s s

s

jm t jm t

m

jm t

t

F t e dt t e dt

e

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Therefore,

+

n=-

1( ) (t-n )= sjm t

m

e

where,

22s sf

Hence,

*m -

1( ) sX j X j m

For simplicity, let x(t) be such that X(j

) is a +ve real-valued

function of

.

i.e. x(t) has only amplitude spectrum and no phase spectrum.

Then

X j X

and

* *1

s s sm m

X j X X m f X m

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sX m has same waveform as X except that it is

centred at sm .

*X is sum of all repetitions of )(Xfs centred at

sm . 0, 1, 2,.....m Consider x(t) as bandlimited signal, with bandwidth W

r/s ; i.e. the frequency spectrum is 0 for

> W.

W

. . in Hz =2

mf B W

Ws 2 i.e., ms ff 2 repetitions do not overlap

theoretically possible to exactly reconstruct x(t) from x*(t) by

passing x*(t) through an ideal (brick-wall type) L.P. filter with cutoff

freq. 2

s

& pass band gain

1

sf(i.e.

).

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If 2s W i.e. 2s mf f adjacent replications of X*()

overlap & resulting X*() will no longer preserve

information of X .

If we try to reconstruct x(t) from x*(t) by L.P. filtering,

resulting signal will be x(t) contaminated by higher frequency

components. This phenomenon is known as aliasing.

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Sampling Theorem :If a Band-limited continuous time

signal x(t) containing no frequency component greater than

mf , is sampled at a frequency sf , then x(t) can be uniquely

reconstructed from its sampled versionx*(t)if 2s mf f .

Sampling rate 2N mf f is known as Nyquist rate.

ANTI-ALIAS FILTER

Practically no analog signal is naturally band limited. So, prior to

sampling (i.e. A/D conversion) the analog signal is made

bandlimited ( as far as practicable ) by passing it through an analog

filter.

Thus, to get rid of aliasing, the C.T. x t is processed (prior to

sampling) by analog L.P. filter with cut-off frequency < 2sf .

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Pass band of anti-alias filter depends on type of analog signal

to be processed.

Frequency spectra of human voices contain useful info up to

about 4 KHz. So for processing speech signals, cutoff freq. fcis

selected a little bit more than 4 KHz. The sampling is carried

out at 2s cf f .

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THE Z-TRANSFORM

What is Z-Transform ?

x(t)

sampled at a rate 1Sf

.

Sampled version ofx(t) is

n

ntnxtx )()()(* (1)

where n = 0, 1, 2,

or,

n

n ntxtx )()(* (1a)

where ( ) [ ]n t nx x n x n x n x t L.T of both sides of equation (1a) yields

n

n ntxLtxL )]([)](*[ (2)

or,

sn

n

nexsX

)(*

(3)

X*(s) Discrete Laplace Transform or Starred Laplace

Transform.

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Letse z . Then z is a complex variable.

Substitution of

s

z e

in (3) gives

)()(* zXzxsX n

n

n

(4)

X(z) is known as the Z-transform of sequencexn

In operational form,

[ ] ( ) nn nn

Z x X z x z

(5)whereZ[ . ] stands for Z-transform operator.

Ifxnrepresents a causal sequence, then,

0[ ] ( )

n

n nn

Z x X z x z

(6) Eq. (5) gives 2-sided or bilateral Z-transform a series

in both +ve and -ve powers of z-1.

Eq. (6) gives one sidedor unilateral Z-transforma

series in +ve power ofz-1

.

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The Complex-Variable z

3 5

1ln

1 1 1 1 1 ln 2 ..............

1 3 1 5 1

sz e

s z

z z zz

z z z

1

1

1ln( ) 2 1

zz z

Hence,

1

1

2 1

1

zs

z

Bilinear transformationor Tustin transformation

s j

s jz e e e

or,(in polar form)z e

cos sin (in cartesian form)e j

u jv

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z e

if 0, 1

if 0, 1

if 0, 1

z

z

z

Z TRANSFORMS OF SOME ELEMENTARY

SEQUENCES:

1.

Unit Step Sequence

1 for 0

0 for 0

nu n

n

-2 -1 01 2 3 4 5 6

n

Un

1

0

1 2

1

1 1 ........

1 1

;ROC : 1

n n

n n

n n

Z u u z z

zz z

z z

z

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1

L u ts

j

s-plane

jv

z-plane

u

2

. Causal Exponential Sequence:

n

an

n uex

01 2 3 3

1

1 .........

1 ;ROC :

1

an n

nn

a a n a

a

a a

Z x e z

e z e z e z

zz e

e z z e

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Alternatively,

an

n nx e u ; where 1a = time const. in terms of

number of samples.

Then,

11

( )1

n aX z Z x

e z

;

ROC : z> e-a

If a is real:

3

. Unit Discrete Impulse Sequence or Kronecker Delta

Sequence or Unit Sample Sequence.n = 1 for n=0

=0 for n 0

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n

1

n

-5 -4 -3 -2 -10

1 2 3 4 5

1 ; ROC : Entire z-plane.n on nn

Z z z

4

. Causal Sinusoidal Sequence:

0 for 0

sin for 0

n

o

x n

A n n

. . = sinn o ni e x A n u x

n

n

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0

sin[ ]

o o

on

j j

AzZ A Sin n u

z e z e

A causal sinusoidal sequence can be also expressed as :

n 0 0x = A Sin ( ) ; where is in radian.nn u ,

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then,

1

0

-1 20

Sin( ) ; ROC : z 1

1-2z

AzX z

Cos z

Some Important Properties of Z-Transform

Linearity of z-transform :

If [ ]nF z Z f , [ ]nz Z , & are constants and

n n nx f , then

[ ]nX z Z x F z z

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Proof:

0

0 0

n n n

n

n nn

n n

n nn n

n n

X z Z x Z f

f z

f z z

Z f ZF z z

Multiplication by an(scaling in the z-domain)

If X(z) = Z[ xn] ; 2 1:ROC r z r ,

then ,

1n nZ a x X a z ; 2 1:ROC a r z a r n n n

n nn

Z a x a x z

1

1

n

nn

x a z

X a z

Example : 1

1 1

121

2

n

nZ u

z

Real translation property (shifting properly)

If nX z Z x & K is a +ve integer, then,

Kn kZ x z X z

(1)

ROC:same as of X(z) except z = 0

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and K

n kZ x z X z (2)

ROC: same as of X(z) except for z =

.

Thus, multiplication of a Z-transform byKz has the effect of

delaying the time function *x t by K time units, i.e. xn by

K samples. also known as right shift or backward time

shift.

Multiplication of a Z-transform by zk has the effect of

advancing time function *x t by K time units, i.e. by K

samples.

Also known as left shiftor forward time shift.

Real Convolution Property

D.T. linear convolution or convolution sum of sequences nx &

ng is given by

n n n k n k k

y x g x g

(1)

for 0, 1, 2, 3n etc.

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If , andY z X z G z are Z-transforms of ny , nx

& ng , then,

Y z X z G z ;

ROC: At least the intersection of those for

andX z G z .

That is, D.T. convolution of 2 sequences results in

multiplication in z-domain.

Proof:

[ ]n nn K n K n n K

Y z y z x g z

or,

Y(z) =

n

K n KK n

n KK

K n KK n

x g z

x z g z

m=n-K

( ) ( ). ( )K mK mK m

Let

Y z x z g z X z G z

Note:

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If nx and ng are causal sequences, then

0

n

n n n k n k ky x g x g for n = 0, 1, 2,

etc.

Energy and Power Signals

Normalized energy:

2 2Nn n

N n N n

E im x x

Normalized power:

21

2 1

N

nN n N

P im xN

If E is finite, obviously P = 0

In this case, nx is an energy signal.

If E is infinite, P may be finite or infinite.

When P is finite and E = , then nx is a power signal.

Correlation Property

The cross-correlation of finite energy sequences x(n) is y(n) is

given by

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( ) ( ) ( ) ( ) ( )xyn n

r l x n y n l x n l y n

(1)

for 0, 1, 2, 3l etc.and,

( ) ( ) ( ) ( ) ( )yxn n

r l y n x n l y n l x n

(2)for 0, 1, 2, 3l etc.

Then,

1xy xyS z Z r l X z Y z ; ROC is at leastthe intersection of those for 1and YX z z

Note:

( ) ( )xy yxr l r l

Autocorrelation of a finite-energy sequence x(n) is

( ) ( ) ( ) ( ) ( ) ( )x xxn n

r l r l x n x n l x n l x n

1( ) ( ) ( ) ( )x xS z Z r l X z X z

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Time Multiplication Property ( Differentiation in Z

Domain)

( ) [ ]

dX(z)[ ] -z ; both transforms have the same ROC

dz

n

n

If

X z Z x

Z nx

n 2

[ ] ( ) ;1

dU(z)

Z[nu ] -z dz

:

1 ( 1)

n

Example

zZ u U z

z

d z z

zdz z z

Initial and Final Value Theorems

If nZ x X z , nx being a causal sequence,

Then,

0 zx im X z

IVT

11

1lim( 1) ( )zz

zx im X z z X z

z

.FVT

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Mapping Between s-plane and z-plane:

Importance of the study

Design & analysis of C.T. system often rely on pole-zero

configuration of system T.F. in the s-plane.

Similarly, poles & zeros of Z-transform of the system transfer

function determine response of a D.T. system at sampling

instants.

Hence the study of mapping between s-plane and z-plane is

important.

cos sinz e e j

j

S-plane

Z

Trnasform

jv

Z-Plane

u

1

For

= 0, |z| = 1.

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Thus, the jaxis in the s-plane corresponds to the circumference

of unit circle centred at origin in the z-plane.

For

< 0 , |z| < 1

So entire left half of s-plane is mapped into interior of unit

circle centred at origin, in z-plane.

For

> 0, |z| > 1.

Hence entire right half of s-plane is mapped into exterior of

unit circle centred at origin, in z-plane.

:

Mapping of Left half of the s-plane into the z-plane:

s-plane divided into infinite no. of horizontal periodic strips,

each of width

2 s

in direction of jaxis.

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Each strip extends from s)2

1j(Nto)

2

1( sNj in direction

of jaxis, where0, 1, 2, 3,...........N

j

N = 1

N = 0

N = - 1Complementary strip

Complementary strip

Primary strip

32

sj

2

sj

32

sj

2

sj

0

N = 0primary strip,

Others are complementary strips.

Mapping of left half of Primary Strip

B

E

C

D A

jv

Z-plane

u

2

sj

2

sj

BC

D E

A

Primary

strip

S-plane

j

- 0

1

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In s-plane, AB C DEA representing boundary of

L.H. of primary strip, is traversed in counter clockwise

direction.

During traversal, z varies in following manner.

At , 1 0

Interval [A,B]

oA z

1 , 0 1802os

z

At , 1 180

Interval ,

oB z

B C

180 , 1 0 0oz z z l

At , 0 180

Interval , ,

o

C z

C D

0 0

,

180 180 ; 0

z z

z

At , 0 180 .

Interval , :

oD z

D E

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180 , 1 0oz z z

At , 1 180

Interval , :

oE z

E A

01 , 180 0oz

Entire L.H. of primary strip is mapped into interior of unit

circle centred at origin in z-plane.

Mapping of left halves of complementary strips:

32

sj

2

sj

32

sj

2

sj

0

j

j

sS

c

Sp

Gc

p

sc pt.in L.H. of strip with N = 1.

where p c ps j is point in

L.H. of primary.

In general, any point sc in L.H. of any compl. strip can be

represented by

c p ss s jN

where 1, 2,........N & spis a point in primary strip.

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Hence, in z-domain,

2

p s pc s

p p

s jN ss jN

c

s sj N

p

z e e e e

e e e z

Inference:

Correspondence between z-plane & s-plane is not

unique.

A point in z-plane corresponds to infinite no. of points in

s-plane, although a point in s-plane corresponds to a

single point in z-plane.

So L.H. of every complementary strip is mapped into unit circle

in z-plane.

Aliasing of complementary strips into primary strip .

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j

N = 2

N = 1

Z

Trnasform

Primary

Strip (N=0)

N = - 1

N = - 2

S-Plane

Z-Plane

jv

u

= 0, +

s, + 2

s------

Entire L.H. of s-plane is mapped into interior of unit circle in z-

plane, & entire R.H. of s-plane is mapped into exterior of unit

circle in z-plane.

j axis in s-plane maps into circumference of unit circle in z-

plane.

THE REGION OF CONVERGENCE

ROC) OF Z-TRANSFORM

Consider bilateral Z-transform

nn nn

X z Z x x z

(1)

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where,jz re and is in radians.

Now,

r z e

n j nnn

X z x r e

(2)

The region of convergence (ROC) of X(z)set of all values of z

for which X z attains a finite value.

Now,

n j n n j n nn n nn n n

X z x r e x r e x r

or, 1

0

n n

n nn n

X z x r x r

or,

1 0

n nn n

n n

xX z x r

r

(3)

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If X(z) converges in some region of z-plane, both

summations in (3) must be finite in this region.

1st sum converges for values of r small enough such that

product sequence , (1 )n

nx r n

is absolutely

summable, i.e.1

n

nn

x r

.

ROC of 1st sum consists of all points within a circle of

radius r < r1.

2nd

sum converges, for values of r large enough such that

product sequence , (0 n )n

n

x

r is absolutely summable, i.e.

0

n

nn

x

r

ROC of 2nd

sum consists of all points outside a circle of radius

2.r r

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Therefore, ROC of X(z) can be represented in general by

an annular region 2 1r r r , i.e. 2 1r z r , which is a

common region where both of above mentioned sums are

< .

If r1> r2, there is no common region of convergence

of the 2 sums & hence X(z) does not exist.

Example:

n

n nx a u , an infinite duration causal sequence

10 0

nn n

n n

X z a z az

If 1 1, i.e. ,az z a power series converges to

11

1 az .

11

; ROC :1

X z z aaz

ROC is exterior of a circle of radius |a|.

If, a = 1,

1:;1

1][)(

1

zROC

zuzX n

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jv

ROC

ua

Example:

1

n

n nx a u , anti-causal infinite duration sequence.

1

1

1

mn n

n m

X z a z a z

where, m =n .

Now,2 3 ..........

1

CC C C

C

where, |C| < 1.

1

1 1

1,

1 1

a zX z

a z az

if,1 1, i.e.a z z a

That is, 1 1

1 ; ROC :

1

n

nZ a u z aaz

.

ROC is interior of a circle of radius |a| & centred at origin in

z-plane.

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Observation:

Closed form expressions of Z-transforms of

sequences anun and -anu-n-1 are identical but

their ROCs are different.

Above examples reveal 2 important facts:-

1stuniqueness of Z-transform.

2nd

non-uniqueness of inverse Z-transform in absence of

citation of the ROC.

Thus a uniformly sampled signal xn is uniquely determined

by its Z-transform X(z) and the ROC of X(z).

Also seenROC of Z-tr of a causal signal is exterior

of a circle centred at origin & that of anti-causal

sequence is interior of a circle centred at origin in z-

plane.

Example:

1

n n

n n nx a u b u ; two sided infinite duration sequence.

1

0

1 1

0 1

n n n n

n n

n m

n m

X z a z b z

az b z

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The first sum converges if z a . The second sum converges if

z b .CaseI:

If b a , the two ROC do not overlap. Hence X(z) doesnot exist.

Example :

1 1,3 2

b a

CaseII:

If b a , there is a ring in z-plane with inner radius |a| and

outer radius |b| where both power series converge simultaneously.

Thus,

1

1 1 1 2

1 1; ROC:

1 1 1

a b zX z a z b

az bz a b z abz

This example shows that if there is a ROC for an infinite

duration two-sided time series, it is an annular region

centred at origin in the z-plane.

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jv

ub

a

X(z) does not exist

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4 3 29 5 3 2 1X z z z z z ;ROC: entire z-plane except

z = . [r1=]

NOTE :

1.

A rational Z-transform always has at least one pole

located on any boundary separating its regions of

convergence and divergence.

2.

There is never a pole located inside the ROC.

INVERSE Z-TRANSFORM

If ,nZ x X z then,

1nx Z X z

xn is inverse Z-tr of X(z)

POWER SERIES EXPANSION

LONG DIVISION (EXAMPLE- 1)

)(ZFind

1:;)2.0)(1(

)(

1-

2

zX

zROCzz

zzX

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2

2 1 2

1

1.2 0.2 1 1.2 0.2

zX z

z z z z

1

n

( )

ROC is z x1; is causal

nx Z X z

We seek power series expansion in -ve powers of z.

1 2

11 1

11

1 1.2 0.2

1 1 11.25

1 0.2 0.8

oz z

z z

x Lim X z Limz z

zx Lim X z Lim

z z

The long division is illustrated below:

1 2 3

1 2

1 2

1 2

1 1.2 1.24 1.248 .......

1 1.2 0.2 1

1 1.2 0.2

1.2 0.2

1.

z z z

z z

z z

z z

3

1 2 3

2 3

-2 3 4

3 4

1.248 ...................................

2 1.44 0.24

1.24 0.24

1.24z 1.488 0.2481.248 0.248

z

z z z

z z

z zz z

1 2 31 1.2 1.24 1.248 .............X z z z z

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But, 1 2 3

1 2 3 ..............oX z x x z x z x z

1

2

3

1

1.2

1.24

1.248

ox

x

x

x

div. can be contd. to give as many terms as desired.

Disadvantage not yielding xnin closed form.

Example 2.

If 2

2 1 2

1

1.2 0.2 1 1.2 0.2

zX z

z z z z

; ROC : |z| < 0.2,

It is clear thatxnis anti-causal .

So power series expansion in +ve powers of z reqd.

2 3 4 5

2 1

2

2

5 30 155 780 .......

0.2 1.2 1 1

1 6 5

6 5

z z z z

z z

z z

z z

2 3

2 3

2 3 4

3 4

3

6 36 30

31 30

31 186 155

156 155

156

z z z

z z

z z z

z z

z

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2 3 4 5

1 2 3 4 5

5 30 155 780 ..............

0, 5, 30, 155, 780

..........780,155,30,5,0,0n

X z z z z z

x x x x x

x

INVERSE TRANSFORM OF NON-RATIONAL FUNCTION

Example

1log 1 ;X z az z a Using power series expansion for log (1+x) with |x|< 1,

1

1

1log 1

n n

n

xx

n

Here,

1

x az

1

1

1

1

1

1

1

1for 1

0 for 01

or,

n n n

n

n

n

n

n

n

n n

a zX z

n

x Z X z a nn

n

x a un

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Alternatively

1

2

1

log 1 ;

1

X z az z a

dX z az

dz az

Let, 1

nx Z X z

From the differentiation property,

1

1

1

1

1

Z1

1

n

n

n

n

n n

dX z azn x z

dz az

aZ a a u

az

Z nx Z a a u

1

1

1

1

[ ]

1or,

n

n n

n

n

n n

Z nx a a u

x a un

Problem:

sin , ROC includes 1X z z z .

Expanding X(z) in a Taylor series about z = 0,

22

20

0 0 0

3 5 2 1

0

...... .....2! !

..... 13! 5! 2 1 !

KK

Kz

z z z

nn

n

dX z d X z d X z z zX z X z zdz dz K dz

z z zz

n

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11 , 1, 3, 5,.........

2 1 !

n

nn

n

n

X z x z

x nn

PARTIAL FRACTION METHOD

1

11

1

ln

n

n

at an

n a

t n

n

x t X s X zx

zu t u

s zz

e u t e us a z e

za u t a u

s a z a

Examination of 1st -order functions in table for Z-

transforms shows that a factor z is required in the numerator

of these terms.

Hence we expand

X z

z into partial fractions & then

multiply by z to obtain expansion of X(z).

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Problem:

2 1z

X zz z

is Z-transform of causal sequence xn.

Determine xn.

Solution:

2

2

3 3

1

1

1 3 1 3

1 2 2 2 2

j j

X z

z z z

z z z j z j

z e z e

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1 2

3 3

13 3

23 3

1 1

2sin3

1 1

2sin3

j j

j j

j j

X z K K

z z e z e

Ke e j

K

e e j

3 3

3 3

1

3 3

1 1

2sin 2sin3 3

1 1

3 3( ) ( )

1

3

2 sin

33

j j

j j

n

jn jn

n

n

z zX z

z e z ej j

z z

j jz e z e

x Z X z

e e uj

nu

Problem:

3z:ROC;34

)(2

23

zz

zzzX

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Solution:

2

1 3

X z z z

z z z

Converting above improper rational function M N into sum

of a constant & a proper rational function,

1 2

5 31

1 3

5 3

1 3 1 3

X z z

z z z

K Kz

z z z z

11

5 31 1

1 3z

zK z

z z

2 3

5 3

3 61 3z

z

K z z z

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1 1

1

1 61

1 3

1 6or,1 1 3

6 3 n

n n n n

X z

z z z

X z zz z

x u u

CONTOUR INTEGRATION METHOD

1 11

2

n

n

nC

X z Z x

x Z X z X z z dzj

where C is a closed contour within ROC of X(z) that encircles

origin in z-plane in a counter clockwise direction.

According to Cauchys residue theorem,

1 11 Residues of at the poles inside C2

n nn

C

x X z z dz X z zj

For a simple pole z = zo ,residue at

1 iso

n

o o z zz z z z X z z

For a pole of multiplicity m > 1 at oz z ,

Res.

11

1

1( )

1 !o

mm n

om

z z

dz z X z z

m dz

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Proof:

n

nn

X z x z

Multiplying both sides by Zk-1 and integrating w.r.t. Z about

closed contour in the ROC of X(z),

1 1

1

k n k

nnC C

k n

nn C

X z z dz x z dz

x z dz

Now,

1 2k n knC

z dz j

By Cauchy

1 if

0 if

kn

k n

k n

Hence

1 2

2

K

n knnC

k

X z z dz j x

jx

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1

1

1

2

12

K

KC

n

nC

x X z z dzj

x X z z dzj

C is a contour in the ROC of X(z) traversed about origin in the

counter clockwise direction/

DISCRETE-TIME LTI SYSTEMS

Discrete-time

System

x*(t)

input

y*(t)

output

Discrete-time

System

xn

yn

input output

1.

Discrete-Time Differentiator

Differentiator x t

dx ty t

dt

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Derivative of C.T. signal x(t) :

dx t

y tdt

(1)

( )( ) slope of x(t) at t=t

t t

dy ty t

dt

Discrete-time

Differentiator

y(n)x(t) x(n)

t n

dx ty n

dt

n=0,1,2,..

y n =slope of x(t) at t n ; n = 0,1,2,

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Actual slope

Approximate

slope

t

x(t)

x n

x n

1n n

x n x n

0

2

slope approx. by backward difference relation.

dx(t) x(n )- x(n - ) ( ) ( 2)

dt t ny n

smaller, approximation better.

1 1

y n x n x n (3)

or, 1 1n o n ny a x a x (3)

where 11 1

, andoa a

D.T differentiation algorithm can be pictorially represented

as

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+

xn

ao

a1

yn

xn-1

Unit

delay

Taking Z-transforms , transfer fn of D.T. differentiator is

1

1

11 1

o

Y zG z a a Z

X z

z

(4)

D.T. Integrator

Integral of C.T. fn x(t) over interval 0 to t is:

t

o

y t x t dt (1)

Value of y(t) = area under x(t) over 0 tt.

Integratorx(t) y(t)

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Discrete-time

Integrator

y(n)x(t) x(n)

Then,

n

o

y n x t dt

(3)

or,

n n

o ny n x t dt x t dt

or

n

n

y n y n x t dt

(4)

2nd

term of R.H.S. of relation (4) = area under x(t) over

n t n .

area shaded rectangleshown below.

Smaller better approx..

x n

x n

x(t)

n x n t'

0

2

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Hence,

( ) dt x(n ) (5)

n

nx t

Then, equation (4) becomes,

y n y n x n

or, 1n n ny y x (6)

or, 1n o n ny a x y (6a)

where, oa

Equation (6a) gives the 1storder difference equation for D.T.

integrator using rectangular integration.

Algo can be pictorially represented as:

+

xn

ao

yn

yn-1

unit delay

Taking Z-transforms of both sides,

)(z)()( -1 zYzXazY o (5)

Z-transfer-function of the D.T. integrator is

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1 1

Y(z) z( )

X(z) 1 1 z-1o

RI

aG z

z z

(8)

X(Z) Y(Z)

1

zG z

z

j

S-plane

True integrator

jv

Z-plane

1

u

Approximate discrete-time

integrator

For more accurate integration, n

nx t dt

approx. by area of the shaded trapezium.

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Hence,

(9)

From equation (4) and (9),

2

y n y n x n x n

or, 1 12 2

n n n ny x x y

or, 1 1 1n o n n ny a x a x y (10)

where, 1,2 2oa a

Eq. (10) 1st order difference equation representing algo for

D.T. integrator based on trapezoidal integration.

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+ +

xn

ao y

n

a1x

n-1 yn-1

Taking Z-transforms :

1 1

0 1( ) ( ) ( ) ( )Y z a X z a z X z z Y z

The Z-transfer function is

1 1

1

1

1

1 2 1

1

2 1

oTI

Y z a a z zG z

X z b z

z

z

(11)

X(Z) Y(Z)1

2 1

z

z

j

S-plane

jv

Z-plane

1

u

-1

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1

1

1 1

2 1

2 1 2 1 Tustin Transformation relation1 1

z

s z

z zsz z

Observations:

DTLTISystem

S(t) S(n)or

Sn

Z(n)or

Zn

* If

t n

ds tz n

dt

1 11 1

1 11

1 1,

n o n n

o

z a s a s

a a

* If

2

2

t n

d s tz ndt

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2

t n t n

ds t ds t

dt dt z n

x n x n x n x n

2 12 1 22 2

2 22 122 2

1 2,

n o n n n

o

z a s a s a s

a a a

* Similarly, if

3

3

t n

d s tz n

dt

3 13 1 23 2 33 3n o n n n nz a s a s a s a s

* Hence, if

i

i

t n

d s tz n

dt

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1 1 2 2 ........n oi n i n i n ii n iz a s a s a s a s

Representing a discrete-time LTI system by a constant-

coefficient difference equations:

Continuous-time

LTI System

x(t) y(t)

The most basic mathematical model of a stable causal

continuous-time LTI system is the differential equation

representation, wherein i/p x(t) & o/p y(t) are related by the

linear differential equation.

For BIBO stability

0 0

;

i iN M

i ii ii i

d y t d x t a b M N

dt dt

(1)

{ ai} & {bi} are constants.

N = order of the system.

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Discrete-time

LTI System

x(t) xn yn

Causal

Signal

0 0

i iN M

i i ii i

t n t n

d y t d x t a

dt dt

Hence i/p & o/p of a causal DTLTI system can be related

by constant coeff. linear difference equation.

1 10 0

N M

i n i ni i

y x

(2)

If N 0

Equation (2) can be expressed as

10 1

1 M Nn i n i i n

i io

y x y

Directly expresses o/p at instant n in terms of previous

values of i/p & o/p.

Taking Z-transform of both sides of eq. (2) ,

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0 0

N Mi i

i ii i

z Y z z X z

Z- T.F. of the system is :

00

0 1

1

MM ii ii

i oi

N Ni iii

i io

zzY zG z

X z z z

(3)

Properties of Discrete-time Convolution and

Interconnection of LTI systems

gn

n

X Z

x

n

Y Z

y

G Z

1

ng z G Z

Y Z G Z X Z

O/p ynof a DTLTI system excited by i/p sequence xn,

can be expressed as convolution of xnand gn, where gnis

the weighting sequenceof the system.

n n n m n mm

y x g x g

(1)

Also, n n n m n mm

y g x g x

(2)

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If n nx

n n n m n mm

y g g

Now 1 forn m n m

and 0 for n m

n ny g Hence gnis the impulse response of the system.

Following are the properties of convolution:-

1.

Commutative Property:-

n n n nx g g x (3)

i.e.0 0

m n m m n mm m

x g g x

(4)

gn

xn

xn

yn gn

yn

In the Z-domain,

G(Z) X(Z)X(Z) Y(Z) G(Z) Y(Z)

2.

Associative Law:

nnnnn ggggx 21n21 x (5)

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1 1

1 2 1 2

&

n n n

n n n n n n

x g y

y g x g g y

From a physical point of view, xn may be interpreted asthe input signal to an LTI system with impulse sequence

g1n . The output of this system denoted as y1n , becomes

the input to a second LTI system with impulse sequence

g2n. Then the output is

1 2 1 2n n n n n ny y g x g g

which is the left hand side of equation (5).

Thus, the LHS of equation (5) corresponds to having two

LTI systems in cascade. RHS of equation (5) indicates that

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xn is applied to an equivalent system having an impulse

response 1 2n n ng g g

and n n ny x g

From commutative law we also have

nn gg 12ng

As generalization of associative law, if there are L

DTLTI systems in cascade, with impulse response

sequence g1n , g2n ,.., gln, then equivalent system will

have an impulse-response

1 2 ..........n n n Lng g g g

g1n

g2n

gLn

ynxnynxn

1 2,.......,

n n n Lng g g g

In the Z-domain

G1

(Z) G2

(Z) G3

(Z) GL

(Z)X(Z)

Y(Z)G(Z) = G

1(Z) G

2(Z)

GL(Z)

X(Z) Y(Z)

3.

Distributive to Property:

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G1(Z)

GL(Z)

G2(Z)

X(Z)Y(Z) X(Z) G(Z) = G

1(Z) + G

2(Z)

+ + GL(Z)

Y(Z)

CONCEPT OF CAUSALITY OF DTLTI SYSTEMS

A system is said to be causal if the output ynat any

instant ndepends on present and past inputs xn, xn-1, xn-2,

etc. and does not depend on future inputs xn+1, xn+2etc.

If a system does not satisfy this condition, it is non-

causal.

Examples of non-causal systems

2

3

2

2 2

# 2 1.5

#

#

#

n n n

n n

nn

n n

y x x

y x

y x

y x y x

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From the definition of causality it is clear that for a causal system,

the impulse response sequencegnshould be a causal sequence.

Constraint on Transfer Function

Example:

3 2

2

1 2

1 2

1 2 1 2

1 1 2 1 2

2 1

2 4 1

2

2 4

0.5 0.5 0.5 2 0.5

0.5 0.5 0.5 2 0.5n n n n n n n

Y zz z zG z

z z X z

Y zz z zG z

z z X z

Y z zX z X z z X z z X z z Y z z Y z

y x x x x y y

So system is non-causal.

So, order of numerator polynomial in z Order of denominator

polynomial in z for causality.

STABILITY OF DTLTI SYSTEMS

An LTI system is BIBO stable if and only if its output

sequenceynis bounded for every bounded inputxn.

Ifxnis bounded there exists a constant Bxsuch that

for alln xx B n

Similarly ifynis bounded, there exists a constantBysuch that,

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for alln yy B n Constraint on impulse response

The response of a DTLTI system (with weighting sequence gn) to

a bounded inputxn, is expressed as

; 0, 1, 2 etc.

n m n mm

n m n m m n mm m

n m x

n x mm

y g x n

y g x g x

x B

y B g

Forynto be bounded, we should have

or,

mm

nn

g

g

Thus the DTLTI system is BIBO stable if its impulse

response is absolutely summable.

Constraint on transfer function

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n

n nn

n nn nn n

G z Z g g z

G z g z g z

when evaluated on the unit circle (i.e. 1z )

nn

G z g

If the system is BIBO stablen

n

g

should be finite.

Hence, G z evaluated on the unit circle, should be finite.

G z can only be finite in the ROC of G(Z).

A DTLTI system is BIBO stable if and only if the ROC of thesystem transfer function includes the unit circle centred at origin in

the Z-plane.

The above is valid for both causal and non-causal systems.

For a causal system, however, the condition for stability can be

narrowed to a certain extent.

For a causal system ROC of G(z) is |z| > r

For a stable system ROC of G(z) must include the

unit circle.

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For a causal and stable system the ROC of the

system function is |z| > r < 1.

Since the ROC can not contain any pole of G(z), it can be seen that

a causal DTLTI system is BIBO stable if and only if all the poles

of G(z) are inside the unit circle.

Example: A DTLTI system has transfer function

11

3 4

1 1 21

4

G zz

z

Determine the ROC of G(z) and findgnfor

i)

a stable system.

ii)

a causal system.

iii)

Purely anti-causal system.

Solution:Poles of G(z) are at1

4z and z = 2

i)

ROC must include unit circle and poles can not be within

ROC. So ROC is1

2.

4

z

1

is non-causal

1 3 4 2

4

n

nn

n n n

g

g u u

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ii)

For causality ofgn, ROC of G(z) is |z| > 2

13 4 2

4

n n

n n ng u u

Since ROC does not include unit circle,

G(z) is unstable.

iii)

gnis purely anti-causal. Hence ROC of G(z) is1

4z

1 11

3 4 2

4

nn

n n ng u u

Since ROC does not include unit circle,

G(z) is unstable.

Systems with Finite-Duration and

Infinite-Duration Impulse Response:

It is convenient to subdivide the class of discrete-time LTI

systems into two types, those that have a finite-duration

impulse response (FIR) and those that have an infinite-

duration impulse response (IIR).

An FIRsystem has an impulse response that is zero outside

some finite interval of time. For a causal FIR system,

0 for 0 andng n n K

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The convolution formula for such a system is

1

0 0

K

n m n m m n m m n m

m m m K

y g x g x g x

or,

1

0

K

n m n mm

y g x

It is worth noting that that output of the system at any instant n,

is simply a weighted linear combination of input signal samples

xn, xn-1 ,., xn-K+1.The system acts as a window that views

only the most recent K input signal samples and sums the K

products.It neglects all prior input samples i.e., xn-K, xn-K-1, ..

Thus it is said that an FIR system has a finite memory of length

of K samples.

A IIR linear time-invariant system has an infinite-duration

impulse response. Output of a causal IIR system, based on the

convolution formula, is

0n m n m

m

y g x

Here, the system output is a weighted linear combination of

input signal samples xn , xn-1 , xn-2 , ., xo . Since thisweighted sum involves the present and all the past input

samples,it is said that the system has an infinite memory.

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The convolution summation expression suggests a means for

ready implementation of FIR systems involving additions,

multiplications, and a finite number of memory locations. If thesystem is IIR, its practical realization using convolution is

clearly impossible, since it requires an infinite number of

memory locations, multiplications and additions. A question

that naturally arises, then, is whether or not it is possible to

realize IIR systems other than in the form suggested by the

convolution summation. Fortunately there is a practical and

computationally efficient means for implementing a family of

IIR systems.

This family of IIR systems are more conveniently described

by difference equations.This subclass of IIR systems is veryuseful in a variety of practical applications, including the

implementation of digital filters.

RECURSIVE AND NON-RECURSIVE SYSTEMS:

Definitions: A system whose output ynat instant n depends on

any number of past output values yn-1 , yn-2 ,.. is called a

recursivesystem.

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A system whose output ynat any instant n depends only on the

present and past inputs xn , xn-1, ., but not on any past

output value, is called a non-recursive system.Discussions: Let us consider a discrete-time system

represented by a linear constant coefficient difference equation

0 0

N M

i n i i n ii i

y x

(1)

If N 0,On rearranging the terms, we have

0 0

1 M Nn i n i i n i

i io

y x y

(2)

If N 0,This equation directly expresses the output at time n in terms of

previous values of input and output. It can be immediately

seen that in order to calculate yn, we need to know yn-1, yn-2

,, yn-N. An equation of this form is called a recursive

equation, since it specifies a recursive procedure for

determining the output in terms of the input and previous

outputs. The system is hence, called a recursive system.

In the special case, when, in equation (1), N is zero, the

equation reduces to

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0

Mi

n n ii

oy x

(3)

In this case, ynis a function of the present and previous values

of the input. For this reason, equation (3) is called a non-

recursive equation, since we do not recursively use previously

computed values of the output to compute the present value of

the output. The system represented by the equation is known as

non-recursive system.The impulse response of the system is

given by

0

Mi

n n ii

o

g

will be 0 forn i n i

which is obtained as

, for 0

0 , otherwise

for , can never be equal to

n

on

n Mg

n M i n

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Note that the impulse response has finite duration, i.e. it is non-

zero only over a finite time interval. Because of this property

the system is an FIR system.

The basic differences between non-recursive and

recursive discrete-time systems are illustrated below.

xn

yn 1, ,.....,n n n M F x x x

Non-recursive System

yn

xn

1 ,

1

,....,

, ,.....,

n n N

n n n M

F y y

x x x

Recursive System

delay (s)

The fundamental difference between the two systems is the

presence of feedback loop in the recursive system, which feed

back the output sample(s) of the system through a delay

element. The presence of this delay element is crucial for the

realizability of the system, since its absence would force the

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system to compute yn in terms of yn, which is not possible for

discrete-time systems.

There is another important difference between recursive andnon-recursive systems. Suppose we wish to compute the output

ynoof a system when it is excited by an input applied at time n =

0. if the system is recursive, to compute yno, all the previous

values yo , y1 ,., yno-1 should be computed first. If the

system is non-recursive, we can compute yno immediately

without computing yno-1, yno-2, It implies that the output of

a recursive system should be computed in order [i.e. yo, y1, y2,

.], whereas for a non-recursive system, the output can be

computed in any order [e.g. y100, y25, y2, y150, etc.]. This feature

is desirable in some practical applications.

Impulse Response of Recursive Systems:

The general form of difference equation for recursive systems is

repeated below

0 1

M N

n i n i i n i

i i

y b x a y

(4)

By substituting xn= n , the impulse response of the system is

obtained as

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1

N

n n i n ii

g b a g

(5)

No simple interpretation for this result can be given.

However it can be noted that the impulse response can remain

non-zero for even large indices. The recursive portion

continues to generate an output long after the bns

are zero. Therefore recursive discrete-time systems have

infinite impulse response. The point can be illustrated by

considering the following simple causal recursive system.

1 1n o n ny b x a y (6)The response of the filter to a unit sample sequence is obtained

by substituting xn= n and is given by

1 1

2

2 1

3

3 1

1 1

o o

o

o

o

n n

n o o n

g b

g a b

g a b

g a b

g a b a b u

The response has clearly infinite duration.

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gn

bo

a1

>1

a1= 1

0

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1

1 *

*

1 over the range to 2

1

2

1

2

s

s

Wj t

Ws

Wjn j t

nWs

x t X

X W W W f

f

X e df

x n e e df

F

F

where1

sf = sampling period.

Inter-changing the order of integration and summation and

considering that 2 m sW f f

sin sin sin

t n

W t n W t t W t t t n dt

W t n W t t W t t

1

2

2

sin

Wj t n

n Ws

jw t n jw t n

n

n

x t x n e df

x n e e

W j t n

W t nx n

W n

(1)

This gives the ideal interpolation formula for reconstructing x(t)

from x*(t).

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Therefore, in equation (1),

*

*

sin( )

n

i

i

W t tx t x n t n dtW t t

x t g t t dt

x t g t

where sin

sin sin 2s

i

tWt Wt

g tWt t t

is the impulse

response function of the ideal-interpolator or ideal discrete-time

to continuous-time converter or ideal digital-to-analog

converter.The frequency response function of the ideal-interpolator is

sfor

2 2

0 otherwise.

s

i iG j g t

Hence the ideal interpolator is nothing but an ideal low pass filter

with pass band of2

s and pass band gain of .

Hence the ideal interpolation formula may be used to

reconstruct x(t) from x*(t) if fs> 2fm.

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There are however two important problems associated with the

ideal interpolation formula.

Firstly it requires infinite numbers of additions and

multiplications. This is not feasible in practice.

.

Secondly, the computation can not be carried out in real time.

By this we mean that the computation of x(to) requires not only

x(n) for ntobut also x(n) for n> to.

Hence we can start to compute x(t) only after all x(n) (n = 0,

1, 2, .) are known.

Because of these two reasons, equation (1) is not very useful

in the reconstruction of x(t) from x*(t).

PRACTICAL DAC

In practice a D/A converter employs a zero order hold (ZOH) for

reconstructing the analog signal from its sampled version.

The function of the DAC is to decode the input digital words

(binary numbers) and then convert the digital signal into an analog

signal. Hence a DAC can be represented by a decoder followed by

a ZOH.

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Decoder ZOHx*(t) xr(t)

DAC

If we consider a DAC as a system, the decoder does not affect the

system transfer function. So we can only consider the ZOH.

ZOHx*(t)

xr(t)

A zero-order hold (ZOH) holds a sample value constant

until receiving the next sample value.

xr(t)

Samples

0 t

There are discontinuities in the analog signal constructed from the

digital signal by a ZOH. This implies the presence of high

frequency components.

The output staircase waveform of the ZOH can be expressed

as

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0, 1, 2, 3, ..............

r

n

x t x n u t n u t n

n

Taking the Laplace Transform of both sides,

1

1

n sn s

rn

sn s

n

e eX s x n

se

x n es

Now,

* *( ) ( ) [ ( )]n s

n

x n e X s L x t

-s

ho 1-e G ( )s

s

[A]

is the transfer function of the zero-order hold.

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Now,

[B]

Thus it is often said that the ZOH introduces a time delay of

approximately half the sampling period.

Frequency response function of Z.O.H

s

e-1)(G

-s

ho

s

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Sin 2 ( )

2

hoG j

[C]

( ) Sin -2 2hoG j

or, ( ) m - 2hoG j [D]

m = 0,1,2,..

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G j

ideal filter iG j

ZOH hoG j

0.637

3 s 2 s s s 2 s 3 s2

s

2

s0

The nature of ( )hoG j reveals that the ZOH allows high

frequency components of x*(t) to pass through.

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Hence the ZOH (i.e. DAC) should be followed by a low pass

analog filter (post-filter) with cut off frequency 2s to remove

these high frequency components.

xr(t)

0 t

ZOHPost

filter

x*(t) xr(t) x

o(t)

xo(t)

0 t

Hence a complete DSP system can be represented as

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Pre-Filter

ADCSignal

ProcessingAlgorithm

DACPostFilter

x(t) xb(t) x*(t) y*(t) yr(t) yo(t)

dsp - sgm

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