circuits ii: phasors & complex numbersbisonacademy.com/ece111/videos/09 circutis ii.pdf · j50...

Circuits II: Phasors

& Complex NumbersECE 111 Introduction to ECE

Jake Glower - Week #9

Please visit Bison Academy for corresponding

lecture notes, homework sets, and solutions

Numbers make a difference. They can create and end empires.

The number zero is an odd concept.

It isn't necessary: Romans had an extensiveeconomy without the number zero

Addition and multipliation is much easier withthe number zero

Roman Numbers Arabic Numbers

XXVII+ CIX

--------------?

27+ 106

------------?

Negative NumbersWeird concept: negative one apple?

The Dutch invented the double-entrybookkeeping system

Keep track of expenses (debits)

Keep track of assets (credits)

Only invest in profitable ventures

With the double-entry bookkeeping systemthe Dutch were able to challenge England,France, and Spain for world dominance.

"We will bury you"Nikita Khrushchev

Accounting can decide the fate of empires

In the 1960's, the Soviet Union thought theireconomy was growing 20% per year

vs. 4% for the United States

At that rate, by 2000 the Soviet Union would crushthe U.S. ecomomically

Exponential growth is a powerful thing...

Actually, the Soviet economy was shrinking 1-2%per year

Eventually, the size of the goverment (based upon20% growth) could not be supported

Real Numbers & DC Signals1st half of Circuits I

At DC, real numbers canrepresent

Resistors

Voltages

Currents

When working with DC,real numbers are all youneed.

AC Signals2nd half of Circuits I

All of Circuits II

You need 2 terms to represent a sine wave

Frequency is usually known

x(t) = a cos (ωt) + b sin (ωt)

or

x(t) = r cos (ωt + θ)

Real numbers don't work well

Only one degree of freedom

Two real numbers are needed to represent asine wave 0 1 2 3 4 5 6 7 8 9 10

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

Time

cos(t)sin(t)

Complex Numbers

Complex numbers have two terms

Real Part & Complex Part

Allows you to represent cosine() and sine() terms with a single number

Let

j = −1

A complex number can be written as

x = a + jb = c∠θ

a2 + b2 = c2

tan θ =ba

a + jb

b

a

c

real

imag

Complex Number Math

Addition:

The real parts add and the complex parts add.

(a1 + jb1) + (a2 + jb2) = (a1 + a2) + j(b1 + b2)

Subtraction:

The real parts subtract and the complex parts subtract

(a1 + jb1) − (a2 + jb2) = (a1 − a2) + j(b1 − b2)

Multiplication:

Multiply out: results in a complex number

(a1 + jb1)(a2 + jb2) = a1a2 + ja1b2 + ja2b1 + j2b1b2

= (a1a2 − b1b2) + j(a1b2 + a2b1)

Division: Division also results in a complex number but takes even morecomputations:

a1+jb1

a2+jb2

=

a1+jb1

a2+jb2

a2−jb2

a2−jb2

=

(a1a2+b1b2)+j(−a1b2+a2b1)

a22+b2

2

=

a1a2+b1b2

a22+b2

2

+ j

−a1b2+a2b1

a22+b2

2

HP Calculators to the Rescue!Free42 is a free app on your cell phone

HP35s is a $60 from Amazon

Download it

Get familiar with it

Use it

HP calculators are worth 10 points on midtermsfrom my experience

Handout (part 1)

Determine the result of the following operations with complex numbers

2 + j3 2 + j3 2 + j3 2 + j3

+ 9 + j4 - 9 + j4 * 9 + j4 ------------

----------- ---------- ----------- 9 + j4

Phasor Representation of cosine()

Euler's identity states that the complex exponential is

ejωt = cos (ωt) + j sin (ωt)

Real part gives you cosinecos (ωt) = real(ejωt

)

Phasor representation of cosine is 1 + j0

1 ↔ cos (ωt)

Phasor representation of a sine wave:

If you multiply by (a + jb)

(a + jb)ejωt = (a + jb)(cos (ωt) + j sin (ωt))

= (a cos (ωt) − b sin (ωt)) + j(. .. )

Take the real part

a + jb ↔ a cos (ωt) − b sin (ωt)

You can also represent voltages in polar form:

r∠θ ↔ r cos (ωt + θ)

Example: Determine the phasor representation for x(t)

The frequency comes from the period

T = 6.28ms

f =1

T= 159.2Hz = 159.2

cycles

second

ω = 2πf = 1000radsec

The amplitude is 5.000 (polar form)

The delay is 57 degrees

θ = −1ms delay

6.28ms period 2π

θ = −1.0 radian = -57.30

meaning

X = 5∠ − 57.30

x(t) = 5 cos (1000t − 57.30)

0 1 2 3 4 5 6 7 8 9 10

-6

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

Time (ms)

Vp = 5.0V

Delay = 57 deg

= 1 radian

Period = 6.28ms

x(t)

Phasor Representation for Impedance's

Resistors: V = IR

R → R

Inductors: V = LdI

dt

V = Ld

dt(ejωt

) = jωL ejωt = jωL ⋅ I

L → jωL

Capacitors: V =1

C ∫ I dt

V =1

C ∫ (e jωt); dt =

1

jωC e jωt =

1

jωC I

C →1

jωC

Simplification of RLC circuitsSame as resisitor circuits

Now with complex numbers

Example: Determine the impedance Zab.

-j100 + 300 = 300 - j100

(200) (300 − j100) =

1

200+

1

300−j100

−1

= 123.07 − j15.38

(200) + (123.07 − j15.38) = 323.07 − j15.38

(j300) (323.08 − j15.38) = 156.85 + j161.83

(100) + (156.85 + j161.83) = 256.85 + j161.83

Answer:

Zab = 256.85 + j161.83

100 200 300

-j100200j300

a

b

Example: Determine Zab

Solution

(10) + (−j60) = 10 − j60

(10 − j60) (j50) = 125.00 + j175.00

(125.00 + j175.00) + (40) = 165.00 + j175.00

(165.00 + j175.00) (30 − j20) = 31.42 − j14.98

answer:

Zab = 31.42 − j14.98

-j20 30

40

j50

10-j60

a b

Handout

Problem #2

a

b

20 j30 40

50 j60-j70

Circuit Analysis with Phasors

Same as DC

Now with complex numbers

Example: Determine y(t)

V0 = 10 sin (628t)

Step 1: Replace each term with phasor value

V0 → 0 − j10

R → 400

C →1

jωC= −j159Ω

Step 2: Solve ( Voltage division )

Y =

−j159

−j159+400 (0 − j10) = −3.436 − j1.368

y(t) = −3.436 cos(628t) + 1.368 sin(628t)

ValidationUse CircuitLab

Input the circuit

Run a Time Domain simulation

The output (orange line) is easier tosee in polar form

Y = −3.436 − j1.368

= 3.698∠ − 1580

Example 2: 3-Stage RC Circuit

Step 1: Convert to phasor form

V0 → 0 − j10

ω = 100

C1 →1

jωC= −j50Ω

C2 →1

jωC= −j40Ω

C3 →1

jωC= −j33.33Ω

5 10 15

100 150 200

200uF 250uF 300uF+

-10 sin(100t)

V0 V1 V2 V3

-j50 -j40 -j33.33

Step 2: Write N equations for N unknowns

V0 = 0 − j10

V1−V0

5 +

V1

100 +

V1

−j50 +

V1−V2

10 = 0

V2−V1

10 +

V2

150 +

V2

−j40 +

V2−V3

15 = 0

V3−V2

15 +

V3

200 +

V3

−j33.33 = 0

5 10 15

100 150 200

200uF 250uF 300uF+

-10 sin(100t)

V0 V1 V2 V3

-j50 -j40 -j33.33

Step 3: Solve. First, group terms

V0 = −j10

−1

5V0 +

1

5+

1

100+

1

−j50+

1

10V1 +

−1

10V2 = 0

−1

10V1 +

1

10+

1

150+

1

−j40+

1

15V2 +

−1

15V3 = 0

−1

15V2 +

1

15+

1

200+

1

−j33.33V3 = 0

Place in matrix form

1 0 0 0

−1

5

1

5+

1

100+

1

−j50+

1

10

−1

10 0

0

−1

10

1

10+

1

150+

1

−j40+

1

15

−1

15

0 0

−1

15

1

15+

1

200+

1

−j33.33

V0

V1

V2

V3

=

−j10

0

0

0

Put into MATLAB and solve

a1 = [1,0,0,0];

a2 = [-1/5,1/5+1/100+1/(-j*50)+1/10,-1/10,0];

a3 = [0,-1/10,1/10+1/150+1/(-j*40)+1/15,-1/15];

a4 = [0,0,-1/15,1/15+1/200+1/(-j*33.33)];

A = [a1;a2;a3;a4]

1.0000 0 0 0

-0.2000 0.3100 + 0.0200i -0.1000 0 0 -0.1000 0.1733 + 0.0250i -0.0667

0 0 -0.0667 0.0717 + 0.0300i

B = [-j*10;0;0;0];

V = inv(A)*B

V0 0 -10.0000i

V1 -1.6314 - 8.0724i

V2 -3.4430 - 5.3506i

V3 -4.4982 - 3.0942i

V0 0 -10.0000i

V1 -1.6314 - 8.0724i

V2 -3.4430 - 5.3506i

V3 -4.4982 - 3.0942i

meaning

V0 = 10 sin(100t)

V1 = −1.6314 cos (100t) + 8.0742 sin (100t)

V2 = −3.4430 cos (100t) + 5.5306 sin (100t)

V3 = −4.4982 cos (100t) + 3.0942 sin (100t)

CircuitLab Simulation

|Vin| |V1| |V2| |V3|

Calculated 10.0000V 8.2356V 6.3627V 5.4596V

CircuitLab 10.00V 8.231V 6.348V 5.435V

Handout

Problem #3:

Convert to phasors

Write the voltage node equations

20 40

50

0.1H

0.2H

0.01F

8 cos(100t)

V0 V1 V2 V3

+

-