ECEN 301Discussion #13 – Phasors1 Imaginary JS-H 1: 16 16 But, exerting all my powers to call upon...

ECEN 301 Discussion #13 – Phasors 1

ImaginaryJS-H 1: 16 16 But, exerting all my powers to call upon God to deliver me

out of the power of this enemy which had seized upon me, and at the very moment when I was ready to sink into despair and abandon myself to destruction—not to an imaginary ruin, but to the power of some actual being from the unseen world, who had such marvelous power as I had never before felt in any being—just at this moment of great alarm, I saw a pillar of light exactly over my head, above the brightness of the sun, which descended gradually until it fell upon me.


Lecture 13 – Network Analysis with Capacitors and Inductors

PhasorsA “real” phasor is NOT the same thing used in Star Trek


Euler’s Identity

• Appendix A reviews complex numbers

Complex exponential (ejθ) is a point on the complex plane

A

jAAAe

je

je

j

j

j

sincos

1sincossincos1

sincos

22

1

1-1

j

-j

Re

Im

sinθ

cosθ

θ

ejθ

Euler’s equation


Phasors

• Rewrite the expression for a general sinusoid signal:

}Re{)cos( )( tjeAtA

Angle (or argument)magnitude

Complex phasor notation for the simplification:

jAeAtA )cos(

NB: The ejwt term is implicit (it is there but not written)

ECEN 301 Discussion #16 – Frequency Response 5

Frequency Domain

Graphing in the frequency domain: helpful in order to understand Phasors

-1.5

0.0

1.5

0.00 2.00 4.00

time

x(t)

0.00

3.50

-600.0 0.0 600.0

w

X(jw)

-ω ω

π π

cos(ω0t)π[δ(ω – ω0) + δ(ω – ω0)]

Time domain Frequency domain

The Electromagnetic Spectrum

Visiblelight

UltravioletXray

Gamma rays InfraredMicrowaves

Radio waveswavelength

0.03

Å

0.3Å

3Å 30Å

300Å

1020

Hz

0.3 m

m

3 m

m

30 m

m

0.3

mm

3 m

m

30 m

m

0.3

m

3 m

30 m

300

m

3 km

30 k

m

……

1019

Hz

1018

Hz

1017

Hz

1016

Hz

1015

Hz

1014

Hz

1013

Hz

1012

Hz

1011

Hz

1010

Hz

109 H

z

108 H

z

107 H

z

106 H

z

105 H

z

104 H

z

300

km10

3 Hz

3000

km

102 H

z


Phasors

1. Any sinusoidal signal can be represented by either:• Time-domain form: v(t) = Acos(ωt+θ)

• Frequency-domain form: V(jω) = Aejθ = A θ

2. Phasor: a complex number expressed in polar form consisting of:

• Magnitude (A)

• Phase angle (θ)

3. Phasors do not explicitly include the sinusoidal frequency (ω) but this information is still important


Phasors• Example 1: compute the phasor voltage for the equivalent voltage vs(t)

• v1(t) = 15cos(377t+π/4)• v2(t) = 15cos(377t+π/12)

v1(t)+–~

v2(t)+–~

vs(t)+–~



• v1(t) = 15cos(377t+π/4)• v2(t) = 15cos(377t+π/12)

v1(t)+–~

v2(t)+–~

vs(t)+–~

1. Write voltages in phasor notation

V

ejV

V

ejV

j

j

1215

15)(

415

15)(

12/2

4/1



• v1(t) = 15cos(377t+π/4)• v2(t) = 15cos(377t+π/12)

Vj

jjV

VjV

88.349.14

12sin15

12cos15)(

:rrectangula Convert to12

15)(

2

2

v1(t)+–~

v2(t)+–~

vs(t)+–~

1. Write voltages in phasor notation2. Convert phasor voltages from polar to

rectangular form (see Appendix A)

Vj

jjV

VjV

61.1061.10

4sin15

4cos15)(

:rrectangula Convert to4

15)(

1

1



• v1(t) = 15cos(377t+π/4)• v2(t) = 15cos(377t+π/12)

v1(t)+–~

v2(t)+–~

vs(t)+–~


rectangular form (see Appendix A)3. Combine voltages

49.1410.25

)()()( 21

j

jVjVjVS



• v1(t) = 15cos(377t+π/4)• v2(t) = 15cos(377t+π/12)

v1(t)+–~

v2(t)+–~

vs(t)+–~


rectangular form (see Appendix A)3. Combine voltages4. Convert rectangular back to polar

698.28)(

6

10.25

49.14tan

98.28

(14.49)(25.10) r

:polar toConvert

49.1410.25)(

1

22

jV

jjV

S

S



• v1(t) = 15cos(377t+π/4)• v2(t) = 15cos(377t+π/12)

v1(t)+–~

v2(t)+–~

vs(t)+–~


rectangular form (see Appendix A)3. Combine voltages4. Convert rectangular back to polar5. Convert from phasor to time domain

6377cos98.28)(

698.28)(

ttv

jV

S

S

Bring ωt back

NB: the answer is NOT simply the addition of the amplitudes of v1(t) and v2(t) (i.e. 15 + 15), and the addition of their phases (i.e. π/4 + π/12)



• v1(t) = 15cos(377t+π/4)• v2(t) = 15cos(377t+π/12)

v1(t)+–~

v2(t)+–~

vs(t)+–~

6377cos98.28)(

698.28)(

ttv

jV

S

S

Re

Im

14.49 π/6

25.10

Vs(jω)


Phasors of Different Frequencies

Superposition of AC signals: when signals do not have the same frequency (ω) the ejωt term in the phasors can no longer be implicit

i1(t) Load

+

v

–

I

i2(t)0

20

1

02

01

2211

21

21

)()()(

)()()(

jj

tjjtjj

eAeA

eeAeeA

jIjIjI

tititi

NB: ejωt can no longer be implicit



Superposition of AC signals: when signals do not have the same frequency (ω) solve the circuit separately for each different frequency (ω) – then add the individual results

R1

R2

i1(t) vs(t)+–



+R1

–

+ R2 –

i1(t) vs(t)+–

• Example 2: compute the resistor voltages

• is(t) = 0.5cos[2π(100t)] A

• vs(t) = 20cos[2π(1000t)] V

• R1 = 150Ω, R2 = 50 Ω



+R1

–

+ R2 –

i1(t)

1. Since the sources have different frequencies (ω1 = 2π*100) and (ω2 = 2π*1000) use superposition • first consider the (ω1 = 2π*100) part of

the circuit• When vs(t) = 0 – short circuit

• Example 2: compute the resistor voltages• is(t) = 0.5cos[2π(100t)] A• vs(t) = 20cos[2π(1000t)] V• R1 = 150Ω, R2 = 50 Ω

+R1

–

+ R2 –

i1(t) vs(t)+–


Phasors of Different Frequencies• Example 2: compute the resistor voltages

• is(t) = 0.5cos[2π(100t)] A• vs(t) = 20cos[2π(1000t)] V• R1 = 150Ω, R2 = 50 Ω

+R1|| R2

–

i1(t)


the circuit

075.18

)150()50(

)150)(50(05.0

||)()(

05.0)(

21

21

2121

RR

RRI

RRIjVjV

jI

s

sII

s




5

05

)150()50(

)50(020

)(

020)(

21

22 RR

RVjV

jV

sV

s


the circuit• Next consider the (ω2 = 2π*1000) part

of the circuit+R1

–

+ R2 –

vs(t)+–

015

)150()50(

)150(020

)(

020)(

21

11

RR

RVjV

jV

sV

s

+R1

–

+ R2 –

i1(t) vs(t)+–




1. Since the sources have different frequencies (ω1 = 2π*100) and (ω2 = 2π*1000) use superposition • first consider the (ω1 = 2π*100) part of the

circuit• Next consider the (ω2 = 2π*1000) part of

the circuit• Add the two together

)]1000(2cos[15)]100(2cos[75.18)(

015075.18

)()()(

1

111

tttv

jVjVjV VI

)]1000(2cos[5)]100(2cos[75.18)(

05075.18

)()()(

1

222

tttv

jVjVjV VI

+R1

–

+ R2 –

i1(t) vs(t)+–


ImpedanceImpedance: complex resistance (has no physical significance)

• Allows us to use network analysis methods such as node voltage, mesh current, etc.

• Capacitors and inductors act as frequency-dependent resistors

vs(t)+–~ R

+vR(t)

–

i(t)

vs(t)+–~ C

+vC(t)

–

i(t)

vs(t)+–~ L

+vL(t)

–

i(t)

Vs(jω) +–~

+VZ(jω)

–

I(jω)

Z


Impedance – Resistors

Impedance of a Resistor:• Consider Ohm’s Law in phasor form:

0)(

0)(

R

AjI

AjV

Z

Z

Vs(jω) +

–~

+VZ(jω)

–

I(jω)

Z

)cos(

)()(

)cos()(

tR

AR

tvti

tAtv

s

s

Re

Im

I V

Phasor

Phasor domain

RjI

jVjZ

Z

ZR

)(

)()(

NB: Ohm’s Law works the same in DC and AC


Impedance – Inductors

Impedance of an Inductor:• First consider voltage and current in the time-domain

)sin(

)cos(1

)(1

)(

)()(

)(

tL

A

dAL

dvL

ti

tvdt

tdiLtv

LL

SL

L

vs(t)+–~ L

+vL(t)

–

i(t)

NB: current is shifted 90° from voltage

2cos

)sin()(

)cos()()(

tL

A

tL

Ati

tAtvtv

L

LS


Impedance – Inductors

Impedance of an Inductor:• Now consider voltage and current in the phasor-domain

2)(

0)(

L

AjI

AjV

Z

Z

vs(t)+–~ L

+vL(t)

–

i(t)

2cos

)sin()(

)cos()()(

tL

A

tL

Ati

tAtvtv

L

LS

Vs(jω) +–~

+VZ(jω)

–

I(jω)

Z

Phasor

Phasor

LjjI

jVjZ

Z

ZL

)(

)()(

Phasor domain

Re

Im

I

V

-π/2


Impedance – Capacitors

Impedance of a capacitor:• First consider voltage and current in the time-domain

2cos

)]sin([

)]cos([

)()(

)()(1

)(

tCA

tAC

tAdt

dC

dt

tdvCti

tvdiC

tv

CC

SCC

vs(t)+–~ C

+vC(t)

–

i(t)

Vs(jω) +–~

+VZ(jω)

–

I(jω)

Z

Phasor

Phasor

2)(

0)(

CAjI

AjV

Z

Z


Impedance – Capacitors

Impedance of a capacitor:• Next consider voltage and current in the phasor-domain

2)(

0)(

CAjI

AjV

Z

Z

Vs(jω) +–~

+VZ(jω)

–

I(jω)

Z

Cj

C

jC

jI

jVjZ

Z

ZL

1

2

1

)(

)()(

Phasor domain

Re

Im

I Vπ/2

jej

j 12


Impedance

LjjZL )(RjZR )(

+L–

+C–

Re

Im

-π/2

π/2R

-1/ωC

ωL

ZR

ZC

ZL

Phasor domain

+R–

CjjZC 1

)(

Vs(jω) +–~

+VZ(jω)

–

I(jω)

Z

Impedance of resistors, inductors, and capacitors


Impedance

)()()(

:generalin Impedence

jjXjRjZ

Re

Im

-π/2

π/2R

-1/ωC

ωL

ZR

ZC

ZL

Phasor domainVs(jω) +

–~

+VZ(jω)

–

I(jω)

Z

Impedance of resistors, inductors, and capacitors

AC resistance reactanceNot a phasor but a complex number


ImpedancePractical capacitors: in practice capacitors contain a real component

(represented by a resistive impedance ZR)• At high frequencies or high capacitances

• ideal capacitor acts like a short circuit

• At low frequencies or low capacitances• ideal capacitor acts like an open circuit

+C–

+R–

+C–Re

Im

-π/2

-1/ωC

ZC

Re

Im

-π/2

R

-1/ωC

ZR

ZC

Ideal Capacitor Practical Capacitor

NB: the ratio of ZC to ZR is highly frequency dependent


ImpedancePractical inductors: in practice inductors contain a real component (represented

by a resistive impedance ZR)• At low frequencies or low inductances ZR has a strong influence

• Ideal inductor acts like a short circuit

• At high frequencies or high inductances ZL dominates ZR• Ideal inductor acts like an open circuit• At high frequencies a capacitor is also needed to correctly model a practical inductor

Re

Im

π/2

ωLZL

Re

Im

π/2 R

ωL

ZR

ZL

Ideal Inductor Practical Inductor

NB: the ratio of ZL to ZR is highly frequency dependent

+L–

+R–

+L–


Impedance• Example 3: impedance of a practical capacitor

• Find the impedance• ω = 377 rads/s, C = 1nF, R = 1MΩ

+R–

+C–

+Z–


Impedance• Example 3: impedance of a practical capacitor

• Find the impedance• ω = 377 rads/s, C = 1nF, R = 1MΩ

+R–

+C–

+Z–

)36.0(1036.9

377.01

10

)10)(10)(377(1

10

1

)/1(

)/1(

||

5

6

69

6

j

j

CRj

R

CjR

CjR

ZZZ CR


Impedance

• Example 4: find the equivalent impedance (ZEQ)

• ω = 104 rads/s, C = 10uF, R1 = 100Ω, R2 = 50Ω, L = 10mH

ZEQ

R2 C

R1

L


Impedance



ZEQ

R2 C

R1

L

)37.1(81.9

62.992.1

51

50

)50)(1010)(10(1

50

1

)/1(

)/1(

||

64

2

2

2

2

21

j

j

j

CRj

R

CjR

CjR

ZZZ CREQ


Impedance



723.02.136

38.9092.101

62.992.1)10)(10(100

)37.1(81.924

1

11

j

jj

LjR

ZZZZ EQLREQ

ZEQ

R1

L

ZEQ1

)37.1(81.91EQZNB: at this frequency (ω) the circuit has an inductive impedance (reactance or phase is positive)

ECEN 301Discussion #13 – Phasors1 Imaginary JS-H 1: 16 16 But, exerting all my powers to call upon...

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Transcript of ECEN 301Discussion #13 – Phasors1 Imaginary JS-H 1: 16 16 But, exerting all my powers to call upon...