ELEG 205Fall 2017

Lecture #15

Mark Mirotznik, Ph.D.Professor

The University of DelawareTel: (302)831-4221

Email: mirotzni@ece.udel.edu

Chapter 10: Steady-StateSinusoidal Analysis

)cos()( φω += tAtV

Review of Complex NumbersHow can we convert a complex number from rectangular to polar notation?

22ir AAA +=

)(tan 1

r

i

AA−=φ

Review of Complex NumbersHow can we convert a complex number from polar to rectangular notation?

)sin(

)cos(

φ

φ

AA

AA

i

r

=

=

Review of Complex NumbersAj

ir eAjAAA φ⋅=+=Bj

ir eBjBBB φ⋅=+=Summary

( )BAjeBA

C φφ −⋅=

( )BAjeBAC φφ +⋅=( ) ( )iirr BAjBAC +++=

( ) ( )iirr BAjBAC −+−=

Addition and Subtraction Multiplication and Division

Review of Complex Numbers: ExampleConvert to single complex number in polar notation

485

832π

π

j

j

je

jeC−

+

+−=

Review of Complex Numbers: ExampleConvert to single complex number in polar notation

485

832π

π

j

j

je

jeC−

+

+−=

42185

83)sin(2)cos(2ππ

ππjj

ee

jjC−

+

+−+=

Review of Complex Numbers: ExampleConvert to single complex number in polar notation

42185

83)sin(2)cos(2ππ

ππjj

ee

jjC−

+

+−+=

485

83)0(2)1(2πj

e

jjC+

+−⋅+−⋅=

Review of Complex Numbers: ExampleConvert to single complex number in polar notation

)4

sin(8)4

cos(85

85ππ j

jC++

+−=

485

83)0(2)1(2πj

e

jjC+

+−⋅+−⋅=

Review of Complex Numbers: ExampleConvert to single complex number in polar notation

657.5657.1085

228

2285

85j

j

j

jC++−

=++

+−=

)4

sin(8)4

cos(85

85ππ j

jC++

+−=

Review of Complex Numbers: ExampleConvert to single complex number in polar notation

657.5657.1085j

jC++−

=

−

−

−

⋅+

⋅+=

657.10657.5tan

22

58tan

22

1

1

657.5657.10

85j

j

e

eC

Review of Complex Numbers: ExampleConvert to single complex number in polar notation

−

−

−

⋅+

⋅+=

657.10657.5tan

22

58tan

22

1

1

657.5657.10

85j

j

e

eC

488.0

0122.1

065.12434.9

j

j

eeC⋅⋅

=−

Review of Complex Numbers: ExampleConvert to single complex number in polar notation

5.17819.0 jeC −⋅=

488.0

0122.1

065.12434.9

j

j

eeC⋅⋅

=−

Phasors

)sin()cos( xjxe jx +=Recall:( )[ ]

[ ][ ]

φ

ω

ωφ

φωφω

jm

tj

tjjm

tjmm

eVV

eV

eeVeVtVtV

⋅=

⋅=

⋅⋅=

⋅=+= +

~

~Re

ReRe)cos()(

Where is called a Phasor

Phasors

φjm eVV ⋅=~

What does the Phasor tell us about the sinusoidal voltage or current

Phasors

φjm eVV ⋅=~

What does the Phasor tell us about the sinusoidal voltage or current

For example if a Phasor voltage is ojeV 305~ ⋅=

and the frequency was known as kHzf 5=could you tell me what the voltage waveform was in time?

PhasorsFor example if a Phasor voltage is

ojeV 305~ ⋅=and the frequency was known as kHzf 5=could you tell me what the voltage waveform was in time?

)3050002cos(5)( ottv +⋅= π

ojeV 305~ ⋅= kHzf 5=

Phasor Transform of Sinusoidal Sources

)cos()( φω += tVtV m

Time-Domain Phasor-Domain

φjm eVV ⋅=~

Phasor Transform of Sinusoidal Sources

)cos()( φω += tIti m

Time-Domain Phasor-Domain

φjm eII ⋅=~

Phasor Transform of Sinusoidal Sources: Examples

)3

1000cos(10)( π+= ttV

Time-Domain Phasor-Domain

310~ πjeV ⋅=

Phasor Transform of Sinusoidal Sources: Examples

Time-Domain Phasor-Domain

)40000,10cos(5.1)( otti +=ojeI 405.1~ ⋅=

Phasor Transform of Sinusoidal Sources: Examples

)4

1000sin(7)( π−= ttV

Time-Domain Phasor-Domain

?~ =V

Inverse Phasor Transform

?)( =tV

Time-Domain Phasor-Domain

58~ πjeV ⋅=

kHzf 5=

Inverse Phasor Transform

)5

50002cos(8)( ππ +⋅⋅= ttV

Time-Domain Phasor-Domain

58~ πjeV ⋅=

kHzf 5=

Inverse Phasor Transform

?)( =tV

Time-Domain Phasor-Domain

jV 15~ =MHzf 1=

Inverse Phasor Transform

)2

102cos(15)( 6 ππ +⋅=tV

Time-Domain Phasor-Domain

215

15~

πje

jV

⋅=

=

MHzf 1=

Phasors with Circuit Components

=== tjtjtjjm e

RVeIeeIti I ωωωφ~

Re~ReRe)(

R)(ti tj

tjjm

vm

eV

eeV

tVtvv

ω

ωφ

φω

~Re

Re

)cos()(

=

=

+=

= tjtj eRVeI ωω~

Re~Re

Phasors with Circuit Components

R)(ti tj

tjjm

eV

eeVtv v

ω

ωφ

~Re

Re)(

=

=

= tjtj eRVeI ωω~

Re~Re

= tjtj eRVeI ωω~

Re~Re

Phasors with Circuit Components

R)(ti tj

tjjm

eV

eeVtv v

ω

ωφ

~Re

Re)(

=

=

= tjtj eRVeI ωω~

Re~Re

RVI~~ =

IVR ~~

=

Phasors with Circuit Components

R tjeVtv ω~Re)( =

IVR ~~

=

The ratio of the phasor voltage to phasor current is called the impedance (symbol Z)

tjeIti ω~Re)( =

Phasors with Circuit Components

IVRZ ~~

==

The ratio of the phasor voltage to phasor current is called the impedance (Z)

IVZ ~~

=

In general For a resistor

C

[ ] tjtjtj eCjVeVdtdCeI

dttdvCti

ωωω ω ⋅⋅==

=

~Re~Re~Re

)()(

tjeVtv ω~Re)( = tjeIti ω~Re)( =

Phasors with Circuit Components

C

tjtj

tjtj

eCjVeI

eCjVeIωω

ωω

ω

ω

⋅⋅=

⋅⋅=~Re~Re

~Re~Re

tjeVtv ω~Re)( = tjeIti ω~Re)( =

Phasors with Circuit Components

C

CjVI ω⋅= ~~

tjeVtv ω~Re)( = tjeIti ω~Re)( =

CjIVZ

ω1

~~==

Phasors with Circuit Components

C tjeVtv ω~Re)( =

tjeIti ω~Re)( =

CjIVZ

ω1

~~==I

VZ ~~

=

In general For a capacitor

Phasors with Circuit Components

L tjeVtv ω~Re)( =

tjeIti ω~Re)( =

[ ]

⋅==

=

∫

∫tjtjtj eV

LjdteV

LeI

dttvL

ti

ωωω

ω~1Re~Re1~Re

)(1)(

Phasors with Circuit Components

L tjeVtv ω~Re)( =

tjeIti ω~Re)( =

⋅=

⋅=

tjtj

tjtj

eVLj

eI

eVLj

eI

ωω

ωω

ω

ω

~1Re~Re

~1Re~Re

Phasors with Circuit Components

L tjeVtv ω~Re)( =

tjeIti ω~Re)( =

⋅= tjtj eVLj

eI ωω

ω~1Re~Re

VLj

I ~1~ω

= LjIVZ ω== ~~

Phasors with Circuit Components

L tjeVtv ω~Re)( =

tjeIti ω~Re)( =

LjIVZ ω== ~~

IVZ ~~

=

In general For an inductor

Phasors with Circuit Components

Phasors with Circuit ComponentsSummary

Impedance, Z, Ω

Resistor

Inductor

Capacitor

LjZ ω=

CjZ

ω1

=

RZ =

Phasors with Circuit ComponentsSummary

Impedance,Z, Ω

Admittance,Y, sieman

Resistor

Inductor

Capacitor

LjIVZ ω== ~~

CjIVZ

ω1

~~==

RIVZ == ~~

RVIY 1~~==

LjVIY

ω1

~~==

CjVIY ω== ~~

Combining Impedances

1 mH

0.5 mH

1 µF

10 ΩeqZ

Convert all of the components to their impedance values and determine the equivalent impedance of the entire network

kHzf 10=

1 mH

0.5 mH

1 µF

Z=10 ΩeqZ

Convert all of the components to their impedance values and determine the equivalent impedance of the entire network

kHzf 10=

jjCj

Z 9.15101000,102

116 −=

⋅⋅⋅== −πω

jjLjZ 8.62101000,102 3 =⋅⋅⋅== −πω

jj

LjZ

4.31105.0000,102 3

=⋅⋅⋅=

=−π

ω

Combining Impedances

62.8j Ω

31.4j Ω

-15.9j Ω

10 ΩeqZ

Convert all of the components to their impedance values and determine the equivalent impedance of the entire network

kHzf 10=

jjjjjZeq 8.4908.99.158.62

4.31104.3110

+=−++⋅

=

Combining Impedances

62.8j Ω

31.4j Ω

-15.9j Ω

10 ΩeqZ

Convert all of the components to their impedance values and determine the equivalent impedance of the entire network

kHzf 10=

Ω==+= ⋅⋅ ojjeq eejZ 7.7939.1 6.506.508.4908.9

Combining Impedances

Solving Sinusoidal Steady State Circuits

STEP #1: Transform all sources to the phasor domain using the phasor transform

Solving Sinusoidal Steady State CircuitsSTEP #2: Transform all components (i.e. resistors, inductors and capacitors) into their complex impedance values.

Solving Sinusoidal Steady State Circuits

STEP #3: Using Kirchhoff's laws and Ohm’s law for impedances solve for the unknown phasor quantities.

I~

Solving Sinusoidal Steady State Circuits

STEP #4: Transform phasor results back into the time domain.

tjeIti ω~Re)( =

Solving Sinusoidal Steady State CircuitsExamples

)cos()( tAtv ω= CR

Find vc(t)

)(tvc

Solving Sinusoidal Steady State CircuitsExamples

STEP #1 and #2: Transform all sources to the phasor domain using the phasor transform and transform all components to their impedances

)cos()( tAtv ω= CR

)(tvc

Time-Domain Phasor-Domain

0~ jAeV = Cjω1

R

cV~

Solving Sinusoidal Steady State CircuitsExamples

)cos()( tAtv ω= CR

)(tvc

Time-Domain Phasor-Domain

0~ jAeV = Cjω1

R

cV~

STEP #3: Using Kirchhoff's laws and Ohm’s law for impedances solve for the unknown phasor quantities.

Solve this circuit like you would if it was only just filled with resistors and DC sources.

Solving Sinusoidal Steady State CircuitsExamples

Phasor-Domain

0~ jAeV = Cjω1

R

cV~

STEP #3: Using Kirchhoff's laws and Ohm’s law for impedances solve for the unknown phasor quantities.

CjR

CjVVc

ω

ω1

1~~

+=Voltage divider

Solving Sinusoidal Steady State CircuitsExamples

Phasor-Domain

0~ jAeV = Cjω1

R

cV~

STEP #3: Using Kirchhoff's laws and Ohm’s law for impedances solve for the unknown phasor quantities.

11

1

1~~

+⋅=⋅

+=

CRjA

CjCj

CjR

CjVVc ωωω

ω

ωVoltage divider

Solving Sinusoidal Steady State CircuitsExamples

Phasor-Domain

0~ jAeV = Cjω1

R

cV~

STEP #3: Using Kirchhoff's laws and Ohm’s law for impedances solve for the unknown phasor quantities.

Voltage divider)

1(tan22

0

1

)(11

~RCj

j

c

eRC

eACRjAV ω

ωω −

⋅+

⋅=

+=

Solving Sinusoidal Steady State CircuitsExamples

Phasor-Domain

0~ jAeV = Cjω1

R

cV~

STEP #3: Using Kirchhoff's laws and Ohm’s law for impedances solve for the unknown phasor quantities.

Voltage divider)

1(tan

2

1

)(1~ RCj

c eRC

AVω

ω

−−⋅

+=

Solving Sinusoidal Steady State CircuitsExamples

)cos()( tAtv ω= CR

)(tvc

Time-Domain Phasor-Domain

0~ jAeV = Cjω1

R

cV~

STEP #4: Transform phasor results back into the time domain.

)1

(tan

2

1

)(1~ RCj

c eRC

AVω

ω

−−⋅

+=( )

))(tancos(1

)( 1

2RCt

RC

Atvc ωωω

−−+

=

Solving Sinusoidal Steady State CircuitsExamples

)cos()( tAtv ω= CR

)(tvc

( )))(tancos(

1)( 1

2RCt

RC

Atvc ωωω

−−+

=

Lets look at this solution at different frequencies.

Solving Sinusoidal Steady State CircuitsExamples

)cos()( tAtv ω= CR

)(tvc

( )))(tancos(

1)( 1

2RCt

RC

Atvc ωωω

−−+

=

Lets look at this solution at different frequencies.

1. At DC (ω=0) what does the capacitor voltage look like?

Solving Sinusoidal Steady State CircuitsExamples

)cos()( tAtv ω= CR

)(tvc

( )))(tancos(

1)( 1

2RCt

RC

Atvc ωωω

−−+

=

Lets look at this solution at different frequencies.

1. At DC (ω=0) what does the capacitor voltage look like?

( )ARCt

RC

Atvc =⋅−⋅⋅+

= − ))0(tan0cos(01

)( 1

2 Does this make sense?

Solving Sinusoidal Steady State CircuitsExamples

)cos()( tAtv ω= CR

)(tvc

( )))(tancos(

1)( 1

2RCt

RC

Atvc ωωω

−−+

=

Lets look at this solution at different frequencies.1. At very very high frequencies (ω=infinity) what does the capacitor voltage look like?

Solving Sinusoidal Steady State CircuitsExamples

)cos()( tAtv ω= CR

)(tvc

( )))(tancos(

1)( 1

2RCt

RC

Atvc ωωω

−−+

=

Lets look at this solution at different frequencies.1. At very very high frequencies (ω=infinity) what does the capacitor voltage look like?

( )0))(tancos(

1)( 1

2=⋅∞−⋅∞

⋅∞+= − RCt

RC

Atvc Does this make sense?

Solving Sinusoidal Steady State CircuitsExamples

Lets now plot the magnitude of the capacitor sinusoid as a function of frequency

ω

( )21 RC

A

ω+

A

0 0 RC20

As the frequency increases the amplitude of the output voltage (vc(t)) gets smaller.

RC10

Solving Sinusoidal Steady State CircuitsExamples

Lets now plot the magnitude of the capacitor sinusoid as a function of frequency

ω

( )21 RC

A

ω+

A

0 0 RC20

As the frequency increases the amplitude of the output voltage (vc(t)) gets smaller.

RC10

THIS PLOT IS CALLED A MAGNITUDE FREQUENCY RESPONSE PLOT

Solving Sinusoidal Steady State CircuitsExamples

As the frequency increases the amplitude of the output voltage (vc(t)) gets smaller.

( )))(tancos(

1)( 1

2RCt

RC

Atvc ωωω

−−+

=

Solving Sinusoidal Steady State CircuitsExamples

Lets now plot the phase angle of the capacitor sinusoid as a function of frequency ω

)(tan 1 RCω−−

o90−0 RC20

As the frequency increases the phase of the output voltage (vc(t)) goes from 0 degrees to -90 degrees.

RC10

THIS PLOT IS CALLED THE PHASE FREQUENCY RESPONSE PLOT

o0