ELEG 205 Fall 2017 Lecture #14mirotzni/ELEG205/Lecture15.pdf · Chapter 10: Steady-State....

ELEG 205Fall 2017

Lecture #15

Mark Mirotznik, Ph.D.Professor

The University of DelawareTel: (302)831-4221

Email: [email protected]

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−

+

+−=


485

832π

π

j

j

je

jeC−

+

+−=

42185

83)sin(2)cos(2ππ

ππjj

ee

jjC−

+

+−+=


42185

83)sin(2)cos(2ππ

ππjj

ee

jjC−

+

+−+=

485

83)0(2)1(2πj

e

jjC+

+−⋅+−⋅=


)4

sin(8)4

cos(85

85ππ j

jC++

+−=

485

83)0(2)1(2πj

e

jjC+

+−⋅+−⋅=


657.5657.1085

228

2285

85j

j

j

jC++−

=++

+−=

)4

sin(8)4

cos(85

85ππ j

jC++

+−=


657.5657.1085j

jC++−

=

−

−

−

⋅+

⋅+=

657.10657.5tan

22

58tan

22

1

1

657.5657.10

85j

j

e

eC


−

−

−

⋅+

⋅+=

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⋅⋅

=−


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


φjm eII ⋅=~

Phasor Transform of Sinusoidal Sources: Examples

)3

1000cos(10)( π+= ttV


310~ πjeV ⋅=



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


)4

1000sin(7)( π−= ttV


?~ =V

Inverse Phasor Transform

?)( =tV


58~ πjeV ⋅=

kHzf 5=


)5

50002cos(8)( ππ +⋅⋅= ttV


58~ πjeV ⋅=

kHzf 5=


?)( =tV


jV 15~ =MHzf 1=


)2

102cos(15)( 6 ππ +⋅=tV


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


R)(ti tj

tjjm

eV

eeVtv v

ω

ωφ

~Re

Re)(

=

=

= tjtj eRVeI ωω~

Re~Re

= tjtj eRVeI ωω~

Re~Re


R)(ti tj

tjjm

eV

eeVtv v

ω

ωφ

~Re

Re)(

=

=

= tjtj eRVeI ωω~

Re~Re

RVI~~ =

IVR ~~

=


R tjeVtv ω~Re)( =

IVR ~~

=

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

tjeIti ω~Re)( =


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)( =


C

tjtj

tjtj

eCjVeI

eCjVeIωω

ωω

ω

ω

⋅⋅=

⋅⋅=~Re~Re

~Re~Re



C

CjVI ω⋅= ~~


CjIVZ

ω1

~~==


C tjeVtv ω~Re)( =

tjeIti ω~Re)( =

CjIVZ

ω1

~~==I

VZ ~~

=

In general For a capacitor


L tjeVtv ω~Re)( =

tjeIti ω~Re)( =

[ ]

⋅==

=

∫

∫tjtjtj eV

LjdteV

LeI

dttvL

ti

ωωω

ω~1Re~Re1~Re

)(1)(


L tjeVtv ω~Re)( =

tjeIti ω~Re)( =

⋅=

⋅=

tjtj

tjtj

eVLj

eI

eVLj

eI

ωω

ωω

ω

ω

~1Re~Re

~1Re~Re


L tjeVtv ω~Re)( =

tjeIti ω~Re)( =

⋅= tjtj eVLj

eI ωω

ω~1Re~Re

VLj

I ~1~ω

= LjIVZ ω== ~~


L tjeVtv ω~Re)( =

tjeIti ω~Re)( =

LjIVZ ω== ~~

IVZ ~~

=

In general For an inductor


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


kHzf 10=

jjCj

Z 9.15101000,102

116 −=

⋅⋅⋅== −πω

jjLjZ 8.62101000,102 3 =⋅⋅⋅== −πω

jj

LjZ

4.31105.0000,102 3

=⋅⋅⋅=

=−π

ω


62.8j Ω

31.4j Ω

-15.9j Ω

10 ΩeqZ


kHzf 10=

jjjjjZeq 8.4908.99.158.62

4.31104.3110

+=−++⋅

=


62.8j Ω

31.4j Ω

-15.9j Ω

10 ΩeqZ


kHzf 10=

Ω==+= ⋅⋅ ojjeq eejZ 7.7939.1 6.506.508.4908.9


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.


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

I~


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


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


0~ jAeV = Cjω1

R

cV~


)cos()( tAtv ω= CR

)(tvc


0~ jAeV = Cjω1

R

cV~


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


Phasor-Domain

0~ jAeV = Cjω1

R

cV~


CjR

CjVVc

ω

ω1

1~~

+=Voltage divider


Phasor-Domain

0~ jAeV = Cjω1

R

cV~


11

1

1~~

+⋅=⋅

+=

CRjA

CjCj

CjR

CjVVc ωωω

ω

ωVoltage divider


Phasor-Domain

0~ jAeV = Cjω1

R

cV~


Voltage divider)

1(tan22

0

1

)(11

~RCj

j

c

eRC

eACRjAV ω

ωω −

⋅+

⋅=

+=


Phasor-Domain

0~ jAeV = Cjω1

R

cV~


Voltage divider)

1(tan

2

1

)(1~ RCj

c eRC

AVω

ω

−−⋅

+=


)cos()( tAtv ω= CR

)(tvc


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

−−+

=


)cos()( tAtv ω= CR

)(tvc

( )))(tancos(

1)( 1

2RCt

RC

Atvc ωωω

−−+

=

Lets look at this solution at different frequencies.


)cos()( tAtv ω= CR

)(tvc

( )))(tancos(

1)( 1

2RCt

RC

Atvc ωωω

−−+

=


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


)cos()( tAtv ω= CR

)(tvc

( )))(tancos(

1)( 1

2RCt

RC

Atvc ωωω

−−+

=


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

( )ARCt

RC

Atvc =⋅−⋅⋅+

= − ))0(tan0cos(01

)( 1

2 Does this make sense?


)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?


)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?


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


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

ω

( )21 RC

A

ω+

A

0 0 RC20


RC10

THIS PLOT IS CALLED A MAGNITUDE FREQUENCY RESPONSE PLOT



( )))(tancos(

1)( 1

2RCt

RC

Atvc ωωω

−−+

=


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

ELEG 205 Fall 2017 Lecture #14mirotzni/ELEG205/Lecture15.pdf · Chapter 10: Steady-State....

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