Sinusoidal Steady-state Analysiseestaff.kku.ac.th/~jamebond/182304/chapter2.pdf · Complete response and sinusoidal steady-state response Superposition in the steady state If a linear

Sinusoidal Steady-state Analysis

Complex number reviews

Phasors and ordinary differential equations

Complete response and sinusoidal steady-state response

Concepts of impedance and admittance

Sinusoidal steady-state analysis of simple circuits

Resonance circuit

Power in sinusoidal steady –state

Impedance and frequency normalization

Complex number reviews

1, −=+= jjyxz

yzxz == )Im()Re(

| |j

z z eθ=

Complex number

In polar form

2 2 1/ 2 1| | ( ) tan

yz x y

xθ −= + =

| |z z θ= ∠or | | cos | | sinx z y zθ θ= =

The complex number can be of voltage, current, power, impedance etc..in any circuit with sinusoid excitation.

z

Magnitude

Phase or angle

Operations: Add, subtract, multiply, divide, power, root, conjugate

Phasors and ordinary differential equationsA sinusoid of angular frequency is in the formω

)cos( φω +tAm

Theorem

The algebraic sum of sinusoids of the same frequency and of their derivatives is also a sinusoid of the same frequency

Example 1

( ) 2cos(2 60) 4sin 2 2sin 2

2cos 2 cos 60 2sin 2 sin 60 4sin 2 4cos 2

cos 2 3 sin 2 4sin 2 4cos 2

5cos 2 (4 3)sin 2 7.6cos(2 48.8)

df t t t t

dt

t t t t

t t t t

t t t

= + − +

= − − +

= − − +

= − + = +


jmA A e

φ@

)3/602cos(2110)( ππ += ttv

phasor form

Example 2

/ 3110 2

jA e

π=(120 )

( ) Re(110 2 )j t

v t eπ=

phasor

( )Re( ) Re( )

Re( cos( ) sin( ))

cos( ) ( )

j t j tm m

m m

m

A e A e

A t jA t

A t x t

ω ω φ

ω φ ω φ

ω φ

+=

= + + +

= + =


Ordinary linear differential equation with sinusoid excitation

1

0 1 1.. cos( )........(1)

n n

n mn n

d x d xa a a x A t

dt dtω φ

−

−+ + + = +

Lemma: Re[.. ] is additive and homogenous

)](Re[)](Re[)]()(Re[ 2121 tztztztz +=+

1 1 2 2 1 1 2 2Re[ ( ) ( )] Re[ ( )] Re[ ( )]z t z t z t z tα α α α+ = +

Re( ) Re Re( )j t j t j td d

Ae Ae j Aedt dt

ω ω ωω = =


Application of the phasor to differential equation

,j j

m mA A e X X eφ ψ

@ @Let

substitute ( ) Re( )j t

x t Xeω= in (1) yields

0 Re( ) .. Re( ) Re( )

nj t j t j t

nn

dXe Xe Ae

dt

ω ω ωα α+ + =

0Re( ) .. Re( ) Re( )

nj t j t j t

nn

dXe Xe Ae

dt

ω ω ωα α+ + =

0Re( ( ) ) .. Re( ) Re( )n j t j t j t

nj Xe Xe Aeω ω ωα ω α+ + =


10 1 1Re [ ( ) ( ) .. ( ) ] Re( )

n n j t j tn nj j j Xe Ae

ω ωα ω α ω α ω α−−+ + + + =

12

10 1 1

10 1 1

2 2 3 22 1 3

[ ( ) ( ) .. ( ) ]

[ ( ) ( ) .. ( ) ]

[( .. ) ( ..) ]

n nn n

n nn n

mm

n n n n

j j j X A

AX

j j j

AX

α ω α ω α ω α

α ω α ω α ω α

α α ω α ω α ω

−−

−−

− − −

+ + + + =

=+ + + +

=− + + + − +

even powerodd power

31 1 3

22

..tan

..

n n

n n

α ω α ωψ φ

α α ω− − −

−

− += −

− +


)cos(||)Re()( φωω +== tEEetetj

s

Example 3

From the circuit in fig1 let the input be a sinusoidal voltage source and the output is the voltage across the capacitor.

( ) | | cos( )se t E tω φ= + ( ) | | cos( )c cv t V tω ψ= +

( )se t( )cV t

Fig1


)()()()( tetvtRitidt

dL sc =++

2

2

( ) ( )( ) ( )c c

c s

d v t dv tLC RC v t e t

dtdt+ + =

)cos(||)Re()( ψωω +== tVeVtv ctj

cc

KVL

2[ )( ) ( ) 1] (2)cLC j RC j V Eω ω+ + =

Particular solution


2

2 2 2 1/ 2

1

2

1

| || |

[(1 ) ( ) ]

tan1

c

c

EV

LC j RC

EV

LC RC

RC

LC

ω ω

ω ωω

ψ φω

−

=− +

=− +

= −−

Complete response and sinusoidal steady-state

responseComplete reponse

)()()( tytyty ph +=

)(ty p = sinusoid of the same input frequency (forced component)

( )hy t =solution of homogeneous equation (natural component)

1

( ) i

ns t

h i

i

y t k e

=

=∑ (for distinct frequencies)


responseExample 4

)(2cos)( tuttes =For the circuit of fig 1, the sinusoid input is appliedto the circuit at time . Determine the complete response of theCapacitor voltage. C=1Farad, L=1/2 Henry, R=3/2 ohms.

0=t

1)0(,2)0( 0 === cL vIi

2

2

( ) ( )1 3( ) cos 2 ( )

2 2

c cc

d v t dv tv t t u t

dtdt+ + =

From example 3

Initial conditions

2)0()0(

,1)0( ===−−

−

C

i

dt

dvv Lc

c


responseCharacteristic equation

2,1,01232

21 −−==++ sss

tth ekektv

221)(

−− +=2

( ) Re( ) | | cos(2 )j t

pv t Ve V t ψ= = +

2( ) Re( ) cos 2

j tse t Ee t= =

From (2) 2 312 2[ ( ) ( ) 1]j j V Eω ω+ + =

108.4

2 312 2

10.316

1 31

jEV e

jjω ω−= = =

− +− +

o

Natural component

Forced component


responseThe complete solution is

21 2

( ) ( ) ( )

0.316cos(2 108.4 )

c h p

t t

v t v t v t

k e k e t− −

= +

= + + − o

1 2

1 2

(0) 1 0.316cos( 108.4 )

1.1

cv k k

k k

= = + + −

+ =

o

1 2

1 2

(0) 2 2 0.316 2sin( 108.4 )

2 1.4

c

dv k k

dt

k k

= = − − − × −

+ = −

o

6.31 =k 2 2.5k = −


response

2( ) 3.6 2.5 0.316cos(2 108.4 )

t tcv t e e t

− −= − + − o

The complete solution is


responseSinusoidal steady-state response

)(ty

In a linear time invariant circuit driven by a sinusoid source, the response

Is of the form

1 21 2( ) .. cos( )ns ts t s t

n my t k e k e k e A tω ψ= + + + + +

Irrespective of initial conditions ,if the natural frequencies lie inthe left-half complex plane, the natural components converge

to zero as and the response becomes close to a sinusoid. The sinusoid steady state response can be calculated

by the phasor method.

∞→t


response

40

220

4220

22)( ωωω ++=+ sss

1 2 0 3 4 0,s s j s s jω ω= = = = −

0 0

1 2 3 4( ) ( ) ( )j t j t

hy t k k t e k k t eω ω−= + + +

Example 5

Let the characteristic polynomial of a differential a differential equationBe of the form

The characteristic roots are

and the solution is of the form

1 0 1 2 0 2( ) cos( ) cos( )hy t k t k t tω φ ω φ= + + +In term of cosine

The solution becomes unstable as ∞→t


response

0 01 2 0( ) cos( )

j t j thy t k e k e K t

ω ω ω φ−= + = +

∞→t0ωω =

220s ω+

1 0 2 0,s j s jω ω= = −

( ) ( ) ( )h py t y t y t= +

Example 6

Let the characteristic polynomial of a differential a differential equationBe of the form

The characteristic roots are

and the solution is of the form

( ) cos( )py t B tω ψ= +and

The solution is oscillatory at different frequencies. If the output isunstable as ( ) cos( )y t At tω θ= +


responseSuperposition in the steady state

If a linear time-invariant circuit is driven by two or more sinusoidalsources the output response is the sum of the output from each source.

Example 7

The circuit of fig1 is applied with two sinusoidal voltage sources and the output is the voltage across the capacitor.

1 1 1cos( )mA tω φ+( )cV t

2 2 2cos( )mA tω φ+


1 2( ) ( ) ( ) ( ) ( )c s s

dL i t Ri t v t e t e t

dt+ + = +

2

1 1 1 2 2 22

( ) ( )( ) cos( ) cos( )c c

c m m

d v t dv tLC RC v t A t A t

dtdtω φ ω φ+ + = + + +

21 1

1 1 1 12

( ) ( )( ) cos( )

p p

p m

d v t dv tLC RC v t A t

dtdtω φ+ + = +

KVL

Differential equation for each source

22 2

2 2 2 22

( ) ( )( ) cos( )

p p

p m

d v t dv tLC RC v t A t

dtdtω φ+ + = +


)cos()cos(

)()()(

22221111

21

θφωθφω +++++=

+=

tVtV

tvtvtv

mm

ppp

The particular solution is

1

1 1

2

2 2

( ) 11 2

1 1

( ) 22 2

2 2

1

1

jj m

m

jj m

m

A eV e

LC j RC

A eV e

LC j RC

φφ θ

φφ θ

ω ω

ω ω

+

+

=− +

=− +

where


response

Summary

A linear time-invariant circuit whose natural frequencies are all withinthe open left-half of the complex frequency plane has a sinusoid steady state response when driven by a sinusoid input. If the circuit has Imaginary natural frequencies that are simple and if these are differentfrom the angular frequency of the input sinusoid, the steady-stateresponse also exists.

The sinusoidal steady state response has the same frequencyas the input and can be obtained most efficiently by the phasor method


Properties of impedances and admittances play important roles in

circuit analyses with sinusoid excitation.

Linear time-invariant

circuit in sinusoid

steady stateElement

( ) Re( )j t

i t Ieω=

( ) Re( )j t

v t Veω=

+

-

Phasor relation for circuit elements

( ) Re( ) | | cos( )j t

v t Ve V t Vω ω= = + R

( ) Re( ) | | cos( )j t

i t Ie I t Iω ω= = + R

Fig 2


( ) ( ), ( ) ( )

,

v t Ri t i t Gv t

V RI I GV

I V

= =

= =

=R R

Resistor

The voltage and current phasors are in phase.

Capacitor

1,

90 , 90 ,

dvi C

dt

I j CV V Ij C

V I I V

ωω

=

= =

∠ = ∠ − ∠ = ∠ +o o

The current phasor leads the voltage phasor by 90 degrees.


Inductor

1,

90 , 90 ,

div L

dt

V j L I I Vj L

V I I V

ωω

=

= =

= + = −o oR R R R

The current phasor lags the voltage phasor by 90 degrees.


Definition of impedance and admittance

ℜThe driving point impedance of the one port at the angular frequency is the ratio of the output voltage phasor V to the input current phasor Iω

| || ( ) | , ( )

| |

VZ j Z j V I

Iω ω= = −R R R

( ) | ( ) || | cos( )s sv t Z j I t Z Iω ω= + +R Ror

ℜThe driving point admittance of the one port at the angular frequency is the ratio of the output current phasor I to the input voltage phasor Vω

| || ( ) | , ( )

| |

IY j Y j I V

Vω ω= = −S R R

( ) | ( ) || | cos( )si t Y j V t Y Vω ω= + +R Ror

Concepts of impedance and admittance1

| ( ) | , ( ) ( )| ( |

Z j Z j Y jY j

ω ω ωω

= = −R R

Inductor

Capacitor

Resistor

YZAngular frequency ω

RG

1=

1

j Cωj Cω

j Lω1

j Lω

R


0)()()( 321 =++ tvtvtv

ω

0)cos()cos()cos( 332211 =+++++ φωφωφω tVtVtV mmm

In the sinusoid steady state Kirchhoff’s equations can be written directlyin terms o voltage phasors and current phasors. For example:

If each voltage is sinusoid of the same frequency

1 2 3

1 2 3

1 2 3

1 2 3

Re( ) Re( ) Re( ) 0

Re( ) 0

Re[( ) ] 0

0

j t j t j t

j t j t j t

j t

V e V e V e

V e V e V e

V V V e

V V V

ω ω ω

ω ω ω

ω

+ + =

+ + =

+ + =

+ + =


n

n

VVVV

IIII

....

....

21

21

++=

====

Series parallel connections

In a series sinusoid circuit

I1I 2I nI

1V 2V nV

V

1

( ) ( )

n

i

i

Z j Z jω ω=

=∑

Fig 3


1 2

1 2

....

....

n

n

V V V v

I I I I

= = = =

= + +

In a parallel sinusoid circuit

Y1 YnY2

I

1I 2I nI

1V

+

-

2V

-

++

nV

-

+

-

V

1

( ) ( )

n

i

i

Y j Y jω ω=

=∑

Fig 4


Node and mesh analyses

)302cos(10)(ο+= ttis

Node and mesh analysis can be used in a linear time-invariant circuit to

determine the sinusoid steady state response. KCL, KVL and the concepts

of impedance and admittance are also important for the analyses.

Example 8

Fig 5

In figure 5 the input is a current source

Determine the sinusoid steady-state voltage at node 3

1 32

2FCi

-

+++

--1v 2v 3v1

1 1

2H

Lisi

Node and mesh analyses2 30

( ) 10cos(2 30 ) Re( ) 10j t j

s s si t t I e or I e= + = =o

o

)Re()(),Re()(),Re()(2

332

222

11tjtjtj eVtveVtveVtv ===

1 1 3 1 24( ) ( ) sV j V V V V I+ − + − =

1 3 1 3 2 1 3( ) ( ) 4( )C CI Y V V j C V V V j V Vω= − = − = −

2224

11V

jV

LjVYI LL ===

ω

KCL at node 1

KCL at node 2 22 1 2 3( ) ( ) 0

4

VV V V V

jω+ − + − =

KCL at node 33

3 1 3 24( ) ( ) 02

Vj V V V V+ − + − =

Node and mesh analysesRearrange the equations

1 2 3(2 4) 4 sj V V j V I+ − − =

1 2 3

1(2 ) 0

4V V V

j− + + − =

1 2 3

34 ( 4) 0

2j V V j V− − + + =

By Crammer’s Rule

3

32

2 4 1

11 2 0

4

4 1 0 2 8

6 11.252 4 1 4

11 2 1

4

4 1 4

s

s

j I

j

j jV I

jj j

j

j j

+ −

− +

− − += =

++ − −

− + −

− − +


3010

jsI e=

o44

3 6.45j

V e=o

3 ( ) 6.45cos(2 44 )v t t= + o

Since

and the sinusoid steady-state voltage at node 3 is

Then

Example 9

Solve example 8 using mesh analysis

2i

Lv3v

sv 1i 3i

Fig 6

Node and mesh analyses2 30

( ) 10cos(2 30 ) Re( ) 10j t j

s s sv t t V e or V e= + = =o

o

2 2 21 1 2 2 3 3( ) Re( ), ( ) Re( ), ( ) Re( )

j t j t j ti t I e i t I e i t I e= = =

1 1 2 1 3( ) 4( ) sI I I j I I V+ − + − =

3 3

1C CV Z I I

j Cω= =

1 24( )L L L LV Z i j LI j I Iω= = = −

KVL at mesh 1

KVL at mesh 2

3 3 2 3 12 ( ) 4( ) 0I I I j I I+ − + − =KVL at mesh 3

2 2 3 3 1

1( ) 4( ) 0

4I I I j I I

j+ − + − =


1 2 3(2 4) 4 sj I I j I V+ − − =

1 2 3

1(2 ) 0

4I I I

j− + + − =

1 2 34 (3 4) 0j I I j I− − + + =

By Crammer’s Rule

3

2 4 1

11 2 0

4

4 1 0 2 8

12 22.52 4 1 4

11 2 1

4

4 1 3 4

s

s

j V

j

j jI V

jj j

j

j j

+ −

− +

− − += =

++ − −

− + −

− − +

Rearrange the equations


303 310 2

jsV e and V I= =

o

443 6.45

jV e=

o

3 ( ) 6.45cos(2 44 )v t t= + o

Since

and the sinusoid steady-state voltage at node 3 is

Then

The solution is exactly the same as from the node analysis

Resonance circuit

Resonance circuits form the basics in electronics and communications. It is

useful for sinusoidal steady-state analysis in complex circuits.

( ) Re( )j t

s si t I eω= C L

G=1/R

+

-

V

Y

CI LI RI

Impedance, Admittance, Phasors

Fig 7

Figure 7 show a simple parallel resonant circuit driven by a sinusoid source.

( ) Re( ) | | cos( )j t

s s s si t I e I t Iω ω= = + R

Resonance circuit

1( )

1( )

Y j G j Cj L

G j CL

ω ωω

ωω

= + +

= + −

ω

( )Y jω

The input admittance at the angular frequency is

The real part of is constant but the imaginary part varies with frequency

1( )B j C

Lω ω

ω= −

At the frequency 0 00

1

2 2 2f

LC

ω ωπ π π

= = =

the susceptance is zero. The frequency is called the resonant frequency.0f

Resonance circuit

LCjY

GjY

ωωω

ω1

)](Im[

)](Re[

−=

=

2 2

2 2

Re[ ( )]( )

( )Im[ ( )]

( )

GZ j

G B

BZ j

G B

ωω

ωω

ω

=+

−=

+

2 2 2 2

1 1( )

( ) ( )

( )

( ) ( )

Z jY j G jB

G Bj

G B G B

ωω ω

ω

ω ω

= =+

−= +

+ +

The admittance of the parallel circuit in Fig 7 is frequency dependant

ω

B

Cω

1

Lω

0ω

Susceptance plot

Fig 8

Resonance circuit

0ω =

GRe( )Y

Im ( )Y

ω = ∞

| |Y

Y0ω ω=

Admittance Plane

1( ) ( )Y j G j C

Lω ω

ω= + −

1( )

1( )

Z j

G j CL

ωω

ω

=+ −

Locus of Y

Re( )Z

ω

0ω ω=

Im( )Z

1

2G

1

G

0ω =ω = ∞

Locus of Z

Fig 9

Resonance circuit

CVjIVLj

IGVI CLR ωω

=== ,1

,

SCLR IIII =++

)Re(cos)(jt

ss eItti ==

The currents in each element are

and

If for example1

1 , , 14

R L H C F= Ω = =

0, 1 /

jsI e rad sω =

The admittance of the circuit is

71.6( 1) 1 (1 4) 1 3 10

jY j j j e

−= + − = − =o

The impedance of the circuit is71.61 1

( 1)( 1) 10

jZ j eY j

= =o

Resonance circuit

71.61( 1)

10

jsV Z j I e= =

o

71.6 18.4 161.61 1 1, ,

10 10 10

j j jR L CI e I e I e

−= = =o o o

The voltage phasor is

Thus

Re

ImV

LI

CIRI

SI

Fig 10

Resonance circuit

sradLC

/21

0 ==ω2

( ) cos 2 Re( )j t

s si t t I e= =

0, 2 /

jsI e rad sω =

Similarly if and

( 2) 1, ( 2) 1, 1Y j Z j V= = =90 90

1, 2 , 2j j

R L CI I e I e−= = =

o o

The voltage and current phasors are

LI

CI

R sI I=

SI

Note that it is a resonance and

,L s C sI I I I> >

Fig 11

Resonance circuit

| || |

| | | |

pCLp

s s

RIIQ

I I Lω= = @

1>>Q

)1

()(C

LjRjZω

ωω −+=

The ratio of the current in the inductor or capacitor to the input current is the quality factor or Q-factor of the resonance circuit.

Generally and the voltages or currents in a resonance circuit is very large!

Analysis for a series R-L-C resonance is the very similar

( )( )

ss

VI j

Z jω

ω=

S R L CV V V V= + + | || |

| | | |

CLs

s s s

VV LQ

V V R

ω= = @

Power in sinusoidal steady-state

)()()( titvtp −ℜ

ℜ

The instantaneous power enter a one port circuit is

The energy delivered to the in the interval is

∫=t

t

o

o

dttpttW ')'(),(

),( tto

ℜ

( )i t

( )v t

Fig 12


Instantaneous, Average and Complex power

ℜIn sinusoidal steady-state the power at the port is

( ) cos( ) Re( )j t

mv t V t V Veωω= + =R

, | |j V

m mV V e V V=R@

where

If the port current is

( ) cos( ) Re( )j t

mi t I t I Ie ωω= + =R

, | |j I

m mI I e I I=R@

where


m

1 1m m2 2

( ) ( ) ( )

cos( )cos( )

cos( ) cos(2 )

m

m m

p t v t i t

V I t V t I

V I V I V I t V I

ω ω

ω

=

= + +

= − + + +

R R

R R R R

1m2

0

1( ') ' cos( )

T

av mP p t dt V I V It

= −∫@ R R

)(tp

Then

avP)(tv

)(ti

Fig 13

Power in sinusoidal steady-state Remarks

The phase difference in power equation is the impedance

angle

Pav is the average power over one period and is non negative. But p(t) may be negative at some t

The complex power in a two-port circuit is 1

2 rms rmsP V I V I=@( )1

2

1 1

2 2

| || |

| || | cos( ) | || | sin( )

j V IP V I e

V I V I j V I V I

−=

= − + −

R R

R R R R

1

2Re( ) Re( ) Re( )av rms rmsP P V I V I= = =

2

2 21 12 2

2

| | Re[ ( )] | | Re[ ( )]

Re[ ( )] Re[ ( )]rms rms

avP I Z j V Y j

I Z j V Y j

ω ω

ω ω

= =

= = Average power is additive


Maximum power transfer

SL ZZ =

The condition for maximum transfer for sinusoid steady-state is thatThe load impedance must be conjugately matched to the source imedance

s

sav

R

VP

8

||2

max=

2| |

4

ss

s

VP

R=

Q of a resonance circuit

0

0

212

0 212

| |

| |

RQ CR

L

C V

G V

ωω

ω

=

=

@

For a parallel resonance circuit

0

energy storedQ

average power dissipated in the resistorω=

(Valid for both series and parallel resonance circuits)


In designing a resonance circuit to meet some specification component

values are usually express in normalized form.

L

R

Qbandwidthdb

LCRZ ==∆−== 0

00 3,1

,ω

ωωFrom

200

1,

RL C and

QR

ω ωω ωω

∆ ∆= = =∆

Let the normalized component values are

0 0 0

11, ,R L Q C

Q= = =

Then

0 00 2

0 0 0 0 0

1, ,

QZ ZR Z L C

Q Z Z

ωω ω ω ω

∆= = = = =

∆


Popularity of normalized design: The circuit design can be made at any impedance level and

center frequency

Well-known solutions exist

n

desired impedance levelr

impedance level of normalized design=

Let

n

desired typical frequency

typical frequency of normalized designΩ =

Then

nnn

nn

r

CCL

rLRrR

Ω=

Ω== 0

00 ,,


Example

6

2

1

2

1

1

ω+=

I

E

Fig. 14 shows a low pass filter whose transfer impedance

1I1.5F 0.5F

1.333H

1

+

-

E2

ω

2

1

E

I

1

2

10Rad/s

1

The gain of the filter is 1 at 0ω = And at 1ω =

Design the circuit to have an impedance of 600 ohms at 0ω =and equal to 1/ 2 at 3.5 kHz then

600nr = 3 42 3.5 10 2.199 10n πΩ = × ⋅ = ×and

2/1


0

0 4

011 4

022 4

600

6001.333 36.37

2.199 10

1.50.1137

600 2.199 10

0.50.0379

600 2.199 10

n

n

n

n n

n n

R r R

rL L mH

C FC F

r

C FC F

r

µ

µ

= =

= = =Ω ×

= = =Ω ⋅ ×

= = =Ω ⋅ ×


1I6 00

+

-

E2

ω

2

1

E

I

600

2

2 x35000Rad/s

600

0.11378 F0.0379 F

36.37mH

Designed circuit

Sinusoidal Steady-state Analysiseestaff.kku.ac.th/~jamebond/182304/chapter2.pdf · Complete response and sinusoidal steady-state response Superposition in the steady state If a linear

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