Circuit and System Analysis EHB 232EThe current leads the voltage across the terminals of a capacitor by 90 . Prof. Dr. Mu ˘stak E. Yal˘c n (_IT U) Circuit and System Analysis Spring,

Circuit and System AnalysisEHB 232E

Prof. Dr. Mustak E. Yalcın

Istanbul Technical UniversityFaculty of Electrical and Electronic Engineering

[email protected]

Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 1 / 34

Outline I

1 SinusoidaI Steady-State AnalysisPhasor analysisProperties of phasorsRepresentation of state-space equationsTransfer functionKirchhoff’s Laws in the Frequence DomainThe Passive Circuit Elements in the Frequency DomainThe Concept of Impedance and Admittance


SinusoidaI Steady-State Analysis

If a system is exponentially stable

limt→∞

Φ(t) = 0

Forced response is the complete response.

x(t) = Φ(t)x0 − Φ(t)xp(t0) + xp(t)

Sinusoidal steady-state behavior

Sinusoidal steady-state behavior of linear time-invariant circuits when thecircuits are driven by one or more sinusoidal sources at some frequency wand when, after all ”transients” have died down, all currents and voltagesare sinusoidal at frequency w .

Electric Circuits, James W. Nilsson and Susan A. Riedel, Ch.9 and 10


Phasor

The idea is to associate with each sine wave (of voltage or current) acomplex number called the phasor.x(t) is a complex variable and in polar coordinate

x(t) = Xmej(wt+θ).

where j =√−1, Xm (|X | ) is called the magnitude of x(t) and θ (∠) is

called the phase of x(t).Rectangular representation of complex x(t) is

x(t) = Xm cos(wt + θ) + jXm sin(wt + θ)

Real partXm cos(wt + θ) = Rex(t)

Imaginary partXm sin(wt + θ) = Imx(t)

from Euler’s identity.Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 4 / 34

The quantity ( phasor)X = Xme

jθ

is a complex number that carries the amplitude and phase angle of thegiven sinusoidal function.This complex number is by definition the phasorrepresentation of the given sinusoidal function.

Using phasor representation, a complex variable is given

x(t) = Xewtj

Examples: Electric Circuits, James W. Nilsson and Susan A. Riedel, pp. 334


Example

Phasor of the sinusoidal function

x(t) = 110√

2 cos(wt +π

2)

usingx(t) = Re110

√2e j

π2︸︷︷︸

phasor

e jwt

is obtainX = 110

√2e j

π2

Xm = 100√

2 (or |X | = 100√

2), θ = π2 (or ∠π

2 ). |X |rms = 100A complex number (phasor) in rectangular coordinate Z = a + bj ,magnitude

Zm =√

a2 + b2

phase

θ = arctan(b

a)


3 + 4j = 5e j0.927 3 = 5 cos(0.927) 4 = 5 sin(0.927)

5 =√

32 + 42 0.927 = arctan4

3

1 + j =√

2e j0.785 1 =√

2 cos(0.785) 1 =√

2 sin(0.785))√

2 =√

12 + 12 0.785 = arctan1

1

5e j0.927 +√

2e j0.785 =?

3 + 4j + 1 + j = 4 + 5j = 6.403e j0.896

(3 + 4j)× (1 + j) =?

5e j0.927 ×√

2e j0.785 = 5√

2e j1.71


Properties of phasors

y(t) = α1x1(t) + α2x2(t)

Y = α1X1 + α2X2

d

dtXme

θje jwt = ddt

Xmeθje jwt = jwAXme

θj︸︷︷︸phasor

e jwt

X is phasor of x(t)

y(t) =dnx(t)

dtn

then phasor of y(t)Y = (jw)nX .

A are B phasorIf ReAe jwt = ReBe jwt then A = B.If A = B then ReAe jwt = ReBe jwt.


Phasor & State-space equation

Lets find the sinusoidal particular solution (Xe jwt) of linear time invariantstate equation

x = Ax + Be

for a sinusoidal input Ee jwt . Substituting the solution and input

j(w)X = AX + BE

The sinusoidal solution is then

X = (jwI − A)−1BE

The solution is defined for det(jwI − A) 6= 0 which means jw 6= λ.Inputfrequency is equal the natural frequency.


Transfer function

A linea time invariant single input single output system (e, y ∈ R) isdefined by

x = Ax + Be

y = Cx + De

what if e ∈ Rm, y ∈ R l ...

e yx = A x + B ey= C x + D e

.


Using phasors, the output of the system

Y = (C (jwI − A)−1B + D)︸︷︷︸transfer function

E

Transfer function

H(jw) = (C (jwI − A)−1B + D)

from input E to output Y


Example

dx

dt=

[−√

2 −11 0

]x +

[10

]e(t); y = [0 1]x

Transfer function (which is a function of w !)

H(jw) =1

1− w2 + j√

2w

For the input e(t) = sin(wt)(phasor of input signal E = 1) phasor of theoutput is

Y = H(j1)E =−j√

2=

1√2e−jπ/2

Hence output signal is

y(t) =1√2

sin(t − π/2)


0 0.5 1 1.5 2 2.5 3 3.50.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

w

|H(jw

)|

X: 1Y: 0.7071

Linear Simulation Results

Time (sec)A

mp

litu

de

0 5 10 15 20 25 30−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

System: sysTime (sec): 15.7Amplitude: 0.707


0 0.5 1 1.5 2 2.5 3 3.5−3

−2.5

−2

−1.5

−1

−0.5

0

w

∠ H

(jw

)

Linear Simulation Results

Time (sec)A

mp

litu

de

0 5 10 15 20 25 30−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

System: sysTime (sec): 15.7Amplitude: 0.707


Kirchhoff’s Laws in the Frequence Domain

Lets assuming that v1, v2...vne ,represent voltages arourd a closed path in acircuit.KVL requires that

ne∑k=1

vk(t) = 0

We assume that the circuit is operating in a sinusoidal steady statetherefore

ne∑k=1

ReVke

jwt

= 0

Factoring the term e jwt from each term yields

ne∑k=1

Vk = 0.


A similar derivation applies to a set of sinusoidal currents (KCL). Thus if

ne∑k=1

ik(t) = 0

We assume that the circuit is operating in a sinusoidal steady statetherefore

ne∑k=1

ReIkeθk ) = 0

Factoring the term e jwt from each term yields

ne∑k=1

Ik = 0.

Question: Four branches terminates at a common node. The referencedirection of each branch current (i1, i2, i3, i4, is toward the node ifi1 = 100 cos(wt + 25)A i2 = 100 cos(wt + 145)Ai3 = 100 cos(wt − 95)A, find i4


The Passive Circuit Elements in the Frequency Domain

Resistors From Ohm’s law, if the current in a resistor varies sinusoidallywith time, the voltage at the terminals of the resistor

v(t) = R ReIRe jwt

v(t) = ReRIRe jwt

ReVRejwt = ReRIRe jwt

from the properties of phasor

VR = RIR

orIG = GVG

There is no phase shift between the current and voltage of resistor. Thesignals of voltage and current are said to be in phase.



Capacitor Substituting the phasor representation of the current andphasor voltage at the terminals of a capacitor into i = C dv

dt )

ReICe jwt = CdReVCe

jwtdt

using the properties of phasor

ReICe jwt = ReCVCde jwt

dt = ReCVC jwe

jwt

we getIC = jwCVC

The current leads the voltage across the terminals of a capacitor by 90.



InductorVL = jwLIL

The current lags the voltage by 90.Independent current and voltage sources

Ik = Imkejθk

andVk = Vmke

jθk


The Concept of Impedance and Admittance

The driving-point impedance of the one-port N (is formed by an arbitraryinterconnection of linear time-invariant elements) at the frequency w to bethe ratio of the port-voltage phasor V and the input-current phasor I thatis,

Z (jw) =V

IThus the amplitude of the port voltage is the product of the currentamplitude times the magnitude of the impedance. Z represents theimpedence of the circuit element

Z =V

I= R + jL

R, is called resistance and L, is called reactance.Y represents the admittance of the circuit element

Y =I

V= G + jB

G , is called conductance and B, is called susceptance.Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 20 / 34

Element Impedance Reactance Admintance Susceptance

Resistor R - G -Capacitor -j/wC -1/wC jwC wCInductor jwL wL -j/wL -1/wL


Combining Impedance in Series and Parallel

Impedances in series can be combined into a single impedance by simplyadding the individual impedances.

Z = Z1 + Z2 + ...+ Zn

when they in parallel

Z =

1

Z1+

1

Z2+ ...+

1

Zn

−1

admittances in parallel:

Y = Y1 + Y2 + ...+ Yn


Mutual Inductance

φ1 = L1i1 + Mi2

φ2 = L2i2 + Mi1

v1 = L1di1dt

+ Mdi2dt

v2 = L2di2dt

+ Mdi1dt


Example

Find the complete solution for R = 1/3Ω, C = 1F , L = 1/2H,VC (0) = 1V ve iL(0) = 1A and i(t) = cos(wt) .

Yeq(jω) =1

jωL+

1

R+ jωC =

R + jωL− ω2RLC

jωRL

VC = Zeq(jω) · IK =IK

Yeq(jω)=

1

Yeq(jω)=

jωRL

R − ω2RLC + jωL


Magnitude of VC (jω) is maximun when ω = 1√LC

!

0 0.5 1 1.5 2 2.5 3 3.50

0.05

0.1

0.15

0.2

0.25

0.3

0.35

w

abs

Z (j1√LC

) = R

Resonance

Resonance occurs at a particular resonance frequency when the imaginaryparts of impedances or admittances of circuit elements cancel each other.

ω = 1√LC

is the resonance frequency.Prof. Dr. Mustak E. Yalcın (ITU) Circuit and System Analysis Spring, 2020 25 / 34

Example

R

L C

Find the impedance Z between the terminals:

Y = Cjw +1

Ljw

Z = R +Ljw

1− LCw2

Z =R − RLCw2 + Ljw

1− LCw2


Example

+

− e

R

L C

State equation for R = 1/5, C = 1F ve L = 1/6

d

dt

[VC1

iL

]=

[− 1

RC − 1C

1L 0

] [VC1

iL

]+

[1RC0

]e

which isd

dt

[VC1

iL

]=

[−5 −16 0

] [VC1

iL

]+

[50

]e


The roots of the function

det

λI −

[−5 −16 0

]= λ(λ+ 5) + 6

are the eigenvalues of A which λ1 = −3 and λ2 = −2. Correspondingeigenvalues are [1 − 2]T and [1 − 3]T . Fundamental matrix

M =

[e−3t e−2t

−2e−3t −3e−2t

]The homogeneous solution

xh(t) =

[e−3t e−2t

−2e−3t −3e−2t

] [α1

α2

]The state transition matrix of the circuit

φ(t) =

[e−3t e−2t

−2e−3t −3e−2t

] [1 1−2 −3

]−1


Example

Using the The Concept of Impedance, lets find the VR

VR = Re

Z

= R(1− LCw2)e

Ljw + R(1− LCw2)e


Using the state-equation

VC = [1 0]

[jw + 1

RC1C

− 1L jw

]−1 [ 1RC0

]e

=jwRC

1LC − w2 + jw

RC

VR = e − VC = e

1−

jwRC

1LC − w2 + jw

RC

= e

1LC − w2

1LC − w2 + jw

RC

=R(1− LCw2)e

Ljw + R(1− LCw2)


Example

Sekil 7.2

v1 1 0 sin(0 .1 10) dc 0 ac 1

r 1 2 4k

l 2 0 2m

c 2 0 2m

.control

ac lin 1000 .1 100

plot v(1,2)/4k

.endc

.end


Example

frequency

voltage

0 20 40 60 80 100

Hz

205.0

210.0

215.0

220.0

225.0

230.0

235.0

240.0

245.0

250.0

uV v(1,2)/4k

YouTube Video: Example with two sources


https://youtu.be/ZbvkNNrbkcY

Phasor diagram

A phasor diagram shows the magnitude and phase angle of each phasorquantity in the complex number plane.

VR

VS

VRS

120

VR = V , VS = Ve−j120 , VRS =√

3Ve j30 , VRS?f (VS ,VR)


Documents

Circuit and System Analysis EHB 232EThe current leads the voltage across the terminals of a capacitor by 90 . Prof. Dr. Mu ˘stak E. Yal˘c n (_IT U) Circuit and System Analysis Spring,