Circuit Analysis Using Fourier and Laplace Transforms

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Circuit Analysis Using Fourier and Laplace Transforms

EE2015: Electrical Circuits and Networks

Nagendra Krishnapura

https://www.ee.iitm.ac.in/nagendra/

Department of Electrical Engineering

Indian Institute of Technology, Madras

Chennai, 600036, India

July-November 2017

Nagendra Krishnapura https://www.ee.iitm.ac.in/nagendra/ Circuit Analysis Using Fourier and Laplace Transforms ..Circuit Analysis Using Fourier and Laplace Transforms

Based on

exp(st)being an eigenvector of linear systems Steady-state response toexp(st)isH(s)exp(st)whereH(s)is some scaling factor Signals being representable as a sum(integral) of exponentialsexp(st) Nagendra Krishnapura https://www.ee.iitm.ac.in/nagendra/ Circuit Analysis Using Fourier and Laplace Transforms ..Fourier series

Periodicx(t)can be represented as sums of complex exponentialsx(t)periodic with periodT0Fundamental (radian) frequency!0=2=T0

x(t) =1∑ k=1a kexp(jk!0t) x(t)as a weighted sum of orthogonal basis vectors exp(jk!0t)

Fundamental frequency!0and its harmonics

a k: Strength ofkthharmonic Coefficientsakcan be derived using the relationship a k=1 T

0∫

0x(t)exp(jk!0t)dt

Inner product" ofx(t)with exp(jk!0t)

Nagendra Krishnapura https://www.ee.iitm.ac.in/nagendra/ Circuit Analysis Using Fourier and Laplace Transforms ..Fourier series

Alternative form

x(t) =a0+1∑ k=1b kcos(k!0t) +cksin(k!0t)Coefficientsbkandckcan be derived using the relationship b k=2 T

0∫

0x(t)cos(k!0t)dt

c k=2 T

0∫

0x(t)sin(k!0t)dt

Another alternative form

x(t) =a0+1∑ k=1d kcos(k!0t+ϕk) Coefficientsbkandckcan be derived using the relationship d b

2k+c2k

k=tan1(ck b k) Nagendra Krishnapura https://www.ee.iitm.ac.in/nagendra/ Circuit Analysis Using Fourier and Laplace Transforms ..Fourier series Ifx(t)satisfies the following (Dirichlet) conditions, it can be represented by a Fourier series x(t)must be absolutely integrable over a period T0

0jx(t)jdtmust exist

x(t)must have a finite number of maxima and minima in the interval[0;T0] x(t)must have a finite number of discontinuities in the interval[0;T0] Nagendra Krishnapura https://www.ee.iitm.ac.in/nagendra/ Circuit Analysis Using Fourier and Laplace Transforms ..Fourier transform Aperiodicx(t)can be expressed as an integral of complex exponentials x(t) =1

2∫

1 1 X !(!)exp(j!t)d! x(t)as a weighted sum(integral) of orthogonal vectors exp(j!t)

Continuous set of frequencies!

X !(!)d!: Strength of the component exp(j!t) X !(!): Fourier transform ofx(t) X !(!)can be derived using the relationship X 1 1 x(t)exp(j!t)dt

Inner product" ofx(t)with exp(j!t)

Nagendra Krishnapura https://www.ee.iitm.ac.in/nagendra/ Circuit Analysis Using Fourier and Laplace Transforms ..Fourier seies Ifx(t)satisfies either of the following conditions, it can be represented by a Fourier transform

FiniteL1norm∫1

1 jx(t)jdt<1

FiniteL2norm∫1

1 jx(t)j2dt<1 Many common signals such as sinusoids and unit step fail these criteria

Fourier transform contains impulse functions

Laplace transform more convenient

Nagendra Krishnapura https://www.ee.iitm.ac.in/nagendra/ Circuit Analysis Using Fourier and Laplace Transforms ..Fourier transform x(t)in volts)X!(!)has dimensions of volts/frequencyX

!(!): Density over frequencyTraditionally, Fourier transformXf(f)defined as density per Hz"(cyclic frequency)

Scaling factor of 1=2when integrated over!(radian frequency) x(t) =∫ 1 1

Xf(f)exp(j2ft)df

2∫

1 1 X !(!)exp(j!t)d! X !(!) =Xf(!=2) X f(f): volts/Hz(density per Hz) ifx(t)is a voltage signal X f(f) =∫ 1 1 x(t)exp(j2ft)dt X 1 1 x(t)exp(j!t)dt Nagendra Krishnapura https://www.ee.iitm.ac.in/nagendra/ Circuit Analysis Using Fourier and Laplace Transforms ..Fourier transform as a function ofj!

Ifj!is used as the independent variable

x(t) =1 j2∫ j1 j1X(j!)exp(j!t)d(j!)

X(j!) =X!(!)

Same function, butj!is the independent variable

Scaling factor of 1=j2

Withj!as the independent variable, the definition is the same as that of the

Laplace transform

Nagendra Krishnapura https://www.ee.iitm.ac.in/nagendra/ Circuit Analysis Using Fourier and Laplace Transforms ..Fourier transform pairs

Signals in1 t 1

1$2(!)

exp(j!0t)$2(!!0) cos(!0t)$(!!0) +(!+!0) sin(!0t)$ j (!!0) j (!+!0) exp(ajtj)$2a a 2+!2

Not very useful in circuit analysis

Nagendra Krishnapura https://www.ee.iitm.ac.in/nagendra/ Circuit Analysis Using Fourier and Laplace Transforms ..Fourier transform pairs

Signals in 0t 1

u(t)$(!) +1 j! exp(j!0t)u(t)$(!!0) +1 j(!!0) cos(!0t)u(t)$(!!0) +(!+!0) +j! 20!2 sin(!0t)u(t)$ j (!!0) j (!+!0) +!0 20!2 exp(at)u(t)$1 j!+a Useful for analyzing circuits with inputs starting att=0 Nagendra Krishnapura https://www.ee.iitm.ac.in/nagendra/ Circuit Analysis Using Fourier and Laplace Transforms ..Circuit analysis using the Fourier transform

For an input exp(j!t), steady state output isH(j!)exp(j!t)A general inputx(t)can be represented as a sum(integral) of complex

exponentials exp(j!t)with weightsX(j!)d!=2 x(t) =1

2∫

1 1

X(j!)exp(j!t)d!

Linearity)steady-state outputy(t)is the superposition of responses

H(j!)exp(j!t)with the same weightsX(j!)d!=2

y(t) =1

2∫

1Y(j!)z

X(j!)H(j!)exp(j!t)d!

Therefore,y(t)is the inverse Fourier transform ofY(j!) =H(j!)X(j!) Nagendra Krishnapura https://www.ee.iitm.ac.in/nagendra/ Circuit Analysis Using Fourier and Laplace Transforms ..Circuit analysis using the Fourier transformFourier transform

Inverse

Fourier

transform circuit analysisx(t)y(t) X(j!)

Y(j!) =H(j!)X(j!)

H(j!)

CalculateX(j!)

CalculateH(j!)

Directly from circuit analysis

From differential equation, if given

Calculate(look up) the inverse Fourier transform ofH(j!)X(j!)to gety(t) Nagendra Krishnapura https://www.ee.iitm.ac.in/nagendra/ Circuit Analysis Using Fourier and Laplace Transforms ..Circuit analysis using the Fourier transform In steady state with an input of exp(j!t), Ohms law" also valid for L, C+ v R+ v C+ v Li Ri Ci L R C L v(t) i(t) v(t)=i(t)

Resistor

v R=RiR RI

Rexp(j!t)

Inductor

L=L(diL=dt)

j!LILexp(j!t) I

Lexp(j!t)

j!L

Capacitor

C=C(dvC=dt)

Cexp(j!t)

j!CVCexp(j!t)

1=(j!C)

R,IL,VC: Phasors corresponding toiR,iL,vC

Use analysis methods for resistive circuits with dc sources to determineH(j!)asratio of currents or voltages

e.g. Nodal analysis, Mesh analysis, etc.

No need to derive the differential equation

Nagendra Krishnapura https://www.ee.iitm.ac.in/nagendra/ Circuit Analysis Using Fourier and Laplace Transforms ..Example: Calculating the transfer function+- V sRL2 C 1 C 3V 1V 3I 2I 0

Mesh analysis with currentsI0,I2

2 6 64R+1
j!C11 j!C1 1 j!C1j!L2+1 j!C1+1 j!C33 7 75[
I0I 2] =[Vs 0] I 0(j!) V s(j!)=(j!)3C1C3L2+ (j!) (C3+C1) (j!)3C1C3L2+ (j!)2C3L2+ (j!) (C3+C1)R+1 I 2(j!) V s(j!)=(j!)C3 (j!)3C1C3L2+ (j!)2C3L2+ (j!) (C3+C1)R+1 V

1= (I0I2)=(j!C1),V3=I2=(j!C3)

Nagendra Krishnapura https://www.ee.iitm.ac.in/nagendra/ Circuit Analysis Using Fourier and Laplace Transforms ..Example: Calculating the response of a circuit+-R C v o(t) v i(t) v i(t) =Vpexp(at)u(t) From direct time-domain analysis, with zero initial condition v o(t) =Steady-state response z V p

1aCRexp(at)u(t)Transient response

z V p

1aCRexp(t=RC)u(t)

Nagendra Krishnapura https://www.ee.iitm.ac.in/nagendra/ Circuit Analysis Using Fourier and Laplace Transforms ..Example: Calculating the response of a circuit+-R C+ vo(t)vi(t)H(j!)Vi(j!)Vo(j!) v i(t) =Vpexp(at)u(t) V i(j!) =Vp a+j!

Using Fourier transforms and transfer function

V o(j!) =Vp a+j!1

1+j!CR

Vp 1aCR1 a+j!Vp

1aCRCR

1+j!CR

From the inverse Fourier transform

v o(t) =Steady-state response z V p

1aCRexp(at)u(t)Transient response

z V p

1aCRexp(t=RC)u(t)

We get both steady-state and transient responses with zero initial condition Nagendra Krishnapura https://www.ee.iitm.ac.in/nagendra/ Circuit Analysis Using Fourier and Laplace Transforms ..Fourier transform of the input signal-20 0 2000.51jVi(j!)j vi(t) =exp(-t)u(t);Vi(j!) = 1=(1 +j!) -20 0 20 !-10001006Vi(j!)[o]

Fourier transform magnitude and phase(Vp=1,a=1)

Shown for20!20

Nagendra Krishnapura https://www.ee.iitm.ac.in/nagendra/ Circuit Analysis Using Fourier and Laplace Transforms ..Fourier transform of the input signal-5 0 5-101

Samples of constituent sinusoids

-5 0 5 t-101 x(t) vi(t) 1

2πR

20 !20Vi(j!)exp(j!t)d! Fourier transform componentsVi(j!)d!exp(j!t): Sinusoids fromt=1to1

A small number of sample sinusoids shown above

The integral is close, but not exactly equal tox(t) Extending the frequency range improves the representation Nagendra Krishnapura https://www.ee.iitm.ac.in/nagendra/ Circuit Analysis Using Fourier and Laplace Transforms ..How do we get the total response by summing up steady-state responses? Fourier transform componentsVi(j!)d!exp(j!t): Sinusoids fromt=1to1

For anyt>1, the output is the

steady-state response

H(j!)Vi(j!)d!exp(j!t)

Sum(integral) of Fourier transform components produces the inputx(t)(e.g. exp(at)u(t)) which starts fromt=0

Sum(integral) of

steady-state responses produces the output including the response to changes att=0, i.e. including the transient response

Inverse Fourier transform ofVi(j!)H(j!)is the

total zero-state response Nagendra Krishnapura https://www.ee.iitm.ac.in/nagendra/ Circuit Analysis Using Fourier and Laplace Transforms ..Accommodating initial conditions+- v C+ v′CA B A B v

C(0) =V0v

′C(0) =0A B iLA B i′Li

L(0) =I0i

′L(0) =0 I 0u(t) V 0u(t) C C L L A capacitor cannot be distinguished from a capacitor in series with a constant voltage source An inductor cannot be distinguished from an inductor in parallel with a constant current source Initial conditions reduced to zero by inserting sources equal to initial conditions Treat initial conditions as extra step inputs and find the solution Step inputs because they start att=0 and are constant afterwards Nagendra Krishnapura https://www.ee.iitm.ac.in/nagendra/ Circuit Analysis Using Fourier and Laplace Transforms ..Accommodating initial conditions+- +-v

C(0) =V0R

C+ v o(t)v i(t) v i(t) =Vpexp(at)u(t)v ′C(0) =0R C+ v o(t)vi(t)+ v ′C(t) v x(t) =V0u(t) v x(t) V o(j!) =Vi(j!)H(j!)z 1

1+j!CR+Vx(j!)H

x(j!)z j!CR

1+j!CR

Vp a+j!1

1+j!CR+V0(

(!) +1 j!) j!CR

1+j!CR

Vp 1aCR( 1 a+j!CR

1+j!CR)

+V0CR

1+j!CR

v o(t) =Vp

1aCRexp(at)u(t) +(

V oVp 1aCR)quotesdbs_dbs8.pdfusesText_14

[PDF] Circuit Analysis Using Fourier and Laplace Transforms - EE2015

EE2015: Electrical Circuits and Networks

Nagendra Krishnapura

Department of Electrical Engineering

Indian Institute of Technology, Madras

Chennai, 600036, India

July-November 2017

Based on

Fundamental frequency!0and its harmonics

0∫

0x(t)exp(jk!0t)dt

Inner product" ofx(t)with exp(jk!0t)

Alternative form

0∫

0x(t)cos(k!0t)dt

0∫

0x(t)sin(k!0t)dt

Another alternative form

2k+c2k

0jx(t)jdtmust exist

2∫

Continuous set of frequencies!

Inner product" ofx(t)with exp(j!t)

FiniteL1norm∫1

FiniteL2norm∫1

Fourier transform contains impulse functions

Laplace transform more convenient

Xf(f)exp(j2ft)df

2∫

Ifj!is used as the independent variable

X(j!) =X!(!)

Same function, butj!is the independent variable

Scaling factor of 1=j2

Laplace transform

Signals in1 t 1

1$2(!)

Not very useful in circuit analysis

Signals in 0t 1

2∫

X(j!)exp(j!t)d!

H(j!)exp(j!t)with the same weightsX(j!)d!=2

2∫

1Y(j!)z

X(j!)H(j!)exp(j!t)d!

Inverse

Fourier

Y(j!) =H(j!)X(j!)

CalculateX(j!)

CalculateH(j!)

Directly from circuit analysis

From differential equation, if given

Resistor

Rexp(j!t)

Rexp(j!t)

Inductor

L=L(diL=dt)

Lexp(j!t)

Capacitor

C=C(dvC=dt)

Cexp(j!t)

1=(j!C)

R,IL,VC: Phasors corresponding toiR,iL,vC

No need to derive the differential equation

Mesh analysis with currentsI0,I2

1= (I0I2)=(j!C1),V3=I2=(j!C3)

1aCRexp(at)u(t)Transient response

1aCRexp(t=RC)u(t)

Using Fourier transforms and transfer function

1+j!CR

1aCRCR

1+j!CR

From the inverse Fourier transform

1aCRexp(at)u(t)Transient response

1aCRexp(t=RC)u(t)

Fourier transform magnitude and phase(Vp=1,a=1)

Shown for20!20

Samples of constituent sinusoids

2πR

A small number of sample sinusoids shown above

For anyt>1, the output is the

Inner product" ofx(t)with exp(jk!0t)

Inner product" ofx(t)with exp(j!t)