Laplace Solution Transient Circuits

Laplace Transform Solutions of Transient Circuits

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CE00436-1 ELECTRICAL PRINCIPLES LAPLACE TRANSFORM SOLUTION OF TRANSIENT CIRCUIT

LEARNING OUTCOMES

Laplace transformProperties of Laplace transformLaplace Circuit Element ModelsApplication of Laplace transform in transient analysis


Introduction

• In a circuit with energy storage elements, voltages and currents are the solutions to linear, constant coefficient differential equations

• Real engineers almost never solve the differential equations directly

• It is important to have a qualitative understanding of the solutions

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Definition of Laplace Transform

• Definition of the unilateral (one-sided) Laplace transform

where s=+j is the complex frequency, and f(t)=0 for t<0

• The inverse Laplace transform requires a course in complex variable analysis

0

dtetfstf stL F


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Laplace Transform Pairs

Most of the time we find Laplace transforms from tables like Table 1 rather than using the defining integral

(t) F(s)

δ(t) 1

u(t) {a constant} s

1

e–at as

1

t 2

1

s

t e–at 2

1

as


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Some Laplace Transform Properties


Laplace Circuit Solutions

• In this section we will use previously established techniques (e.g., KCL, KVL, nodal and loop analyses, superposition, source transformation, Thevenin) in the Laplace domain to analyze circuits

• The primary use of Laplace transforms here is the transient analysis of circuits

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Laplace Circuit Element Models

• Here we develop s-domain models of circuit elements

• DC voltage and current sources basically remain unchanged except that we need to remember that a dc source is really a constant, which is transformed to a 1/s function in the Laplace domain

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Resistor

• We start with a simple (and trivial) case, that of the resistor, R

• Begin with the time domain relation for the element

v(t) = R i(t)

• Now Laplace transform the above expression

V(s) = R I(s)

• Hence a resistor, R, in the time domain is simply that same resistor, R, in the s-domain

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Capacitor



I(s) = s C V(s) – C v(0)

• Interpretation: a charged capacitor (a capacitor with non-zero initial conditions at t=0) is equivalent to an uncharged capacitor at t=0 in parallel with an impulsive current source with strength C·v(0)

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dt

v(t)dCi(t)


Capacitor (cont’d.)

• Rearranging the above expression for the capacitor

• Interpretation: a charged capacitor can be replaced by an uncharged capacitor in series with a step-function voltage source whose height is v(0)

• Circuit representations of the Laplace transformation of the capacitor appear on the next page

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s

v(0)

Cs

I(s)V(s)


s-Domain Model for C

s

vsI

sCsV

CvssCVvssVCsI

dt

tdvCti

)0()(

1)(or

)0()()0()()(

)()(

capacitor, aFor

Time domain s domain s domain


Inductor



V(s) = s L I(s) – L i(0)• Interpretation: an energized inductor (an

inductor with non-zero initial conditions) is equivalent to an unenergized inductor at t=0 in series with an impulsive voltage source with strength L·i(0)

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dt

i(t)dLv(t)


Inductor (cont’d.)

• Rearranging the above expression for the inductor

• Interpretation: an energized inductor at t=0 is equivalent to an unenergized inductor at t=0 in parallel with a step-function current source with height i(0)

• Circuit representations of the Laplace transformation of the inductor appear on the next page

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s

i(0)

Ls

V(s)I(s)


s-Domain Models for an inductor

s

isV

sLsI

LissLIissILsV

dt

tdiLtv

)0()(

1)(or

)0()()0()()(

)()(

inductor,an For

Time domain s domain s domain


Circuit Element Models

• Steps in applying the Laplace transform:

– Transform the circuit from the time domain to the s domain

– Solve the circuit using circuit analysis technique (nodal/mesh analysis, source transformation, etc.)

– Take the inverse Laplace transform of the solution and thus obtain the solution in the time domain


Summary

s

isV

sLsI

LissLIsV

)0()(

1)(

)0()()(

s

vsI

sCsV

CvssCVsI

)0()(

1)(

)0()()(

For inductor: For capacitor:


Summary

Element Z(s)

Resistor R

Inductor sL

Capacitor 1/sC

*Assuming zero initial conditions

• Impedance in the s domain– Z(s)=V(s)/I(s)

• Admittance in the s domain– Y(s)=1/Z(s)=V(s)/I(s)


Transient Analysis

• Sometimes we not only must Laplace transform the circuit, but we must also find the initial conditions

Element DC Steady-State

Capacitor I = 0; open circuit

Inductor V = 0; short circuit

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EEE 202

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Laplace Solution Transient Circuits

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