Chapter 13 The Laplace Transform in Circuit Analysis PDF

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Chapter 13 The Laplace Transform in Circuit Analysis

13.1. Circuit Elements in the s Domain. 13.2-3 Circuit Analysis in the s Domain. 13.4-5 The Transfer Function and Natural Response.

&KDSWHU 2

Key points

How to represent the initial energy of L, C in the s-domain?

Why the

functional forms of natural and steady- state responses are determined by the poles of transfer function H(s) and excitation source X(s), respectively?

Why the output of an LTI circuit is the

convolution of the input and impulse response?

How to interpret the

memory of a circuit by convolution? 3

Section 13.1

Circuit Elements in the s

Domain

Equivalent elements of R, L, C

A resistor in the s domain

iv-relation in the time domain: ).()(tiRtv

By operational Laplace transform:

sIRsVtiLRtiRLtvL

Physical units: V(s) in volt-seconds, I(s) in

ampere-seconds. 5

An inductor in the s domain

).()(tidtdLtv 00

LIsIsLIssILsVtiLLtiLLtvL

initial current iv-relation in the time domain:

By operational Laplace transform:

Equivalent circuit of an inductor

Series equivalent:

Parallel equivalent:

Thévenin

Norton

A capacitor in the s domain

).()(tvdtdCti 00

CVsVsCVssVCsItvLCtvCLtiL

initial voltage iv-relation in the time domain:

By operational Laplace transform:

Equivalent circuit of a capacitor

Parallel equivalent:

Series equivalent:

Norton

Thévenin

Section 13.2, 13.3

Circuit Analysis in the s

Domain1.

Procedures

Nature response of RC circuit

Step response of RLC circuit

Sinusoidal source

5. MCM 6.

Superposition

How to analyze a circuit in the s-domain?

Replacing each circuit element with its s-domain

equivalent . The initial energy in L or C is taken into account by adding independent source in series or parallel with the element impedance. 2.

Writing & solving algebraic equations by the

same circuit analysis techniques developed for resistive networks. 3.

Obtaining the t-domain solutions by inverse

Laplace transform

Why to operate in the s-domain?

It is convenient in solving transient responses of linear, lumped parameter circuits, for the initial conditions have been incorporated into the equivalent circuit.

It is also useful for circuits with

multiple essential nodes and meshes, for the simultaneous ODEs have been reduced to simultaneous algebraic equations. 12

Nature response of an RC circuit (1)

i(t)v(t)=? .)(1)( , 1000
RCsRV

RCsCVsIIRsCI

Replacing the charged capacitor by a Thévenin

equivalent circuit in the s-domain.

KVL, algebraic equation & solution of I(s):

Nature response of an RC circuit (2)

The t-domain solution is obtained by inverse

Laplace transform

).( 1 )(01)( 0 10 1 tue R

VsLeRV

RCsRVLti

RCtRCt

i (0 ) = V 0 /R, which is true for v C (0 ) = v C (0 V 0 i ()= 0,which is true for capacitor becomes open (no loop current) in steady state. 14

Nature response of an RC circuit (3)

To directly solve v(t), replacing the charged

capacitor by a Norton equivalent in the s-domain. 10 0

RCsVsVRVsCVCV

Solve V(s), perform inverse Laplace transform:

01 01 tRitueVRCsVLtv RCt 15

Step response of a parallel RLC (1)

i L (0 ) = 0 v C (0 ) = 0 Q: i L (t)=? 16

Step response of a parallel RLC (2)

KCL, algebraic equation & solution of V(s):

112

LCsRCsCIsVsLV

RVsCVsI

dcdc

Solve I

L (s): .)106.1()104.6(1084.3 )()()()()(

94271121

sssLCsRCss LCI sLsVsI dc L 17

Step response of a parallel RLC (3)

Perform partial fraction expansion and inverse

Laplace transform:

.s)(mA )k24k32(12720 )k24k32(1272024)(jsjsssI L .(mA) )( )k2432sin()k2424cos(24 (mA) )( 127k)24(cos4024 ..)(20)(24)( k)32(k)32()k24(k)32(127 tuttetutecctueeetuti tttjtj L 18

Transient response due to a sinusoidal source (1)

For a parallel RLC circuit, replace the current

source by a sinusoidal one:

The algebraic equation changes:

11222111222222

LCsRCss

sLCI sLVsILCsRCsssCIsVssIIsLV RVsCV m Lm m g ).(cos)(tutIti mg 19 22*
11 jsK jsK jsK jsKsI L

Driving

frequency Neper frequency

Damped

frequency ).( cos2cos2)( 2211
tuKteKKtKti t L

Steady-state

response (source)Natural response (RLC parameters)

Transient response due to a sinusoidal source (2)

Perform partial fraction expansion and inverse

Laplace transform:

Step response of a 2-mesh circuit (1)

i 2 (0 = 0i 1 (0 ) = 0 Q: i 1 (t), i 2 (t)=? 21

Step response of a 2-mesh circuit (2)

)2(0)4810()(42)1(

336)(424.8

212211

IsIIsIIsI

.0336

109042424.842

21
s II ss 124.1

24.87121

21415
0 336

109042424.842

1 2 1 ssssss s ss II 22

Step response of a 2-mesh circuit (3)

.A 15)48//42(336)(1415)( 122
1 tueeti tt .A 748424215)(4.14.87)( 122
2 tueeti tt 23

Use of superposition (1)

Given 2 independent sources v

g i g and initially charged C, L, v 2 (t)=? 24

Use of superposition: V

g acts alone (2) 12 .01, 11 .0 )(,0 2 211
21
1 2 2 1121
211
11

VsCRVsCRVVsCVsCsLR

R V sCVVsCVV sLV RVV gg 25

Use of superposition (3)

.01 11 1 21

221212112

1 21
RV VV YYYYV V sC

RsCsCsC

sLR g 2

122211112

VYYYRYV

For convenience, define admittance matrix:

Use of superposition: I

g acts alone (4) 12 . ,0 2

12221111

2 21
quotesdbs_dbs8.pdfusesText_14

[PDF] Chapter 13 The Laplace Transform in Circuit Analysis

Key points

Why the

Why the output of an LTI circuit is the

How to interpret the

Section 13.1

Circuit Elements in the s

Domain

Equivalent elements of R, L, C

A resistor in the s domain

By operational Laplace transform:

Physical units: V(s) in volt-seconds, I(s) in

An inductor in the s domain

LIsIsLIssILsVtiLLtiLLtvL

By operational Laplace transform:

Equivalent circuit of an inductor

Series equivalent:

Parallel equivalent:

Thévenin

Norton

A capacitor in the s domain

CVsVsCVssVCsItvLCtvCLtiL

By operational Laplace transform:

Equivalent circuit of a capacitor

Parallel equivalent:

Series equivalent:

Norton

Thévenin

Section 13.2, 13.3

Circuit Analysis in the s

Domain1.

Procedures

Nature response of RC circuit

Step response of RLC circuit

Sinusoidal source

Superposition

How to analyze a circuit in the s-domain?

Replacing each circuit element with its s-domain

Writing & solving algebraic equations by the

Obtaining the t-domain solutions by inverse

Laplace transform

Why to operate in the s-domain?

It is also useful for circuits with

Nature response of an RC circuit (1)

RCsCVsIIRsCI

Replacing the charged capacitor by a Thévenin

KVL, algebraic equation & solution of I(s):

Nature response of an RC circuit (2)

The t-domain solution is obtained by inverse

Laplace transform

VsLeRV

RCsRVLti

RCtRCt

Nature response of an RC circuit (3)

To directly solve v(t), replacing the charged

RCsVsVRVsCVCV

Solve V(s), perform inverse Laplace transform:

Step response of a parallel RLC (1)

Step response of a parallel RLC (2)

KCL, algebraic equation & solution of V(s):

LCsRCsCIsVsLV

RVsCVsI

Solve I

94271121

Step response of a parallel RLC (3)

Perform partial fraction expansion and inverse

Laplace transform:

Transient response due to a sinusoidal source (1)

For a parallel RLC circuit, replace the current

The algebraic equation changes:

11222111222222

LCsRCss

Driving

Damped

Steady-state

Transient response due to a sinusoidal source (2)

Perform partial fraction expansion and inverse

Laplace transform:

Step response of a 2-mesh circuit (1)

Step response of a 2-mesh circuit (2)