Laplace Transforms Circuit Analysis

• Passive element equivalents

• Review of ECE 221 methods in s domain

• Many examples

J. McNames Portland State University ECE 222 Laplace Circuits Ver. 1.63 1

Example 1: Circuit Analysis

We can use the Laplace transform for circuit analysis if we can definethe circuit behavior in terms of a linear ODE.

For example, solve for v(t). Check your answer using the initial andfinal value theorems and the methods discussed in Chapter 7.

10 u(t) v(t)

i(0-) = -2 mA

Example 1:Workspace

Hint: � [r,p,k] = residue([-2e-3 2e3],[1 1e6 0])

r = -0.0040, 0.0020,

p = -1000000, 0

k = []

Example 1:Workspace

Laplace Transform Circuit Analysis Overview

• LPT is useful for circuit analysis because it transforms differentialequations into an algebra problem

• Our approach will be similar to the phasor transform

1. Solve for the initial conditions– Current flowing through each inductor

– Voltage across each capacitor

2. Transform all of the circuit elements to the s domain

3. Solve for the s domain voltages and currents of interest

4. Apply the inverse Laplace transform to find time domainexpressions

• How do we know this will work?

Kirchhoff’s Laws

N∑k=1

vk(t) = 0N∑

Vk(s) = 0

M∑k=1

ik(t) = 0M∑

Ik(s) = 0

• Kirchhoff’s laws are the foundation of circuit analysis

– KVL: The sum of voltages around a closed path is zero

– KCL: The sum of currents entering a node is equal to the sumof currents leaving a node

• If Kirchhoff’s laws apply in the s domain, we can use the sametechniques that you learned last term (ECE 221)

• Apply the LPT to both sides of the time domain expression forthese laws

• The laws hold in the s domain

Defining s Domain Equations: Resistors

v(t) -+

i(t) R

V(s) -+

v(t) = R i(t) V (s) = R I(s)

• Generalization of Ohm’s Law

• As with KCL & KVL, the relationship is the same in the s domainas in the time domain

• Note that we used the linearity property of the LPT for bothOhm’s law and Kirchhoff’s laws

Defining s Domain Equations: Inductors

v(t) -+

i(t) L

V(s) -+

I(s) LsL I0

V(s) -+

I(s) Ls

v(t) = Ldi(t)dt

i(t) =1L

v(τ) dτ + I0

V (s) = L [sI(s) − I0] I(s) =1sL

V (s) +1sI0

V (s) = sLI(s) − LI0

Where I0 � i(0-)

Defining s Domain Equations: Capacitors

v(t) -+

V(s) -+

i(t) = Cdv(t)dt

v(t) =1C

i(τ) dτ + V0

I(s) = C [sV (s) − V0] V (s) =1C

[1sI(s)

I(s) = sCV (s) − CV0 V (s) =1

sCI(s) +

Where V0 � v(0-)

s Domain Impedance and Admittance

Impedance: Z(s) =V (s)I(s)

Admittance: Y (s) =I(s)V (s)

• The s domain impedance of a circuit element is defined for zeroinitial conditions

• This is also true for the s domain admittance

• We will see that circuit s domain circuit analysis is easier when wecan assume zero initial conditions

s Domain Circuit Element Summary

Resistor V (s) = RI(s) V = RI R

Inductor V (s) = sLI(s) V = sLI Ls

Capacitor V (s) = 1sC I(s) V = 1

• All of these are in the form V (s) = ZI(s)

• Note similarity to phasor transform

• Identical if s = jω

• Will discuss further later

• Equations only hold for zero initial conditions

10 u(t) v(t)

i(0-) = -2 mA

Solve for v(t) using s-domain circuit analysis.

Example 2: Workspace

+1 kΩ

sin(1000t) 1 μF

Given vo(0) = 0, solve for vo(t) for t ≥ 0.

Hint: � [r,p,k] = residue([1e6],conv([1 0 1e6],[1 1e3]))

r = [ 0.5000, -0.2500 - 0.2500i, -0.2500 + 0.2500i]

p = 1.0e+003 *[ -1.0000, 0.0000 + 1.0000i, 0.0000 - 1.0000i]

k = []

� [abs(r) angle(r)*180/pi]

ans = [ 0.5000 0, 0.3536 -135.0000, 0.3536 135.0000]

Example 3: Plot of Results

0 5 10 15 20 25−0.8

−0.6

−0.4

−0.2

1 TotalTransientSteady State

Time (ms)

100 Ω

175 Ω

Solve for v(t).

Hint: � [r,p,k] = residue([1e-3 20 0],[1 21.25e3 10e3])

r = [-1.2496, -0.0004]

p = [-21250,-0.4706]

k = [0.0010]

Example 5: Parallel RLC Circuits

C L R v(t)

Find an expression for V (s). Assume zero initial conditions.

20 kΩ0.125 μF

Given v(0) = 0 V and the current through the inductor isiL(0-) = −12.25 mA, solve for v(t).

Hint: � [r,p,k] = residue([98e3],[1 400 1e6])

r = [ 0 -50.0104i, 0 +50.0104i]

p = 1.0e+002 * [ -2.0000 + 9.7980i, -2.0000 - 9.7980i]

k = []

� [abs(r) angle(r)*180/pi]

ans =[ 50.0104 -90.0000, 50.0104 90.0000]

Example 6: Plot of v(t)

0 5 10 15 20 25 30 35 40

Time (ms)

Example 6: MATLAB Code

t = 0:0.01e-3:40e-3;v = 50*exp(-200*t).*sin(979.8*t);t = t*1000;h = plot(t,v,’b’);set(h,’LineWidth’,1.2);xlim([0 max(t)]);ylim([-23 40]);box off;xlabel(’Time (ms)’);ylabel(’(volts)’);title(’’);

Example 7: Series RLC Circuits

vR(t) -+ vL(t) -+

Find an expression for VR(s), VL(s), and VC(s). Assume zero initialconditions.

10 nF 2.5 nF

20 kΩ

There is no energy stored in the circuit at t = 0. Solve for v2(t).

Hint: � [r,p,k] = residue([320e3],[1 5e3 0])

r = [-64, 64]

p = [-5000,0]

k = []

0.25 v1

0.1 F600 u(t)

140 Ω

Solve for V2(s). Assume zero initial conditions.

9 i(t)

100 Ω

Find the Thevenin equivalent of the circuit above. Assume that thecapacitor is initially uncharged.

R vo(t)

Find an expression for vo(t) given that v(t) = e−αtu(t) andiL(0) = I0 mA.

100 mH

vs(t) vo(t)

200 Ω

1 kΩ1 μF

Find an expression for Vo(s) in terms of Vs(s). Assume there is noenergy stored in the circuit initially. What is vo(t) if vs(t) = u(t)?

Hint: � [r,p,k] = residue([-0.2 0],conv([1 2e3],[1 1e3]))

r = [-0.4000, 0.2000]

p = [-2000, -1000]

k = []

+vs(t)

Find an expression for Vo(s) in terms of Vs(s). Assume there is noenergy stored in the circuit initially.

vs(t)vo(t)

Find an expression for Vo(s) in terms of Vs(s). Assume there is noenergy stored in the circuit initially.

Laplace Transforms Circuit Analysis - Web Services Overview

Documents

Lecture 21 Laplace transforms revisitedlinks.uwaterloo.ca/amath351docs/set9.pdf · Lecture 21 Laplace transforms revisited You have seen Laplace transforms in an earlier course, e.g.,

4 Laplace Transforms (1)

Circuit Analysis with Laplace Transforms - جامعة نزوى · Circuit Analysis with Laplace Transforms his chapter presents applications of the Laplace transform. Several examples

Schaum Laplace Transforms

Introduction to Laplace Transforms

Asymptotic Laplace transforms

Circuit Analysis by Laplace and Fourier Transforms

Presentation on laplace transforms

Unit IV Laplace Transforms

Circuit Analysis by Laplace and Fourier Transforms - Jaynes 1945.pdf

LaPlace Transforms in Design and Analysis of …...LaPlace Transforms in Design and Analysis of Circuits© Part 2 by Tom Bertenshaw Basic Circuit Analysis - Series Circuits Series

Circuit Analysis by Laplace and Fourier Transforms (1945) - Jaynes US NAVY

Laplace Transforms - Maplesoft

Laplace transforms and applications

18686025 Laplace Transforms

Fourier Laplace Transforms

Chapter 6 Laplace Transforms

Introduction to Laplace Transforms. Definition of the Laplace Transform Some functions may not have Laplace transforms but we do not use them in circuit

5 LAPLACE TRANSFORMS

Laplace transforms