Laplace Transforms Circuit Analysis • Passive element equivalents

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)

-

+

5 mH

i(0-) = -2 mA

5 kΩ


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∑

k=1

Vk(s) = 0

M∑k=1

ik(t) = 0M∑

k=1

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

R

v(t) -+

i(t) R

V(s) -+

I(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

I0s

v(t) = Ldi(t)dt

i(t) =1L

∫ t

0-

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) -+

i(t)

V(s) -+

I(s)

V(s) -+

I(s)

CV0C

1sC

V0s 1

sC

i(t) = Cdv(t)dt

v(t) =1C

∫ t

0-

i(τ) dτ + V0

I(s) = C [sV (s) − V0] V (s) =1C

[1sI(s)

]+

1sV0

I(s) = sCV (s) − CV0 V (s) =1

sCI(s) +

V0

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

sC I1

sC

• 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)

-

+

5 mH

i(0-) = -2 mA

5 kΩ

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


Example 2: Workspace



t = 0

vo

-

+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

0

0.2

0.4

0.6

0.8

1 TotalTransientSteady State

Time (ms)

vo(t

)(V

)



40 V

10 mF

10 mH

t = 0

v

-

+

50 Ω

50 Ω

100 Ω

175 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)

-

+

i(t)

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



8 H v

-

+

iL

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

−40

−20

0

20

40

60

80

Time (ms)

(vol

ts)


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

R L

Cv(t)

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

vC(t)

-

+

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





50 nF

10 nF 2.5 nF

80 V

t = 0

v1(t)

-

+

v2(t)

-

+

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

20 H

0.1 F600 u(t)

v1 v2

10 Ω

140 Ω

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







10 mF

9 i(t)

u(t)

a

b

i(t)

100 Ω

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







v(t)

L

R vo(t)

-

+

iL

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 = []





vo(t)

-

+vs(t)

RL

RA

CA RB

CB

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







vs(t)vo(t)

-

+

RL

C

R

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




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Laplace Transforms Circuit Analysis • Passive element equivalents