Laplace Transforms Circuit Analysis • Passive element equivalents

Laplace Transforms Circuit Analysis

•

•

•

Passive element equivalents

Review of ECE 221 methods in

s

Many examples domain

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 define the circuit behavior in terms of a linear ODE.

For example, solve for

v

(

t

) . Check your answer using the initial and final value theorems and the methods discussed in Chapter 7.

i(0-) = -2 mA

10 u(t)

5 k Ω

5 mH

+

v(t)

-


2

Example 1:Workspace

Hint:

[r,p,k] = residue([-2e-3 2e3],[1 1e6 0]) r = -0.0040, 0.0020, p = -1000000, 0 k = []


3

Example 1:Workspace


4

Laplace Transform Circuit Analysis Overview

•

LPT is useful for circuit analysis because it transforms differential equations 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

3. Solve for the

s s

domain domain voltages and currents of interest

4. Apply the inverse Laplace transform to find time domain expressions

•

How do we know this will work?


5

Kirchhoff’s Laws

N v k

(

t

) = 0

k

=1

M k

=1

i k

(

t

) = 0

N

V k

(

s

) = 0

k

=1

M k

=1

I k

(

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 sum of currents leaving a node

If Kirchhoff’s laws apply in the

s

domain, we can use the same techniques that you learned last term (ECE 221)

•

•

Apply the LPT to both sides of the time domain expression for these laws

The laws hold in the

s

domain


6

Defining

s

Domain Equations: Resistors

i(t)

R

+

v(t)

-

I(s)

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 as in the time domain

s

domain

•

Note that we used the linearity property of the LPT for both

Ohm’s law and Kirchhoff’s laws


7

v

(

t

) =

L

d

i

(

t

) d

t

V

(

s

) =

L

[

sI

(

s

)

− I

0

]

V

(

s

) =

sLI

(

s

)

− LI

0

Where

I

0

Defining

s

Domain Equations: Inductors

i(t)

L

+

v(t)

-

I

0 s

L I

0

I(s)

+

Ls

V(s)

-

I(s)

+

Ls

V(s)

-

i

(

t

) =

I

(

s

) =

1

L

1

sL

V

0

-

t v

(

τ

) d

(

s

) +

1

s

τ

I

0

+

I

0

i

(0

-

)


8

Defining

s

Domain Equations: Capacitors

i(t)

C

I(s)

+

v(t)

1

sC

-

V

0 s

+

V(s)

i

(

t

) =

C

d

v

(

t

) d

t

I

(

s

) =

C

[

sV

(

s

)

− V

0

]

-

Where

I

(

s

) =

sCV

(

s

)

− CV

0

V

0

v

(0

-

)

CV

0

1

sC

I(s)

+

V(s)

-

V

V v

(

t

) =

(

(

s s

) =

) =

1

C

1

C

1

sC

I t i

(

τ

) d

τ

0

-

1

s

I

(

s

) +

(

s

) +

V

0

s

+

V

1

s

V

0

0


9

s

Domain Impedance and Admittance

Impedance:

Admittance:

Z

(

s

) =

Y

(

s

) =

V

(

s

)

I

(

s

)

I

(

s

)

V

(

s

)

•

The

s

domain

impedance

initial conditions of a circuit element is defined for zero

•

•

This is also true for the

s

domain admittance

We will see that circuit can assume

s

domain circuit analysis is easier when we

zero initial conditions


10

Resistor

Inductor

s

Domain Circuit Element Summary

V

(

s

) =

RI

(

s

)

V

(

s

) =

sLI

(

s

)

V

V

=

=

RI sLI

Capacitor

V

(

s

) =

1

sC

I

(

s

)

V

=

1

sC

I

•

•

•

•

•

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

R

Ls

1

sC


11

10 u(t)

i(0-) = -2 mA

5 k Ω

5 mH


+

v(t)

-

Solve for

v

(

t

) using

s

-domain circuit analysis.


12

Example 2: Workspace


13


sin(1000

t

)

t = 0

1 k Ω

1

μ

F

+

v o

-

Given

v o

(0) = 0 , solve for

v o

(

t

) for

t ≥

0 .


14


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]


15



16

Example 3: Plot of Results

0

−0.2

−0.4

−0.6

−0.8

0

1

0.8

0.6

0.4

0.2

Total

Transient

Steady State

5 10

Time (ms)

15 20 25


17

50 Ω

10 mF

50 Ω


100 Ω

t = 0

175 Ω

175 Ω

40 V

+

v

-

10 mH

Solve for

v

(

t

) .


18


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]


19



20

Example 5: Parallel RLC Circuits

i(t) C L R

+

v(t)

-

Find an expression for

V

(

s

) . Assume zero initial conditions.


21


0.125

μ

F

i

L

8 H

+

20 k Ω

v

-

i

Given

L

(0

-

v

(0) = 0

) =

−

12

.

25

V and the current through the inductor is mA, solve for

v

(

t

) .


22


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]


23



24

Example 6: Plot of

v

(

t

)

80

60

40

20

0

−20

−40

0 5 10 15 20

Time (ms)

25 30 35 40


25

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(’’);


26

v(t)

Example 7: Series RLC Circuits

+

v

R

(t)

+

v

L

(t)

-

R L

C

+

v

C

(t)

-

Find an expression for conditions.

V

R

(

s

) ,

V

L

(

s

) , and

V

C

(

s

) . Assume zero initial


27



28

t = 0

80 V

+

v

2

(t)

-


20 k Ω

+

v

1

(t)

-

50 nF

10 nF 2.5 nF

There is no energy stored in the circuit at

t

= 0 . Solve for

v

2

(

t

) .


29


Hint:

[r,p,k] = residue([320e3],[1 5e3 0]) r = [-64, 64] p = [-5000,0] k = []


30



31


0.25 v

1

600 u(t)

10 Ω

v

1

20 H

0.1 F

v

2

140 Ω

Solve for

V

2

(

s

) . Assume zero initial conditions.


32



33



34


10 mF

i(t) a

u(t)

9 i(t)

100 Ω b

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


35



36



37

v(t)

i

L

L

R


+

v

o

(t)

-

i


L

(0) =

I

0 mA.

v o

(

t

) given that

v

(

t

) = e

− αt u

(

t

) and


38



39



40

1

μ

F

1 k Ω

v s

(t)


200 Ω

100 mH

+

v

o

(t)

-


V o

(

s

) in terms of

V s

(

s

) . Assume there is no energy stored in the circuit initially. What is

v o

(

t

) if

v s

(

t

) =

u

(

t

) ?


41


Hint:

[r,p,k] = residue([-0.2 0],conv([1 2e3],[1 1e3])) r = [-0.4000, 0.2000] p = [-2000, -1000] k = []


42



43

v s

(t)

C

A

R

A


C

B

R

B

R

L

+

v

o

(t)

-


V o

(

s

) in terms of energy stored in the circuit initially.

V s

(

s

) . Assume there is no


44



45



46


C

R

v s

(t) R

L

+

v

o

(t)

-


V o

(

s

) in terms of energy stored in the circuit initially.

V s

(

s

) . Assume there is no


47



48

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