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Overview of Laplace Transforms for Circuit Analysis

Passive element equivalents

Review of ECE 221 methods in

s

domain

Examples

Portland State University

ECE 222

Laplace Circuits

Ver. 1.

Prerequisite and New Knowledge

Prerequisite knowledge

Ability to find Laplace transforms of signals

Ability to find inverse Laplace transforms

methodsAbility to perform DC circuit analysis using all of the standard

New knowledge

switches, and sources)basic linear elements (resistors, capacitors, inductors, op amps,Ability to solve for any current or voltage in a circuit with the

No longer restricted to DC or sinusoidal steady-state analysis

Portland State University

ECE 222

Laplace Circuits

Ver. 1.

Example 1:Workspace

Hint:



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

k = []p = -1000000, 0r = -0.0040, 0.0020,

Portland State University

ECE 222

Laplace Circuits

Ver. 1.

Example 1:Workspace

Portland State University

ECE 222

Laplace Circuits

Ver. 1.

Kirchhoff’s Laws

N

k ∑

v k ( t ) = 0

N

k ∑

V

k ( s ) = 0

M

k ∑

i k ( t ) = 0

M

k ∑

I

k ( s ) = 0

Kirchhoff’s laws are the foundation of circuit analysis

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

of currents leaving a nodeKCL: The sum of currents entering a node is equal to the sum

If Kirchhoff’s laws apply in the

s

domain, we can use the same

techniques that you learned last term (ECE 221)

these lawsApply the LPT to both sides of the time domain expression for

The laws hold in the

s

domain

Portland State University

ECE 222

Laplace Circuits

Ver. 1.

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

domain

as in the time domain

Ohm’s law and Kirchhoff’s lawsNote that we used the linearity property of the LPT for both

Portland State University

ECE 222

Laplace Circuits

Ver. 1.

Defining

s

Domain Equations: Capacitors

v ( t )

-

+

i ( t ) V ( s ) -

+

I ( s )

V ( s )

-

+

I ( s )

CV

0

1 C

sC

V 0

s

1

sC

i ( t ) =

C d v ( t )

d t

v ( t ) =

C 1

t

0

• i ( τ (^) ) d

τ

V

0

I

s ) =

C

[

sV

( s ) − V 0 ] V ( s

C 1

[

s^1 I ( s ) ] +

s^1 V 0

I

s ) =

sCV

s ) −

CV

0

V

s ) =

sC

I

s ) +

V

0

s

Where

V

0



v (

)

Portland State University

ECE 222

Laplace Circuits

Ver. 1.

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 zero

initial conditions

This is also true for the

s

domain admittance

We will see that circuit

s

domain circuit analysis is easier when we

can assume

zero initial conditions

Portland State University

ECE 222

Laplace Circuits

Ver. 1.

Example 2: Workspace

Portland State University

ECE 222

Laplace Circuits

Ver. 1.

Example 3: Circuit Analysis

t = 0

v

- o +

1 k

Ω

sin(

t )

1 μ F

Given

v o (0) = 0

, solve for

v o ( t )

for

t

≥

Portland State University

ECE 222

Laplace Circuits

Ver. 1.

Example 3: Workspace

Portland State University

ECE 222

Laplace Circuits

Ver. 1.

Example 3: Plot of Results

0

5

10

15

20

25

−0.8−0.6−0.4−0.

0

0.2 0.4 0.6 0. 1

Steady StateTransientTotal

Time (ms)

vo(t) (V)

Portland State University

ECE 222

Laplace Circuits

Ver. 1.

Example 4: Workspace

Hint:



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

k = [0.0010]p = [-21250,-0.4706]r = [-1.2496, -0.0004]

Portland State University

ECE 222

Laplace Circuits

Ver. 1.

Example 4: Workspace

Portland State University

ECE 222

Laplace Circuits

Ver. 1.