Chapter 5 The Laplace Transform - University of Kentucky



Chapter 5 The Laplace Transform

5-1 Introduction

(1) System analysis

static system: y(t) = ax(t) => easy (simple processing)

dynamic system:

[pic]

Can we determine y(t) for given u(t) easily?

Easier solution method

[pic]

(1) we have systematic way to obtain H(f) based on the differential equation

(2) we can obtain X(f)

( Fourier transform: an easier way

(2) Problem

Fourier transform of the input signal:

[pic]

if x(t) does not go to zero when [pic] and [pic]

⇨ X(f) typically does not exist! (the existence of X(f) if not guaranteed)

A very strong condition, can not be satisfied by many signals!

(3) Solution:

Why should we care about t [pic]

[pic]

HW#2-2: Assume [pic]. Prove

[pic]

HW#2-3: Express [pic] using [pic]

Example 5-2

Find i(t) using Laplace transform method for t>0

Solution:

1) Before switched from 1 to 2 at t=0

[pic]

(2) System equation (t>0)

KVL: [pic]

(3) Solve system equation using Laplace transform

[pic]

3. Laplace Transform of an integral

Assume [pic]

Then [pic]

where [pic]

Proof :

[pic]

(1)

[pic]

(2)

[pic]

(3)

[pic] Proved!

Example 5.3:

Find I(s) = L(i(t))

Solution:

1) Differential equation

KVL : [pic]

[pic] [pic]

2) Laplace transform

[pic]

[pic]

5-3-4. Complex Frequency shift (s-shift) Theorem

Assume [pic]

[pic]

Then [pic]

• [pic]

• [pic]

=>[pic]

Example 5-4 Find [pic]

Solution:

[pic]

[pic]

4. Delay Theorem

question: How to express delayed function?

Assume [pic]

Then [pic]

(If [pic], it will not be a delay!)

Proof :

[pic]

Question: will [pic] be true if [pic]

No! (it will not be a delay)

Example 5-5: Square wave beginning at t = 0

[pic]

5. Convolution

Signal 1: [pic] Signal 2 : [pic]

[pic]

if [pic]

[pic]

if [pic] [pic]

Therefore, if [pic]

[pic]

[pic]

[pic]

Look at

[pic]

Then

[pic]

7. Product

8. Initial Value Theorem

[pic]

Example: A demonstration where x(0) is obvious

[pic]

It is evident: [pic]

Using Laplace transform

[pic]

[pic]

9. Final Value Theorem: if [pic] and [pic] are Laplace transformable, then

[pic]

(condition: [pic] has no poles on [pic] or in the right-half s-plan or [pic] exists)

10. Scaling

a>0: x(at) ( a times fast (if a>1)

or slow (if a=n [pic]

proper m proper + Polynomial (using long division)

[pic]

[pic]

How to find inverse Laplace transform for polynomials?

[pic]

(consider proper rational functions only!

4) Proper Rational Functions: Partial Fraction Expansion

( sum of [pic]

Let’s look at examples, and then summarize!

Techniques:

Common Denominator Factorize first!

Specific value of s Expand second!

Heaviside’s Expansion Find coefficients third!

Matlab

Example 5-9: Simple Factors

[pic]

Solution:

(1) Factorize and expand [pic]

(2) Common Denominator Methods

[pic]

[pic]

specific values of s

[pic]

Can you solve for A and B?

Heaviside Expansion

[pic]

and [pic]

Example 5-10 Imaginary Roots

[pic]

Solution: what do we have:

[pic]

[pic]

A1+A2 must be real number

(-A1+A2)j must be real number

[pic]

Heaviside Expansion => A3=1 and A4= - 2.

[pic]

s=1 => [pic]

s=2 => [pic]

Can we solve for c1 and c2?

c1=1 c2=1

[pic]

=>[pic]

[pic]Too complex: use MATLAB

Example 5-11 Repeated linear Factors

[pic]

Example 5-12

[pic]

Example 5-13 Complex - Conjugate Factors

[pic]

Example 5-14 Repeated Quadratic Factors

[pic]

* Summary of Partial–Fraction Expansion

1) Expansion Structure:

Simple Roots (including complex conjugate)

=> [pic] could be complex.

Repeated Roots: m multiplicity

=>[pic]

real number or complex number

2) Avoid complex number

For complex conjugates: [pic]

[pic]

3) Inverse Laplace transform

[pic]

[pic]

Matlab use for Partial – Fraction Expansion.

Have to be memorized:

(1) Table 5-2 all except for No. 7

(2) Table 5-3 No. 1-No. 7

Appendix: Partial-Faction Expansion with MatLab

1. Command (MatLab Help)

2. An Introduction

3. An Example: Complex Conjugate

1. Command (MatLab Help)

» help residue

RESIDUE Partial-fraction expansion (residues).

[R,P,K] = RESIDUE(B,A) finds the residues, poles and direct term of

a partial fraction expansion of the ratio of two polynomials B(s)/A(s).

If there are no multiple roots,

B(s) R(1) R(2) R(n)

---- = -------- + -------- + ... + -------- + K(s)

A(s) s - P(1) s - P(2) s - P(n)

Vectors B and A specify the coefficients of the numerator and

denominator polynomials in descending powers of s. The residues

are returned in the column vector R, the pole locations in column

vector P, and the direct terms in row vector K. The number of

poles is n = length(A)-1 = length(R) = length(P). The direct term

coefficient vector is empty if length(B) < length(A), otherwise

length(K) = length(B)-length(A)+1.

If P(j) = ... = P(j+m-1) is a pole of multplicity m, then the

expansion includes terms of the form

R(j) R(j+1) R(j+m-1)

-------- + ------------ + ... + ------------

s - P(j) (s - P(j))^2 (s - P(j))^m

[B,A] = RESIDUE(R,P,K), with 3 input arguments and 2 output arguments,

converts the partial fraction expansion back to the polynomials with

coefficients in B and A.

Warning: Numerically, the partial fraction expansion of a ratio of

polynomials represents an ill-posed problem. If the denominator

polynomial, A(s), is near a polynomial with multiple roots, then

small changes in the data, including roundoff errors, can make

arbitrarily large changes in the resulting poles and residues.

Problem formulations making use of state-space or zero-pole

representations are preferable.

See also POLY, ROOTS, DECONV.

»

2. An Introduction

[pic]

[pic]

[pic]

[pic]

3. An Example: Complex Conjugate

[pic]

-----------------------

Dynamic

System

x(t) input

y(t) output

A processor which processes the input signal to produce the output

H(f )

Y(f )

X(f)

Dynamic

System

Algebraic equation, not differential equation

Inverse Laplace transform

G(s )

Y(s )

X(s)

Dynamic

System

0

1

[pic]

Same as real roots!

t

[pic]

[pic]

[pic]

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