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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