Laplace is no more difficult!

2. There are techniques for finding the system response of a system
described by a differential equation, based on the replacement of
functions of a real variable (usually time or distance) by certain
frequency-dependent representations, or by functions of a complex
variable dependent upon frequency. The equations are converted from
the time or space domain to the frequency domain through the use of
mathematical transforms.

3. Time Domain Frequency Domain
Laplace Transform
Differential Algebraic
equations equations
Input Input
excitation e(t) excitation E(s)
Output Output
response r(t) response R(s)
Inverse Laplace Transform

4. Let f(t) be a real function of a real variable t (time) defined for t>0. Then
is called the Laplace transform of f(t). The Laplace transform is a
function of a complex variable s. Often s is separated into its real and
imaginary parts: s= σ +jω , where σ and ω are real variables.

5. After a solution of the transformed problem has been obtained in terms
of s, it is necessary to "invert" this transform to obtain the solution in
terms of the time variable, t. This transformation from the s-domain into
the t-domain is called the inverse Laplace transform.

6. Let F(s) be the Laplace transform of a function f(t), t>0. The contour integral
where c> σ 0 (σ 0 as above) is called the inverse Laplace transform of F(s).

7. It is seldom necessary to perform the integration in the Laplace transform
or the contour integration in the inverse Laplace transform. Most often,
Laplace transforms and inverse Laplace transforms are found using tables
of Laplace transform pairs.

8. Time Domain Frequency Domain
f(t), t> 0 F(s)
1. δ 1
2. K K/s
3. Kt K/s2
4. Ke-at K/(s+ a)
5. Kte-at K/(s+ a) 2
6. Ksinωt Kω/(s2+ ω2)
7. Kcosωt Ks/(s2+ ω2)
8. Ke-at sinωt Kω/((s+ a) 2+ ω2))
9. Ke-at cosωt K(s+ a)/ ((s+ a) 2+ ω2))

9. Time Domain Frequency Domain
f(t), t> 0 F(s)
10. t s
11. f(t) F(s)
12. L-1{ F(s)} = f(t) L{ f(t )} = F(s)
13. Af 1(t) + Bf 2(t) AF1(s)+ BF2(s)
14.
15.

10. The inverse Laplace transform is usually more
difficult than a simple table conversion.
8( s + 3)( s + 8)
X ( s) =
s( s + 2)( s + 4)

11. If we can break the right-hand side of the
equation into a sum of terms and each term is in a
table of Laplace transforms, we can get the
inverse transform of the equation (partial fraction
expansion).
8( s + 3)( s + 8) K1 K2 K3
X ( s) = = + +
s( s + 2)( s + 4) s s+2 s+4

12. In general, there will be a term on the right-hand
side for each root of the polynomial in the
denominator of the left-hand side. Multiple roots
for factors such as (s+2)n will have a term for each
power of the factor from 1 to n.
8( s + 1) K1 K2
Y ( s) = = +
( s + 2) 2
s + 2 ( s + 2) 2

13. Complex roots are common, and they always
occur in conjugate pairs. The two constants in
the numerator of the complex conjugate terms
are also complex conjugates.
5.2 K K*
Z ( s) = 2 = +
s + 2 s + 5 ( s + 1 − j 2) ( s + 1 + j 2)
where K* is the complex conjugate of K.

14. The solution of each distinct (non-multiple)
root, real or complex uses a two step process.
The first step in evaluating the constant is to
multiply both sides of the equation by the factor
in the denominator of the constant you wish to
find.
The second step is to replace s on both sides of
the equation by the root of the factor by which
you multiplied in step 1

15. 8( s + 3)( s + 8) K1 K2 K3
X ( s) = = + +
s( s + 2)( s + 4) s s+2 s+4
8( s + 3)( s + 8) 8(0 + 3)(0 + 8)
K1 = = = 24
( s + 2)( s + 4) s=0
(0 + 2)(0 + 4)
8( s + 3)( s + 8) 8( −2 + 3)( −2 + 8)
K2 = = = −12
s( s + 4 ) s =−2
−2( −2 + 4 )

16. 8( s + 3)( s + 8) 8( −4 + 3)( −4 + 8)
K3 = = = −4
s( s + 2 ) s =−4
−4( −4 + 4)
The partial fraction expansion is:
24 12 4
X ( s) = − −
s s+2 s+4

17. The inverse Laplace transform is found from
the functional table pairs to be:
−2 t −4 t
x (t ) = 24 − 12e − 4e

18. Any unrepeated roots are found as before.
The constants of the repeated roots (s-a)m are
found by first breaking the quotient into a
partial fraction expansion with descending
powers from m to 0:
Bm B2 B1
++ +
(s − a) m
(s − a) 2
(s − a)

19. The constants are found using one of the
following:
1 d m −i
 P( s) 
Bi =  m
(m − i )! ds m −i  Q( s ) /( s − a1 )  s = a1
P(a )
Bm =
Q( s) / ( s − a ) m
s=a

20. 8( s + 1) K1 K2
Y ( s) = = +
( s + 2) 2
s + 2 ( s + 2) 2
8( s + 1)( s + 2) 2
K2 = = 8( s + 1) s=−2 = −8
( s + 2) 2
s =−2

21. 1 d  8( s + 1) 
Bi =  ( s + 2) 2 /( s + 2) 2  =8
(2 − 1)! ds   s = −2
The partial fraction expansion yields:
8 8
Y ( s) = −
s + 2 ( s + 2) 2

22. The inverse Laplace transform derived from the functional
table pairs yields:
−2 t −2 t
y (t ) = 8e − 8te

23. 8( s + 1) K1 K2
Y ( s) = = +
( s + 2) 2
s + 2 ( s + 2) 2
8( s + 1) = K1 ( s + 2) + K 2
8s + 8 = K1s + 2 K1 + K 2
Equating like terms:
8 = K1 and 8 = 2 K1 + K 2

24. 8 = K1 and 8 = 2 K1 + K 2
8 = 2 × 8 + K2
8 − 16 = −8 = K 2
Thus
8 8
Y (s) = −
s + 2 ( s + 2) 2
y (t ) = 8e −2t − 8te −2t

25. 8( s + 1) K1 K2
Y (s) = = +
( s + 2) 2
s + 2 ( s + 2) 2
As before, we can solve for K2 in the usual manner.
8( s + 1)( s + 2) 2
K2 = = 8( s + 1) s=−2 = −8
( s + 2) 2
s =−2

26. 8( s + 1) 2 K1 8
( s + 2)
2
= ( s + 2) − ( s + 2) 2
( s + 2) 2
s+2 ( s + 2) 2
d [ 8( s + 1)] d [ ( s + 2 ) K1 − 8]
=
ds ds
8 = K1
8( s + 1) 8 8
Y (s) = = −
( s + 2) 2
s + 2 ( s + 2) 2
y (t ) = 8e −2t − 8te −2t

27. Unrepeated complex roots are solved similar to
the process for unrepeated real roots. That is you
multiply by one of the denominator terms in the
partial fraction and solve for the appropriate
constant.
Once you have found one of the constants, the
other constant is simply the complex conjugate.

28. 5.2 K K*
Z ( s) = 2 = +
s + 2 s + 5 ( s + 1 − j 2) ( s + 1 + j 2)
5.2( s + 1 − j 2)
K= = − j13
.
( s + 1 − j 2)( s + 1 + j 2) s =−1+ j 2
K = j13
*
.

29. 5.2 − j1.3 j1.3
Z ( s) = 2 = +
s + 2 s + 5 ( s + 1 − j 2) ( s + 1 + j 2)
e − j1.3 e j 1. 3
Z ( s) = +
( s + 1 − j 2) ( s + 1 + j 2)