Tutorials - Digital Signal Processing: An Introduction with MATLAB and Applications - page 255

U ð x Þ¼ pd ð x Þþ jx relation d ð 2pf Þ¼ 2p d ð f Þ from

Therefore, [Using the

Tables.]
It is apparent that U(x) = U(s)| s=jx .
3. The function t n u(t) has no Fourier transform, but it has Laplace transform
given by

n!

s n þ 1 .

Tutorial 21

Q: Expand the following expression using partial fraction expansion, then find
y ð t Þ : Y ð s Þ¼ 2s 2 þ 13s þ 12

ð s þ 1 Þð s 2 þ 5s þ 6 Þ :

To find the inverse Laplace transform ðL 1 Þ for an expression of the Solution:

form:

a 0 þ a 1 s þþ a m s m
ð s p 1 Þð s p 2 Þð s p n Þ ;

Y ð s Þ¼ N ð s Þ

D ð s Þ ¼

we apply partial fraction expansion rules as follows:
Y ð s Þ¼ c 1
s p 1

þ c 2
s p 2

þþ c n

s p n :

Notes:

1. If m = degree [N(s)] C n = degree [D(s)], perform long division first.
2. For complex conjugate poles p k+1 = p * we have c k+1 = c * .
3. For a pole of multiplicity r at s = p k there will be r terms:
c k1 /(s - p k )+c k2 /(s - p k ) 2 + + c kr /(s - p k ) r in the expansion.

Now back to the main question.

ð s þ 1 Þð s p 1 Þð s p 2 Þ ) s 2 þ 5s þ 6 ¼ 0 ) p 1 ; 2 ¼ 5

p

2s 2 þ 13s þ 12 5 2 4 6

Y ð s Þ¼

2

¼ 2 ; 3 :

Hence, Y ð s Þ¼ 2s 2 þ 13s þ 12

ð s þ 1 Þð s þ 2 Þð s þ 3 Þ . Let Y ð s Þ¼ a s þ 1 þ b s þ 2 þ c s þ 3 .

2s 2 þ 13s þ 12

ð s þ 1 Þð s þ 2 Þð s þ 3 Þ ¼ a s þ 1 þ b s þ 2 þ c

)

s þ 3

To find a do the following:

ð s þ 1 Þð s þ 2 Þð s þ 3 Þ ¼ a þ b ð s þ 1 Þ

2s 2 þ 13s þ 12

s þ 2 þ c ð s þ 1 Þ

1. Multiply by ð s þ 1 Þ : .

s þ 3

ð 1 þ 2 Þð 1 þ 3 Þ ¼ a þ 0 þ 0 ! a ¼ 2 .

2 13 þ 12

2. Put s =-1: