Derive the transfer function the following differential equation the form

1. Find Laplace transform F(s) of f(t) = e−2tcos(3t) + 5 e−2tsin(3t).

Hint

F(s) = L�e−2tcos(3t)�	+ L�
ω=3	;		k=5, ω=3
a=2	;		a=2
F(s) =		3

n=1

n=2

;

n=3

F(s) = 10s3 + 5s2 + 2s − 24

F(s) = 2L�	�t e3t	�
a=−3	n=1	;	n=2
a=−3	a=−3	;	a=−3
1 1 (s − 3)3				2
1 1 (s − 3)3				(s − 3)3

4. Find Laplace transform F(s) of f(t) = 6 e5tcos(2t) − e7t.

Hint

6. Apply partial fraction expansion to the expression:

	4s + 7
F(s) =	(s + 1)(s2+ 5s + 16)

(s + 1)(s2+ 5s ~~+ 16)~~ = A(s2 + 5s + 16) + (Bs + C)(s + 1)
4s + 7 = A(s2+ 5s + 16) + (Bs + C)(s + 1) (1)

• Apply lims→−1 to find A:
4 × −1 + 7 = A((−1)2+ 5 × (−1) + 16) + 0 A = 1
4

4s + 7 =	�	s2+
7 = 4 + C		∴	C = 3.
				∴
F(s) =	1
F(s) =	4 s + 1 +

Solution

• Take the Laplace transform of the both sides and find Y (s)

sY (s) − y(0) = 3 1 s− 2 1 s2

3s − 2 s3	=A s+ B s2 + C	(1)
3s − 2 = As2+ Bs + C		(1)

In order to find A, we differentiate both sides of equation (2):

Y (s) = 3 s2 − 2 s3.	(3)

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