Consider an LTI system with transfer function H(s) = 1s(s

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Consider an LTI system with transfer function
H(s) = 1
s(s + 4)

If the input to the system is cos(3t) and the steady state output is A sin(3t + a), then the value of A is

1. 1/30
2. 1/15
3. 3/4
4. 4/3

Correct Option: B

H(s) =	1
	s(s + 4)

Input r(t) = cos 3t,

R(s) =	s
	s² + 9

Then C(s) = H(s). R(s)

C(s) =	s	1	=	1
C(s) =	s² + 9	s(s + 4)	=	(s + 4)(s² + 9)

Using partial fraction, we have

C(s) =	A	+	B	+	C
	s + 4		s - 3j		s + 3j

A =	1			=	1
	s² + 9		_s=-4		25

B =	1			=	1
	(s + 4)(s + 3j)		_s=-3j		(-6j)(4 - 3j)

C =	1			=	1
	(s + 4)(s - 3j)		_s=-3j		(-6j)(4 - 3j)

Now, C(s) =	1	+	1	+	1
	25(s + 4)		(4 + 3j)(-6j)(s - 3j)		(4 - 3j)(-6j)(s + 3j)

Inverse Laplace gives,

C(t) =	1	e^-4t +	1	[4e^3jt - 4e^-3jt - 3e^3jt - 3e^-3jt]
	25		150j

=	1	e^-4t +	1	[8 sin 3t + 6 cos 3t]
	25		150

=	1	e^-4t +	1		4	sin 3t -	3	cos 3t
	25		150		5		5

=	1	e^-4t +	1	sin (3t + α)
	25		15

where α = tan^-1 (3/4)
At steady-state t → ∞

and C(t) =	1	sin (3t + α)
	15

gives A =	1
	15