121. An input signal x(t) = 2 + 5sin(100πt) 2 is sampled with a sampling frequency of 400 Hz and applied to the system whose transfer function is represented by
     

     Y(z)	=	1		(1 - z-N)
     Z(z)		N		(1 - z-1)

     
     where, N represents the number of samples per cycle. The output y(n) of the system under steady state is

1. 1. 0

   
   
   

   2. 1

   
   
   

   3. 2

   
   
   

   4. 5

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

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122. A discrete system is represented by the difference equation
     

     	X1 (k + 1)			a	a - 1			X1(k)
     	X2 (k + 1 )			a + 1	a			X2(k)

     
     It has initial conditions X_{1 (0) = 1; X2 (0) = 0. The pole locations of the system for a = 1, are}

1. 1. 1 ± j0

   
   
   

   2. – 1 ± j0

   
   
   

   3. ± 1 + j0

   
   
   

   4. 0 ± j1

   
   
   

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1. View Hint View Answer Discuss in Forum

   From the given difference equation,
   

   A =		a	a - 1
   		a + 1	a

   
   The pole locations of the system for a = 1.
   

   Then A =		1	0
   		2	1

   
   SI – A| ⇒ (s – 1)² = 0
   S = 1 ± j0

   Correct Option: A

   From the given difference equation,
   

   A =		a	a - 1
   		a + 1	a

   
   The pole locations of the system for a = 1.
   

   Then A =		1	0
   		2	1

   
   SI – A| ⇒ (s – 1)² = 0
   S = 1 ± j0

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123. Consider an LTI system with impulse response h(t) = e^–5t u(t). If the output of the system is y(t) = e^–3t u(t) then the input, x(t), is given by

1. 1. e^–3t u(t)

   
   
   

   2. 2^–3t u(t)

   
   
   

   3. e^–5t u(t)

   
   
   

   4. 2^–5t u(t)

   
   
   

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1. View Hint View Answer Discuss in Forum

   h(t) = e^–5t u(t)
   Laplace transform gives,
   

   H(s) =	1
   	s + 5

   
   y(t) = e^–3t u(t) – e^–5t u(t)
   Laplace transform gives,
   

   Y(s) =	1	-	1	=	2
   	s + 3		s + 5		(s + 3)(s + 5)

   
   

   Then, X(s) =	Y(s)	=	2(s + 5)	=	2
   	H(s)		(s + 3)(s + 5)		s + 3

   
   Inverse Laplace transform gives,
   x(t) = L^–1 X (s) = 2e^–3t u(t)

   Correct Option: B

   h(t) = e^–5t u(t)
   Laplace transform gives,
   

   H(s) =	1
   	s + 5

   
   y(t) = e^–3t u(t) – e^–5t u(t)
   Laplace transform gives,
   

   Y(s) =	1	-	1	=	2
   	s + 3		s + 5		(s + 3)(s + 5)

   
   

   Then, X(s) =	Y(s)	=	2(s + 5)	=	2
   	H(s)		(s + 3)(s + 5)		s + 3

   
   Inverse Laplace transform gives,
   x(t) = L^–1 X (s) = 2e^–3t u(t)

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124. 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. 1/30

   
   
   

   2. 1/15

   
   
   

   3. 3/4

   
   
   

   4. 4/3

   
   
   

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1. View Hint View Answer Discuss in Forum

   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
   	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	[4e3jt - 4e-3jt - 3e3jt - 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

   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
   	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	[4e3jt - 4e-3jt - 3e3jt - 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

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125. Let ƒ(t) be a continuous time signal and let F(ω) be its Fourier Transform defined by
     
     What is the relationship between ƒ(t) and g(t)?

1. 1. g(t) would always be proportional to ƒ(t)

   
   
   

   2. g(t) would be proportional to ƒ(t) if ƒ(t) is an even function

   
   
   

   3. g(t) would be proportional to ƒ(t) only if ƒ(t) is a sinusoidal function

   
   
   

   4. g(t) would never be proportional to ƒ(t)

   
   
   

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1. View Hint View Answer Discuss in Forum

   The relation of time domain with frequency domain is given by fourier transform.
   

   Correct Option: B

   The relation of time domain with frequency domain is given by fourier transform.
   

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