106. The response h(t) of a linear time invariant system to an impulse δ(t), under initially relaxed condition is h(t) = e^–t + e^–2t. The response of this system for a unit step input u(t) is

1. 1. u(t) + e^–t + e^–2t

   
   
   

   2. (e^–t + e^–2t) u(t)

   
   
   

   3. (1.5 – e^–t – 0.5e^–2t) u(t)

   
   
   

   4. e^–t δ(t) + e^–2t u(t)

   
   
   

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1. View Hint View Answer Discuss in Forum

   Impulse response, H(s) = α h(t)
   

   =	1	+	1
   	(s + 1)		(s+ 2)

   
   

   Step response = L-1		1	+	1
   		(s + 1)		(s+ 2)

   
   

   = L-1			1	-	1		+		0.5	-	0.5
   			s		s + 2				s		(s + 2)

   
   = (1 – e^–t + 0.5 – 0.5e^–2t)u(t)
   = (1.5 – e^–t – 0.5e^–2t)u(t)

   Correct Option: C

   Impulse response, H(s) = α h(t)
   

   =	1	+	1
   	(s + 1)		(s+ 2)

   
   

   Step response = L-1		1	+	1
   		(s + 1)		(s+ 2)

   
   

   = L-1			1	-	1		+		0.5	-	0.5
   			s		s + 2				s		(s + 2)

   
   = (1 – e^–t + 0.5 – 0.5e^–2t)u(t)
   = (1.5 – e^–t – 0.5e^–2t)u(t)

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107. A zero mean random signal is uniformly distributed between limits – a and +a and its mean square value is equal to its variance. Then the r.m.s value of the signal is

1. 1. a

      √3

   
   
   

   2. a

      √2

   
   
   

   3. a√2

   
   
   

   4. a√3

   
   
   

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1. View Hint View Answer Discuss in Forum

   Variance =	(a(-a))²	=	4a²	=	a²
   	12		12		3

   
   R.M.S. value = √variance = a/√3

   Correct Option: A

   Variance =	(a(-a))²	=	4a²	=	a²
   	12		12		3

   
   R.M.S. value = √variance = a/√3

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108. Given two continuous time signals x(t) = e^–t and y(t) = e^–2t which exist for t > 0, the convolution z(t) = x(t) * y(t) is

1. 1. e^–t – e^–2t

   
   
   

   2. e^–3t

   
   
   

   3. e^+t

   
   
   

   4. e^–t + e^–2t

   
   
   

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1. View Hint View Answer Discuss in Forum

   z(t) = x(t) * y(t), Taking laplace transform both side
   Z(s) = X(s).Y(s)
   

   =	1	.	1	=	1	.	1
   	(s + 1)		(s + 2)		(s + 1)		(s + 2)

   
   L^–1{z(s)} = z(t) = e^–t – e^–2t

   Correct Option: A

   z(t) = x(t) * y(t), Taking laplace transform both side
   Z(s) = X(s).Y(s)
   

   =	1	.	1	=	1	.	1
   	(s + 1)		(s + 2)		(s + 1)		(s + 2)

   
   L^–1{z(s)} = z(t) = e^–t – e^–2t

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109. A low-pass filter with a cut-off frequency of 30Hz is cascaded with a high-pass filter with a cutt-off frequency of 20 Hz. The resultant system of filters will function as

1. 1. an all-pass filter

   
   
   

   2. an all-stop filter

   
   
   

   3. a band stop (band-reject) filter

   
   
   

   4. a band-pass filter

   
   
   

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1. View Hint View Answer Discuss in Forum

   

   Correct Option: D

   

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110. The Fourier transfom of asignal h(t) i s H(jω) = (2 cos ω) (sin 2 ω)/ω. The value of h(0) is

1. 1. 1/4

   
   
   

   2. 1/2

   
   
   

   3. 1

   
   
   

   4. 2

   
   
   

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1. View Hint View Answer Discuss in Forum

   

   Correct Option: C

   

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