31. The raised cosine pulse x(t) is defined as
    

    x(t) =		1/2(cos ωt + 1),		- π/ω ≤ t ≤ π/ω
    		0,		otherwise

    
    The total energy of x(t) is

1. 1. 3π

      4ω

   
   
   

   2. 3π

      8ω

   
   
   

   3. 3π

      ω

   
   
   

   4. 2π

      2ω

   
   
   

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   Correct Option: A

   

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32. The relationship between the input x(t) and output y(t)of a causal system is defined as
    

    d²y(t)	-	dy(t)	- 2y(t) = 4x(t) + 5	dx(t)
    dt²		dt		dt

    
    The impulse response of system is

1. 1. 3e^{– t} u(t) + 2e^2t u(– t)

   
   
   

   2. (3e^{– t} + 2e^2t) u(t)

   
   
   

   3. 3e^{– t} u(t) – 2e^2t u(– t)

   
   
   

   4. (3e^{– t} – 2e^2t) u(– t)

   
   
   

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   Y(s) (s² – s – 2) = X (s) (5s – 4)
   

   H(s) =	Y(s)	=	5s - 4
   	X(s)		s ² - s - 2

   
   

   =	3	+	2
   	s + 1		s - 2

   
   h(t) = 3e^{– t} u(t) + 2e^2t u(t).

   Correct Option: B

   Y(s) (s² – s – 2) = X (s) (5s – 4)
   

   H(s) =	Y(s)	=	5s - 4
   	X(s)		s ² - s - 2

   
   

   =	3	+	2
   	s + 1		s - 2

   
   h(t) = 3e^{– t} u(t) + 2e^2t u(t).

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33. The two inputs to an analogue multiplier are x(t) and y(t) with Fourier transform X(ƒ) and Y(ƒ) respectively. The output z(t) will have a transform Z(ƒ) given by

1. 1. X(ƒ).Y(ƒ)

   
   
   

   2. X(ƒ) + Y(ƒ)

   
   
   

   3. X(ƒ) /Yƒ

   
   
   

   4. 

   
   
   

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   Multiplication of two signals in frequency domain is equivalent to their convolution in time domain.
   

   Correct Option: D

   Multiplication of two signals in frequency domain is equivalent to their convolution in time domain.
   

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34. Consider a continuous-time signal x(t) whose magnitude and phase spectra are as follows:
    | X (jω)| = 2{u(ω + 3) – u(ω – 3)},
    

    ∠X (jω) =	- 3ω	+ π
    	2

    
    The value of t, for which x(t) = 0, is

1. 1. kπ	+	3	,for = 0, 1, 2, 3,...
      3		2

   
   
   

   2. kπ	+	3	,for = 1, 2, 3,...
      3		2

   
   
   

   3. kπ	+	2	,for = 0, 1, 2, 3,...
      3		3

   
   
   

   4. kπ	+	2	,for = 1, 2, 3,...
      3		3

   
   
   

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   Correct Option: B

   

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35. A real and non-negative signal x(t) has Fourier transform X(jω). The following facts are given 1. F^{– 1}{(1 + jω) X(jω)} = Ae^{– 2t}u(t), where A is constant
    2.
    The signal x(t) may be

1. 1. √12(e^-2t - e^-t)u(t)

   
   
   

   2. √12(e^-t - e^-2t)u(t)

   
   
   

   3. √3(e^-t - e^-2t)u(t)

   
   
   

   4. √3(e^-t - e^-2t)u(t)

   
   
   

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   Y(jω) =	1	-	1	=	1
   	3 + jω		4 + jω		(3 + jω)(4 + jω)

   
   

   X(jω) =	Y(jω)	=	1
   	H(jω)		4 + jω

   
   ⇒ x(t) = e^{– 4t}u(t)

   Correct Option: B

   Y(jω) =	1	-	1	=	1
   	3 + jω		4 + jω		(3 + jω)(4 + jω)

   
   

   X(jω) =	Y(jω)	=	1
   	H(jω)		4 + jω

   
   ⇒ x(t) = e^{– 4t}u(t)

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