81. The mean square power of the power spectral density given is R_x(τ)—

1. 1. NB

   
   
   

   2. 1

      NB

   
   
   

   3. NB cos 2πBτ

   
   
   

   4. 	1	cos 2πBτ
      	NB

   
   
   

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1. View Hint View Answer Discuss in Forum

   The mean square power,
   R_x(0) = NB sin c 2πBτ
   

   = NB	sin 2πBτ
   	2πBτ

   
   

   = NB	cos 2πBτ.2πB	⎪τ = 0
   	2πB

   
   = NB.

   Correct Option: A

   The mean square power,
   R_x(0) = NB sin c 2πBτ
   

   = NB	sin 2πBτ
   	2πBτ

   
   

   = NB	cos 2πBτ.2πB	⎪τ = 0
   	2πB

   
   = NB.

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82. The power spectral density is shown below then R_x(τ) is—
    
    
    

1. 1. NB sin c (2πBτ)

   
   
   

   2. 	1	sin c (2πBτ)
      	NB

   
   
   

   3. NB sin c (πBτ)

   
   
   

   4. 	1	sin c (πBτ)
      	NB

   
   
   

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1. View Hint View Answer Discuss in Forum

   Rxx(τ) =	1		∞	Sx(ω) ejωt .dω
   	2π		0

   
   

   Given that Sx(ω) =		N	;|ω| = 2πB
   		2
   		0; otherwise

   
   

   so, Rx(τ) =	1		2πB		N	· ejωt dω
   	2π		– 2πB		2

   
   

   =	N			ejωt			2πB
   	4π			jτ			– 2πB

   
   

   =	N			eej2πBτ – e– j2πBτ
   	2πτ			2τ

   
   

   =	N	sin 2πBτ
   	2πτ

   
   

   = NB	sin (2πBτ)	2πBτ {∵sin x/x = sinc x}
   	2πτ

   
   = NB sinc (2πBτ)

   Correct Option: A

   Rxx(τ) =	1		∞	Sx(ω) ejωt .dω
   	2π		0

   
   

   Given that Sx(ω) =		N	;|ω| = 2πB
   		2
   		0; otherwise

   
   

   so, Rx(τ) =	1		2πB		N	· ejωt dω
   	2π		– 2πB		2

   
   

   =	N			ejωt			2πB
   	4π			jτ			– 2πB

   
   

   =	N			eej2πBτ – e– j2πBτ
   	2πτ			2τ

   
   

   =	N	sin 2πBτ
   	2πτ

   
   

   = NB	sin (2πBτ)	2πBτ {∵sin x/x = sinc x}
   	2πτ

   
   = NB sinc (2πBτ)

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83. The joint pdf of random variable x and y is given by f_xx(x, y) = ke^{– (ax + by)} u(x).u(y)

1. 1. K = ab

   
   
   

   2. K =	a
      	b

   
   
   

   3. K =	1
      	ab

   
   
   

   4. K =	b
      	a

   
   
   

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1. View Hint View Answer Discuss in Forum

   Fxy(x, y) =		∞		∞	fxy(x, y)dx dy
   		-∞		-∞

   
   

   1 =		∞		∞	ke–(ax + by) u(x).u(y)dy.dx
   		-∞		-∞

   
   

   1 =		∞		∞	ke–ax.e–by.dydx
   		0		0

   
   

   1 =	k
   	ab

   
   or k = ab

   Correct Option: A

   Fxy(x, y) =		∞		∞	fxy(x, y)dx dy
   		-∞		-∞

   
   

   1 =		∞		∞	ke–(ax + by) u(x).u(y)dy.dx
   		-∞		-∞

   
   

   1 =		∞		∞	ke–ax.e–by.dydx
   		0		0

   
   

   1 =	k
   	ab

   
   or k = ab

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84. If the variance of the random variable X is σ2 x then the variance of – kx is—

1. 1. Kσ²_x

   
   
   

   2. – Kσ²_x

   
   
   

   3. K²σ²_x

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   We know that
   σ²_x = E[x²] – [μ_x]²
   

   =		∞	x2 fx(x)dx -			∞	xfx(x)dx		2
   		-∞				-∞

   
   Let, new variance,
   

   σ2 x′ =		∞	(– kx)2fx(x)dx -			∞	(– kx)fx(x)dx		2	{∵ x′ = – kx}
   		-∞				-∞

   
   or
   

   σ2 x′ = k2		∞	x2fx(x)dx - k2			∞	xfx(x)dx		2
   		-∞				-∞

   
   σ² x′ = k²σ²_x

   Correct Option: C

   We know that
   σ²_x = E[x²] – [μ_x]²
   

   =		∞	x2 fx(x)dx -			∞	xfx(x)dx		2
   		-∞				-∞

   
   Let, new variance,
   

   σ2 x′ =		∞	(– kx)2fx(x)dx -			∞	(– kx)fx(x)dx		2	{∵ x′ = – kx}
   		-∞				-∞

   
   or
   

   σ2 x′ = k2		∞	x2fx(x)dx - k2			∞	xfx(x)dx		2
   		-∞				-∞

   
   σ² x′ = k²σ²_x

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85. The spectral density of random process, where
    R_xx(τ) = Ke^{– 3|τ|}

1. 1. K

      3 + ω2

   
   
   

   2. 6K

      9 + ω2

   
   
   

   3. 3K

      6 + ω2

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

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