51. Calculate y(t) =	d2
    	x(t – 1) dt2

1. 1. ejω ω5

      1 + ω2

   
   
   

   2. ejω·ω

      1 + ω2

   
   
   

   3. – je-jω·ω5

      1 + ω2

   
   
   

   4. – ejω·ω

      1 + ω2

   
   
   

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1. View Hint View Answer Discuss in Forum

   If x(t) ← F.T.→ X(jω)
   

   then	d	x(t) ←F.T.→ jω X(jω)
   	dx

   
   

   and	d2	x(t) ←F.T.→ (jω)2 X(jω)
   	dt2

   
   

   and	d2	x(t – 1) ←F.T.→ (jω)2 X(jω) e– jω
   	dt2

   
   

   Therefore, Y(jω) = – ω2·	e– jω.jω3
   	1 + ω2

   
   

   or Y(jω) =	– je– jω.ω5
   	1 + ω2

   
   Hence, alternative (C) is the correct choice.

   Correct Option: C

   If x(t) ← F.T.→ X(jω)
   

   then	d	x(t) ←F.T.→ jω X(jω)
   	dx

   
   

   and	d2	x(t) ←F.T.→ (jω)2 X(jω)
   	dt2

   
   

   and	d2	x(t – 1) ←F.T.→ (jω)2 X(jω) e– jω
   	dt2

   
   

   Therefore, Y(jω) = – ω2·	e– jω.jω3
   	1 + ω2

   
   

   or Y(jω) =	– je– jω.ω5
   	1 + ω2

   
   Hence, alternative (C) is the correct choice.

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52. If x(t) ←F.T→	2 sin ω	= X(jω), when
    	1 + ω

    
    

    x(t) =		1, |t| < 1
    		0, otherwise

    
    then the Fourier transform of signal y(t)
    
    
    

1. 1. 4 sin2 ω

      ω2

   
   
   

   2. 8π sin2 ω

      ω2

   
   
   

   3. 2 sin2 ω

      πω2

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   First approach:
   y(t) = x(t) * x(t)
   Y(jω) = X(jω) X(jω)
   

   =	2 sin ω	.	2 sin ω
   	ω		ω

   
   

   =	4 sin2 ω
   	ω2

   
   Second approach
   Given, waveform
   

   slope =	2	= 1
   	2

   
   If
   y(t) ←F.T.→ Y(jω)
   then
   

   d2	y(t) ←F.T.→ (jω)2 Y(jω)
   dt2

   
   y′′(t) = δ(t + 2) – 2δ(t) + δ(t – 2)
   F{y′′(t)} = e^jω2 – 2 + e^–jω2
   

   = 2		e2jω + e–2jω		- 2
   		2

   
   = 2.cos 2ω – 2
   = 2(cos 2ω – 1)
   = 2(1 – 2 sin² ω – 1)
   = 2(– 2 sin² ω)
   = – 4 sin² ω
   Now, (– jω)² Y(jω) = – 4 sin² ω
   or
   

   Y(jω) =	4	sin2 ω
   	ω2

   
   Hence, alternative (A) is the correct choice.

   Correct Option: B

   First approach:
   y(t) = x(t) * x(t)
   Y(jω) = X(jω) X(jω)
   

   =	2 sin ω	.	2 sin ω
   	ω		ω

   
   

   =	4 sin2 ω
   	ω2

   
   Second approach
   Given, waveform
   

   slope =	2	= 1
   	2

   
   If
   y(t) ←F.T.→ Y(jω)
   then
   

   d2	y(t) ←F.T.→ (jω)2 Y(jω)
   dt2

   
   y′′(t) = δ(t + 2) – 2δ(t) + δ(t – 2)
   F{y′′(t)} = e^jω2 – 2 + e^–jω2
   

   = 2		e2jω + e–2jω		- 2
   		2

   
   = 2.cos 2ω – 2
   = 2(cos 2ω – 1)
   = 2(1 – 2 sin² ω – 1)
   = 2(– 2 sin² ω)
   = – 4 sin² ω
   Now, (– jω)² Y(jω) = – 4 sin² ω
   or
   

   Y(jω) =	4	sin2 ω
   	ω2

   
   Hence, alternative (A) is the correct choice.

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53. In case of man-made noise—

1. 1. receiver noise decreases as the receiver bandwidth is increased

   
   
   

   2. receiver noise increases as the receiver bandwidth is increased

   
   
   

   3. receiver noise increases as the receiver bandwidth is increased

   
   
   

   4. receiver noise is independent of bandwidth.

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

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54. Mean of X if
    

    fx(x) =		3(1 – x)2; for 0 ≤ x ≤ 1
    		0; otherwise

1. 1. 1

      3

   
   
   

   2. 1

      2

   
   
   

   3. 1

      4

   
   
   

   4. 1

      5

   
   
   

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1. View Hint View Answer Discuss in Forum

   μx = E[X] =		∞	xfx(x)dx
   		– ∞

   
   

   =		1	x.3(1 – x) 2dx
   		0

   
   

   = 3		1	(x – 2x2 + x3)dx
   		0

   
   

   = 3		x2	-	2x3	+	x4		1
   		2		3		4		0

   
   

   = 3		1	-	2	+	1
   		2		3		4

   
   

   = 3		6 - 8 + 3		=	1
   		12			4

   
   

   Correct Option: C

   μx = E[X] =		∞	xfx(x)dx
   		– ∞

   
   

   =		1	x.3(1 – x) 2dx
   		0

   
   

   = 3		1	(x – 2x2 + x3)dx
   		0

   
   

   = 3		x2	-	2x3	+	x4		1
   		2		3		4		0

   
   

   = 3		1	-	2	+	1
   		2		3		4

   
   

   = 3		6 - 8 + 3		=	1
   		12			4

   
   

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55. Variance of X if
    

    fx(x) =		3(1 – x)2; for 0 ≤ x ≤ 1
    		0; otherwise

1. 1. 3

      80

   
   
   

   2. 1

      30

   
   
   

   3. 2

      50

   
   
   

   4. 1

      6

   
   
   

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1. View Hint View Answer Discuss in Forum

   Variance, σ² x = E[x²] – (μx)²
   

   E[x2] = 3		1	x2 (1 – x)2dx
   		a

   
   

   = 3		1	(x2 – 2x3 + x4)dx
   		a

   
   

   E[x2] = 3		x3	-	2x4	+	x5		1
   		3		4		5		0

   
   

   = 3		1	-	1	+	1
   		3		2		5

   
   

   = 3		10 - 15 + 6
   		30

   
   = 3 × 1/30
   = 1/10
   Now,
   σ²_x = 1/10 – (1/4)²
   

   =	1	–	1	=	16 - 10
   	10		16		160

   
   = 6/160
   = 3/80

   Correct Option: A

   Variance, σ² x = E[x²] – (μx)²
   

   E[x2] = 3		1	x2 (1 – x)2dx
   		a

   
   

   = 3		1	(x2 – 2x3 + x4)dx
   		a

   
   

   E[x2] = 3		x3	-	2x4	+	x5		1
   		3		4		5		0

   
   

   = 3		1	-	1	+	1
   		3		2		5

   
   

   = 3		10 - 15 + 6
   		30

   
   = 3 × 1/30
   = 1/10
   Now,
   σ²_x = 1/10 – (1/4)²
   

   =	1	–	1	=	16 - 10
   	10		16		160

   
   = 6/160
   = 3/80

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