31. In NBFM, the maximum modulation frequency is 3 kHz and maximum deviation is 5 kHz. The modulation index is—

1. 1. less than 1

   
   
   

   2. equal to 1

   
   
   

   3. greater than 1

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   For NBFM modulation index < 1

   Correct Option: C

   For NBFM modulation index < 1

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32. The disadvantage of FM over AM is that—

1. 1. Noise is very high for high frequency signals

   
   
   

   2. Larger bandwidth is required

   
   
   

   3. High modulating power is required

   
   
   

   4. None of these
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   Refer synopsis.

   Correct Option: B

   Refer synopsis.

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33. The power in normalized message signal m_n (t) would be—

1. 1. 0.693

   
   
   

   2. 0.542

   
   
   

   3. 0.254

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given message signal
   m(t) = 9 cos 20πt + 7 cos 60πt
   |m(t )|_max = 9 + 7 = 16 at t = 0
   Therefore, the power in normalized message signal
   

   m(t) =	Power in the signal m(t)
   	Max. power in signal m(t)

   
   

   =	92/2 + 72/2	= 0.254
   	(16)2

   Correct Option: C

   Given message signal
   m(t) = 9 cos 20πt + 7 cos 60πt
   |m(t )|_max = 9 + 7 = 16 at t = 0
   Therefore, the power in normalized message signal
   

   m(t) =	Power in the signal m(t)
   	Max. power in signal m(t)

   
   

   =	92/2 + 72/2	= 0.254
   	(16)2

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34. The double-sided spectrum of x_c(t) would be—

1. 1. A
      

   
   
   

   2. B
      

   
   
   

   3. C
      

   
   
   

   4. None of the above
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   Try yourself.

   Correct Option: B

   Try yourself.

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35. An AM modulator has output x_c (t) = 40 cos 400 πt + 4 cos 360 πt + 4 cos 440 πt The AM modulation efficiency is—

1. 1. 0.01

   
   
   

   2. 0.02

   
   
   

   3. 0.03

   
   
   

   4. 0.04

   
   
   

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1. View Hint View Answer Discuss in Forum

   The given modulator output
   x_c(t) = 40 cos 400 πt + 4 cos 360 πt + 4 cos 440πt x_c(t) may be written as
   

   xc(t) = 40		1 +	1	cos40πt		cos400πt
   			5

   ...(1)
   On comparing equation (1) with standard equation
   x_c(t )= A_c (1 + μ cos ω_mt) cos ω_ct
   

   μ =	1	,Ac = 40
   	5

   
   
   
   
   

   Efficiency =	μ2	=		(1/5)2		=	1/25	=	1
   	2 + μ2			2 + (1/5)2			50 + 1/25		51

   
   = 0.02
   Hence alternative (B) is correct choice.
   

   Correct Option: B

   The given modulator output
   x_c(t) = 40 cos 400 πt + 4 cos 360 πt + 4 cos 440πt x_c(t) may be written as
   

   xc(t) = 40		1 +	1	cos40πt		cos400πt
   			5

   ...(1)
   On comparing equation (1) with standard equation
   x_c(t )= A_c (1 + μ cos ω_mt) cos ω_ct
   

   μ =	1	,Ac = 40
   	5

   
   
   
   
   

   Efficiency =	μ2	=		(1/5)2		=	1/25	=	1
   	2 + μ2			2 + (1/5)2			50 + 1/25		51

   
   = 0.02
   Hence alternative (B) is correct choice.
   

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