141. The biggest advantage of VSB system is—

1. 1. that it conserves bandwidth and overcomes the problem of low video frequency attenuation

   
   
   

   2. lesser power requirement.

   
   
   

   3. Its bandwidth conservation

   
   
   

   4. simplicity of transmitter circuitry.

   
   
   

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1. View Hint View Answer Discuss in Forum

   Refer synopsis

   Correct Option: A

   Refer synopsis

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142. In the above question, the total bandwidth required to transmit the AM wave is—
     

1. 1. 1.5 kHz
      

   
   
   

   2. 3 kHz
      

   
   
   

   3. 100 kHz
      

   
   
   

   4. 10 kHz

   
   
   

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1. View Hint View Answer Discuss in Forum

   Total bandwidth,
   BW = 2f_m = 2 × 1500 = 3000 = 3 kHz

   Correct Option: B

   Total bandwidth,
   BW = 2f_m = 2 × 1500 = 3000 = 3 kHz

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143. An audio signal 15 sin 2π(1500t) amplitude modulates 60 sin 2π(10⁶t) with carrier wave. The modulation in percentage is—

1. 1. 20%
      

   
   
   

   2. 25%
      

   
   
   

   3. 50%
      

   
   
   

   4. 75%

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given m(t) = 15 sin 2π (1500t, i.e., A_m = 15
   c(t) = 60 sin 2π (10⁶t), i.e., A_c = 60
   Modulation index,
   

   μ =	Am	=	15	=	1	= 25%
   	Ac		60		4

   
   

   Correct Option: B

   Given m(t) = 15 sin 2π (1500t, i.e., A_m = 15
   c(t) = 60 sin 2π (10⁶t), i.e., A_c = 60
   Modulation index,
   

   μ =	Am	=	15	=	1	= 25%
   	Ac		60		4

   
   

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144. In AM, if the modulation index is more than 100% then—
     

1. 1. Power of the wave increases
      

   
   
   

   2. Efficiency of transmission increases
      

   
   
   

   3. The wave gets distorted
      

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   Refer synopsis.
   

   Correct Option: C

   Refer synopsis.
   

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145. The number of AM broadcast stations that can be accommodated on a 100 kHz bandwidth for the highest modulating frequency of 5 kHz will be—

1. 1. 5
      

   
   
   

   2. 10
      

   
   
   

   3. 20
      

   
   
   

   4. 50
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   f_s = 2f_m = 2 × 5 = 10 kHz
   

   No. of stations  ≤	BW	=	100	= 10
   	fs		10

   Correct Option: B

   f_s = 2f_m = 2 × 5 = 10 kHz
   

   No. of stations  ≤	BW	=	100	= 10
   	fs		10

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