146. The AM broadcast band is given by—

1. 1. 10 kHz to 30 kHz
      

   
   
   

   2. 500 kHz to 1500 kHz
      

   
   
   

   3. 2 kHz to 30 kHz
      

   
   
   

   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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147. In an AM wave, the total power content is 600 W and that of each sideband is 75 W. The modulation index is—

1. 1. 53.3%
      
      

   
   
   

   2. 60.7%
      

   
   
   

   3. 81.6%
      

   
   
   

   4. 40.3%

   
   
   

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1. View Hint View Answer Discuss in Forum

   We know that
   

   PT = Pc		1 +	μ2
   			2

   
   

   PT = Pc + Pc.	μ2
   	2

   
   Where,
   P_T = Total Power
   P_c = Carrier Power
   

   Pc.	μ2
   	2

   = Power of both sideband
   600 = P_c + 2 × 75
   or
   P_c = 600 - 150 = 450W
   Now,
   

   150 = Pc.	μ2
   	2

   
   

   150 = 450.	μ2
   	2

   
   or
   

   μ2 =	2
   	3

   
   or
   

   μ =	√	2	= 0.816
   		3

   
   or
   μ = 81.6 %

   Correct Option: C

   We know that
   

   PT = Pc		1 +	μ2
   			2

   
   

   PT = Pc + Pc.	μ2
   	2

   
   Where,
   P_T = Total Power
   P_c = Carrier Power
   

   Pc.	μ2
   	2

   = Power of both sideband
   600 = P_c + 2 × 75
   or
   P_c = 600 - 150 = 450W
   Now,
   

   150 = Pc.	μ2
   	2

   
   

   150 = 450.	μ2
   	2

   
   or
   

   μ2 =	2
   	3

   
   or
   

   μ =	√	2	= 0.816
   		3

   
   or
   μ = 81.6 %

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148. The modulation index of over-modulated wave is—
     

1. 1. < 1
      

   
   
   

   2. > 1
      

   
   
   

   3. infinity
      

   
   
   

   4. 1

   
   
   

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1. View Hint View Answer Discuss in Forum

   Refer synopsis.
   

   Correct Option: B

   Refer synopsis.
   

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149. The intermediate frequency of a superheterodyne receiver is 150 kHz. If the image frequency of a station is 2100 kHz, its actual frequency is—

1. 1. 750 kHz
      

   
   
   

   2. 900 kHz
      

   
   
   

   3. 1200 kHz
      

   
   
   

   4. 1010 kHz

   
   
   

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1. View Hint View Answer Discuss in Forum

   Image frequency f_img = fc + 2f_IF
   2100 kHz = f_c + 2 × 450 kHz
   or f_c = 2100 – 900 = 1200 kHz

   Correct Option: C

   Image frequency f_img = fc + 2f_IF
   2100 kHz = f_c + 2 × 450 kHz
   or f_c = 2100 – 900 = 1200 kHz

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150. An AM wave is given by
     e_AM = 10(1 + 0·4 cos 10³t + 0·3 cos 10⁴t) cos 10⁶t
     The modulation index is—

1. 1. 0.4
      

   
   
   

   2. 0.5
      

   
   
   

   3. 0.3
      

   
   
   

   4. 0.9

   
   
   

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1. View Hint View Answer Discuss in Forum

   Try yourself.

   Correct Option: B

   Try yourself.

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