Amplitude Modulation and Hz

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  • Topic: Amplitude modulation, Frequency modulation, Baseband
  • Pages : 5 (1126 words )
  • Download(s) : 1074
  • Published : August 29, 2009
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1. A 1.4M Hz carrier is modulated by a music signal that has a frequency components from 20 Hz to 10k Hz. Determine the range of frequencies generated for the upper and lower sidebands.

Solution:
Upper Side Band = 1, 400, 000 Hz + 20 Hz = 1,400, 020 Hz
         Up to  = 1, 400, 000 Hz + 10, 000 Hz = 1, 410, 000 Hz        
Lower Side Band = 1, 400, 000 Hz – 10, 000 Hz = 1, 390, 000 Hz          Up to  = 1, 400, 000 Hz – 20 Hz = 1, 399, 980 Hz

2. Determine the %m for the following conditions for an un modulated carrier of 80V peak to peak.

Given:
Maximum Peak to Peak Carrier = 100
Minimum Peak to Peak Carrier = 60
Solution:    %m = [(B – A) / (B + A)] x 100%
        %m = [(100 - 60) / (100 + 60)] x 100% = 25%

3.    A 500W carrier is to be modulated to a 90% level. Determine the total transmitted power.
Solution:    Pt = Pc [ 1 + (m2/2) ]
            = 500W [ 1 + (0.92/2) ] = 702.5W

4.    An intelligence signal is amplified by a 70% efficient amplifier before being combined with a 10kW carrier to generate the AM signal. If it is desired to operate at 100% modulation, what is the dc input power to the final intelligence amplifier?

Solution: The efficiency of an amplifier is the ratio of AC output power to DC input power. To fully modulate a 10kW carrier requires 5kW of intelligence. Therefore, to provide 5kW of sideband (intelligence) power through a 70% efficient amplifier requires a dc input of:         5kW / 0.70 = 7.14kW

5.    A transmitter with a 10kW carrier transmits 11.2kW when modulated with a single sine wave. Calculate the modulation index. If the carrier is simultaneously modulated with another sine wave at 50% modulation, calculate the total transmitted power.

Solution:    Pt = Pc [ 1 + (m2/2) ]
       11.2kW = 10kW [ 1 + (m2/2) ]
         m = 0.49
              meff = ²√ (m1² + m2²)
             = ²√ (o.49² + 0.5²)
             = 0.7
        Pt = Pc [ 1 + (m2/2) ]
            = 10kW [ 1 + (0.7 ²/2) ]
            = 12.45kW

6.    A TRF receiver is to be designed with a single tuned circuit using a 10μH inductor. The ideal 10k Hz BW is to occur at 1100k Hz. Determine the required Q.
Solution:    Q = fr / BW
            = 1100k Hz / 10k Hz = 110

7.    The antenna receives an 8μV signal into its 50Ω input impedance. Calculate the input power in watts.

Solution: P = [(V²)/ R] = (8μV²) / 50Ω = 1.28x10-12W
8.    Determine the bandwidth required to transmit an FM signal with fi = 10k Hz and a maximum deviation δ = 20k Hz.

Solution:     mf = δ/fi = 20k Hz / 10k Hz = 2

9.    Determine the relative total power of the carrier and side frequencies when mf = 0.25 for a 10kW FM transmitter.
Solution: For mf = 0.25, the carrier is equal to 0.98 times its un modulated amplitude and the only significant sideband is J1, with a relative amplitude of 0.12 therefore, since power is proportional to the voltage squared, the carrier power is:

    (0.98)² * 10kW = 9.604kW
and the power of each sideband is:
    (0.12)² * 10kW = 144 W
The total power is:
    9604 W + 144 W + 144 W = 9.892 kW

10.    Determine the worst case output S/N for a broadcast FM program that has a maximum intelligence frequency of 5k Hz. The input S/N is 2.

Solution: The input S/N = 2 means that the worst case deviation is about ½ rad. Therefore:     δ = Φ * fi = 0.5 * 5k Hz = 2.5k Hz     Since full volume in broadcast FM corresponds to a 75k Hz  deviation, this 2.5k Hz worst case noise deviation means that the output S/N is         75k Hz / 2.5k Hz = 30

11.    A certain FM receiver provides a voltage gain of 200,000 (106 dB) prior to its limiter. The limiter’s quieting voltage is 200mV. Determine the receiver’s sensitivity.
Solution: To reach quieting, the input must be...
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