# Amplitude Modulation and Hz

**Topics:**Amplitude modulation, Frequency modulation, Baseband

**Pages:**5 (1126 words)

**Published:**August 29, 2009

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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