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)