Consider a message signal with bandwidth 50KHZ. The amplitude of the message signal is uniformly distributed on [-2, 2]. This message signal is modulated and transmitted through a channel with 40DB attenuation and additive white noise with psd N, /2=10-12 W/Hz. If we require that the output SNR (the SNR after demodulation) be 50dB, in each of the following cases find the necessary transmitted power. DSB-SC amplitude modulation Conventional amplitude modulation with modulation index of 0.6. Phase modulation with phase deviation constant k, = 2 rad/V. Frequency modulation with modulation index of 3. а. b. с. d. Note 10log (P, / PR)= 40DB

Introductory Circuit Analysis (13th Edition)
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Author:Robert L. Boylestad
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Consider a message signal with bandwidth 50KHZ. The amplitude of the
message signal is uniformly distributed on [-2, 2]. This message signal is
modulated and transmitted through a channel with 40DB attenuation and additive
white noise with psd N, /2=10-12 W/Hz. If we require that the output SNR (the
SNR after demodulation) be 50dB, in each of the following cases find the
necessary transmitted power.
DSB-SC amplitude modulation
Conventional amplitude modulation with modulation index of 0.6.
Phase modulation with phase deviation constant k, = 2 rad/V.
Frequency modulation with modulation index of 3.
а.
b.
с.
d.
Note 10log (P, / PR)= 40DB
Transcribed Image Text:Consider a message signal with bandwidth 50KHZ. The amplitude of the message signal is uniformly distributed on [-2, 2]. This message signal is modulated and transmitted through a channel with 40DB attenuation and additive white noise with psd N, /2=10-12 W/Hz. If we require that the output SNR (the SNR after demodulation) be 50dB, in each of the following cases find the necessary transmitted power. DSB-SC amplitude modulation Conventional amplitude modulation with modulation index of 0.6. Phase modulation with phase deviation constant k, = 2 rad/V. Frequency modulation with modulation index of 3. а. b. с. d. Note 10log (P, / PR)= 40DB
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