*Drawback of using PAM method is .17 a. Bandwidth is very large as compared to modulating...

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*Drawback of using PAM method is .17 a. Bandwidth is very large as compared to modulating signal b. Varying amplitude of carrier varies the peak power required for transmission O c. Due to varying amplitude of carrier, it is difficult to remove noise at receiver O d. All of the above O Calculate the Nyquist rate for sampling when a continuous time signal is given 18 *by x(t) = 5 cos 100nt +10 cos 200nt - 15 cos 300mt a. 300Hz O b. 600Hz O c. 150Hz O d. 200Hz O The desired signal of maximum frequency Wm centered at frequency W=0.19 *may be recovered if a. The sampled signal is passed through low pass filter O b. Filter has the cut off frequency Wm c. Both a and b d. None of the above O A distorted signal of frequency fm is recovered from a sampled signal if the .20 sampling frequency fs is a. fs> 2fm O b. fs < 2fm O c. fs = 2fm O d. fs 2 2fm O *Drawback of using PAM method is .17 a. Bandwidth is very large as compared to modulating signal b. Varying amplitude of carrier varies the peak power required for transmission O c. Due to varying amplitude of carrier, it is difficult to remove noise at receiver O d. All of the above O Calculate the Nyquist rate for sampling when a continuous time signal is given 18 *by x(t) = 5 cos 100nt +10 cos 200nt - 15 cos 300mt a. 300Hz O b. 600Hz O c. 150Hz O d. 200Hz O The desired signal of maximum frequency Wm centered at frequency W=0.19 *may be recovered if a. The sampled signal is passed through low pass filter O b. Filter has the cut off frequency Wm c. Both a and b d. None of the above O A distorted signal of frequency fm is recovered from a sampled signal if the .20 sampling frequency fs is a. fs> 2fm O b. fs < 2fm O c. fs = 2fm O d. fs 2 2fm O