A signal has a single-sided bandwidth of W = 35 kHz. If this signal is sampled a

Question

A signal has a single-sided bandwidth of W = 35 kHz. If this signal is sampled at the Nyquist rate (f_s = 2W samples/sec) and PAM encoded what is the mini mu m required bandwidth needed for a channel to transmit this signal? B = 70 KHz B = 140 KHz B = 52.5 KHz B = 35 KHz B= infinity B= 17.5 KHz A signal has a single-sided bandwidth of W = 35 kHz. If this signal is sampled at the Nyquist rate (f_s = 2W samples/sec) and is PCM encoded with 2 bits. What is the mini mu m required bandwidth needed for a channel to transmit this new signal? B= infinity B= 140 KHz B = 35 KHz B = 70 KHz B= 17.5 KHz B = 52.5 KHz

Explanation / Answer

given signal bandwidth is 35 khz it is sampled by twice the nuquist frequency

hence total bandwidth is 4 times of signal bandwidth

hence band width is 4*35=140khz

4.)

given signal bandwidth is 35 khz it is sampled by twice the nuquist frequency

hence total bandwidth is 4 times of signal bandwidth

but pcm is encoded with 2 bits hence total bandwidth is 35*4*2=280 kbps

but option is not there

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