Four signals each band limited to 5 kHz are sampled at twice the Nyqui...
Fm = 5 kHz, Nyquist Rate 2 x 5 = 10 kHz Since signal are sampled at twice the Nyquist rate so sampling rate 2 x 10 = 20 kHz. Total transmission bandwidth 4 x 20 = 80 kHz.
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Four signals each band limited to 5 kHz are sampled at twice the Nyqui...
Given information:
- Four signals, each band-limited to 5 kHz
- Sampled at twice the Nyquist rate
- PAM (Pulse Amplitude Modulation) samples are transmitted after time-division multiplexing
- Find the theoretical minimum transmission bandwidth of the channel
Nyquist rate:
- The Nyquist rate is the minimum sampling rate required to accurately reconstruct a signal from its sampled values
- It is given by the formula: Nyquist rate = 2 x maximum signal frequency
Maximum signal frequency:
- Each signal is band-limited to 5 kHz
- Therefore, the maximum signal frequency for each signal is 5 kHz
Sampling rate:
- The signals are sampled at twice the Nyquist rate
- Therefore, the sampling rate for each signal is 2 x 2 x 5 kHz = 20 kHz
Time-division multiplexing:
- The PAM samples from each of the four signals are transmitted over a single channel using time-division multiplexing
- This means that the samples from each signal are transmitted in a cyclic manner, with each sample taking a fixed amount of time
Minimum transmission bandwidth:
- The theoretical minimum transmission bandwidth of the channel is the sum of the individual bandwidths of each signal
- Each signal has a bandwidth of 5 kHz
- Therefore, the total bandwidth required for all four signals is 4 x 5 kHz = 20 kHz
However, since the signals are sampled at twice the Nyquist rate, the minimum transmission bandwidth of the channel is doubled:
- Minimum transmission bandwidth = 2 x 20 kHz = 40 kHz
Therefore, the correct answer is option D (80 kHz).
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