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Four signals each band limited to 5 kHz are sampled at twice the Nyquist rate. The resulting PAM samples are transmitted over a single channel after time division multiplexing. The theoretical minimum transmissions bandwidth of the channel should be equal to
  • a)
    5 kHz
  • b)
    20 kHz
  • c)
    40 kHz
  • d)
    80 kHz
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
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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Four signals each band limited to 5 kHz are sampled at twice the Nyquist rate. The resulting PAM samples are transmitted over a single channel after time division multiplexing. The theoretical minimum transmissions bandwidth of the channel should be equal toa)5 kHzb)20 kHzc)40 kHzd)80 kHzCorrect answer is option 'D'. Can you explain this answer?
Question Description
Four signals each band limited to 5 kHz are sampled at twice the Nyquist rate. The resulting PAM samples are transmitted over a single channel after time division multiplexing. The theoretical minimum transmissions bandwidth of the channel should be equal toa)5 kHzb)20 kHzc)40 kHzd)80 kHzCorrect answer is option 'D'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about Four signals each band limited to 5 kHz are sampled at twice the Nyquist rate. The resulting PAM samples are transmitted over a single channel after time division multiplexing. The theoretical minimum transmissions bandwidth of the channel should be equal toa)5 kHzb)20 kHzc)40 kHzd)80 kHzCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Four signals each band limited to 5 kHz are sampled at twice the Nyquist rate. The resulting PAM samples are transmitted over a single channel after time division multiplexing. The theoretical minimum transmissions bandwidth of the channel should be equal toa)5 kHzb)20 kHzc)40 kHzd)80 kHzCorrect answer is option 'D'. Can you explain this answer?.
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