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The telemetry signals, each of bandwidth 2 kHz are to be transmitted simultaneously by binary PCM. The maximum tolerable error in sample amplitudes is 0.2% of the peak signal amplitude. The signals must be sampled at least 20% above the Nyquist rate. Framing and synchronizing requires an additional 1% extra bits. The minimum possible data rate must be—
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- 273.64 kbits/sec
- 436.32 kbits/sec
- 936.64 kbits/sec
- None of the above
- 273.64 kbits/sec
Correct Option: B
If Vp is the peak sample amplitude, then
| error ≤ | = | ||
| 100 | 500 |
If L level are used, then each sample is separated by
| δ = | |
| L |
| = | = | = | ||||
| 2 | 2L | L | 500 |
⇒ L = 500 = 2n
n = 8.96 ≈ 9
This require 9 bits binary code per sample.
Nyquist rate. = 2kHz × 2 = 4 kHz
The actual sampling rate = 4 kHz X 120 = 4.8 kHz (˙.˙ 20% greater than the sampling rate)
Thus each signal has 4800 samples/sec, and each sample is coded by 9 bits. Therefore, each signal uses 9 × 4800 = 43.2 kbits/sec.
10 signals are multiplexed. Hence we need 43.2 × 10 = 432 k bit/sec.
Framing and synchronization bits
| = | = 4320 additional bits | |
| 100 |
= 432 k + 4320 = 436·32 k bits/sec.
