Consider an audio signal with spectral components in the range 300 to
3000 Hz. Assume that a sampling rate of 7000 samples per second will
be used to generate a PCM signal.

a) For SNR = 30 dB, what is the number of uniform quantization levels
needed?
b) What data rate is required?

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Clarification of Question by kosa-ga on 06 Dec 2002 05:44 PST

I am not sure what clarification is needed. But, this is a question on
data transmission and transmission media. The two major equations in
this topic are the shannon capacity formula and the nyquist formula
given respetively as;
C = B log (to base 2){1+S/N) and D=2Blog (base 2)K. 
I hope this helps.

Thank you.

Hi Kosa,

Does the SNR of 30dB relate to quantization noise or to channel noise?

Regards,  West

Well, the question is not specific on that, but I would think the dB
is related to  quantization noise.

Hi again Kosa,

Thank you for your answering comment. I agree that the context of the
question indicates that quantization noise is the only form of noise
to be considered in the problem.

The main reference for the following comment/answer is the text book
"Introduction to Communication Systems", third edition (1990)
published by Addison-Wesley.

1. SAMPLING RATE: The minimum Nyquist sampling rate is approximately
3000 * 2 = 6000 samples per second for the band-pass signal in
question. It can be assumed that the stated sampling rate of 7000
samples per second is sufficiently high to enable adequately low
aliasing error performance provided suitable filtering (band-limiting)
is applied before sampling.

2. QUANTIZATION NOISE: The random errors introduced in encoding the
signal are called quantization noise. It can be reduced by increasing
the number of quantizing levels. It can be shown that the peak
signal-to-noise (power) ratio in decibels is as follows:
S/N (pk qnt) = 4.8 + 20 log (to base 10) n, where n is the number of
quantizing levels.
If the binary code is used to code the digits corresponding to
quantizing levels, the equation for the peak signal-to-noise (power)
ratio in decibels is as follows:
     S/N (pk qnt) = 4.8 + 6m, where m is the number of bits in the
binary code.

For the case in question:
   30 dB = 4.8 + 6m
      6m = 30 - 4.8
       m = (30 - 4.8) / 6
       m = 5 bits (rounding up)
A 5-bit binary code will define 32 quantizing levels.

3. DATA RATE: The clock rate is as follows (ignoring overhead bits
such as those for clock and frame synchronization):
          Sampling rate * bits per sample = 7000 * 5 = 35000 bits per
second

The minimum bandwidth required for a data rate of 35000 bits per
second is 35000 / 2 = 17.5 kHz.

Note that the Hartley-Shannon law, C = B log (to base 2){1 + S/N} bps,
states the theoretical maximum information transmission rate for a
given channel. It does not enable the design of systems that will meet
the theoretical ideal.


I hope that you find the above answers your question.

Regards,   West