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A signal has a bandwidth of 4 KHz. If every sample is digitized using 16 bits and the digital
speech is to be transmitted over a communication channel, what is the minimum
bandwidth requirement of the channel?
Answers
Answer:
Simplest calculation will be twice the frequency of the signal to be sent.
If BW of signal to be transmitted is 4 KHz then BW of the channel should be 8 KHz.
It means that the signal will be sampled twice only. Lot of information would be lost in this process. For speech to be transmitted with significant amount of information is retrieved, the speech signal should be sampled atleast 10 to 12 times the frequency range.
if frequency BW is 4 KHz, than channel BW should be in the range of 4x10x8- 320 Kbps.
For a high fidelity music signal, sampling rate would be higher and thus maximum information in the original signal shall be reproduced at the receiving end.
please mark as brilliant answer
256 kbps
Given:
- A signal has a bandwidth of 4 KHz,
- every sample is digitized using 16 bits and the digital.
- speech is transmitted over the communication channel.
To Find:
- To find the minimum bandwidth requirement of a channel .
Solution:
Here the sampling frequency is not given,
So, we'll assume it as the sampled at the Nyquist rate,
that is : f.s = 2 f.m
fm : The Maximum frequency present at modulating signal.
Thus,
For given band--limited signal with the frequency of 4 kHz,
so, the sampling frequency will be:
fs = 2 x 4 = 8 kHz
L = 256, Number of bits are :
n = log2 256 = log2 28
n = 8 bits
Now,
Rb = 4 x n x fs = 4 x 8 x 8000
Rb = 256 kbps
Therefore, the answer is 256 kbps
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