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I am confused with regard to how bandwidth increases with increase in number of bits transmitted for given duration.

1) Usually say in PCM system, when we increase the number of bits (n) for a given sampling frequency (fs), the bitrate increases and minimum transmission bandwidth thus increases.

Rb = nFs and Bandwidth = Rb/2.

2)While in M-ary schemes,when the number of bits is increased the bandwidth decreases because in time domain the signal is expanding,so obviously in frequency domain the overall width should reduce.

I am not able to understand the context in which either is true!

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    $\begingroup$ Looks like you started work on a homework problem set in a course on digital communication systems. Four questions today! When is the homework due? $\endgroup$
    – Dilip Sarwate
    Dec 7 '15 at 20:58
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The required bandwidth depends on the symbol rate (or baud rate). If you have a binary system (i.e. two symbols) then the bit rate and the symbol rate are equal. If you use $M>2$ symbols, you can transmit more than 1 bit per symbol and your symbol rate, hence also the required bandwidth, is decreased. The problem with increasing the number of symbols is that you need more energy to keep the average distance between the symbols (i.e., the robustness against noise) the same. As pointed out by @vaz in a comment, the latter is only true for passband PAM systems, such as QAM or PSK.

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  • $\begingroup$ Just an observation: the sentence "the problem with increasing the number of symbols is that you need more energy to keep the average distance between the symbols" is true for power-limited modulations. This does not hold for the bandwidth-limited case, such as M-ary FSK. $\endgroup$
    – vaz
    Dec 7 '15 at 8:57
  • $\begingroup$ @vaz: Yes, thanks. I added an explanation to my answer. $\endgroup$
    – Matt L.
    Dec 7 '15 at 9:02

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