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[SOLVED] (HELP!!)Linear PCM BitRate and analog signal BW Relation

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Hi Guys, appreciate any help and input on the following problem
:

An analog signal is to be converted to digital signals using linear PCM with 256 quantization levels and transmitted over an 8 kHz band-limited channel. Assume that binary baseband transmission is employed. Ans: (a) Rb = 16 kbps. (b) B = 1 kHz.

(a)Find the maximum PCM output bit rate.

Rb(max)= 2*BT
= 2*8KHZ
=16kbps

=> how do you know that this is the max bitrate?
=>why can't i use Rb=n*fs=n*2*B
=8*2*8KHZ??

(b)What is the corresponding maximum analog signal bandwidth that can be accommodated?
=>Here is the problem that i am facing, how do you get the analog Signal BW from the Bit Rate?

Thanks for your time!
 

Nyquist gives us that a 8KHz channel bandwidth supports 16k baud (from his famous telegraphy eqn) and we are using a 2 level (binary) code, so 16 k baud = 16kbps.

You get the analogue bandwidth from the bit rate by noting that 256 level quantisation is 8 bit quantisation, and that 16kbps/8 bits = 2k samples/s

As you require the sample rate to be strictly greater then twice the channel bandwidth, 2k samples/s the channel bandwidth must be limited to just less then 2k samples/s / 2 = 1khz.

HTH.

Regards, Dan.
 

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