THIS IS A SAMPLE PROBLEM WITH THE SOLUTION EXPLAINED PLEASE

THIS IS A SAMPLE PROBLEM WITH THE SOLUTION EXPLAINED.

PLEASE HELP EXPLAIN THE ANSWER MORE IN DETAIL. EXPLAIN WITH AS MUCH DETAIL AS POSSIBLE PLEASE.. DO NOT JUST REWRITE THE SOLUTION I PROVIDED. I NEED MORE DETAIL AS TO HOW THE SOLUTION WAS DERIVED.

Problems 2-3 What is the bandwidth needed in a TDM system to transmit 4 measurements, 16-bits each, in 1ms? Assume a very noisy environment where only a 0-1 transmission is possible. In 1ms we should transmit 4 channels + sync x16 bits 80 pulses. Therefore, the bandwidth of the system should be at least 80kHz g bandwidth in a FDM system, discounting any noise problems? nt at 1mslimplies that the signal maximum frequency cannot be larger than -The-measureme 500Hz. 4 channels of 500Hz each requires a total bandwidth of 2kHz. (The comparison is not fair anyway!)

Solution

Q2-3)

a)

TDM system:

In TDM,Time duration is shared between different transmitting system(measurement).

Here 4 measurements each of 16 bits are transmitted during 1 msec time duration.

So total number of bits transmitted during 1 msec =4*16 = 64 bits

As each bit is represented by 1 pulse,we can say

total number of pulses transmitted during 1 msec = 64

Bandwidth =Total number of pulses transmitted during 1 sec=64/1msec =64 KHz

[Note:answer here is as per details provided in this question]

b)

FDM System:

As total time duration t=1 msec

Sampling rate fs=1/1msec = 1KHz

fs=2*fm

where fm:signal maximum frequency

so fm=0.5*1000 =500 Hz each signal.

So total bandwidth for 4 measuremetn (or channel) =4*fm =4*500 =2KHz [answer]