Electronics and Communication Engineering - Communication Systems - Discussion
Discussion Forum : Communication Systems - Section 2 (Q.No. 22)
22.
Three analog signals, having bandwidths 1200 Hz, 600 Hz, 600 Hz are sampled at their Nyquist rates, encoded with 12 bit words, and time division multiplexed. The bit rate for the multiplexed signal is
Answer: Option
Explanation:
The total message bandwidth = 1200 Hz + 600 Hz + 600 Hz = 2400 Hz
Sampling frequency = 4800 samples/sec.
Hence, Bit rate = 4800 x 12
57.6 kbps.
Discussion:
4 comments Page 1 of 1.
Rohit prakash said:
1 decade ago
The maximum frequency component is 1200 hz.
So, fm = 1200 Hz.
fs = 2*1200 Hz = 2400 Hz.
B.W = n*fs = 12*2400.
Rb/2 = B.W.
Hence, Rb = 12*2400*2 = 57.6 kbps.
So, fm = 1200 Hz.
fs = 2*1200 Hz = 2400 Hz.
B.W = n*fs = 12*2400.
Rb/2 = B.W.
Hence, Rb = 12*2400*2 = 57.6 kbps.
(1)
Swetha said:
9 years ago
You are right @Rohit.
AdarshDM said:
8 years ago
You are wrong @Rohit Prakash.
The relation rb=2(BW) is a completely different Question, that's channel bandwidth and maximum possible bit rate relation.Here we need to find sampling frequency which is the 2xsum of all frequencies(not twice the highest) and then Ts=nTb or Rb=nFs.
The relation rb=2(BW) is a completely different Question, that's channel bandwidth and maximum possible bit rate relation.Here we need to find sampling frequency which is the 2xsum of all frequencies(not twice the highest) and then Ts=nTb or Rb=nFs.
Pankhuri Saxena said:
6 years ago
True @Adarshdm.
Here signals are being time division multiplexed. No need to choose higher frequency component!
Here signals are being time division multiplexed. No need to choose higher frequency component!
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