Electronics and Communication Engineering - Communication Systems - Discussion
Discussion Forum : Communication Systems - Section 4 (Q.No. 1)
1.
24 telephone channels, each band limited to 3.4 kHz are to be time division multiplexed using PCM. If sampling frequency is 10 kHz and number of quantization levels is 128, the required bandwidth of PCM is
Answer: Option
Explanation:
Bandwidth ≈ 24 (ln2 128) 10 kHz = 1.68 MHz.
Discussion:
5 comments Page 1 of 1.
ARKADEEPDEB said:
6 years ago
What about bandwidth limit = 3.4kHz?
Md. Juwel Rana, Bangladesh said:
9 years ago
Given,
Sampling Frequency, Fs = 10 KHz.
Quantization Level, L = 128, So 2n = 128 or n = 7.
Now, Band width for 1 channel =nFs = 7 * 10 KHz = 70 KHz.
Band width for 24 channel = 24 * 70 = 1680 KHz = 1.68 MHz.
Sampling Frequency, Fs = 10 KHz.
Quantization Level, L = 128, So 2n = 128 or n = 7.
Now, Band width for 1 channel =nFs = 7 * 10 KHz = 70 KHz.
Band width for 24 channel = 24 * 70 = 1680 KHz = 1.68 MHz.
(1)
Chinna said:
9 years ago
Can any one explain it briefly?
Jobon123 said:
9 years ago
It is log not ln.
Gelectronics said:
9 years ago
Please, can anyone give the simple solution?
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