Electronics and Communication Engineering - Measurements and Instrumentation - Discussion
Discussion Forum : Measurements and Instrumentation - Section 2 (Q.No. 38)
38.
PCM employing 4 bit code is used to send data having frequency range from 0 to 2 kHz. The minimum required bandwidth of carrier channel is
Answer: Option
Explanation:
Bandwidth required is twice the required frequency range.
Discussion:
5 comments Page 1 of 1.
Saurabh kumar said:
5 years ago
In pcm system minimum bandwidth = n * fm =2 * 2khz = 4khz.
Here q = 2^n.
q given as 4 bit.
The right answer is 4khz -> option B.
Here q = 2^n.
q given as 4 bit.
The right answer is 4khz -> option B.
Dayal bhai said:
7 years ago
8 kHz is the right answer.
Sunil said:
9 years ago
Yes, you are right @Ayush.
Ayush kumar said:
9 years ago
The given answer is wrong because minimum bandwidth required in PCM is ((n * sampling frequency)÷2) and also we know( sampling frequency is = 2 * modulating frequency ) so in this condition the right solution is (4 *2 * 2)÷2 = 8KHz.
Amandeep said:
1 decade ago
This is wrong in pcm as we know bandwidth is equal to no. of bits multiplied by sampling rate here sampling rate is 2 fm and no. of bits are 4 but in question we have to found minimum bandwidth.
So as we use 2 bits per symbol in pcm so right way is no. of bits into fs devided by 2. So answer will be 8.101% sure.
So as we use 2 bits per symbol in pcm so right way is no. of bits into fs devided by 2. So answer will be 8.101% sure.
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