Electronics and Communication Engineering - Communication Systems - Discussion

Discussion Forum : Communication Systems - Section 5 (Q.No. 24)
24.
In a PCM system, if the code word length is increased from 6 to 8 bits, the signal to quantization noise ratio improves by the factor.
8/6
12
16
8
Answer: Option
Explanation:

Given n1 = 6, n2 = 8,

then (SNR)Q = 22(n2 - n1) 16.

Discussion:
10 comments Page 1 of 1.

Preeti said:   1 decade ago
How 16? Do calculations again?

Shadab Akhtar said:   9 years ago
The answer should be 12.

SNR = (1.7 + 6n)

SNR1 = (1.7 + 6*8)
=> 49.7

SNR2 = ( 1.7 + 6*6)
=> 37.7

SNR2 - SNR1 = 49.7 - 37.7 = 12

SHailesh said:   9 years ago
Are you sure @Shadab Akhtar.

Thanks. It's Good Method it is right!

Shailesh Patel said:   9 years ago
Theoretical SNR is 36dB for a 6-bit sample depth.

Simply add or subtract 6dB for each bit difference.

Here for 8 bit 48db and 36db for 6 bit.

Now, 48db - 36db =12.

So 12 is the right answer.

Jyotirmay nath said:   9 years ago
How 16? I didn't get it, please help me.

Pratik said:   8 years ago
16 is right.

(3/2 * 2^2n1)/(3/2*2^2n2).
now you get 2^16/2^12.
2^4= 16 ans.

Shadan said:   8 years ago
S/N = 2^2n= 2^12.
S/N =2^2n= 2^16.
So,
= 2^16/2^12.
=2^4 = 16.
(1)

#JustAsking said:   7 years ago
@Shadan.

Your computation is in dB? Because it seems like just a normal ratio and it doesn't even represent a quantization error as what is being asked in the problem.

Zulu Chungtia said:   6 years ago
Correct @Shadab.

Because the question isn't asked in dB.

Chandni said:   4 years ago
Agree, 16 is the correct answer.

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