Electronics and Communication Engineering - Exam Questions Papers - Discussion

3. 

A certain 8 bit uniform quantization PCM system can accommodate a signal ranging from - 1 V to + V. The rms value of the signal is V . The signal to quantization noise ratio is:

[A]. 30 dB
[B]. 46.91 dB
[C]. 40 dB
[D]. 49.92 dB

Answer: Option D

Explanation:

(SNR)dB = 1.76 + 6.02n

n = 8

Thus (SNR)dB = 1.76 + 6.02 x 8 = 49.92 dB.


Santhosh said: (Mar 30, 2016)  
Why did you apply 1.76 + 6.02? I want clear explanation. Please help me.

Pranali said: (May 4, 2016)  
I didn't understand the concept.

Please elaborate it.

Jiten said: (Sep 9, 2016)  
Please explain this clerly.

Vineeta Pal said: (Oct 9, 2016)  
Hi, if anyone knows the solution explain this question.

Avinash Bhagawat said: (Nov 8, 2016)  
Why did you apply 1.76 + 6.02?

I want a clear explanation. Please help me.

M Chinna said: (Jan 25, 2017)  
I didn't understand the answer, please explain me.

Govind said: (Feb 13, 2017)  
SQNR= 1.5*(2^2n) by converting this factor in db will get SQNR IN DB = 1.8+6n where n is no of bits per sample.

Anil Gautam said: (Jun 12, 2018)  
(SNR)dB = 1.76 + 6.02n.

Please explain this equation.

Singamayum Naeema said: (Jul 12, 2018)  
Please explain the answer clearly.

Subham Subhrajeet Parida said: (Jul 19, 2018)  
Please explain this equation clearly.

Gayathri said: (Sep 4, 2018)  
Please anyone solve this problem.

Dimpu said: (Mar 4, 2019)  
I am not getting. Please explain it clearly.

Viresh Madgundi said: (Jun 18, 2019)  
Why you apply 1.76+6.02?

Please give an explanation.

Pavan Kumar said: (Jul 8, 2019)  
Not understanding this, could anyone explain more?

Aditya. said: (Feb 16, 2020)  
Snr = 1.5 x 2^2n.

In db, snr (dB) = 10log 1.5 + 2n x 10log 2.

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