Electronics and Communication Engineering - Exam Questions Papers - Discussion
Discussion :: Exam Questions Papers - Exam Paper 1 (Q.No.3 )
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
[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.
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?
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.
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?
Pavan Kumar said:
(Jul 8, 2019)
Not understanding this, could anyone explain more?
Aditya. said:
(Feb 16, 2020)
Snr = 1.5 x 2^2n.
V . The signal to quantization noise ratio is: