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Electronics and Communication Engineering - Exam Questions Papers - Discussion

Discussion Forum : Exam Questions Papers - Exam Paper 1 (Q.No. 3)
Exam Questions Papers
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:
30 dB
46.91 dB
40 dB
49.92 dB
Answer: Option
Explanation:

(SNR)dB = 1.76 + 6.02n

n = 8

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

Discussion:
18 comments Page 1 of 2.
Argha said:   5 years ago
Answer given is the wrong answer is option D.

Signal power is (1/2)^2=1/4 and noise power is 1^2/(3*2^16). Then after signal power /noise power expressed in dB answer will be 46.91approx.
Kaveri Ammu said:   4 years ago
(SNR) db = 10 log 1.5 + 2 n*10 log 2.
Log1.5 value = 0.176,
Log 2 value = 0 3010,
N= 8.

10 * 0.176 + 2*8*0.3010* 10.
1.76 + 4.816*10,
1.76 + 48.16,
49.92 db.
(1)
Govind said:   9 years ago
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.
Avinash Bhagawat said:   9 years ago
Why did you apply 1.76 + 6.02?

I want a clear explanation. Please help me.
Santhosh said:   10 years ago
Why did you apply 1.76 + 6.02? I want clear explanation. Please help me.
Aditya. said:   6 years ago
Snr = 1.5 x 2^2n.

In db, snr (dB) = 10log 1.5 + 2n x 10log 2.
Sowmya said:   5 years ago
How come 1.76+6.02?

Can anyone please solve this clearly?
Vineeta pal said:   9 years ago
Hi, if anyone knows the solution explain this question.
Viresh Madgundi said:   6 years ago
Why you apply 1.76+6.02?

Please give an explanation.
Pranali said:   9 years ago
I didn't understand the concept.

Please elaborate it.
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