Electronics and Communication Engineering - Exam Questions Papers - Discussion

Discussion Forum : Exam Questions Papers - Exam Paper 1 (Q.No. 3)

Exam Questions Papers

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.

Santhosh said: 10 years ago

Why did you apply 1.76 + 6.02? I want clear explanation. Please help me.

Pranali said: 9 years ago

I didn't understand the concept.

Please elaborate it.

Jiten said: 9 years ago

Please explain this clerly.

Vineeta pal said: 9 years ago

Hi, if anyone knows the solution explain this question.

Avinash Bhagawat said: 9 years ago

Why did you apply 1.76 + 6.02?

I want a clear explanation. Please help me.

M chinna said: 9 years ago

I didn't understand the answer, please explain me.

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.

Anil gautam said: 7 years ago

(SNR)dB = 1.76 + 6.02n.

Please explain this equation.

Singamayum naeema said: 7 years ago

Please explain the answer clearly.

Subham Subhrajeet Parida said: 7 years ago

Please explain this equation clearly.

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