# Electronics and Communication Engineering - Exam Questions Papers - Discussion

Discussion Forum : 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 V . The signal to quantization noise ratio is:

Answer: Option

Explanation:

(SNR)_{dB} = 1.76 + 6.02*n*

*n* = 8

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

Discussion:

18 comments Page 1 of 2.
Kaveri Ammu said:
2 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.

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)

Argha said:
4 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.

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.

Sowmya said:
4 years ago

How come 1.76+6.02?

Can anyone please solve this clearly?

Can anyone please solve this clearly?

Aditya. said:
4 years ago

Snr = 1.5 x 2^2n.

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

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

Pavan kumar said:
5 years ago

Not understanding this, could anyone explain more?

Viresh Madgundi said:
5 years ago

Why you apply 1.76+6.02?

Please give an explanation.

Please give an explanation.

Dimpu said:
5 years ago

I am not getting. Please explain it clearly.

Gayathri said:
6 years ago

Please anyone solve this problem.

Subham Subhrajeet Parida said:
6 years ago

Please explain this equation clearly.

Singamayum naeema said:
6 years ago

Please explain the answer clearly.

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