For a. PCM signal, determine L if the compression parameter $μ$ = 100 and the minimum SNR required is 45 dB. Determine the output SNR with this value of L. Remember that L must be a power of 2, that is, L = 2ⁿ for a binary PCM.
ক জন্য. PCM সংকেত, L বের করুন যদি কম্প্রেশন প্যারামিটার $μ$ = 100 এবং ন্যূনতম SNR 45 dB হয়। L-এর এই মান দিয়ে আউটপুট SNR নির্ণয় করুন। মনে রাখবেন L অবশ্যই 2 এর একটি পাওয়ার হতে হবে, অর্থাৎ, একটি বাইনারি PCM এর জন্য L = 2ⁿ ।

Eastern Refinery Limited (ERL) - Probationary Engineer (Electrical) - (30-12-2022) ইলেকট্রিক্যাল এন্ড ইলেকট্রনিক্স ইঞ্জিনিয়ারিং (EEE)

Created: 7 months ago | Updated: 7 months ago

Updated: 7 months ago

Ans :

For μ - Law Compression,

SNR in watt = $\frac{3 L^{2}}{{[\ln (1 + μ)]}^{2}}$

31622.78 = $\frac{3 L^{2}}{{[\ln (1 + 100)]}^{2}}$

$L = \sqrt{\frac{31622.78 \times {[\ln (1 + μ)]}^{2}}{3}} = 473.83$

Compression parameter $μ$ = 100

Minimum SNR_in _dB = 45

SNR_{in dB} =10 $\frac{S N R_{i n d B}}{10}$

SNR_in _watt = $10 \frac{45}{10}$ $= 10^{4.5}$ = 31622.78W

Number of Levels, L =?

As, Number of Levels L will be the power of 2, let

we select L = 512 = $2^{9}$ Ans.

$∴$ SNR for this value of L = 512 is

$S N R_{i n w a t t} = \frac{3 L^{2}}{{[\ln (1 + μ)]}^{2}} = \frac{3 \times 542^{2}}{{[\ln (1 + 100)]}^{2}}$

=31622.78 W

SNR _ln(dB) = 10 log(SNR _{ln watt})

=10 log 36922.84

=45.67 dB

7 months ago

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Question:
For a. PCM signal, determine L if the compression parameter $μ$ = 100 and the minimum SNR required is 45 dB. Determine the output SNR with this value of L. Remember that L must be a power of 2, that is, L = 2ⁿ for a binary PCM.
ক জন্য. PCM সংকেত, L বের করুন যদি কম্প্রেশন প্যারামিটার $μ$ = 100 এবং ন্যূনতম SNR 45 dB হয়। L-এর এই মান দিয়ে আউটপুট SNR নির্ণয় করুন। মনে রাখবেন L অবশ্যই 2 এর একটি পাওয়ার হতে হবে, অর্থাৎ, একটি বাইনারি PCM এর জন্য L = 2ⁿ ।

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