Question

Consider a message signal m(t) with power of 20 mW that is applied to an analog-to-digital converter, with dynamic range of -1 to 1 volt. a) If uniform quantization is utilized, and the minimum signal-to-quantization-noise (SQNR) is 43 dB, determine the minimum number of bits required, and the corresponding SQNR obtained with this quantizer. b) Repeat Part (a) if a u-law compander is used with u-100. c) If the power of the signal m(t) is reduced to 5 mW, repeat Parts (a) and (b), and comment on the results

Please explain part 2 thoroughly. Thanks!

Answer 1

* case o case Signal power = 20 mwi © SQNR signal to Quantisation Noise ratio (SQNR ) = 403 dB 10 log(sQNR) – 43 dB Linear 4.3 SONR = lo * 19952.623 S 19952.623 Q NR (QNR) € 1x10 c w Scanned with CamScanner

(200 5 5 .me a 2 to stop te 5136 - 1x10-6 12 as 12 xlo-6 A [ 3.46x16 3. A = Vmax - Vmin 2 m -4 = 3:46x103 3:46X163 € 3146X163 3:46 X 16-3 an m 2 X1600 m 2 2 < 7 2600 3:46 3.46 2n,5,78.03 n=107 na no. of bits bits m = 10 n=no.of M = 100 L→ No of quantisation Levels Game) = 312 [in (1+x)] L=27 Cam - Senec Scanned with CamScanner

312 QNR 22.179 4,3 - 10 x 22:179 a 12 384.06 LL 384.06 4.2" 2"Y1 384.06 n=9] na noof bits n = 9 na no of bits Case: Power 5 mwi 7. (SNR) = 10 ** QNR= 10-403x5 xlo 3 (QNR) = 2.50 x10 9 x103 A2 = 122.50 X10-7 (QNR )= A2 = 3 x106 Scanned with CamScanner

A 21,73 x163. A Vmax - Vmin 173x102 2n bits 2000 1.73 ☺ = 100 uloo 342 (in (1+r)) ? Ų.3 31 10 x 22:17 104.3 < 36² 1. (In (III) L 383.99 2" > 983.99 [n=q bits * It the signal power using the uniform changes quantisation the mo. of bits changes, but there is no change in M-Law comp and Scanned with CamScanner