Question

A video signal has bandwidth of 4 MHz. This signal is sampled, quantized, and binary encoded using the PCM technique. The video signal amplitude is uniformly distributed in the range (-2, 2). If the signal is sampled at the Nyquist rate and quantized into 512 levels: (a) Determine the bit rate. (b) Determine the signal-to-quantization-noise ratio. (©) If the signal-to-quantization-noise ratio must be increased by a factor of 16, how you can achieve that? Will this affect the bit rate? How?

Answer 1

Msg signal fm = 4MHz. Sampling frequency [&s=2fm s2x4 M H22 8 MHZ. Quantization Voltage #L) = 512 L=2n.. 512=2" In=9 bits No of a Bit Rate (Rb ³ nfs R6=9*8 MH2 = ALARM 72 Mbps R6 2 725X bbs

S 2 © Signal to Quantization Noise Rah'on' = 3x225 Na 2 2x9 3x2 → 393216. S 2393216 cor) 6 Qon 2 + = 10 log10 ( 3 ) tologio ( haven ~ (1.8 +6xn) dB tologist = 1:76 +6.02n. Zlco S 2 1.8 +6x9 + 55-8dB NGdB So, signal to Noise ratio is 393216 or 65.8dB © Com - 1209 for n bits 2(n+K) - for (nek) bits. 이도 이얼 3 X2 2 ) en (3/99) nak a nik 26+K) 2xK 2 2 time increases. 22inde (mign Ini

Hence from n to oth era) if no. no. of bits increased increases by gek times. Given that s Na → Increased by 16 times. 2xk 16 z 2 K22 No.q bit increased from 9 bits to il bits. Bit Rate Rb Rb = nxfs 2 11%8MHZ 788 Mibbs. | R6=88 Mbps

If you have any query comment down below I will surely help you as soon as possible please upvote my answer thanks in advance...!