Question

part c wasnt answered

Question 2 (25 Marks) Consider the following analog waveform that is to be digitized using PCM The resulting PCM code is to be sent at a bit rate of 240000 bps over a communication medium having an SNR of 15 dB. Also, the sampling rate used is 40000 samples/sec. Answer the following questions: a. What is the number of quantizing levels? b. If the sampling time is doubled: 1. What is the number of bits per sample? ii. What is the number of quantizing levels? c. Using the above illustration, assume that the maximum signal intensity is approximately 34 millivolts. If the sampling time is halved, answer the following questions: i. Draw on the illustration above the quantization levels needed. ii. If the "quantizing step" generated the following quantized code numbers: 0,0,3,3,3,4,6,7,7,7,7,6,5,4,2 What will be the output of the "encoding step"?

Answer 1

Answer:

Ansi- Теги . Given a anlog signal to be digitized using PCM → The pcm ocetpeet bit rate is = 240 kbps → and Communication medicem having an SNR of 15dB. + Gives the sampling rate is 40,000 Samples/sec. a) ansi No of bits cused for Sampling = bit rate Sample rate = 240 kbps 40,000 Samples/sec = 240 K 40 K - 6 bits/samples i No of Quantization levels = 26 = 64 b) ans,- If i sampling rate is doubled No of samples /sec (Sampling rate) decreases log & = 20 kbps Inorder to maintain. 240 Kbps bit rate No of bits per sample = bitrate Sample rate No of Quantization levels = 12 Sampling rate is clockbled = 22 :21:

C) ansi . If the sampling time is halved, then Sampling rate increases - 8000 Samples/sec No of bits per Sample = 240 k - 3 bils/sample 80 K No of quantization levels = 23 = 8 Step size = Dynamic range 1. No of levels 34m = 4.25mv Range Quantization level 2.125 m 0-4.25 m 6.375 m 4:25-8.5m 10.625 3 8.5m-12.75m 14.875 m § 12.75m - 17 m o at d m tisoi 19.125 m 3 17m - 21.25m 21.25m – 25.5m 23.375 m 3 27. 625 m. 25.5m - 29.75 m 31.575 m 3 29.75m - 34 m

Given quantization code, then acetpeet of enoding 0000 > 000 > 011 > 011 00 110 ↑ ↑ ↑ ↑ ↑ co 3 3 46 777 76 s 4 2 ↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑ 111 111 >111 -111 lo 104 4 → 400 2010

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