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Solution Quiz

This document discusses digital transmission and PCM coding. It contains examples of: - Calculating Nyquist sampling rates for analog signals - Determining dynamic range and minimum bits for a given dynamic range using PCM codes - Calculating voltages for linear sign-magnitude PCM codes - Finding resolution and quantization error for PCM codes - Mapping PCM codes to voltage ranges - Calculating output voltages for μ-law and linear compression encoding schemes.

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djun033
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367 views3 pages

Solution Quiz

This document discusses digital transmission and PCM coding. It contains examples of: - Calculating Nyquist sampling rates for analog signals - Determining dynamic range and minimum bits for a given dynamic range using PCM codes - Calculating voltages for linear sign-magnitude PCM codes - Finding resolution and quantization error for PCM codes - Mapping PCM codes to voltage ranges - Calculating output voltages for μ-law and linear compression encoding schemes.

Uploaded by

djun033
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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Homework 5_1

DIGITAL TRANSMISSION

1. Determine the Nyquist sample rate for a maximum analog input frequency of:

(a) 4 KHz
2(4 kHz) = 8 kHz samples/sec
(b) 10 KHz
2(10 kHz) = 20 kHz samples/ sec

2. Determine the dynamic range for a 10-bit sign-magnitude PCM code.

2n = DR + 1
20log (29) = 54.17 dB

3. Determine the minimum number of bits required in a PCM code for a


dynamic range of 80 dB. What is the coding efficiency?

20 log (DR+1) = 80 dB
log (DR+1) = 4 dB
DR = 104 + 1 = 10001
log(2n) = log (10001)
N * .301 = 4.00
N = 13.29
Coding efficiency = (13.29/14)*100 = 95%

4. For a resolution of 0.04 V, determine the voltages for the following linear
seven-bit sign magnitude PCM codes.

(a) 1001011
23+ 21+20 = -11
-11 * 0.04 = -0.44 V

(b) 0101101
25+ 23+22+20= 45
45 * 0.04 = 1.8 V

1
5. Determine the resolution and quantization error for an eight-bit linear sign
magnitude PCM code for a maximum decoded voltage of 1.27 V.

Resolution = 1.27 / (28+1) = .01V


Q error = .01 / 2 = .005

6. For a 12-bit linear PCM code with a resolution of 0.02 V, determine the voltage
range that would be converted to the following PCM codes.

Q error = .02/ 2 = 0.01

(a) 1 0 0 0 0 0 0 0 0 0 0 1 = -1
-1 * 0.02 = -0.02
Range = (-0.03, -0.01)
(b) 000000000000=0
0 * 0.02 = 0
Range = (-0.03, -0.01)
(c) 1 1 0 0 0 0 0 0 0 0 0 0 = -1024
-1024 * 0.02 = -20.48
Range = (-20.49, -20.47)
(d) 0 1 0 0 0 0 0 0 0 0 0 0 = 1024
1024 * 0.02 = 20.48
Range = (20.47, 20.49)
(e) 1 0 0 1 0 0 0 0 0 0 0 1 = -257
-257 * 0.02 = -5.14
Range = (-5.15, -5.13)
(f) 1 0 1 0 1 0 1 0 1 0 1 0 = -682
-682 * 0.02 = -13.64
Range = (-13.65, -13.63)

7. A µ-law compression encoder has:

Vi(max) = 8 V, vi(max) = 1V, µ = 255

Determine the output voltage for an input voltage of 2 V.

µ𝑉𝑖𝑛
𝑉𝑚𝑎𝑥 ×ln(1+ )
𝑉𝑖𝑛 𝑚𝑎𝑥
𝑉0 =
ln(1+µ)

(255)(2𝑉)
1𝑉 × ln(1+ )
8𝑉
𝑉0 = = .75V
ln(1+255)

2
8. Repeat problem 7 for µ = 0.
ln(1)
𝑉0 = 1𝑉 ( )=0V
ln(1)

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