TDC462 Data Communications
Homework 2, due July 1, 2002 at 5:59:59pm
Jeffrey Absher – 0673398
Q1. A transmission channel between two communicating devices is made up of three
sections. The first introduces a signal power change of -16dB, the second of 20dB and the
third of –10dB. Assuming a mean transmitted power level of 400mW, determine the
mean output power level of the channel.
-16 = 10 log10(x/400) … x = 3989.95
20 = 10 log10(x/3989.95)….x = 398995.25
-10 = 10 log10(x/398995.25)…x = 39899.52 mW
Q2. The achievable data rates of modem connections over voice-grade lines vary greatly
with the signal-to-noise ratio of the lines and the line bandwidth. In the US, the signal-tonoise ratio varies from 24dB to 38dB and the line bandwidth varies from 2400Hz to
3800Hz. Given the best possible modem technology, what is the difference between the
highest achievable modem data rates across the US within the framework of the
parameters given above?
C = B log2 (1 + SNR)
(SNR)db = 10 log10(SNR)
24dB@2400Hz = …SNR = 1024/10…C = 2400 log2(1 + 102.4) = 19148.063 bps
38dB@3800Hz = …SNR = 1038/10… C = 3800 log2(1 + 103.8) = 47969.510 bps
That is a difference of 28821.447 bps
Q3. Determine the bandwidth requirement for a modem whose data rate is 19,800 bps
and which uses 4 bits per signal element.
19800 bits/sec 1 sigEl/4 bits = 4950 SigEl/Sec = 4950 Hz.
Q4. Problem 6.2 on page 191
A datasource produces 7-bit IRA chars. Derive an expression of the maximum effective datarate (rate of
IRA databits) over a B-bps line for the following
a)Asynch 1.5 unit stop element and parity bit
B bits/1 sec * 1 frame/(9.5 bits) * 1 char/1 frame * 7 bits/1 char = B * 7/9.5 = 0.74 B
b)synch w/ frame of 48 control bits and 128 information bits, the info field contains 8bit (paritiy included
IRA chars
B bits/1 sec * 1 frame/(48 + 128) bits * 16 chars/1 frame * 7 bits/1 char = B * 112/176 = 0.64 B
c) same as b) w/ info field of 1024 bits.
B bits/1 sec * 1 frame/(48+1024) bits * 128 chars/1 frame * 7 bits/1 char = B * 896/1072 = 0.84 B
Q5. A PCM encoder accepts a signal with a full-scale of 10V (0 to 10V) and generates 8bit codes using uniform quantization.
1) How many different discrete values are used to represent the analog signal
voltage level?
8
2 =256 values
2) What is the quantization step size (range of each interval)
10/256 = 0.0390625 V
3) Give the actual range of voltage values that are represented by the following
discrete PCM pulses:
(a) 27 = [1.0546875 V, 1.09375 V)
(b) 34 = [1.328125 V, 1.3671875 V)
(c) 129 = [5.0390625 V, 5.078125 V)
4) What is the value of the highest quantization noise?
0.0390625/2 =0.01953125 V