Infsci 1004 - Homework Assignment
Homework Assignment #2 Solutions
Questions
1. Explain the significance of the Shannon-Hartley theorem.
 Given a S/N ratio over a communication channel and the channel bandwidth for a
particular distance, there is a maximum rate or capacity at which one can transmit
data over that channel.
 The capacity is given by
C = W log2 (!+S/N)

where, C = channel capacity in bits per second.
W = channel bandwidth in Hertz.
S/N = signal-to-noise power ratio.
One can transmit at rates higher than C if the extra errors are tolerable.
2. Compare and contrast the time domain with the frequency domain.


The time-domain characteristics give the amplitude, frequency and phase of a signal with
respect to time.
The frequency-domain characteristics give the frequency components that make up any
arbitrary signal in the time-domain. It gives us the bandwidth that a particular type of
signal will need.
Problems
1. A telephone channel has a bandwidth of 3100 Hz.
1. If the minimum frequency is 300Hz, what is the maximum frequency?
Maximum Frequency = Minimum Frequency + Bandwidth = 3400 Hz.
2. What is the maximum capacity of the channel if the signal to noise ratio is 20dB?
30dB?
C = W log (1+S/N)
2
S/N = 10 ** (SNR/10)
SNR(dB)
20
30
S/N
100
1000
Capacity (bits per sec)
20,600 (approx)
30,800 (approx)
2. Most stereo systems reproduce sound from about 20Hz to about 20KHz, based on the
hearing ability of the human ear, which is much higher than the bandwidth of the
telephone channel. Speculate why the telephone channel's bandwidth is so much lower
than the human ear's audible bandwidth.
Human voice generates frequencies in the range of 300 to 3400 Hz. Hence, if we provide
about 4 kHz of BW (including a guard band)per telephone channel, it would be sufficient
for voice quality. Hence the main reasons are:
 Acceptable Voice Quality when reproduced at the receiving end.
 Low Cost.
 Savings in Bandwidth.
3. There are many tradeoffs inherent in the design of codes for information transmission.
Explore these tradeoffs by answering the following:
1. Compute the number of bits that would be necessary to transmit the uppercase
alphabet, digits from 0 to 9, and the punctuation marks (.,'!)?.
There are 44 symbols in all. So it would require 6 bits per character since
2 ** 6 =64 > 44.
2. Compare the bit efficiency of this code with the Baudot code and with the ASCII
code for transmitting the strings DARTH VADER and R2D2.C3P0 (for Baudot,
assume that the "." character is a "letter")?
Bit
Efficiency
6-bit code
ASCII
Baudot
DARTH VADER
Information Bits Transmitted bits
10
10
10
10
10
11
Bit Efficiency
6-bit code
ASCII
Baudot
DARTH VADER
100
100
91
R2D2.C3PO
Information Bits
Transmitted bits
9
9
9
9
9
16
R2D2.C3PO
100
100
56
3. If the bit rate of a channel is 150bps, what is the difference in transmission times
for the three codes? At 1200bps?
Transmission
DARTH VADER
R2D2.C3PO
time
150 bps
1200 bps
150 bps
1200 bps.
6-bit code
60/150=0.4 sec
60/1200=0.05 sec
54/150=0.36 sec 54/1200=0.045 sec
ASCII
70/150=0.46 sec
70/1200=0.06 sec
63/150=0.42 sec 63/1200=0.052 sec
Baudot
55/150= 0.36 sec 55/1200= 0.045 sec 80/150= 0.53 sec
80/1200= 0.06 sec
4. What do you conclude about the different strategies used in each of the codes
from the previous calculations?
 The lesser the number of bits, the smaller the transmission time.
 The lesser the overhead the higher the efficiency.