# Rick Wagoner - DOC by pengtt

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```									Rick Wagoner
IET680 Digital Communication Network – Homework 2

Chapter 2 Problem set

Problem 2
A noiseless 4-KHz channel is sampled every 1 msec. What is the maximum data rate?
This channel will have an approximate maximum data rate of 2408 bps.

Problem 3
Television channels are 6 MHz wide. How many bits/sec can be sent if four-level digital
signals are used? Assume a noiseless channel.
This configuration can send approximately 14,449 bits/sec.

Problem 8
It is desired to send a sequence of computer screen images over an optical fiber. The
screen is 480 X 640 pixels, each pixel being 24 bits. There are 60 screen images per
second. How much bandwidth is needed, and how many microns of wavelength are
needed for this band at 1.30 microns?
The transmission will require approximately 442 Mb of bandwidth which will
require 0.4 microns. (I think).

Problem 11
Radio Antennas often work best when the diameter of the antenna is equal to the
wavelength of the radio wave. Reasonable antennas range from 1 cm to 5 meters in
diameter. What frequency range does this cover?
The frequency range would be from 5 MHz to 30,000 MHz.

Problem 19
A regional telephone company has 10 million subscribers. Each of their telephones is
connected to a central office by a copper twisted pair. The average length of these
twisted pairs is 10 km. How much is the copper in the local loops worth? Assume that
the cross section of each strand is a circle 1 mm in diameter, the density of copper is 9.0
grams/cm3, and that copper sells for 3 dollars per kilogram.
The value of the copper in the local loops would be \$4,241,150,082.35.

Problem 22
A modem constellation diagram similar to Fig. 2-25 has data points at eh following
coordinates: (1, 1), (1,-1), (-1, 1), and (-1,-1). How many bps can a modem with these
parameters achieve at 1200 baud?
This configuration called Quadrature Phase Shift Keying can achieve 2400 bps.

Problem 23
A modem constellation diagram similar to Fig.2-25 has data points at (0,1) and (0.2).
Does the modem use phase modulation or amplitude modulation?
This modem is using a phase modulation scheme.

Problem 29
Why has the PCM sampling time been set at 125 µsec?
The telephone channel bandwidth is 4-KHz. Thus according to the Nyquist
theorem 2H is sufficient to capture all the information from this channel. 2H is
8000 samples per second or 125 µsec.

Problem 30
What is the percent overhead on a T1 carrier; that is, what percent of the 1.544 Mbps are
not delivered to the end user?
Approximately 12.9% of each frame is used for overhead.

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