ECES 490 Homework #2 Due April 17, 2001
1. A digital cellular system operator has purchased a license for 10 MHz of spectrum.
The geographical region covered by this license is metropolitan Philadelphia. The
operator’s business plan calls for 125,000 customers to be supported by June 30,
2002, when the system will be at maximum utilization for its first phase of operation
(i.e., when the forecasted 125,000 customers have been “signed up” for the service).
The efficiency of the modulation scheme employed by the equipment which is used
by this system operator is 2 bits per second/Hz
Scenario A: All of the customers use the service only to make wireless telephone calls.
During the busy hour, each customer will originate an average of 4 calls; and an average
call will last for 90 seconds. When active, a two-way telephone call uses 16 kilobits per
second of communication capacity.
What frequency re-use factor is required to provide enough total system capacity to
support Scenario A in the first phase of operation?
Solution: The required number of bits per hour during the busy hour is 125,000
customers x 4 calls per hour x 90 seconds per average call x 16 kilobits per second = 7.2
x 10**11 bits per hour. The number of bits per second during the busy hour is 7.2 x
10**11 bits per hour/ 3600 seconds per hour = 200,000,000 bits per second. With 10
MHz of bandwidth and 2 bits per second per Hz of modulation efficiency, a single use of
the available frequencies will support 20,000,000 bits per second. Thus we need a
frequency reuse factor of 10.
Scenario B: In addition to the usage described in scenario A, above, 35% of the users
utilize the system to send and receive E-mail, and 25% of the users utilize the system to
access information from web servers. An average E-mail user downloads 10 E-mail
messages during the busy hour, each of which contains 10,000 bytes of information (on
average). An average E-mail user uploads 10 E-mail messages during the busy hour, each
of which contains 500 bytes of information. An average web user downloads 250,000
bytes of information during the busy hour.
How much additional, aggregate traffic (% increase), is generated by the data users?
[Make any additional assumptions you wish, to answer this question; but state what they
are].
Solution: a) E-mail: 0.35 x 125,000 users x [10 messages per hour x 10,000 bytes per
message x 8 bits per byte + 10 messages per hour x 500 bytes per message x 8 bits per
byte ] = 3.675 x 10**10 bits per hour
b) Web access: 0.25 x 125,000 users x 250,000 bytes per hour x 8 bits per byte = 6.25 x
10**10 bits per hour
Total of a) + b) = 9.925 x 10**10 bits per hour….. compared to 7.2 x 10**11 bits per
hour for voice. The added data traffic corresponds to a 13.78% increase in traffic over
the voice traffic alone.
2. A cellular radio handset delivers 100 mW to its antenna, and the antenna is 70%
efficient in radiating that power (omni-directionally). The receiver in the radio port
(base station) has a worst-case noise temperature of 300K. The nominal frequency of
operation is 2 GHz. The required received signal-to-noise ratio at the radio port is 20
dB (minimum). The effective bandwidth utilized by a single radio channel (for the
purposes of SNR calculations) is 12 kHz. The receiving antenna at the radio port has
an effective area of 0.5 square meters. How much loss can be allowed between the
handset and the radio port.
Solution: The SNR of 20 dB corresponds to a SNR of 100. The radio port requires kTB
x 100 watts of received power = 1.38 x 10**-23 J/Kelvin x 300 Kelvins x 12,000 Hz x
100 = 4.968 x 10**-15 watts. The handset radiates 0.1 watts x 0.7 = 0.07 watts. The
allowable loss between the handset and the base station is 0.07 watts / (4.968 x 10**-15
watts) = 1.409 x 10**13 => ~131.5 dB.
3. In problem 2., above, assume that the maximum distance of the handset from the
radio port is 3 km. How much attenuation will result from free-space path loss
(1/r**2 loss)? What will be the dominant causes of the additional losses that might
use up the link loss budget? (List at least two causes)
Solution: The free space path loss is equal to the area of the base station antenna / area of
a sphere 3 km in radius = 0.5 meters**2 / 4 x pi x 3,000 x 3,000 meters**2 = 4.423 x
10**-9 => ~83.5 dB of loss. Other possible causes of loss include: rain attenuation or
other absorption by the atmosphere, absoption by obstacles in the path between the
handset and the base station (walls of buildings), multipath fading, interference from
other distant handsets using the same frequency, handset power level lower than
nominal, mis-adjustments of the base station receiver that reduce its sensitivity below
nominal, deviations of the radiation pattern of the handset antenna from omni-diectional.
4. A digital wireless local area network is set up in a room that is 15 feet wide x 20 feet
long x 10 feet high. Reflections from metal objects in the room and in the
walls/ceiling will cause interference to the direct-path signal between a given
transmitter and a given receiver. The excess delay (v. the delay associated with the
direct path signal) associated with a reflected, received signal resulting from a single
reflection must be less than 0.5 x the spacing between symbols in the digital symbol
stream.
What is the maximum allowable symbol rate, considering all possible positions of a
transmitter/receiver pair in the room? (Assume that the transmitting and receiving
antennas are omni-directional)
Solution: The longest diagonal in the room is [15**2 + 20**2 + 10**2]**0.5 feet = 26.93
feet. The worst case situation would be one in which the transmitter and the receiver were
both in 1 corner of the room. The direct path would be ~0 feet. The longest reflected path
would be 53.86 ft (round trip) ~ 16.42 meters (round trip) => 54.72 ns of delay. Thus the
spacing between symbols in the symbol stream must be at least 2 x 54.72ns = 109.44 ns.
This corresponds to a maximum symbol rate of 1/[109.44 ns] = ~9.14 Mbps.
5. A free space digital optical LAN uses an LED transmitter mounted on the ceiling of a
room. The room measures 12 feet wide x 15 feet long x 8 feet high. The LED
transmitter emits light uniformly over a solid angle of 2 pi steradians ((i.e., all angles
below the ceiling level). A receiver placed on a table in the room, directly under the
transmitter contains an optical detector that captures all light that falls on its aperture.
The aperture has a diameter of 0.2 inches, and is facing up toward the transmitter. The
aperture of the detector is 30 inches above the floor. The receiver requires the
detector to capture 30,000 photons per received pulse in order to operate with a low
error rate. The LED transmitter’s total optical output power is 1mW.
How high a data rate can the LAN operate at…neglecting any effects of reflections from
the walls of the room?
Solution: The distance between the LED and the receiving aperture is (8-2.5) feet. At a
distance of 5.5 feet, the light from the LED is spread out over a surface having area 2 x pi
x 5.5 x 5.5 square feet = 190.07 square feet = 27,370 square inches. The area of the
aperture is pi x 0.2 x 0.2/4 square inches = 0.0314 square inches. The fraction of the light
captured by the aperture is 0.0314/27,370 = ~ 1.15 x 10**-6. The power captured by the
aperture is 1.15 x 10**-6 mW [-59.4 dBm]. The receiver requires 30,000 photons per
received pulse. If we assume that the photon energy is ~1.5 x 10**-19 J, then the receiver
requires ~ 4.5 x 10**-15 J per received pulse. Thus the duration of a single pulse must be
at least 4.5 x 10**-15 Joules / 1.15 x 10**-9 watts = 3.913 microseconds. Thus the
maximum data rate is 1/[3.913 microseconds] = 256 kbps.
6. A one-way wireless messaging system uses a single omni-directional transmitting
antenna, and operates at 850 MHz. It provides messaging service to the city of
Philadelphia. Each message contains 256 bytes of payload information and 3 bytes of
address information (identifying the customer to whom the message is directed). Each
messaging customer may receive up to 3 messages per hour during the busy hour.
Assume a modulation efficiency of 2 bits per Hz.
How much bandwidth is needed to serve 125,000 customers?
Solution: The required number of bits per hour during the busy hour is 125,000
customers x 3 messages per hour x (256 + 3) bytes x 8 bits per byte = 7.77 x 10**8 bits
per hour. The number of bits per second during the busy hour is 7.77 x 10**8 bits per
hour/ 3600 seconds per hour = 2.15 x 10**5 bits per second. With 2 bits per Hz of
modulation efficiency we require 2.15 x 10**5 / 2 Hz = 108 KHz of bandwidth.