ECE6610: Wireless Networks Fall 2016 Homework 1

ECE6610: Wireless Networks
Fall 2016
Homework 1

Given: August 30, 2016
Due: September 23, 2016 (Midnight) + 1 week for “off Campus” students

Submission Instructions:
Submit your homework as a DOC or PDF file to
No hardcopies will be accepted. Scanned pages are fine.

Dr. Ian F. Akyildiz
Ken Byers Chair Professor in Telecommunications
Broadband Wireless Networking Laboratory
School of Electrical and Computer Engineering
Georgia Institute of Technology, Atlanta, GA 30332
Tel.: 404-894-5141; Fax: 404-894-7883; E-Mail:


Question 1 (Basics of Wireless Communications):
a) A telephone line is known to have a loss of 20 dB. The input signal power is measured at 1
Watt, and the output signal noise level is measured at 1 mW. Using this information, calculate
the output signal to noise ratio in dB.

Input signal power = 1 W = 0 dB
Output signal power = Input signal power - Path loss = 0-20 = -20 dB
Output noise level = 1 mW = -30 dB
Output SNR = Output signal power – Output noise level = -20-(-30)= 10 dB



b) What is the maximum data rate that can be supported on a 10 MHz noise-less channel if
the channel uses eight-level digital signals?

From Nyguist formula,
C=2B log2M = 2  10M  log28 = 60 Mbps




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, c) A sine wave is used for two different signaling schemes (i) BPSK and (ii) QPSK. The duration
of a signal element is 10-5 s. If the received signal is of the form s(t)=0.005 sin(2π106t+) volts
and if the measured noise power at the receiver is 2.510-8 watts, determine the Eb/N0 in dB for
each modulation.

Duration of signal element T=10-5 s
1 (.((*+
Signal power 𝑃 = 𝑇 = = 0.0000125 = 12.5 µ𝑊
𝑇 ( 𝑠&
&
𝑡

i) BPSK:
Energy per bit Eb=PT=
𝐸𝐵 𝑃 12.510–:
T
= = 500 = 𝟐𝟕 𝒅𝑩
𝑁( 2.410–9T 2.510–9
=

ii) QPSK:
𝐸𝐵 𝑃/2T 6.2510–:
= = 250 = 𝟐𝟒 𝒅𝑩
𝑁( 2.410–9T 2.510–9
=




d) Determine the mean received power at the receiver node. The distance between the transmitting
station and the receving nodes is 500m. The transmitter and receiver antenna gains are 10dB
and 5 dB respectively. Use pass loss exponent of 4. The transmitted power is 30 dBm. Do all
calculations using dB.

𝜆 = 𝑣/𝑓

3×109
𝜆= = 𝟎. 𝟏𝟐𝟓 𝐦
2.4×10G




𝑃𝑟𝑥 𝜆
= 𝑃𝑡𝑥 + 𝐺𝑡𝑥 + 𝐺𝑟𝑥 + 10𝛾 log
4𝜋𝑑

𝑃𝑟𝑥
0.125
= 30 + 10 + 5 + 10×4× log
4𝜋×500

𝑃𝑟𝑥 = 45 + 40×−4.70127


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