NAME ____________________________________________
ES 442 Homework # 8
(Spring 2016 – Due May 11, 2016 )
Print out homework and do work on the printed pages.
Textbook: B. P. Lathi & Zhi Ding, Modern Digital and Analog Communication
Systems, 4th edition, Oxford University Press, New york, 2009.
Problem 1 Bandwidth Efficiency (16 points)
We define the bandwidth efficiency (or spectral efficiency) of a communication channel
as the maximum bit rate (bits/second) divided by the bandwidth (in hertz). In this
problem you are asked to compare the three versions of Wi-Fi, namely 802.11g/n/ac for
specified bandwidth efficiency. From the technical specifications for these three
versions of Wi-Fi we know:
802.11g uses a 25 MHz band with a maximum date rate of 54 Mbps (megabits/second)
802.11n uses a 20 MHz band with a maximum date rate of 300 Mbps (bits/second
802.11ac uses a 160 MHz band with a maximum date rate of 6933 Mbps (bits/second
All three Wi-Fi versions operate in the 2.4 GHz ISM band.
Wi-Fi Version
802.11g
802.11n
802.11ac
Bandwidth Efficiency
Don’t forget to include the units for bandwidth efficiency in your answer.
Problem 2 M-ary QAM Symbol Error Rates (16 points)
In slide set 14 on “Digital Carrier Systems” we discussed the symbol error rates versus
Eb/N0 per symbol for several M-ary Quadrature Ampltude Modulation schemes.
Remember that Eb/N0 is the energy per symbol (or bit) divided by the noise power per
hertz (the units for Eb/N0 is [joule/(watts/Hz)] and often expressed in dB).
(b) How does your answer from part (a) above depend upon the actual value of the
symbol error rate (SER)?
Homework 8
(a) Considering 16-QAM, 64-QAM, 256-QAM and 1024-QAM, by how much does Eb/N0
change for an increase in M by a factor of four (e.g., going from 64-QAM to 256-QAM)?
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(c) At a symbol error rate of 10-4, an engineer suggests using 4096-QAM. In that case,
what value of Eb/N0 would we require? [Express your answer in dB.]
Problem 3 US Digital Cellular Standard Data Rate (18 points)
The bandwidth efficiency  is computed from the data rate R divided by the channel
bandwidth B. Remember from an earlier homework assignment that Shannon’s
channel capacity equation is given by

 S 
C  B log2  1   
 N 

in bits/second
where C is the capacity, B is bandwidth and (S/N) is the signal-to-noise ratio as usual.
Capacity C may be thought of as the maximum data rate estimate for a communication
channel (or maximum bandwidth efficiency max for the channel.
The US Digital Cellular Standard (established 1991) calls for a data rate of 48.6 kbps
with a channel bandwidth of 30 kHz and with a signal-to-noise ratio of 20 dB. By the
way, US Digital Cellular operates from 824 MHz to 894 MHz (both uplink and downlink).
(a) Calculate Shannon’s capacity C for the US Digital Cellular Standard with the above
parameters.
Homework 8
(b) How does your answer from part (a) above compare to 48.6 kbps?
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Problem 4 Non-coherent FSK Receiver (20 points)
The block diagram shows a proposed non-coherent binary FSK (binary frequency shift
keying) receiver as shown by a prior student while at SSU.. Two frequencies (fH & fL)
are used to represent a 1 and a 0.
(a) Will this circuit actually demodulate a BFSK encoded signal? If yes, explain why it
will work. If, no, explain why it will not work.
(b) If you do not know the answer for part (a) above, explain why you don’t know.
Problem 5 Serial to Parallel Parser (16 points)
Homework 8
When an input bit stream drives a quadrature modulator, we need to parse the input bit
stream into two bit sequences intended to drive the in-phase and quadrature mixers.
Thus, each pair of bits representing a symbol must be separated into an I-sequence of
bits and a Q-sequence of bits. Let mk(t) be the input bit stream of data and you are to
separate it into the in-phase data sequence mI(t) and quadrature sequence mQ(t). Be
sure to label each in the in-phase and quadrature bit sequences. Illustrate them on the
graph provide below.
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Problem 6 PN Codes in Spread Spectrum (14 points)
Let
m(t) be the digital message to be communicated,
c(t) is the PN code used to spread m(t), and
n(t) is random noise added as the modulated signal travels across the channel
The transmitter multiplies m(t) and c(t) together to form ss(t) which is the signal
launched onto the channel. The received signal from the channel is ss(t) + n(t)
Homework 8
In direct sequence spread spectrum a pseudo-noise bit code (also known as pseudorandom or chip code or bit sequence; PN code for brevity) is used to spread the
spectrum. The PN code is quasi-random and has an autocorrelation function with a
value of unity when it is coincident with itself and of negligible value when shifted in time
over it full length.
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representing the modulated signal plus added noise. In some cases the noise n(t) may
be larger than the signal ss(t).
Let the received signal be r(t) = ss(t) + n(t). The receiver takes r(t) and multiplies it by
an exact replica of c(t).
Homework 8
Given that ss(t) = m(t)c(t), show that the data output of the receiver is m(t).
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