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Sirindhorn International Institute of Technology
Thammasat University at Rangsit
School of Information, Computer and Communication Technology
ECS 455: Problem Set 3
Semester/Year: 2/2012
Course Title: Mobile Communications Instructor: Dr. Prapun Suksompong ([email protected]) Course Web Site: http://www2.siit.tu.ac.th/prapun/ecs455/
Due date: Jan 14, 2013 (Monday), 10:30AM
1. Complete the following M/M/m/m description with the following terms:
(I) Bernoulli (II) binomial (III) exponential
(IV) Gaussian (V) geometric (VI) Poisson
The Erlang B formula is derived under some assumptions. Two important assumptions
are (1) the call request process is modeled by a/an ______(A)________ process and (2)
the call durations are assumed to be i.i.d. ______(B)________ random variables. For the
call request process, the times between adjacent call requests can be shown to be i.i.d.
______(C)________ random variables. On the other hand, if we consider non-
overlapping time intervals, the numbers of call requests in these intervals are
______(D)________ random variables.
In order to analyze or simulate the system described above, we consider slotted time
where the duration of each time slot is small. This technique shifts our focus from
continuous-time Markov chain to discrete-time Markov chain. In the limit, for the call
request process, only one of the two events can happen during any particular slot:
either (1) there is one new call request or (2) there is no new call request. When the
slots are small and have equal length, the numbers of new call requests in the slots can
be approximated by i.i.d. _____(E)_________ random variables. In which case, if we
count the total number of call requests during n slots, we will get a/an
Page 2 of 4
_____(F)_________ random variable because it is a sum of i.i.d. _____(E)_________
random variables.
When we consider a particular time interval I (not necessarily small), the number of
slots in this interval will increase as the slots get smaller. In the limit, the number of call
requests in the time interval I which we approximated by a _____(F)_________ random
variable before will approach a/an ______(D)________ random variable.
Similarly, if we consider the numbers of slots between adjacent call requests, these
number will be i.i.d. ______(G)________ random variables. These random variables can
be thought of as discrete counterparts of the i.i.d. ______(C)________ random variables
in the continuous-time model.
Some term(s) above is/are used more than once. Some term(s) is/are not used.
2. (Markov Chain) The Land of Oz is blessed by many things, but not by good weather. They
never have two nice days in a row. If they have a nice day, they are just as likely to have
snow as rain the next day. If they have snow or rain, they have an even chance of having the
same the next day. If there is change from snow or rain, only half of the time is this a
change to a nice day. [Grinstead and Snell, Ex 11.1][Kemeny, Snell, and Thompson, 1974]
a. Draw the Markov chain corresponding to how the weather in the land of Oz changes
from one day to the next.
Hint: This Markov chain will have three states: nice (N), snow (S), and rain (R).
b. Find the steady-state probabilities.
c. Suppose it is snowing in the land of Oz today. Estimate the chance that it will be a
nice day next year (365 days later)?
Page 3 of 4
3. Consider a system which has 3 channels. We would like to find the blocking probability via
the Markov chain method. For each of the following models, draw the Markov chain via
discrete time approximation. Don’t forget to indicate the transition probabilities on the
arrows. Assume that the duration of each time slot is 1 millisecond. Then, find (1) the
steady-state probabilities and (2) the long-term call blocking probability.
a. Erlang B model: Assume that the total call request rate is 10 calls per hour and the
average call duration is 12 mins.
b. Engset model: Assume that there are 5 users. The call request rate for each user is 2
calls per hour and the average call duration is 12 mins.
c. Engset model: Assume that there are 100 users. The call request rate for each user is
0.1 calls per hour and the average call duration is 12 mins.
4. (Difficult) In class we have seen that the steady-state probabilities for the Engset model are
given by
0
,
i i
u u
i mk
u
k
n nA A
i ip
n z m nA
k
, 0 i m ,
where 0
,m
k
k
nz m n A
k
.
a. Express mp (time congestion) in the form
,1
,m
z m c np
z m n
.
What is the value of c?
Hint: c is an integer.
b. The blocked call probability is given by
0
m
u
b mk
u
k
nn m A
mP
nn k A
k
which can be rewritten
in the form
1 2
3 4
,1
,b
z m c n cP
z m c n c
.
Find 1 2 3 4, , ,c c c c .
Hint: 1 2 3 4, , ,c c c c are all integrers.
c. Suppose 1m n . Simplify the expression for bP .
Hint: Your answer should be of the form m
g A for some function g of A.
Page 4 of 4
5. Consider another modification of the M/M/m/m (Erlang B) system. (There are infinite users)
Assume that there is a queue that can be used to hold all requested call which cannot be
immediately assigned a channel. This is referred to as an M/M/m/ or simply M/M/m
system. We will define state k as the state where there are k calls in the system. If k ≤ m,
then all of these calls are ongoing. If k > m, then m of them are ongoing and k-m of them are
waiting in the queue.
Assume that the total call request process is Poisson with rate and that the call durations
are i.i.d. exponential random variables with expected value 1/.
Also assume that .m
a. Draw the Markov chain via discrete time approximation. Don’t forget to indicate the
transition probabilities on the arrows.
Hint: there are infinite number of states. The transition probabilities for state k
which is < m are the same as in the M/M/m/m system. For k m , the transition
probabilities are given below:
k
m 1 m
Explain why the above transition probabilities make sense.
b. Find the steady-state probabilities
c. Find the long-term delayed call probability (the probability that a call request occurs
when all m channels are busy and thus has to wait).
Hint: This will be a summation of many steady-state probabilities. When you simplify
your answer, the final answer should be
1
0
! 1!
m
kmm
k
A
A AA m
m k
.
ECS 455: Problem Set 3 Solution
1. Complete the following M/M/m/m description with the following terms:
(I) Bernoulli (II) binomial (III) exponential
(IV) Gaussian (V) geometric (VI) Poisson
The Erlang B formula is derived under some assumptions. Two important assumptions
are (1) the call request process is modeled by a/an Poisson process and (2) the call
durations are assumed to be i.i.d. exponential random variables. For the call request
process, the times between adjacent call requests can be shown to be i.i.d. exponential
random variables. On the other hand, if we consider non-overlapping time intervals, the
numbers of call requests in these intervals are Poisson random variables.
In order to analyze or simulate the system described above, we consider slotted time
where the duration of each time slot is small. This technique shifts our focus from
continuous-time Markov chain to discrete-time Markov chain. In the limit, for the call
request process, only one of the two events can happen during any particular slot:
either (1) there is one new call request or (2) there is no new call request. When the
slots are small and have equal length, the numbers of new call requests in the slots can
be approximated by i.i.d. Bernoulli random variables. In which case, if we count the
total number of call requests during n slots, we will get a/an binomial random variable
because it is a sum of i.i.d. Bernoulli random variables.
When we consider a particular time interval I (not necessarily small), the number of
slots in this interval will increase as the slots get smaller. In the limit, the number of call
requests in the time interval I which we approximated by a binomial random variable
before will approach a/an Poisson random variable.
Similarly, if we consider the numbers of slots between adjacent call requests, these
number will be i.i.d. geometric random variables. These random variables can be
thought of as discrete counterparts of the i.i.d. exponential random variables in the
continuous-time model.
Some term(s) above is/are used more than once. Some term(s) is/are not used.
HW3 Q2Tuesday, January 11, 2011 8:52 AM
HW 3 Page 1
HW 3 Page 2
HW3 Q3Tuesday, February 14, 2012 1:04 PM
HW 3 Page 3
HW 3 Page 4
HW 3 Page 5
HW3 Q4Wednesday, January 05, 2011 3:21 PM
HW 3 Page 6
HW 3 Page 7
HW3 Q5Sunday, January 20, 2013 12:44 PM
HW 3 Page 8
HW 3 Page 9
HW 3 Page 10
Page 1 of 3
Sirindhorn International Institute of Technology
Thammasat University at Rangsit
School of Information, Computer and Communication Technology
ECS 455: Problem Set 4
Semester/Year: 2/2012
Course Title: Mobile Communications Instructor: Dr. Prapun Suksompong ([email protected]) Course Web Site: http://www2.siit.tu.ac.th/prapun/ecs455/
Due date: Feb 4, 2013 (Monday), 8:50AM
Instructions 1. ONE sub-question will be graded (5 pt). Of course, you do not know which part will be
selected; so you should work carefully on all of them.
2. It is important that you try to solve all problems. (5 pt)
3. Late submission will be heavily penalized.
4. Write down all the steps that you have done to obtain your answers. You may not get
full credit even when your answer is correct without showing how you get your answer.
1. Consider Global System for Mobile (GSM), which is a TDMA/FDD system that uses 25 MHz
for the forward link, which is broken into radio channels of 200 kHz. If 8 speech channels are
supported on a single radio channel, and if no guard band is assumed, find the number of
simultaneous users that can be accommodated in GSM.
2. Draw the complete state diagrams for linear feedback shift registers (LFSRs) using the
following polynomials. Does either LFSR generate an m-sequence?
a. 2 51 x x
b. 2 51 x x x
c. 2 4 51 x x x x
3. Use any resource, find all primitive polynomials of degree 6 over GF(2). Indicate your
reference.
4. We have seen in class that the following waveforms are orthogonal.
Page 2 of 3
Suppose we want to have four orthogonal waveforms on [0,T]. Find two more nonzero
waveforms, 3c t and 4c t , that are time-limited to [0,T]. Make sure that all four
waveforms are mutually orthogonal.
5. In CDMA, each bit time is subdivided into m short intervals called chips. We will use 8
chips/bit for simplicity. Each station is assigned a unique 8-bit code called a chip-sequence.
To transmit a 1 bit, a station sends its chip sequence. To transmit a 0 bit, it sends the one’s
complement1 of its chip sequence.
Here are the binary chip sequences for four stations:
A: 0 0 0 1 1 0 1 1
B: 0 0 1 0 1 1 1 0
C: 0 1 0 1 1 1 0 0
D: 0 1 0 0 0 0 1 0
For pedagogical purposes, we will use a bipolar notation with binary 0 being -1 and binary 1
being +1. In which case, during each bit time, a station can transmit a 1 by sending its chip
sequence, it can transmit a 0 by sending the negative of its chip sequence, or it can be silent
and transmit nothing. We assume that all stations are synchronized in time, so all chip
sequences begin at the same instant.
When two or more stations transmit simultaneously, their bipolar signals add linearly.
a. Suppose that A, B, and C are simultaneously transmitting 0 bits. What is the resulting
(combined) bipolar chip sequence?
b. Suppose the receiver gets the following chips: (-1 +1 -3 +1 -1 -3 +1 +1).
Which stations transmitted, and which bits did each one send?
c. One of your friends wants to work on part (a) and (b) using MATLAB. Here is his code
with two incomplete lines.
1 You should have seen the “one’s complement” operation in your “digital circuits” class.
t
t
1c t
2c t
T
T
Page 3 of 3
%Chip sequences
C = [0 0 0 1 1 0 1 1; 0 0 1 0 1 1 1 0; 0 1 0 1 1 1 0 0; 0 1 0 0 0 0 1 0];
C = 2*C-1; %Change to bipolar form
% Part a
m = [-1 -1 -1 0] %message to transmit
x = %%%%%%%%%HELP ME%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Part b
r = [-1 1 -3 1 -1 -3 1 1] ;
m_decoded = 1/8* %%%%%%%%%HELP ME%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%This gives [mA mB mC mD]' in bipolar form;
%The value is 1 if 1 was transmitted. The value is 0 if nothing was
%transmitted. The value is -1 if 0 was transmitted.
Help him find the expression for “x” and “m_decoded” in the code above. Note that
the expression for x should be in terms of C and m. The expression for m_decoded
should be in terms of C and r.
6. Consider the list of Walsh sequence of order 64 provided in [Lee and Miller, 1998, Table
5.2].
In class, we observed that one of the sequenced is missing. Find the content of that
sequence.
Hint: Use MATLAB.
Page 1 of 4
Sirindhorn International Institute of Technology
Thammasat University at Rangsit
School of Information, Computer and Communication Technology
ECS 455: Problem Set 4
Semester/Year: 2/2012
Course Title: Mobile Communications Instructor: Dr. Prapun Suksompong ([email protected]) Course Web Site: http://www2.siit.tu.ac.th/prapun/ecs455/
Due date: Feb 4, 2013 (Monday), 8:50AM
1. Consider Global System for Mobile (GSM), which is a TDMA/FDD system that uses 25 MHz
for the forward link, which is broken into radio channels of 200 kHz. If 8 speech channels are
supported on a single radio channel, and if no guard band is assumed, find the number of
simultaneous users that can be accommodated in GSM.
Solution
6
3
25 108
200 11000
0
simultaneous users.
2. Draw the complete state diagrams for linear feedback shift registers (LFSRs) using the
following polynomials. Does either LFSR generate an m-sequence?
Page 2 of 4
Solution
(a) 2 5
1 x x (b) 2 5
1 x x x (c) 2 4 5
1 x x x x
(a) The LFSR will cycle
through the following states:
1 0 0 0 0 0 1 0 0 0
1 0 1 0 0
0 1 0 1 0
1 0 1 0 1
1 1 0 1 0
1 1 1 0 1
0 1 1 1 0
1 0 1 1 1
1 1 0 1 1
0 1 1 0 1
0 0 1 1 0
0 0 0 1 1
1 0 0 0 1
1 1 0 0 0
1 1 1 0 0
1 1 1 1 0
1 1 1 1 1
0 1 1 1 1
0 0 1 1 1
1 0 0 1 1
1 1 0 0 1
0 1 1 0 0
1 0 1 1 0
0 1 0 1 1
0 0 1 0 1
1 0 0 1 0
0 1 0 0 1
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1
(b) The LFSR will cycle
through one of the cycles of
states below. The initial state
determine which cycle it will
go through.
Cycle #1: 1 0 0 0 0
1 1 0 0 0 0 1 1 0 0
1 0 1 1 0
1 1 0 1 1 1 1 1 0 1
1 1 1 1 0
0 1 1 1 1 0 0 1 1 1
1 0 0 1 1
0 1 0 0 1 0 0 1 0 0
0 0 0 1 0
0 0 0 0 1
Cycle #2: 0 0 0 1 1
1 0 0 0 1 0 1 0 0 0
1 0 1 0 0
1 1 0 1 0 0 1 1 0 1
0 0 1 1 0
Cycle #3: 0 0 1 0 1
1 0 0 1 0
1 1 0 0 1 1 1 1 0 0
0 1 1 1 0
1 0 1 1 1 0 1 0 1 1
Cycle #4: 0 1 0 1 0 1 0 1 0 1
Cycle #5: 1 1 1 1 1
(c) The LFSR will cycle
through the following states:
1 0 0 0 0
1 1 0 0 0
0 1 1 0 0 1 0 1 1 0
0 1 0 1 1
1 0 1 0 1 0 1 0 1 0
0 0 1 0 1 1 0 0 1 0
0 1 0 0 1
0 0 1 0 0 0 0 0 1 0
1 0 0 0 1
0 1 0 0 0 1 0 1 0 0
1 1 0 1 0
1 1 1 0 1 1 1 1 1 0
1 1 1 1 1
0 1 1 1 1 1 0 1 1 1
1 1 0 1 1
0 1 1 0 1 0 0 1 1 0
1 0 0 1 1
1 1 0 0 1 1 1 1 0 0
0 1 1 1 0
0 0 1 1 1
0 0 0 1 1
0 0 0 0 1
The polynomial 2 51 x x and 2 4 5
1 x x x x from part (a) and (c) generate m-
sequences. (Their states go thorough cycle of size 25-1)
Page 3 of 4
3. Use any resource, find all primitive polynomials of degree 6 over GF(2). Indicate your
reference.
Solution
Primitive Polynomials
x6 + x1 + 1 x6 + x5 + x2 + x1 + 1 x6 + x5 + x3 + x2 + 1 x6 + x4 + x3 + x1 + 1 x6 + x5 + x4 + x1 + 1 x6 + x5 + 1
Source: http://www.theory.cs.uvic.ca/~cos/gen/poly.html
4. See the hand-written solution at the end.
5. In CDMA, each bit time is subdivided into m short intervals called chips. We will use 8
chips/bit for simplicity. Each station is assigned a unique 8-bit code called a chip-sequence.
To transmit a 1 bit, a station sends its chip sequence. To transmit a 0 bit, it sends the one’s
complement1 of its chip sequence.
Here are the binary chip sequences for four stations:
A: 0 0 0 1 1 0 1 1
B: 0 0 1 0 1 1 1 0
C: 0 1 0 1 1 1 0 0
D: 0 1 0 0 0 0 1 0
For pedagogical purposes, we will use a bipolar notation with binary 0 being -1 and binary 1
being +1. In which case, during each bit time, a station can transmit a 1 by sending its chip
sequence, it can transmit a 0 by sending the negative of its chip sequence, or it can be silent
and transmit nothing. We assume that all stations are synchronized in time, so all chip
sequences begin at the same instant.
When two or more stations transmit simultaneously, their bipolar signals add linearly.
a. Suppose that A, B, and C are simultaneously transmitting 0 bits. What is the resulting
(combined) bipolar chip sequence?
b. Suppose the receiver gets the following chips: (-1 +1 -3 +1 -1 -3 +1 +1).
Which stations transmitted, and which bits did each one send?
Solution
%Chip sequences
C = [0 0 0 1 1 0 1 1; 0 0 1 0 1 1 1 0; 0 1 0 1 1 1 0 0; 0 1 0 0 0 0 1 0];
C = 2*C-1; %Change to bipolar form
1 You should have seen the “one’s complement” operation in your “digital circuits” class.
Page 4 of 4
% Part a
m = [-1 -1 -1 0] %message to transmit
x = %%%%%%%%%HELP ME%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Part b
r = [-1 1 -3 1 -1 -3 1 1] ;
m_decoded = 1/8* %%%%%%%%%HELP ME%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%This gives [mA mB mC mD]' in bipolar form;
%The value is 1 if 1 was transmitted. The value is 0 if nothing was
%transmitted. The value is -1 if 0 was transmitted.
Use the above MATLAB code with x = m*C; and m_decoded = (C*r')/8;
(a) [3 1 1 -1 -3 -1 -1 1]
(b) [1 -1 0 1]'; Hence, A and D sent 1 bits, B sent a 0 bit, and C was silent.
HW4 Q4 SolutionThursday, March 04, 2010 2:22 PM
HW Part 2 Page 1
HW Part 2 Page 2
HW Part 2 Page 3
Page 1 of 6
Sirindhorn International Institute of Technology
Thammasat University at Rangsit
School of Information, Computer and Communication Technology
ECS 455: Problem Set 5
Semester/Year: 2/2011
Course Title: Mobile Communications Instructor: Dr. Prapun Suksompong ([email protected]) Course Web Site: http://www2.siit.tu.ac.th/prapun/ecs455/
Due date: Not due 1. Consider the list of Walsh sequence of order 64 provided in [Lee and Miller, 1998, Table
5.2].
In class, we observed that one of the sequenced is missing.
Find the content of that sequence.
Hint: Use MATLAB.
Page 2 of 6
2. Select the terms (provided at the end of the problem) to complete the following description
of OFDM systems:
Wireless systems suffer from ______________ problem. Equalization can be used to
mitigate this problem. Another important technique that works effectively in wireless
systems is OFDM. The general idea is to ______________ the symbol or bit time so that
it is ______________ compared with the channel delay spread. To do this, we separate
the original data stream into multiple parallel substreams and transmit the substreams
via different carrier frequencies, creating parallel subchannels. This is called
______________. In such direct implementation, there are two new problems to solve:
bandwidth inefficiency and complexity of the transceivers. The inefficient use of
bandwidth is caused by the need of ______________ between adjacent subchannels.
Bandwidth efficiency can be improved by utilizing ______________. The computational
complexity of the transceivers is solved by the use of ______________.
Here are the terms to use. Some term(s) is/are not used.
FFT and IFFT
FDM
multipath fading
local oscillators
guard bands
guard times
reduce
increase
small
large
spectral efficiency
orthogonality
3. Recall that the baseband OFDM modulated signal can be expressed as
0,
1
0
1 2( ) e1 xp
sT
N
k
k s
kts t S j
TNt
where 0 1 1, , , NS S S are the (potentially complex-valued) messages.
Let 1sT [ms], N = 8, and
0 1 1, , , 1 ,1 ,1,1 , 1 ,1,1 , 1NS S S j j j j j j
a. Use MATLAB ifft command to plot Re s t for 0 st T .
b. Let
i. 1
0
1 2Re cos
N
k
k s
kta t S
TN
ii. 1
0
1 2Im sin
N
k
k s
ktb t S
TN
Page 3 of 6
What is the relationship between a t , b t , and Re ( )s t ?
c. Let
0,
1*
2
0
1 2( ) exp1
sT
N
k
k s
kts t S j
TNt
.
Note the extra conjugation in 2 ( )s t .
What is the relationship between a t , b t , and
2Re ( )s t ?
d. Use MATLAB to plot a t and b t for 0 st T .
4. Evaluate the following expressions by hand. Show your calculation.
a. DFT 3 1
b. DFT 1 0 0
c. IDFT 1 0 0
d. DFT 1 0 0 0 0
e. 1 2 1 2 1 2
f. 1 2 1 2 1 2
g. 1 2 1 0 2 1 2 0
h. 1 2 1 0 0 2 1 2 0 0
5. Consider the discrete-time complex FIR channel model
2
0
*m
y n h x n w n h m x n m w n
where w n is zero-mean additive Gaussian noise.
In this question, assume that h[n] has unit energy and that H z has two zeros at
2
31
j
z e
and 2
1z
where 1 .
a. The information given above implies that
1
un
un
h
h k h kE
and 1
un
un
h
H z H zE
where
1 2 1 21 2unh k k z z k z z k ,
1 1 1 2
1 2 1 2 1 21 1 1unH z z z z z z z z z z z ,
Page 4 of 6
and
2 2 22
1 2 1 21unhE z z z z .
Plot j
j
z eH e H z
in the range
20 : : 2
80
for = 0.5 and 0.99.
See Appendix A for the discussion on how I get h k and H z .
b. For OFDM system with block size N = 8, find the corresponding channel gains
2j k
Nk z eH H z
, k = 0, 1, 2, …, N-1 for = 0.5 and 0.99. In particular, complete the
following table.
Ch # k
= 0.5 = 0.99
Hk |Hk| Hk |Hk|
0 -0.5455 + 0.1890i 0.5774 -0.0087 + 0.0050i 0.0100
1
2
3
4 0.9820 + 0.5669i 1.1339 0.5860 + 0.9949i 1.1547
5
6
7
6. OFDM simulation: Write a MATLAB code to perform the following operations
a. Generate 10,000 OFDM blocks, each is an 8 dimensional QPSK vector. Each element
of the vector is independently and randomly chosen from the constellation set
1 ,1 , 1 , 1M j j j j .
b. Perform the MATLAB’s IFFT to each vector. Multiply the result by N . (See Q3a for
the reason why we need to multiply by N .)
c. Add the cyclic prefix to each block and transmit over the FIR channel defined in the
previous question. Assume 0w n . Consider two cases: = 0.5 and 0.99.
d. At the receiver, remove the cyclic prefix and perform the FFT to get kR . Scale the
results by 1/ N .
e. Assume that the receiver knows Hk. Detect the transmitted symbols at each channel.
Page 5 of 6
When there is no noise, you have k k kR H S . Therefore, kk
k
RS
H . Record the
symbol error rates (SER) for each channel.
Remark: All of them should be 0 in this question. The goal of this problem is to make
you that you understand the OFDM system enough to write down MATLAB code for
it.
7. (Difficult) Repeat Question 6. However, in this question, the channel noise is generally non-
zero. w n is now i.i.d. complex-valued Gaussian noise. Its real and imaginary parts are i.i.d.
zero-mean Gaussian with variance 0N /2 where N0 = 1.
For the decoder, use the ML (maximum likelihood) detector to decode back kS . To do this,
first calculate k
k
R
H. Then, find ˆkS which is defined to be the closest message s in the
constellation to k
k
R
H.
a. Complete the following table. The Hk can be copied from the Table in Q2b.
= 0.5 = 0.99
Ch # k |Hk| SER |Hk| SER
0
1
2
3
4
5
6
7
Note that a symbol is counted as decoded correctly if both real and imaginary parts
are decoded correctly.
b. What is the relationship between |Hk| and SER.
Page 6 of 6
Appendix A
The channel h has two zeros at 2
31
j
z e
and 2
1z
.
Thus, before normalized to unit energy, we have
1 1 1 2
1 2 1 2 1 21 1 1unH z z z z z z z z z z z .
Inverse Z-transform gives
1 2 1 21 2unh k k z z k z z k .
The energy of unh k is 2 2 22
1 2 1 21unhE z z z z .
So,
1
un
un
h
h k h kE
and 1
un
un
h
H z H zE
.
Appendix B
When there is some zero-mean Gaussian noise, k k k kR H S W where kW is the noise (in the
frequency domain). Therefore,
New Noise
k kk
k k
R WS
H H .
Because the noise is Gaussian and zero-mean, the noise will most-likely not take k
k
R
H too far
from kS . Therefore, the ML detector gives
ˆ arg min arg min kk k k
s M s M k
RS R sH s
H
,
i.e. it detects kS as the closest message s in the constellation to k
k
R
H. Of course, the noise can
be large and shift k
k
R
H too far from the original kS . Therefore, ˆkS may be different from kS .
This is when symbol error occurs.
Page 1 of 5
Sirindhorn International Institute of Technology
Thammasat University at Rangsit
School of Information, Computer and Communication Technology
ECS 455: Problem Set 5 Solution
Semester/Year: 2/2011
Course Title: Mobile Communications Instructor: Dr. Prapun Suksompong ([email protected]) Course Web Site: http://www2.siit.tu.ac.th/prapun/ecs455/
1. In MATLAB, it is easy to generate the Hadamard matrix by the command hadamard(N).
However, note that the Walsh sequences are indexed by zero crossings. We can follow the
recipe for index changing provided in lecture or we can directly generate the Walsh matrix
by the command ifwht(eye(N)).
Here, N = 64. So, we use ifwh(eye(64)). Note however that the outputs will be in the
{1,-1} from. To map them back to the {0,1} form, we calculate
W = (1- ifwht(eye(64)))/2.
Finally, the missing row is W42. Note that the first row is W0. Therefore, W42 is the 43rd row.
Its content is given by W(43,:). From MATLAB, W42 is
[0110 1001 1001 0110 1001 0110 0110 1001
1001 0110 0110 1001 0110 1001 1001 0110]
2. Wireless systems suffer from multipath fading problem. Equalization can be used to
mitigate this problem. Another important technique that works effectively in wireless
systems is OFDM. The general idea is to increase the symbol or bit time so that it is large
compared with the channel delay spread. To do this, we separate the original data stream
into multiple parallel substreams and transmit the substreams via different carrier
frequencies, creating parallel subchannels. This is called FDM. In such direct
implementation, there are two new problems to solve: bandwidth inefficiency and
complexity of the transceivers. The inefficient use of bandwidth is caused by the need of
guard bands between adjacent subchannels. Bandwidth efficiency can be improved by
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utilizing orthogonality. The computational complexity of the transceivers is solved by the
use of FFT and IFFT.
3. Solution
a.
b. Re ( )s t a t b t
c. 2Re ( )s t a t b t
d. From part (b) and (c), we have
2Re ( ) Re ( )
2
s t s ta t
and
2Re ( ) Re ( )
2
s t s tb t
.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
Re{s
(t)}
t
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-1
-0.5
0
0.5
1
1.5
t
a(t)
b(t)
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4.
a. DFT 3 1 2 4
b. DFT 1 0 0 1 1 1
c. 1 1 1
IDFT 1 0 03 3 3
d. DFT 1 0 0 0 0 1 1 1 1 1
e. 1 2 1 2 1 2 2 5 2 5 2
f. 1 2 1 2 1 2 3 7 2
g. 1 2 1 0 2 1 2 0 4 5 2 5
h. 1 2 1 0 0 2 1 2 0 0 2 5 2 5 2
5. Consider the discrete-time complex FIR channel model
2
0
*m
y n h x n w n h m x n m w n
where w n is zero-mean additive Gaussian noise.
In this question, assume that h[n] has unit energy and that H z has two zeros at
2
31
j
z e
and 2
1z
where 1 .
Solution
a. The plots of j
j
z eH e H z
in the range 0 2 for = 0.5 and 0.99 are
shown below:
0 1 2 3 4 5 6
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
H(
)
=0.5
0 1 2 3 4 5 60
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
H(
)
=0.99
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b. For OFDM system with block size N = 8, find the corresponding channel gains
2j k
Nk z eH H z
, k = 0, 1, 2, …, N-1 for = 0.5 and 0.99.
Ch # k
= 0.5 = 0.99 2
j kNH e
2
j kNH e
2
j kNH e
2
j kNH e
0 -0.5455 + 0.1890i 0.5774 -0.0087 + 0.0050i 0.0100
1 0.1407 + 0.6246i 0.6403 0.5170 + 0.1489i 0.5380
2 0.4657 + 0.3858i 0.6047 0.3710 - 0.2026i 0.4227
3 0.4649 + 0.4555i 0.6508 -0.0624 + 0.2716i 0.2787
4 0.9820 + 0.5669i 1.1339 0.5860 + 0.9949i 1.1547
5 1.4881 - 0.1882i 1.4999 1.6375 + 0.4284i 1.6927
6 0.8436 - 1.1417i 1.4196 1.3609 - 0.7973i 1.5773
7 -0.3480 - 0.8919i 0.9574 0.2171 - 0.8489i 0.8762
6. All symbol error rates (SER) should be 0 because there is no noise.
7. In this question, the channel noise is generally non-zero. w n is now i.i.d. complex-valued
Gaussian noise. Its real and imaginary parts are i.i.d. zero-mean Gaussian with variance 0N
/2 where N0 = 1.
Solution
a.
= 0.5 = 0.99
Ch # k |Hk| SER |Hk| SER
0 0.5774 0.3631 0.0100 0.7487
1 0.6403 0.3316 0.5380 0.3919
2 0.6047 0.3601 0.4227 0.4736
3 0.6508 0.3292 0.2787 0.5750
4 1.1339 0.1032 1.1547 0.0973
5 1.4999 0.0328 1.6927 0.0145
6 1.4196 0.0423 1.5773 0.0235
7 0.9574 0.1684 0.8762 0.2028
b. If you try to plot |Hk| vs. SER, you should get something similar to the plot below.
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So, |Hk| and SER go in the opposite directions. The channel that has large value of
|Hk| will have very good SER performance; that is it will has low SER.
Furthermore, the SER of ch#0 when 0H is about 0 should be very close to 0.75. This
is because the channel gain destroys almost all the information contained in the
received signal. Hence, the ML detector will be correct with probability 0.5 for each
dimension. The complex number (QPSK symbol) has two dimensions. Hence, the
chance that it will be decoded correctly is 0.5 0.5 0.25 .
If you are good at digital communications, you may check that the SER is given by
22p p where 0
2kp Q H
N
.
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.80
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
|Hk|
SE
R