ECS 455: Problem Set 3

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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

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http://www2.siit.tu.ac.th/prapun/ecs455/

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_____(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)?

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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.

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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

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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.



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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

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%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.

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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?



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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)

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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.

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% 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

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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.



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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

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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 ,

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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.

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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.

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.

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ECS 455: Problem Set 5 Solution



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



of 5

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

ECS 455: Problem Set 3

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Transcript of ECS 455: Problem Set 3