ECS 455: Problem Set 5 · Page 3 of 4 k OG mPG 1 OG PGm Explain why the above transition probabilities make sense. b. Find the steady-state probabilities using balance equations c

Page 1 of 4

Sirindhorn International Institute of Technology

Thammasat University at Rangsit

School of Information, Computer and Communication Technology

ECS 455: Problem Set 5

Semester/Year: 2/2014

Course Title: Mobile Communications Instructor: Asst. Prof. Dr. Prapun Suksompong ([email protected]) Course Web Site: http://www2.siit.tu.ac.th/prapun/ecs455/

Due date: April 10, 2015 (Friday), 10:30AM

Instructions 1. It is important that you try to solve all problems. (5 pt)

For questions (or parts of questions) that require the use of MATLAB, include printouts of your scripts and the results

from the command window.

2. 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.

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.

Questions

1. (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 to have rain on 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: rain (R), nice (N), and snow (S).

b. Find the corresponding transition probability matrix P.

mailto:[email protected]

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

Page 2 of 4

c. Find the steady-state probabilities

i. by using balance equations

ii. by using the eigen-values & eigen-vectors (in MATLAB)

d. Modify the script MarkovChain_Demo1.m discussed in class to check your answers

in part (b) using simulation in MATLAB.

e. Suppose it is snowing in the Land of Oz today.

i. (*) Calculate the chance that it will be a nice day tomorrow.

ii. (*) Calculate the chance that it will be a nice day the day after tomorrow.

iii. Estimate the chance that it will be a nice day next year (365 days later).

2. 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. 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.

3. 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 (in terms of , , and ) 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:

Page 3 of 4

k

m 1 m

Explain why the above transition probabilities make sense.

b. Find the steady-state probabilities using balance equations

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

.

4. (Optional) 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.

Page 4 of 4

c. Suppose 1m n . Simplify the expression for bP .

Hint: Your answer should be of the form m

ug A for some function g of Au.

P = sym([1/2 1/4 1/4; 1/2 0 1/2; 1/4 1/4 1/2]);

ECS455 HW5 Q1 Markov Chain: Land of OzTuesday, January 11, 2011 8:52 AM

455 2014 HW5 Page 1

% Method 1

[V,D] = eig(P)

Dinf = (abs(D)==1)

Pinf = V*Dinf*(V^(-1))

p = Pinf(1,:)

eval(p)

V =

[ 1, -1, 1]

[ 1, 0, -4]

[ 1, 1, 1]

D =

[ 1, 0, 0]

[ 0, 1/4, 0]

[ 0, 0, -1/4]

Dinf =

1 0 0

0 0 0

0 0 0

Pinf =

[ 2/5, 1/5, 2/5]

[ 2/5, 1/5, 2/5]

[ 2/5, 1/5, 2/5]

p =

[ 2/5, 1/5, 2/5]

ans =

0.4000 0.2000 0.4000

p =

[ 2/5, 1/5, 2/5]

ans =

0.4000 0.2000 0.4000

% Method 2

[V D] = eig(P');

k = find(diag(D)==1);

v = V(:,k);

p = (v./sum(v))'

eval(p)

455 2014 HW5 Page 2

n = 1e4; % The number of slots to be consideredS = [1,2,3]; % Two possible statesP = [1/2 1/4 1/4; 1/2 0 1/2; 1/4 1/4 1/2]; % Transition probability matrixX1 = 2; % Initial state

X = MarkovChainGS(n,S,P,X1);

% Approximate the transition probabilities from the simulationP_sim = []; x = X(1:(n-1)); y = X(2:n);for k = 1:length(S) I = find(x==S(k)); LI = length(I); yc = y(I); cond_rel_freq = hist(yc,S)/LI; P_sim = [P_sim; cond_rel_freq];endP_sim

% Approximate the proportions of time that the states occurp_sim = hist(X,S)./n

>> MarkovChain_2014_HW5_Q1_dP_sim = 0.4995 0.2553 0.2452 0.5177 0 0.4823 0.2583 0.2468 0.4949p_sim = 0.4090 0.2008 0.3902

455 2014 HW5 Page 3

ECS455 HW5 Q2 Call Blocking Probability: Erlang vs. EngsetTuesday, February 14, 2012 1:04 PM

455 2014 HW5 Page 4

455 2014 HW5 Page 5

455 2014 HW5 Page 6

ECS455 HW5 Q3 M/M/m and the Erlang C FormulaSunday, January 20, 2013 12:44 PM

455 2014 HW5 Page 7

455 2014 HW5 Page 8

455 2014 HW5 Page 9

ECS455 HW5 Q4 More on Engset Model (Optional)Wednesday, January 05, 2011 3:21 PM

455 2014 HW5 Page 10

455 2014 HW5 Page 11

Page 1 of 3







Due date: May 8, 2015 (Friday), 10:30AM

Instructions 1. It is important that you try to solve all problems. (5 pt)

For questions (or parts of questions) that require the use of MATLAB, include printouts of your scripts and the results

from the command window.

2. 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.

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.

Questions

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



Page 2 of 3

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.

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 (which is different from the mapping discussed in class). 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?

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

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.

%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.

Page 1 of 4




ECS 455: Problem Set 6 Solution



Due date: May 8, 2015 (Friday), 10:30AM

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 51 x x (b)

2 51 x x x (c) 2 4 51 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 51 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.

(c) 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 3






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 3

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. Evaluate the following expressions by hand. Show your calculation. (You may use MATLAB

to check your answers later.)

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

Page 3 of 3

4. In this question, we will consider an OFDM system in discrete time. The channel is

characterized by 2 1 h . We would like to transmit 1 1 2 1 1 2 1 2 S

of data across this channel using OFDM. For simplicity, we will assume that there is no

noise. Let N = 4 be the length of each OFDM symbol.

a. Find the transmitted vector x. (Apply IFFT with scaling by √𝑁. Then add cyclic prefix.)

To reduce the overhead, the cyclic prefix should be as short as possible.

b. The received vector is y x h . (Note that this is a regular convolution.) Find y .

c. Find H which is the FFT of the zero-padded h .

d. Remove the “irrelevant parts” from y . Then apply FFT with scaling by 1√𝑁⁄ . Finally,

use the corresponding property in frequency domain of the circular convolution (in

time) for DFT to recover the original data S from y .

5. 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 .

Hint: Use oversampling with large value of L.

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

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 .

Use the relationships found in parts (b) and (c).

Page 1 of 5




ECS 455: Problem Set 7 Solution


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



Page 2 of 5

utilizing orthogonality. The computational complexity of the transceivers is solved by the

use of FFT and IFFT.

3.

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

4. In this question, we will consider an OFDM system in discrete time. The channel is

characterized by 2 1 h . We would like to transmit 1 1 2 1 1 2 1 2 S

of data across this channel using OFDM. For simplicity, we will assume that there is no

noise. Let N = 4 be the size of each OFDM symbol.

a. First note that because we have N = 4 but there are 8 numbers that we want to

transmit; we will need two OFDM symbols; one for 1 1 2 1 and another one

for 1 2 1 2 .

To apply IFFT to a (column) vector of length 4, simply multiply it by 1 *

4 4

1

N

Ψ Ψ . In

class, we have already seen that

11 2

2 12 4

1 2 1 1 1

1 1 1 1

1

1

1

N

N N N

NN N N N

N N N N

N N N

Ψ where 2

jN

N e

.

Here, with N = 4, we have 4

1 1 1 1

1 i 1 i

1 1 1 1

1 i 1 i

Ψ and 1

4

1 1 1 1

1 i 1 i

1 1 1 1

1 i 1

1

i

4

Ψ .

Page 3 of 5

So, the IFFT of 1 1 2 1 is

1 1 1 1 1 3

1 1 1 1 2

1 1 1 1 2 3

1 1 1 2

1

4

1

1

4

j j

j j

j

j

and

The IFFT of 1 2 1 2 is

1 1 1 1 1 4

1 1 2 2

1 1 1 1 1 4

1 i 1 2

1 1

4

2

4

j

j

j

.

After scaling by √𝑁 we have

3

11

2

2

3

1 2

j

j

and 42

4

1 2

2

.

Now, to add cyclic prefixes, we note that the memory length of the channel is 1. So,

the cyclic prefix length for each block is 1. This gives

1 3 1 3 1, , , , , 1, 2, 1, 2, 1

2 2 2

1, 3, 1 2 , 3, 1 2 , , 4, 2, 4, 2,

22

2

1 2

2.

j

j

j

j j

j

x

b. 7 5 7 5 3

1 2 j, j, 2 j, j, 2 j, j, 5, 4, 3, 0, 12 2 2 2 2

.

y .

c. To find H which is the FFT of the zero-padded h , we first need to think about how

many 0s should be padded to h . Because N = 4, we need to zero-pad h by two zeros

to bring it to the same size as a data block for OFDM.

So, 4

1 1 1 1

1 i

2 2

1 1

0 0

0 0

1 i

1 1 1 1

1 i 1

3

2i

1

2 j

j

H Ψ .

d. First we remove the “irrelevant parts” from y :

1 2 j 7 5 7 5 3

, j, 2 j, j, 2 j, j2 2 2 2 2 , 5, 4, 3, 0, 1

.

Note that the first two removals correspond to the locations of the cyclic prefixes.

The last removal is the extra junk we get from doing regular convolution.

So, now, we have two received blocks:

Page 4 of 5

7 5 7 5j, 2 j, j, 2 j

2 2 2 2

and 5, 4, 3, 0 .

Note that, earlier, we added the cyclic prefixes so that these received blocks are the

circular convolutions of (the inverse-transform of) our data (in time domain) with

the (zero-padded) channel. This implies that, in the frequency domain, our data are

simply multiplied element-by-element by H .

So, to recover our data, we convert the received blocks back to frequency domain

via FTT. This gives

2 4 2 j 12 4 2 j and 2 8 4 j 6 8 4 j .

Recall that, in the transmission part, we scale our block by √𝑁. To cancel this scaling,

we need to add another scaling of 1√𝑁⁄ which gives

1, 2 j, 6, 2 j and 1, 4 2 j, 3, 4 2 j .

Finally, the circular convolution (in time) for DFT tells us that in the frequency

domain, this received blocks are simply the data blocks multiplied element-wise by

H . Therefore, to recover the data blocks, we divided the elements in the vectors

above by the corresponding elements in H to get

1 1 2 1 and 1 2 1 2 .

Joining these blocks together, we then get back our data stream

1 1 2 1 1 2 1 2 .

5. For MATLAB codes, see ECS455_2014_HW7_Q5.m.

a. Re ( )s t a t b t

b. 2Re ( )s t a t b 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

Page 5 of 5

c. 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 1-1

-0.5

0

0.5

1

1.5

t

a(t)

b(t)

Documents

ECS 455: Problem Set 5 · Page 3 of 4 k OG mPG 1 OG PGm Explain why the above transition probabilities make sense. b. Find the steady-state probabilities using balance equations c