MEDIA

DIGITAL COMMUNICATION OF MEDIA

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QUESTION

Communications- assignment

 

 

Description In this assignment you will simulate a simple end-to-end digital communication system, and compare the performance of two simple conceptual modulation schemes in terms of bit error rate.
 

Procedure

  

The block diagram of the system is given in the following figure:

 

 

 

In each transmission cycle:

 

  1. The “Random Bit Generator” produces a 0 or 1 (by tossing a coin, for example), with equal probability.
  2. The “Signal Constellation” box maps the input bits into complex S0 or S1 vectors according to the following two schemes:

 

 

 

 

 

  1. The channel box produces a zero mean complex Gaussian vector withand adds it to the transmitted vector to produce the received vector Yi.
  2. The detector box implements a MAP detection scheme to detect the transmitted bits from the observed received vectors.

 

At the end of these steps you will need to compare the detected bit with the transmitted bit. If they are the same, the transmission is considered to be successful; otherwise, a bit error will be recorded.

 

You will need to repeat the above procedure for 1 million bits for

 

 

and record the values of BER for each value of energy per bit.

 

 

 

 

Report:you are required to submit a report including two sections:

 

Section 1:

 


  1. The plots the BER against                    in dB for both schemes in two different plots from your simulations.
  2. The plots of BER predicted by the theory for both schemes. These results should be represented on the same plot from the simulation results in order to let us compare the simulation results with the theory on the same figures.
  3. Compare and discuss the performance of Scheme 1 and 2 using a separate plot that shows the BER against                     for the both schemes.

 

 

Useful note: plot theoretical results using solid line and simulation results with only marks (use mark ‘o’, please) on your plot.

 

Section 2:

 

A printed version of your source code.

 

 
Submission A single printed report – pages must be properly together.

 

Date: April15, 2012

 
Questions Please contact me atadammeclearn@ymail.comif you have questions.
Important  Notes
  1. If the evaluations of the submitted materials indicate any kind of plagiarism, the result will be a zero mark for the assignment.
  2. Please note that you must use MATLAB, not Simulink, for your mini-project. I think you will not need to write more than 100 lines of code for this project.
  3. Explanation for the code (result) must be provided with final submission.
  4. Don’t forget to send all pages of the report together.

 

SOLUTION

1.According to queation—SECTION-1

CODE—

SCHEME-1—

No=1;

M=10.^6;

Eb=[0,2,4,6,8,10,12,14,16,18,20];

y=20.*log((log(2)).*(1./2).*(erfc((Eb./No).^(1/2)))./(log(M)));

x=20.*(log(Eb./No));

plot(x,y,’o’,x,y);

xlabel(‘Eb/No’);

ylabel(‘BER’);

 

 

 

 

 

 

 

 

 

 

EXPLANATION OF CODE—

 

The code for scheme-1 is for BPSK as according to it’s geometrical representation given in figure.

 

For BPSK error probability is given by—

 

P(e)=(1/2)erfc((Eb/No)^(1/2))…………………………(1)

 

Where erfc is complementry error function.

 

According to Schemes—

There is a mapping from binary to M-ary symbols is performed in such a way that the binary m-tuples corresponding to any pair of adjacent symbols in the M-arymodulation scheme differ in only one bit position.

 

So Bit Error Rate is –

 

BER=(Pe/logM)*(log2);

 

Where M>=2.

 

Here M=1000000.

 

In BPSK—

 

BER=(1/2)erfc((Eb/No)^(1/2))*log(2)/log(M);

 

On logarithmic scale—

 

BER=20*log(BER)

 

And ratio Eb/No will be —  20*log(Eb/No)

 

We can get plot between BER and Eb/No on logarithmic graph.

 

SCHEME-2—

 

No=1;

M=10.^6;

Eb=[0,2,4,6,8,10,12,14,16,18,20];

y=20.*log((log(2)).*(1./2).*(erfc((Eb./2.*No).^(1/2)))./(log(M)));

x=20.*(log(Eb./No));

plot(x,y,’o’,x,y);

xlabel(‘Eb/No’);

ylabel(‘BER’);

 

 

 

 

 

 

 

 

 

EXPLANATION OF CODE–

 

The second scheme is BFSK.Here the error probability will be,

 

Pe=(1/2)*erfc((Eb/2No)^(1/2));

 

And BER =Pe.log(2)/log(M);

 

M=1000000;

 

We will plot   20* log(BER)  Vs   20*log(Eb/No)

 

 

2.Plots of BER predicted by theory can be found by theory available to the student.

 

3.For both schemes—

 

CODE—

No=1;

M=10.^6;

Eb=[0,2,4,6,8,10,12,14,16,18,20];

y1=20.*log((log(2)).*(1./2).*(erfc((Eb./No).^(1/2)))./(log(M)));

y2=20.*log((log(2)).*(1./2).*(erfc((Eb./2.*No).^(1/2)))./(log(M)));

x=20.*(log(Eb./No));

plot(x,y1,’o’,x,y1,x,y2,’o’,x,y2);

xlabel(‘Eb/No’);

ylabel(‘BER’);

 

 

DISCUSSION–

 

First case is for BPSK and second case is for BFSK.

The generation of BFSK is easier.But bandwidth of BFSK is double the bandwidth of BPSK.The distance between signal points is less in case of BPSK.Therefore the erroe rate of BFSK is more compared to BPSK.

 

SECTION-2—The printed version of codes can be get by giving command “print”  at the end of each code.

JF83

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