EE12M016_assign1

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EE6170 Introduction to Wireless and Cellular Communications

Computer Assignment 1

SOURAV CHATTERJEE

EE12M016

1)

Function for generating SRRC Pulse

function [h]=rrcpulse(lowlim,incre,highlim,r,tsamp) count=1; for n=lowlim:incre:highlim

if (n==0) h(count)=1-r+(4*r/pi);

elseif (n==((8*tsamp)/(4*r))||n==(-(8*tsamp)/(4*r))) h(count)=(r/sqrt(2))*((1+2/pi)*sin(pi/(4*r))+(1-2/pi)*cos(pi/(4*r)));

else h(count)=(sin(pi*(1-r)*(n/(8*tsamp)))+4*r*(n/(8*tsamp))*cos(pi*(1+r)*(n/(8*tsamp))))/(pi*(n/(8*tsamp))*(1-16*r*r*(n/(8*tsamp))*(n/(8*tsamp))));

end count=count+1;end

Code for generating frequency responseclc;

clear all; r1=.35; r2=.7; r3= 1; %srrc roll off factor tsym=1/25000; %symbol time tsamp=tsym/8; %sampling timef=(2*pi)/tsamp; x1=rrcpulse(-5*tsym,tsamp,5*tsym,r1,tsamp); x2=rrcpulse(-5*tsym,tsamp,5*tsym,r2,tsamp); %srrc pulse generated x3=rrcpulse(-5*tsym,tsamp,5*tsym,r3,tsamp); figure(1) h1=fft(x1); h11=fftshift(h1);

h2=fft(x2); h21=fftshift(h2); h3=fft(x3); h31=fftshift(h3); w=-40:1:40; plot((w*f),20*log10(abs(h11(w+41))), 'r*-',(w*f),20*log10(abs(h21(w+41))), 'b-',(w*f),20*log10(abs(h31(w+41))), 'g^-')%plotting of frequency response grid on; title('Frequency response of SRRC Pulse for different roll-off factors'); xlabel('Frequency (Hz)'); ylabel('Magnitude(db)'); legend('rolloff factor=.35','rolloff factor=.7','rolloff facor=1');

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Plot of Frequency Response

2) Continuation of the above code

xrc=conv(x1,x1); %generate rc pulse by convolving two srrc

pulses count1=1; for a=40:(length(xrc)-41)

xrc1(count1)=xrc(a); %truncate for appropriate sampling count1=count1+1;

end d=randsrc(1,20); %take random sequence of 20 bits dup=zeros(1,8*20); %Generate the upsampled array for i=1:20

dup(8*(i-1)+1)= d(i); %Upsampling the signal by factor of 8 end result0=conv(dup,xrc1/8); %Convolving the result with rc pulse count2=1; for a=40:(length(result0)-41)

result1(count2)=result0(a); %Truncate for approproiate sampling count2=count2+1;

end figure(2) plot(result1,'*-') %Plot the generated waveform title('Resultant waveform for 20 samples'); figure(3) for i=1:20

c(i)=result1(8*(i-1)+1); end eyediag(result1,16); %Generate eye diagram' title('Eye diagram(100 samples taken for proper construction)');

Function for eye diagram

-6 -4 -2 0 2 4 6

x 107

-100

-80

-60

-40

-20

0

20Frequency response of SRRC Pulse for different roll-off factors

Frequency (Hz)

M a g n i t u d e ( d b )

rolloff factor=.35

rolloff factor=.7

rolloff facor=1

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function eyediag(x,n) for i=1:(length(x)/n)

for j=1:n k(j)=x(n*(i-1)+j);

end plot(k)

hold on; end

Values obtained at ideal sampling points

Columns 1 through 4

-1.0064 0.9957 -0.9937 -1.0069

Columns 5 through 8

-0.9940 0.9937 0.9954 -1.0054

Columns 9 through 12

1.0044 -1.0068 0.9985 1.0042


0.9944 -1.0051 1.0071 0.9985


-0.9965 0.9978 1.0004 -0.9975

Sent Values

Columns 1 through 7

-1 1 -1 -1 -1 1 1


-1 1 -1 1 1 1 -1


1 1 -1 1 1 -1

Observations

We see that the sampled values are nearly equal to the sent values. So there is

no ISI in the bandlimited channel due to use of RC pulse shaping.

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Transmitted Waveform

0 20 40 60 80 100 120 140 160-1.5

-1

-0.5

0

0.5

1

1.5

2 Resultant waveform for 20 samples

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Eye Diagram

3)a)Continuation of the above code

%Generation of mdulating bits d1=randsrc(1,1024);%QPSK Symbol generation loopcount=1; for i=1:2:length(d1)-1

if(d1(i)==-1&&d1(i+1)==-1)

qpsk(loopcount)=exp(1i*pi/4); elseif(d1(i)==-1&&d1(i+1)==1) qpsk(loopcount)=exp(3*1i*pi/4);

elseif(d1(i)==1&&d1(i+1)==-1) qpsk(loopcount)=exp(-1i*pi/4);

else qpsk(loopcount)=exp(-3*1i*pi/4);

end loopcount=loopcount+1;

end dup2=zeros(1,(8*loopcount)); dup3=zeros(1,(8*loopcount)); %upsampling by factor of 8

for i=1:loopcount-1 dup2(8*(i-1)+1)= real(qpsk(i)); dup3(8*(i-1)+1)= imag(qpsk(i));

0 2 4 6 8 10 12 14 16-2

-1.5

-1

-0.5

0

0.5

1

1.5

2Eye diagram(100 samples taken for proper construction)

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end %Generation of two orthogonal waveforms for real and imaginary part of QPSK %symbol qpsktx1lin=conv(dup2,x1); qpsktx1qu=conv(dup3,x1); %Truncation for appropriate sampling

count3=1; for a=40:(length(qpsktx1lin)-41) qpsktx1in(count3)=qpsktx1lin(a); qpsktx1qua(count3)=qpsktx1qu(a); count3=count3+1;

end

%addition of noise with Eb/No=6db SNR=9; qpskinrec=awgn(qpsktx1in,SNR); qpskqurec=awgn(qpsktx1qua,SNR);

for b=1:loopcount-1

qpskinrec1(b)=qpskinrec(8*(b-1)+1); qpskqurec1(b)=qpskqurec(8*(b-1)+1);

end qpskrec=qpskinrec1+1i*qpskqurec1; %plotting of received symbols scatterplot(qpskrec)

Scatterplot of received symbols

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b,c)Continuation of code

%Calculation of signal power Eb=0; for f=1:length(qpsktx1in)

Eb=Eb+(qpsktx1in(f))^2; end Ps=Eb/length(qpsktx1in); count6=1; Eb_No_db=0:2:12; %BER and SER simulation for Eb_No=0:2:12

Ps_Pn=Eb_No+3; %Calculation of noise variance variance=Ps/(10^(Ps_Pn/10)); %generation of AWGN noise=sqrt(variance)*randn(1,length(qpsktx1in)); %addition of noise qpsktxinrec2=qpsktx1in+noise; qpsktxqurec2=qpsktx1qua+noise;

%Convolution of receives symbols with matched filter qpsktxinrec3=conv(qpsktxinrec2,x1); qpsktxqurec3=conv(qpsktxqurec2,x1); count4=1; %Truncation for appropriate sampling for a=40:(length(qpsktxinrec3)-41)

qpsktxinrec4(count4)=qpsktxinrec3(a); qpsktxqurec4(count4)=qpsktxqurec3(a); count4=count4+1;

end

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

Q u a d r a t u r e

In-Phase

Scatter plot

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%Extraction of received data for b=1:loopcount-1

qpskinrec5(b)=qpsktxinrec4(8*(b-1)+1); qpskqurec5(b)=qpsktxqurec4(8*(b-1)+1);

end qpskrec2=complex(qpskinrec5,qpskqurec5);

phase=angle(qpskrec2); %Maximum likelihood decoding of received samples for z=1:length(qpskrec2)

if phase(z)>0 && phase(z)<=(pi/2) qpskdec(z)=exp(j*pi/4);

end if phase(z)>(pi/2) && phase(z)<=pi

qpskdec(z)=exp(3*j*pi/4); end if phase(z)<=0 && phase(z)>(-pi/2)

qpskdec(z)=exp(-j*pi/4); end if phase(z)<=(-pi/2) && phase(z)>(-pi)

qpskdec(z)=exp(-3*j*pi/4); end end %Decoding of bits from decoded samples count5=1; for z1=1:length(qpskdec)

if qpskdec(z1)==exp(j*pi/4) qpskdec1(count5)=-1; qpskdec1(count5+1)=-1;

end if qpskdec(z1)==exp(3*j*pi/4)

qpskdec1(count5)=-1; qpskdec1(count5+1)=1;

end if qpskdec(z1)==exp(-j*pi/4) qpskdec1(count5)=1; qpskdec1(count5+1)=-1;

end if qpskdec(z1)==exp(-3*j*pi/4)

qpskdec1(count5)=1; qpskdec1(count5+1)=1;

end count5=count5+2;

end %Calculation of number of errors symerror=0; biterror=0; for k=1:length(qpskdec)

result1=abs(qpsk(k)-qpskdec(k)); if result1~=0

symerror=symerror+1; end

end for i=1:length(qpskdec1)

result2=abs(d1(i)-qpskdec1(i)); if result2~=0

biterror=biterror+1; end

end %Calculation of probability of error psymerror(count6)=symerror/length(qpskdec); pbiterror(count6)=biterror/length(d1);

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pbiterrortheo(count6)=qfunc(sqrt(2*(10^(Eb_No/10))));; count6=count6+1;

endfigure(7) semilogy(Eb_No_db,psymerror,'r^-',Eb_No_db,pbiterror,'b*-',Eb_No_db,pbiterrortheo,'g^-')

title('BER and SER plot versus Eb/No'); xlabel('Eb/No(db)'); ylabel('Probability of error'); legend('SER','BER','Theoretical BER');

Plot of BER and SER versus Eb/No

Conclusion

0 1 2 3 4 5 6 7 8 9 1010

-6

10-5

10-4

10-3

10-2

10-1

BER and SER plot versus Eb/No

Eb/No(db)

SER

BER

Theoretical BE R

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The actual BER and SER curves come a little different from the theoretical

because we have not simulated over sufficient number of samples.

Honour Code

I certify that this assignment submission is my own work and notobtained from any other source

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EE12M016_assign1