bai giai truyen thong khong day moi

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Chapter 2Problem 1 Powerlawprofiles.mclear%powerlawprofiles.m; calculates the powerloss as per inverse square law and 1/d^n.close allnu=[2 3 3.5 4];%four values of the loss parameterfor k=1:4; n=nu(k);d=1:50;% distance in Kmd1=d.^n;dn=1./d1;pn=-10.*log10(dn);%inverse lawplot(d,pn,'k');xlabel('distance in Km');ylabel('powerloss in dB');hold onend;

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Problem 2 From the book, eqn. (2.2) is2 2

2 2 2

1000 2 116 16 3 5000

r r td

P G GPL d

.

The wavelength 13

m . 85.6 10 72.5dP dBm .

Problem 3& 4 hataloss.m

clear% hataloss.m Power loss calculationscarf=input('enter carrier freq. in MHZ...>>');lcrf=log10(carf);ht=input('enter the height of BS antenna in meters..>>');%height in meters of the BSlht=log10(ht);hr=input('enter the height of the MU antenna in meters..>>');%MU ant. height in meterslhr=log10(11.75*hr);N=60;%number of increments for distancedistmax=30;%coverage distance in Kmsfor kk=1:1:59;d(kk)=1+0.5*(kk-1);fact1=69.55+26.16*lcrf-13.83*lht;fact2=(44.9-6.55*lht)*log10(d(kk));fftq=fact1+fact2;%antenna correction factor MU...ahr=3.2*(lhr)^2-4.97;% large cities; freq. above 400 MHzamr=(1.1*lcrf-0.7)*hr-(1.56*lcrf-0.8);alar(kk)=fftq-ahr;amed(kk)=fftq-amr;asub(kk)=amed(kk)-2*(log10(carf/28)).^2-5.4;arur(kk)=amed(kk)-4.78*(lcrf.^2)+18.33*lcrf-40.94;end;

subplot(2,2,1); plot(d,alar); xlabel('distance Km'); ylabel('power loss dB'); title('large city');

subplot(2,2,2); plot(d,amed);

%axis([1 30 -140 -60]); xlabel('distance Km'); ylabel('power loss dB'); title('medium city');

subplot(2,2,3);

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plot(d,asub);

%axis([1 30 -140 -60]); xlabel('distance Km'); ylabel('power loss dB'); title('suburban');

subplot(2,2,4); plot(d,arur);

%axis([1 30 -140 -60]); xlabel('distance Km'); ylabel('power loss dB'); title('rural');

%loss at a reference distance of 100 m Wave=3e8/900e6;%wavelength in meters PL=10*log10((4*pi*100)^2/Wave^2);%relative loss at 100 meters d3=d(3)*1000;%distance in meters %use eqn. (2.19) denom=10*log10(d3/100);%denominator of eqn. (2.19) nuLarge=(alar(3)-PL)/denom%nu for large city nuMed=(amed(3)-PL)/denom%nu for med city nuSub=(asub(3)-PL)/denom%nu for suburbs city nuRur=(arur(3)-PL)/denom%nu for rural city

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Large = 4.1865Med = 4.1801Sub =3.4159Rur =1.9890

Problem 5.fadcom.m

clear%fadcom.m simulates Doppler fading including a carrier frequency and then demodulate%uses random phase model for both Rayleigh and Rician fading% also computes the outage probabilityfc=900e3;%carrier frequencyppf=2*pi*fc;N=11;% Number of mutiple pathsfd=input('enter the Doppler shift...')ppw=2*pi*fd;%scattering strengths are considered to be Gamma distributed

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fs=4*fc;%sampling rateTs=1/fs;t = [0:Ts:1999*Ts]; %time arraynn=length(t);ci=zeros(1,nn); %received signal (Rayleigh for m=1:N; a = gamrnd(1.4,1,1,nn); ppwt=ppw*unifrnd(0,2*pi,1,nn); ci = ci + a.*cos((ppf+ppwt).*t+ppwt+unifrnd(0,2*pi,1,nn));end;cr=ci;crr=ci+2*cos(ppf*t);%generates Rician[x,y]=demod(cr,fc,fs,'qam');[xr,yr]=demod(crr,fc,fs,'qam');ae=(1/N)*sqrt(x.^2+y.^2);aer=(1/N)*sqrt(xr.^2+yr.^2); pha=pi+angle(x+i*y);%phase to be centered around 2*pi figuresubplot(2,1,1)plot(t*1e3,cr),title('Rayleigh fading')xlabel('time ms')ylabel('rf signal volt')xlim([0 .1])aes=20*log10(ae);subplot(2,1,2)plot(t*1e3,aes),title('Rayleigh fading')xlabel('time ms')ylabel('relative power dB')xlim([0 .1])figuresubplot(2,1,1)plot(t*1e3,crr),title('Rician fading')xlabel('time ms')ylabel('rf signal volt')xlim([0 .1])

aerr=20*log10(aer);

subplot(2,1,2)plot(t*1e3,aerr),title('Rician fading')xlabel('time ms')ylabel('relative power dB')xlim([0 .1])

fractra=0.0;for mf=1:nn;

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if aes(mf)<=-10 fractra=fractra+1; else end;end;outageRayl=fractra/nnfractrc=0.0;for mf=1:nn; if aerr(mf)<=-10 fractrc=fractrc+1; else end;end;outageRice=fractrc/nn

Problem 6.fadingsim.m

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clear%fadingsim.m simulates fading including a carrier frequency and then demodulate%uses random phase model as well as random delayfc=900e3;%carrier frequencyppf=2*pi*fc;N=11;% Number of mutiple paths%a are considered to be Gamma distributedfs=4*fc;%sampling rate

Ts=1/fs;t = [0:Ts:1999*Ts]; %time arraynn=length(t);ci=zeros(1,nn); %received signal (Rayleighnn=length(t);dist=1000;%distance between transmit and receive in metersavdel=dist/3e8%average delay in seconds ci=zeros(1,nn); cir=zeros(1,nn); cit=0.0;%this uses a delay instead of Uniform phase for m=1:N; phterm=unifrnd(0,2*pi,1,nn); delay=avdel+(-0.5+unifrnd(0,2*pi,1,nn))*avdel;%uniform delay with a mean of avdel a = gamrnd(1.4,1,1,nn); ci=ci+a.*cos(ppf*t+phterm);%Rayleigh %following term generates the received signal with delay format cit=cit+a.*cos(ppf*t+delay); end;crr=cit+2*cos(ppf*t);%rician[x,y]=demod(ci,fc,fs,'qam');[xt,yt]=demod(cit,fc,fs,'qam');[xr,yr]=demod(crr,fc,fs,'qam');ae=(1/N)*sqrt(x.^2+y.^2);aet=(1/N)*sqrt(xt.^2+yt.^2);aer=(1/N)*sqrt(xr.^2+yr.^2);

pha=pi+angle(x+i*y);%phase to be centered around 2*pi phat=pi+angle(xt+i*yt);%phase of the delay versionfiguresubplot(2,2,1)plot(t*1e3,ci),title('random phase: Rayleigh');xlabel('time ms')ylabel('rf signal volt')xlim([0 0.1])

subplot(2,2,2)

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plot(t*1e3,ae),title('random phase:Rayleigh');xlabel('time ms')ylabel('envelope volt')xlim([0 0.1])

subplot(2,2,3)plot(t*1e3,cit),title('random delay:Rayleigh');xlabel('time ms')ylabel('rf signal volt')xlim([0 0.1])

subplot(2,2,4)plot(t*1e3,aet),title('random delay:Rayleigh');xlabel('time ms')ylabel('envelope volt')xlim([0 0.1])

figuresubplot(2,2,1)plot(t*1e3,crr),title('Rician')xlabel('time ms')ylabel('rf signal volt')xlim([0 0.1])

subplot(2,2,2)plot(t*1e3,aer),title('Rician')xlabel('time ms')ylabel('envelope volt')xlim([0 0.1])

subplot(2,2,4)hist(aer),title('envelope histogram-Rician fading')

subplot(2,2,3);hist(aet),title('envelope histogram-Rayleigh(delay)')

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Problem 7.rayexp.m

clear%rayexp.m%compares the Rayleigh and exponential%verifies that if the envelope is Rayleigh, power is exponentially distributedx=raylrnd(1,1,1000);%1000 Rayleigh random numbers%%%It is also possible to generate Rayleigh random numbers directly from uniform randomnumbers%%%%%%%%%%%%xu=rand(1,1000);%1000 Uniform rand numbers%x=1.0*sqrt(-log(xu));%%%%%%%%%%%%xmean=mean(x);%meanxstd=std(x);%stdevy=x.^2;%generates exponentially distributed setsubplot(2,1,1)hist(x),title('histogram of Rayleigh')subplot(2,1,2)

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hist(y),title('histogram of exponential');ymean=mean(y)ystd=std(y)ratio_Rayl=xmean/xstdratio_exp=ymean/ystdratio_Rayl=xmean/xstdratio_exp=ymean/ystd

ratio_Rayl = 1.8818ratio_exp = 0.9684

Problem 8.fadcom

outageRayl =

0.2945

Problem 9.fadcom

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Problem 10.fadcom.moutageRayl =

0.2945

outageRice =

0.2500

Problem 11. If the signal is Rayleigh faded, the power is exponentially distributed.

Therefore, the outage probability Pout is given by0

1 exp Tout

ZP

Z

where Z0 is the average power and ZT is the threshold power.

Z0 =10 dBm 10 mW

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ZT=0 dBm 1 mW

Outage Probability = 11 exp 0.27110

.

Problem 121Prob SNR<0dB 1 exp2

= 03935.

Problem 13. Threshold = 25-10=15 dB or 101.5.

Mean SNR = 25 dB or 102.5.1.5

2.5

101 exp 0.0952

10outP .

Problem 14. Threshold = 5 dB or 10.5.

.5

0

100.02 1 expoutP

Z where Z0 is the average SNR.

.5

010 156.5 22

log .98e

Z dB

Problem 15. fadingsim.m

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Problem 16.

.001 0 1 .1 3 1 4 .1 2.8926.01 .1 1 .1

s .

2 22 2.001 0 1 .1 3 1 4 .1 8.84

.01 .1 1 .1s

22. . 0.6898rmsdelay s

Coherence BW =290 KHz.

Problem 17.

a. 60 1 85 .1 110 .01 62.7.01 .1 1

ns

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2 2 22 260 1 85 .1 110 .01 4000

.01 .1 1ns

22. . 8.48rmsdelay ns

b.

0 .01 4 .1 8 1 7.57.01 .1 1

s

2 2 22 20 .01 4 .1 8 1 59.1

.01 .1 1s

22. . 1.36rmsdelay s

The response represented by (a) is an indoor channel response while the response represented by(b) is an outdoor channel one.

Problem 18. hatalog.m

clear% hatalog.m Power loss calculations%incluse a plot of the lognormal fading suprimposedclose allPt=30;%power in dBm transmitted 1Wcarf=input('enter carrier freq. in MHZ...>>');lcrf=log10(carf);ht=input('enter the height of BS antenna in meters..>>');%height in meters of the BSlht=log10(ht);hr=input('enter the height of the MU antenna in meters..>>');%MU ant. height in meterslhr=log10(11.75*hr);N=60;%number of increments for distancedistmax=30;%coverage distance in Kmsfor kk=1:1:39;d(kk)=1+0.5*(kk-1);fact1=69.55+26.16*lcrf-13.83*lht;fact2=(44.9-6.55*lht)*log10(d(kk));fftq=fact1+fact2;%antenna correction factor MU...ahr=3.2*(lhr)^2-4.97;% large cities; freq. above 400 MHzamr=(1.1*lcrf-0.7)*hr-(1.56*lcrf-0.8);alar(kk)=Pt-(fftq-ahr);

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amed(kk)=Pt-(fftq-amr);asub(kk)=Pt-(fftq-ahr-2*(log10(carf/28)).^2-5.4);arur(kk)=Pt-(fftq-ahr-4.78*(lcrf.^2)+18.33*lcrf-40.94);xlar(kk)=alar(kk)+normrnd(0,10);%10 dB std deviation for big city with mean equal to thereceived powerxmed(kk)=amed(kk)+normrnd(0,8);%8 dB std deviation for med cityxsub(kk)=asub(kk)+normrnd(0,4);%4 dB std deviationxrur(kk)=arur(kk)+normrnd(0,3);%3 dB std deviation

end;

subplot(2,2,1) plot(d,alar,d,xlar); xlabel('distance Km'); xlim([2 20]) ylabel('Received Power dBm'); title('large city');

subplot(2,2,2) plot(d,amed,d,xmed);

%axis([1 30 -140 -60]); xlabel('distance Km'); xlim([2 20]) ylabel('Received Power dBm'); title('medium city');

subplot(2,2,3) plot(d,asub,d,xsub);

%axis([1 30 -140 -60]); xlabel('distance Km'); xlim([2 20]) ylabel('Received Power dBm'); title('suburban');

subplot(2,2,4) plot(d,arur,d,xrur);

%axis([1 30 -140 -60]); xlabel('distance Km'); xlim([2 20])

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ylabel('Received Power dBm'); title('rural');

Problem 19. rice.m

clear;close all%rice.m%generates a Rician distributionfor kr=1:10rpara=input('mean of the Gaussian to create Rician...>');rp2=rpara^2;n=1000;%number of samplesx=normrnd(0,1,1,1000);xx=normrnd(rpara,1,1,1000);ac=x.^2+xx.^2;r=sqrt(ac);%rician variable

ax=0.:0.2:20;% range of values for the histogramfigure

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yy=hist(r,ax);%hist(r,ax),title('histogram of Rician Distribution');ax1=rp2+ax.^2;bes1=besseli(0,ax*rpara);y=ax.*bes1.*exp(-ax1/2);plot(ax,yy/max(yy),'+',ax,y/max(y))legend('histogram','pdf')xf=input('enter 1 for another mean and 0 to stop..>>');if xf==0 breakelseend;end;

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You can verify that the numbers are Rician distributed by conducting a chi-square test using thefollowing MATALB program.

function [q]=chirice(x,mm);%data, and bins%does the chi-square testing of the data to see Ricain is a fit.xx=sort(x);%using the moments of the data estimate the two parametersa4=mean(xx.^4);my=mean(xx.^2);a = (2*my*my-a4)^0.25; %Rician parameterb=sqrt(0.5*(my-a^2));

inter = max(xx) - min(xx); %interval for chi-square testintstep = inter/mm; %size of sub-intervalintval = [min(xx):intstep:max(xx)]; %samples at the sub-intervalsk = [zeros(1,mm)]; %number of samples in each sub-interval

%counting the samples in each sub-intervalfor i = 1:length(xx) for j = 1:mm

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if xx(i) <= intval(j+1) k(j) = k(j) + 1; break; end; end;end;Fy=1-marcumq(a/b,intval/b);%CDF uses marcumq function from Communications Toolboxp = zeros(1,mm);for i = 1:mm p(i) = Fy(i+1) - Fy(i); %probabilities at the sub-interval samplesend;

np = p.*length(xx);

q = 0;for i = 1:mm q = q + ((k(i)-np(i))^2)/np(i); %chi-square testend;

Problem 20 . ricetoNakagami.m

The Nakagami pdf is given by 2 1 22 expm m

mm x m

mf x x U x . When m=1, this equation

becomes 2 22 expx xf x U x , the Rayleigh density function. Rician becomes Rayleighwhen the Rician parameter K(dB) .

clear%ricetoNakagami.m%generates a Rician distribution, estimates the Nakagami parameter to show that%m>1 for Ricianfor kk=1:21rr(kk)=(kk-1)*0.2;%A0 value of the Ricianr0=rr(kk);x=randn(1,1000);%1000 Gaussian random numbersxx=randn(1,1000);%1000 Gaussian random numbers

x1=xx+r0;%Gaussian with different meansa=sqrt(x1.^2+x.^2);%calculate the envelope

hist(a),title('histogram of Rician Distribution');r2=mean(a.^2);

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r4=mean(a.^4);m(kk)=r2^2/(r4-r2^2);end;figure

RRdB=10*log10(rr.^2/2);plot(RRdB,m,'+')xlabel('Ricean parameter dB')ylabel('Nakagami parameter m')

Problem 21. ricetoNakagami.m

Eqn. (2.59) is 2 2

0 02 2 202exp a A aAA

Af a I

.2 2 2

02E A A …….Rician2E A ……..Nakagami

Thus, 2 202 A . Also, using the expression for m (eqn. 2.72), we can write

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220

24 40 0

2 22 22 200

2

21 1 1 12 1

2

AA A

m AA

.

Note that 20

210 210log AK dB

. It is now possible to relate K(dB) to the Nakagami parameter

m. (Shown above).

Problem 22. Doppler shift 6

8

900 10 80000 66.63 10 60 60d

vf f Hzc

2 .361

2

12 .6 66.67d

eav f

e

=.0043 s

2 .362 2 66.67 .6 e 70A dN f e .

Problem 23. Coherence BW = 10 KHzData rate = 5 kbps

Data rate < coherence bandwidth Channel is flat.

Coherence Time = 100 s

Pulse duration = 1 2005000

s

Pulse duration > coherence time fast fading.

Problem 24. freqselresponse.m

The relative attenuation of the frequency components varies with the strength of the two waves.This will change the coherence bandwidth of the channel.

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Figure Ex. 2-24

Problem 25.Gaussexp and chiexp

clearclose all%Gaussexp.m%generates exponential from two Gaussian random numbers%histogram is compared to the exponential pdfx1=normrnd(0,.5,1,10000);%10000 Gaussian random numbersx2=normrnd(0,.5,1,10000);%10000 Gaussian random numbersy=x1.^2+x2.^2;x=min(y):max(y)/80:max(y);hist(y,x);xh=hist(y,x);hold on%plot(x,xh/max(xh))

ymean=mean(y);

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ystd=std(y);ratio=ymean/ystd

exf=exppdf(x,ymean);%generates exponential pdf having the same meanhold onexf=exf*max(xh)/max(exf);plot(x,exf,'r')legend('expoential pdf')xlabel('x')ylabel('pdf or histogram')

Problem 26.pulsedisp.m

%pulsedisp.m%shows the effect of frequency selective effects by adding ten different pulses%shifted and weighted%Gaussian pulses usedclose allclearfor kp=1:3;%plots three separate generation to show the effect of random paths and random weights%w=rand(10,1);%weights for 10 pulses

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%tt=100*rand(10,1);%delay is also considered to be randomNN=1024;sigma2=200;%sigmasquarep0=500;%for p=1:NN;xx=zeros(1,NN);p=1:NN; t(p)=p; p1=p-p0; for kn=1:10; xx=xx+(1/sqrt(2*pi*sigma2))*rand(1)*exp(-(p1-100*rand(1)).^2/(2*sigma2));%10 pulsesaddedend;x1=(1/sqrt(2*pi*sigma2))*rand(1)*exp(-(p1-100*rand(1)).^2/(2*sigma2));%a single pulse xxx=xx/max(xx); x1=x1/max(x1); subplot(2,2,kp+1) plot(t,xxx); xlim([400 800]) %finds the mean and standard deviation of the Gaussian pulse xpp=sum(xxx);%denominator xp=sum(t.*xxx)/xpp;%mean......sum(pi.xi)/sum(pi) t2=t.^2; xp2=sum(t2.*xxx)/xpp;%second moment stdval=sqrt((xp2-xp^2))%standard deviationend;subplot(2,2,1)plot(t,x1),title('Transmitted Pulse') xlim([400 800]) sum2=sum(x1); xy1=sum(t.*x1)/sum2; xy2=sum(t2.*x1)/sum2;stval=sqrt(xy2-xy1^2)%standard deviation ...single pulse

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r.m.s width (transmitted) = 14.1421r.m.s width (received) = 25.2188, 24.2574, 24.1377

Problem 27 (a) ricetoGaussian.m

clear;close all%ricetoGaussian.m ;generates a Rician distribution%compares the histogram to Rician pdf and to a Gaussian pdf having identical%mean and variancest=2;%standared deviation of Gaussian neededst2=st^2;for kr=1:10rpara=input('mean of the Gaussian to create Rician...>');rp2=rpara^2;%use 5000 random variablesx=normrnd(0,st,1,5000);xx=normrnd(0,st,1,5000);%another gaussian random variablex1=xx+rpara;%Gaussian with different meansx2=x.^2;xx2=x1.^2;

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xy=x2+xx2;sst=2*st2;KdB=10*log10(rp2/sst);kkdb=num2str(KdB);titlefigure=strcat('KdB=',kkdb)r=sqrt(xy);%Rician random variable

meanp=mean(r)varp=var(r);%variance

ax=0.:0.2:20;% range of values for the histogramfigureyy=hist(r,ax);%hist(r,ax),title('histogram of Rician Distribution');ax1=rp2+ax.^2;bes1=besseli(0,ax*rpara/st2);y=(1/st2)*ax.*bes1.*exp(-ax1/(2*st2));yg=sqrt(1/(2*pi*varp))*exp(-(ax-meanp).^2/(2*varp));%Gaussian pdfplot(ax,yy/max(yy),'+',ax,y/max(y),ax,yg/max(yg),'*'),title(titlefigure)xlabel('envelope')ylabel('pdf or histogram')legend('hostogram','envelope pdf','Gaussian approx.')xf=input('enter 1 for another mean and 0 to stop..>>');if xf==0 breakelseend;end;

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Another way to explore this relationship is to see if the ratio of the median to mean is equal tounity. This happens when Rician becomes Gaussian.

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Problem 28. nakagenerice.mclear%nakagenerice.m%generates Nakagami distributed random numbers and estimates the parameters%conduct the Chi-square testing for Rayleigh and Nakagami and Rice

mval=input('enter the value of m......>');om=input('enter the value of omega....>');N=input('the number of random numbers to be generated...>');y=gamrnd(mval,om/mval,1,N);%Generates Gamma distributed random variablesx=sqrt(y);%generates Nakagami dsitributed random variable which is the square root of theGamma variablemy=mean(x.^2)%gives the value of omegaomega=myav4=mean((x.^2-my).^2);mn=my*my/av4%m-value directlymm=20;%number of binsxx=sort(x); % random samples

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inter=max(x)-min(x); %interval for chi-square testintstep=inter/mm; %size of sub-intervalintval=[min(x):intstep:max(x)]; %samples at the sub-intervalsk=[zeros(1,mm)]; %number of samples in each sub-interval intializationfor i=1:N for j=1:mm if xx(i)<=intval(j+1) k(j)=k(j)+1; break; end; end;end;p=zeros(1,mm);for i=1:mm yin=intval(i+1)*intval(i+1); zin=intval(i)*intval(i); p(i)=gamcdf(yin*mn,mn,my)-gamcdf(zin*mn,mn,my); %probabilities at the sub-intervalsamplesend;np=p.*N;q=0;for i=1:mm q = q + ((k(i)-np(i))^2)/np(i); %chi-square testend;chisquare_Nakagami=q%chisquare valueq_Rayl=chirayleigh(x,mm)

if mval<=1%do not conduct the Rice test breakend;%%%% testing for Rice

%%from the paper by Nakagami%using the moments of the dataa4=mean(xx.^4);a = (2*my*my-a4)^0.25; %Rician parameterb=sqrt(0.5*(my-a^2));

inter = max(xx) - min(xx); %interval for chi-square testintstep = inter/mm; %size of sub-intervalintval = [min(xx):intstep:max(xx)]; %samples at the sub-intervalsk = [zeros(1,mm)]; %number of samples in each sub-interval

%counting the samples in each sub-intervalfor i = 1:length(xx) for j = 1:mm

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if xx(i) <= intval(j+1) k(j) = k(j) + 1; break; end; end;end;

Fy=1-marcumq(a/b,intval/b);%CDF uses marcumq function from Communications Toolboxp = zeros(1,mm);for i = 1:mm p(i) = Fy(i+1) - Fy(i); %probabilities at the sub-interval samplesend;np = p.*length(xx);q = 0;for i = 1:mm q = q + ((k(i)-np(i))^2)/np(i); %chi-square testend;q_Rice=q

**The above program needs the following segment: chirayleigh

function [q]=chirayleigh(yy,m)%Tests for Rayleigh distribution%[chu)]=chirayleigh(samp,bins)y=sort(yy); % random samplesN=length(yy);my=mean(yy);%meaninter=max(y)-min(y); %interval for chi-square testintstep=inter/m; %size of sub-intervalintval=[min(y):intstep:max(y)]; %samples at the sub-intervalsk=[zeros(1,m)]; %number of samples in each sub-interval intialization

for i=1:N for j=1:m if y(i)<=intval(j+1) k(j)=k(j)+1; break; end; end;end;

myp=my*sqrt(2/pi);%Rayleigh parameterp=zeros(1,m);for i=1:m p(i)=raylcdf(intval(i+1),myp)-raylcdf(intval(i),myp); %probabilities at the sub-interval samples

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end;

np=p.*N;

q=0;for i=1:m q = q + ((k(i)-np(i))^2)/np(i); %chi-square testend;

*****Problem 29. outlognorm.m

%outlognorm.m%computes the outage probability for a given average power and a fixed threshold% as a function of the standard deviation of lognormal fading%also generates the outage from a set of random numbersclearclose all%outage probability calculation for lognormalN=2000;%number of random numbers to be generatedpdb=input('enter the average power in dBm....>>>>');pthr=input('threshold power in dBm... less than the average power >')st=[1:1:15];%standard deviation of fading sq=sqrt(2)*st; sqt=1./sq; ff=(pdb-pthr)*sqt; pout=0.5*erfc(ff);

for k=1:15; x=normrnd(pdb,k,1,N); count=0; for kk=1:N; if x(kk)<=pthr count=count+1; else end; end; pouts(k)=count/N;% outage from simulation..random numbers end;

plot(st,pout,'g',st,pouts,'r') legend('Theoretical Outage','Simulated Outage')

xlabel('standard deviation of fading dB')ylabel('outage probability')

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enter the average power in dBm....>>>>-95threshold power in dBm... less than the average power >-100

Problem 30. 0.52

av tout

P PP erfc

2 12

av tout

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P P dBerfinv P

Problem 31. Gausschiexp.mclearclose all%Gausschiexp.m%generates exponential from two Gaussian random numbers%conducts a chi-square test to validate exponential pdfx1=normrnd(0,.5,1,10000);%10000 Gaussian random numbersx2=normrnd(0,.5,1,10000);%10000 Gaussian random numbersyy=x1.^2+x2.^2;m=20;%number of bins

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y=sort(yy); % random samplesN=length(yy);%size of the samplemy=mean(y);%mean and exponential parameterinter=max(y)-min(y); %interval for chi-square testintstep=inter/m; %size of sub-intervalintval=[min(y):intstep:max(y)]; %samples at the sub-intervalsk=[zeros(1,m)]; %number of samples in each sub-interval intialization


p=zeros(1,m);for i=1:m p(i)=expcdf(intval(i+1),my)-expcdf(intval(i),my); %probabilities at the sub-interval samplesend;

np=p.*N;


chi_squaretest_val=q

Problem 32. unitoGaussian.m

Chi-square values 21.5713 24.3384 19.3305 19.3305

%unitoGaussian.m%test to see whether the sum of uniform random variables tend towards Gaussian%conducts chi-square testsclearclose all

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N=1000x1=rand(1,N);x2=rand(1,N);x3=rand(1,N);x4=rand(1,N);;x5=rand(1,N);x6=rand(1,N);xx2=x1+x2+x3;xx3=xx2+x4;xx4=xx3+x5;xx5=xx4+x6;m=20;%number of binssubplot(2,2,1)hist(xx2), title('sum of three random variables')chigauss(xx2,m)%chi-square valuesubplot(2,2,2)hist(xx3),title('sum of four random variables')chigauss(xx3,m)%chi-square valuesubplot(2,2,3)hist(xx4),title('sum of five random variables')chigauss(xx4,m)%chi-square valuesubplot(2,2,4)hist(xx5),title('sum of six random variables')chigauss(xx4,m)%chi-square value

also requires the following program……

function [q]=chigauss(yy,m)%Tests for Normal distribution%[chu)]=chigauss(samp,bins)y=sort(yy); % sort samplesN=length(yy);mean=mean(y);stdev=std(y);inter=max(y)-min(y); %interval for chi-square testintstep=inter/m; %size of sub-intervalintval=[min(y):intstep:max(y)]; %samples at the sub-intervalsk=[zeros(1,m)]; %number of samples in each sub-interval intialization

for i=1:N for j=1:m if y(i)<=intval(j+1) k(j)=k(j)+1; break; end; end;

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end;

p=zeros(1,m);for i=1:m p(i)=normcdf(intval(i+1),mean,stdev)-normcdf(intval(i),mean,stdev); %probabilities at thesub-interval samplesend;

np=p.*N;


Problem 33. The results show that as the number of random variables increase, the sum ofthe random variables approach Gaussian. The Rayleigh fading is a result of multiple paths andwhen the number of multiple paths increase, the density function of the sum of the phasors

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giving rise to the amplitude (inphase and quadrature components) approach Gaussian and the pdfenvelope approaches Rayleigh.

Problem 34.rayoutage1

clear%rayoutage1close all%generates Rayleigh faded signal with 10 multiple paths%compares the outage to theoretical outagenumpaths = 10; %number of pathsFc = 900e6; %carrier frequencyFs = 4*Fc; %sampling frequencyTs = 1/Fs; %sampling periodt = [0:Ts:1999*Ts]; %time arraywc = 2*pi*Fc; %radian frequencyray = zeros(1,length(t)); %received signal

for i = 1:numpathsa = weibrnd(1,3,1,length(t)); %scattering strengts are Weibull distributedray = ray + a.*cos(wc*t+unifrnd(0,2*pi,1,length(t)));

end;[rayi rayq] = demod(ray,Fc,Fs,'qam'); %demodulated signalenv_ray = sqrt(rayi.^2+rayq.^2); %envelope of received signal

%computation of the outage probabilitiespower_ray=env_ray.^2;powerdB=10*log10(power_ray);mean_power=10*log10(mean(env_ray.^2))MK=length(env_ray)for k=1:20; pow(k)=mean_power-2*k; %threshold power kps=pow(k); ratio=10^(kps/10)/10^(mean_power/10); poutth(k)=1-exp(-ratio);%theoretical outage count=0; for ku=1:MK; power=powerdB(ku);%power in dB if power <= kps count=count+1; else end; end; poutsim(k)=count/MK;%outage probability simulatedend;figureplot(pow,poutth,':',pow,poutsim,'--'),title('Outage ---Rayleigh')xlabel('threshold power dBm')

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ylabel('outage probability')legend('pout_{th}','pout_{sim}')

Problem 35. riceout1.m

clear%riceout1.m%generates a fading signal where a direct (LOS) component is present%generates the RF signal and calculate the outageclose allnumpaths = 10; %number of pathsFc = 900e6; %carrier frequencyFs = 4*Fc; %sampling frequencyTs = 1/Fs; %sampling periodt = [0:Ts:1999*Ts]; %time arraywc = 2*pi*Fc; %radian frequencyrice = zeros(1,length(t)); %received signalfor i = 1:numpaths-1

a = weibrnd(1,3,1,length(t)); %scattering strengtsh are Weibull distributed

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rice = rice + a.*cos(wc*t+unifrnd(0,2*pi,1,length(t)));end;rice = rice + 4.5*cos(wc*t);%adds the direct path[ricei riceq] = demod(rice,Fc,Fs,'qam'); %demodulated signalenv_rice = sqrt(ricei.^2+riceq.^2); %envelope of received signalb = sqrt((std(ricei)^2 + std(riceq)^2)/2); %Rician parametera = mean(ricei); %Rician parameterpower_rice=env_rice.^2;powerdB=10*log10(power_rice);mean_power=10*log10(mean(env_rice.^2))MK=length(env_rice)for k=1:20; pow(k)=mean_power-2*k; %threshold power kps=pow(k); count=0; for ku=1:MK; power=powerdB(ku);%power in dB if power <= kps count=count+1; else end; end; poutsim(k)=count/MK;%outage probability simulated kbs=10^(kps/10);%power in mW poutth(k)=1-marcumq(a/b,sqrt(kbs)/b);%from the communications toolboxend;figureplot(pow,poutsim,pow,poutth),title('Outage ---Rician')xlabel('thrshold power dBm')ylabel('outage probability')legend('poutth','poutsim')

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Problem 36.lognormf.m

clearclose all%exploration of the generation of lognormal%lognormf.m%generates a set of lognormal random numbers based on the principle that the procuct of a set ofrandom%numbers will give rise to lognormalmm=input('number of components for the product operation to check for lognormal..');for k=1:1000; x(k)=prod(raylrnd(1,1,mm));%generates the product of mm Rayleigh random numbersend;subplot(2,1,1)hist(x),title('histogram of the product of Rayleigh numbers')xlabel('volt')yy=10*log10(x.^2);subplot(2,1,2)hist(yy),title('histogram after conversion..10log_{10}x^2..')xlabel('dBm')

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%Tests for Normal distribution m=20;%number of bins for the chi-square testy=sort(yy); % sort samplesN=length(y);mean=mean(y);stdev=std(y);inter=max(y)-min(y); %interval for chi-square testintstep=inter/m; %size of sub-intervalintval=[min(y):intstep:max(y)]; %samples at the sub-intervalsk=[zeros(1,m)]; %number of samples in each sub-interval intialization



np=p.*N;

q=0;for i=1:m q = q + ((k(i)-np(i))^2)/np(i); %chi-square testend;chi_square_val=q

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Problem 37. fadlogfm.m

%fadlogfm.m%explores the generation of lognormal fading. There is multiple scattering. There are alsomultiple paths.%This will lead to Rayleigh envelope where the mean of the envelope will be randomclearclose all%simulates lognormal fadingfc=1e5;%carrier frequencyppf=2*pi*fc;N=11;% Number of mutiple paths%a are considered to be Gamma distributedfs=4*fc;%sampling rateTs=1/fs;t = [0:Ts:1999*Ts]; %time arraynn=length(t);

for kr=1:1000;%1000 rf waveforms are genaretd to see the statistics of the mean and the meadianpowers %multiple reflections

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% %ggm is the result of multiple scattering; thus each multipath is produced by multiple(20)scattering. %the statistics of the mean and median of the Rayleigh have been computed. %NOTE THAT THE MULTIPLE REFLECTIONS OCCUR FIRST AND THEN THISMULTIPLY REFLECTED COMPONENTS %GET REFLECTED/SCATTERED/DIFFRACTED TO TAKE MULTIPLE PATHS TOTHE RECEIVER. %THE MULTIPLE SCATTERING COMPONENTS IDENTIFIED BY EACH OF THEggm %WILL BE RAYLEIGH DUE TO THE PRESENCE OF MUTIPLE SCATTERERS INTHE REGION %IN WHICH THE THE SCATTERING TAKES PLACE. ggm=prod(raylrnd(1.5,1,20));%generates the product of 20 Rayleigh r.v cr=zeros(1,nn); for m=1:N; phterm=2*pi*unifrnd(0,2*pi,1,nn); cr=cr+ggm*cos(ppf*t+phterm);%Rayleighend;[x,y]=demod(cr,fc,fs,'qam');ae=(1/N)*sqrt(x.^2+y.^2);%ae will be Rayleighae=ae;mm(kr)=mean(ae);med(kr)=median(ae);end;mm2=mm.^2;med2=med.^2;mmdB=10*log10(mm2);%mean power in dBmmeddB=10*log10(med2);%median power in dBm

%%% note that both average ( or median) of envelope and average of the power will belognormal

plot(t*1e3,cr/max(abs(cr))),title('rf signal incorporating lognormal fading');xlabel('time ms')ylabel('rf signal volt')xlim([0 0.5])

figuresubplot(2,2,1)hist(mm),title('histogram of the mean power')xlabel('volt')subplot(2,2,2)hist(med),title('histogram of the median power')xlabel('volt')

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subplot(2,2,3)hist(mmdB),title('histogram of the mean power (dBm)')xlabel('dBm')subplot(2,2,4)hist(meddB),title('histogram of the median power (dBm)')xlabel('dBm')

Problem 38. fadlogfm.m (See the previous problem # 37)

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Problem 39.lognormf.mclearclose all%exploration of the generation of lognormal%lognormf.m%generates a set of lognormal random numbers based on the principle that the procuct of a set ofrandom%numbers will give rise to lognormalmm=input('number of components for the product operation to check for lognormal..');for k=1:1000; x(k)=prod(raylrnd(1,1,mm));%generates the product of mm Rayleigh random numbersend;subplot(2,1,1)hist(x),title('histogram of the product of Rayleigh numbers')xlabel('volt')yy=10*log10(x.^2);subplot(2,1,2)hist(yy),title('histogram after conversion..10log_{10}x^2..')xlabel('dBm')

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%Tests for Normal distribution m=20;%number of bins for the chi-square testy=sort(yy); % sort samplesN=length(y);mean=mean(y);stdev=std(y);inter=max(y)-min(y); %interval for chi-square testintstep=inter/m; %size of sub-intervalintval=[min(y):intstep:max(y)]; %samples at the sub-intervalsk=[zeros(1,m)]; %number of samples in each sub-interval intialization


Problem 40. lognormfchi.m

clearclose all%exploration of the generation of lognormal%lognormfchi.m%generates a set of lognormal random numbers based on the principle that the procuct of a set ofrandom%numbers will give rise to lognormal%chi-square tests directly and indirectly using Gaussianmm=input('number of components for the product operation to check for lognormal..');for k=1:1000; x(k)=prod(raylrnd(1,1,mm));%generates the product of mm Rayleigh random numbersend;subplot(2,1,1)hist(x),title('histogram of the product of Rayleigh numbers')xlabel('volt')yy=10*log10(x.^2);subplot(2,1,2)hist(yy),title('histogram after conversion..10log_{10}x^2..')xlabel('dBm')

%%% testing directly for lognormalm=20;

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N=length(x);%Tests for lognormal distribution%[chu)]=chilogn(samp,bins)y=sort(x); % random samplesms=std(log(y));%sigma of the lognormal *** not the standard deviationinter=max(y)-min(y); %interval for chi-square testintstep=inter/m; %size of sub-intervalintval=[min(y):intstep:max(y)]; %samples at the sub-intervalsk=[zeros(1,m)]; %number of samples in each sub-interval intialization

for i=1:N for j=1:m if y(i)<=intval(j+1) k(j)=k(j)+1; break; end; end;end;my=mean(log(y));%parameter mu of the lognormal pdf*** not the mean of lognormalp=zeros(1,m);for i=1:m p(i)=logncdf(intval(i+1),my,ms)-logncdf(intval(i),my,ms); %probabilities at the sub-intervalsamplesend;

np=p.*N;

q=0;for i=1:m q = q + ((k(i)-np(i))^2)/np(i); %chi-square testend;ch_lognorm_val=q

clear qclear y%Tests for Normal distributionm=20;%number of bins for the chi-square testy=sort(yy); % sort samplesN=length(y);mean=mean(y);stdev=std(y);inter=max(y)-min(y); %interval for chi-square testintstep=inter/m; %size of sub-intervalintval=[min(y):intstep:max(y)]; %samples at the sub-intervalsk=[zeros(1,m)]; %number of samples in each sub-interval intialization

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np=p.*N;

q=0;for i=1:m q = q + ((k(i)-np(i))^2)/np(i); %chi-square testend;chi_square_val=q

Problem 41.hatawalf.m

clearclose all% hatawalf.m Power loss calculations%using Hata model and Walfischcarf=880;%carrier frequency of in MHzlcrf=log10(carf);ht=30;%height in meters of the BSlht=log10(ht);hr=1.5;%MU ant. height in meterslhr=log10(11.75*hr);N=60;%number of increments for distancedistmax=30;%coverage distance in Kmsfor kk=1:30;d(kk)=1+0.5*(kk-1);fact1=69.55+26.16*lcrf-13.83*lht;fact2=(44.9-6.55*lht)*log10(d(kk));fftq=fact1+fact2;%antenna correction factor MU...ahr=3.2*(lhr)^2-4.97;% large cities; freq. above 400 MHz

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amr=(1.1*lcrf-0.7)*hr-(1.56*lcrf-0.8);alar(kk)=fftq-ahr;%Hata model

lfree=32.4+20*log10(d(kk))+20*lcrf;%free space losssw=15;%street widthhb=30;%building height=antenna height=separation of buildingsdhb=0;%no difference in heights of bdg and antennadhm=hb-hr;ang=90;%angle philrts=-16.9+10*lcrf-10*log10(sw)+20*log10(dhm)+4-.114*90;lms=0+54+18*log10(d(kk))+(-4+1.5*(carf/925-1))*lcrf-9*log10(hb);alf(kk)=lfree+lrts+lms;%W-I modelend;

plot(d,alar,d,alf); legend('Hata model','Walfish-Ikegami model') %axis([1 30 -140 -60]); xlabel('distance Km'); ylabel('power loss dB'); %for transfering to excel directly as columns d=d'; alf=alf'; alar=alar';

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Problem 42. See the textbook; relevant sections

Problem 43. See the textbook; relevant sections

Problem 44. coheBW.m

%coheBW.m%plot of the correlation of faded channelsclearclose allfor k=1:50; x(k)=0.05*(k-1); x1=x(k); sigma=1; y(k)=1/(1+(2*pi*x1*sigma)^2);%corr coef. freq y1=besselj(0,2*pi*x1); yy(k)=y1^2;%corr. Coeff timeend;

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subplot(2,1,1)plot(x,y), title('frequency correlation')xlabel('\Delta f')ylabel('\rho (\Delta f,0)')subplot(2,1,2);plot(x,yy),title('temporal correlation')xlabel('\Delta t')ylabel('\rho (0,\Delta t)')

The delay can be described using the exponential pdf, 1 expf U

where is the average delay. Making use of the properties of the exponential density function,we have ; 0.5E ms . Thus, the maximum data rate will be approximately equal

to 1

2000bps

.

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Problem 45.fadcomRayl. m and chirayleigh.m (see next problem)

clear%fadcomRayl.m simulates multiptah fading varying the number of paths%Compares the histogram to a Rayleigh pdfclose allfc=900e3;%carrier frequencyppf=2*pi*fc;N1=[3 5 7 10]%a are considered to be Gamma distributedfs=4*fc;%sampling ratedt=1/fs;nn=1e-3/dt;%total number of samples in one m secM=2^(nextpow2(nn))dist=1000;%distance between transmit and receive in metersavdel=dist/3e8;%average delay in secondsfor kmm=1:4 N=N1(kmm)for k=1:nn; t(k)=k*dt; ci=0.0; cir=0.0; for m=1:N; phterm=2*pi*rand(1); ci=ci+gamrnd(1.4,1)*cos(ppf*t(k)+phterm);%Rayleigh including Dopplerend;cr(k)=ci;end;[x,y]=demod(cr,fc,fs,'qam');ae=(1/N)*sqrt(x.^2+y.^2); pha=pi+angle(x+i*y);%phase to be centered around 2*pi numb=num2str(N); fx=strcat('Multipath fading....','N=',numb); figuresubplot(2,2,1)plot(t*1e3,cr),title(fx)xlabel('time ms')ylabel('rf signal volt')xlim([0 .1])aes=20*log10(ae);subplot(2,2,2)plot(t*1e3,aes),title(fx)xlabel('time ms')ylabel('relative power dB')xlim([0 .1])

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subplot(2,2,3)ax=0.:0.1:5;% range of values for the histogramyy=hist(ae,ax);bb=mean(ae)*sqrt(2/pi);%Rayleigh parameter by=raylpdf(ax,bb);%Rayl density function of parameter bplot(ax,yy/max(yy),'+',ax,y/max(y)),title(fx)legend('histogram','Rayleigh pdf')xlabel('envelope')qq=chirayleigh(ae,20)%chi-square valueend;

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Problem 46.fadconRayl (see previous problem) and chirayleigh

function [q]=chirayleigh(yy,m)%Tests for Rayleigh distribution%[chu)]=chirayleigh(samp,bins)y=sort(yy); % random samplesN=length(yy);my=mean(yy);%meaninter=max(y)-min(y); %interval for chi-square testintstep=inter/m; %size of sub-intervalintval=[min(y):intstep:max(y)]; %samples at the sub-intervalsk=[zeros(1,m)]; %number of samples in each sub-interval intialization

for i=1:N for j=1:m if y(i)<=intval(j+1) k(j)=k(j)+1; break; end;

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end;end;

myp=my*sqrt(2/pi);%Rayleigh parameterp=zeros(1,m);for i=1:m p(i)=raylcdf(intval(i+1),myp)-raylcdf(intval(i),myp); %probabilities at the sub-interval samplesend;

np=p.*N;


N = 3; Chi-square value = 4.7997e+003

N = 5; Chi-square value = 380.6133



Problem 47 .pulsefading.m

clear%pulsefading.m fading including a carrier frequency%three delays are created; Pulse stream is delayed and added%the a uniform phase is introduced in the RF pulse stream; three pulses are addedclose all%examines the fading of pulse/shapes%the program prompts the fractional delays

fc=10e3;%carrier frequencyppf=2*pi*fc;fs=12*fc;%sampling rateddt=1/fs;nn=1e-3/ddt;%total number of samples in one m secM=2^(nextpow2(nn))dist=1000;%distance between transmit and receive in meters

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avdel=dist/3e8;%average delay in seconds%kp*ku:kp*ku+1N=8;%number of bitsa=[1 1 -1 -1 -1 1 -1 1];kp=floor(nn/8);%creates the width of the pulsefor ku=1:8;for k=(ku-1)*kp+1:kp*ku tt(k)=k*ddt; s(k)=a(ku);%creates the pulse stream sf(k)=a(ku)*cos(ppf*tt(k));%rf pulse sf1(k)=a(ku)*cos(ppf*tt(k)+2*pi*rand(1));%random phase sf2(k)=a(ku)*cos(ppf*tt(k)+2*pi*rand(1));%random phase end;end;

del=input('enter the fractional (of the duration) delay...between 0.25 and 0.75..');dd1=round(del*kp);for k=1:kp*ku; if k<=dd1 ss1(k)=0;else ss1(k)=s(k-dd1);%creates a delayed pulse streamend;end;del2=input('enter the fractional (of the duration) delay...between 0.25 and 0.75..');dd2=round(del2*kp);for k=1:kp*ku; tt2(k)=k*ddt; if k<=dd2 ss2(k)=0;else ss2(k)=s(k-dd2);%creates a delayed pulseend;end;sum=s+ss1+ss2;%added pulse shapessum=sum/max(abs(sum));%normalizes to unityrfsum=sf+sf1+sf2;%added RF pulsesrfsum=rfsum/max(abs(rfsum));%normalizes to unity

plot(tt*1e3,s,tt*1e3,sf),title('bit stream [1 1 -1 -1 -1 1 -1 1]')ylim([-2 2]);xlabel('time ms')ylabel('volts')legend('input bit stream','input RF bit stream')

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figureplot(tt*1e3,sum,tt*1e3,rfsum),title('fading-effects of delay and random phase')ylim([-2 2]);xlabel('time ms')ylabel('volts')legend('bit stream in fading','RF bit stream in fading')

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Problem 48. pulsefading

See the m file and results of the previous exercise (# 48)

Problem 49. pulsefadingRayl

clear%pulsefadingRayl.m fading including a carrier frequency%three delays are created; Pulse stream is delayed and added%three pulses are added%includes Rayleigh weighting at each time instant%the program prompts the fractional delaysclose all%examines the fading of pulse/shapesfc=10e3;%carrier frequencyppf=2*pi*fc;fs=12*fc;%sampling rateddt=1/fs;

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nn=1e-3/ddt;%total number of samples in one m secM=2^(nextpow2(nn))dist=1000;%distance between transmit and receive in metersavdel=dist/3e8;%average delay in seconds%kp*ku:kp*ku+1N=8;%number of bitsa=[1 1 -1 -1 -1 1 -1 1];kp=floor(nn/8);%creates the width of the pulsefor ku=1:8;for k=(ku-1)*kp+1:kp*ku tt(k)=k*ddt; s(k)=a(ku);%creates the pulse stream ss1(k)=a(ku)*raylrnd(1);%Rayleigh weighted sf(k)=a(ku)*cos(ppf*tt(k));%rf pulse sf1(k)=raylrnd(1)*a(ku)*cos(ppf*tt(k));%rf pulse Rayleigh weighted sf2(k)=raylrnd(1)*a(ku)*cos(ppf*tt(k));%rf pulse Rayleigh weighted sf3(k)=raylrnd(1)*a(ku)*cos(ppf*tt(k));%rf pulse Rayleigh weighted end;end;

del=input('enter the fractional (of the duration) delay...between 0.25 and 0.75..');dd1=round(del*kp);for k=1:kp*ku; if k<=dd1 ss2(k)=0;else ss2(k)=s(k-dd1)*raylrnd(1);end;end;del2=input('enter the fractional (of the duration) delay...between 0.25 and 0.75..');dd2=round(del2*kp);for k=1:kp*ku;

if k<=dd2 ss3(k)=0;else ss3(k)=s(k-dd2)*raylrnd(1);end;end;sum=ss1+ss2+ss3;%added pulse shapes;Rayleigh weightedsum=sum/max(abs(sum));%normalizes to unityrfsum=sf1+sf2+sf3;%added RF pulses Rayleigh weightedrfsum=rfsum/max(abs(rfsum));%normalizes to unity

plot(tt*1e3,s,tt*1e3,sf),title('bit stream [1 1 -1 -1 -1 1 -1 1]')ylim([-2 2]);

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xlabel('time ms')ylabel('volts')legend('input bit stream','input RF bit stream')figureplot(tt*1e3,sum,tt*1e3,rfsum),title('Rayleigh weighted fading')ylim([-2 2]);xlabel('time ms')ylabel('volts')legend('bit stream in fading','RF bit stream in fading')

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Problem 50. pulsefadingRayl

See exercise # 49.

bai giai truyen thong khong day moi

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