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papr reduction using slm and clipping method

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victory.dj96

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i am working on papr reduction in ofdm using slm and clippping method, please help me in understanding the matlab code for these techniques.

thanks!
 

greygoo2012

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Please let us know what exactly you are not able to understand in the code
 

victory.dj96

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i have code for clipping, but its not give the appropriate result..please help me.

Code:
clc;
clear all;
close all;

 L=input('Enter the L factor(1 to 1.5)= ');
 N=input('Enter the number of transmitted symbols= ');
M=input('Enter the alphabet size= ');

k=[0:0.5:10];
k1=[1:N];

r=floor(M*rand(N,1));                          
modulated_data=qammod(r,M);                                 
                                               
LN=floor(L*N);
parallel_data=modulated_data';
zero_padding=[parallel_data(1:N) zeros(1,LN-N)]';



ifft_data=ifft(zero_padding);                                  
x_mag=abs(ifft_data);
papr=max(x_mag.^2)/mean(x_mag.^2);

figure(1)
plot(x_mag),title('Normal OFDM signal');
xlabel('time'),ylabel('amplitude')

y=abs(ifft_data).^2;
y1=sort(y,'descend');
y3=y1./mean(x_mag.^2);
y4=10*log10(y3)


[d,k] = hist(y4,k);
figure(2)
semilogy(k,1-cumsum(d)/max(cumsum(d)),'-r')

CR=5;

x_max=(10^(CR/20))*mean(x_mag);

for j=1:LN                                   
if(x_mag(j,1)>x_max)
    x_mag1(j,1)=x_max;
else
    x_mag1(j,1)=x_mag(j,1);
end;
end;

figure(3)
plot(x_mag1),title('Clipped OFDM signal with CR=5 dB');
xlabel('time'),ylabel('amplitude')


                                           
papr1=max(x_mag1.^2)/mean(x_mag1.^2);



c=abs(x_mag1).^2;
c1=sort(c,'descend');
c2=c1./mean(x_mag1.^2);
c3=10*log10(c2)


[dc,k] = hist(c3,k);


CR1=1;

x_max1=(10^(CR1/20))*mean(x_mag);

for j=1:LN                                   
if(x_mag(j,1)>x_max1)
    x_mag2(j,1)=x_max1;
else
    x_mag2(j,1)=x_mag(j,1);
end;    
end;

c21=abs(x_mag2).^2;
c22=sort(c21,'descend');
c31=c22./mean(c21);
c34=10*log10(c31);

[c35,k]=hist(c34,k);

papr2=max(x_mag2.^2)/mean(x_mag2.^2);

figure(4)
plot(x_mag2),title('Clipped OFDM signal with CR=1 dB');
xlabel('time'),ylabel('amplitude')


figure(5)
semilogy(k,1-cumsum(dc)/max(cumsum(dc)),'-k')

figure(6)
semilogy(k,1-cumsum(c35)/max(cumsum(c35)),'-k')

figure(7)
semilogy(k,1-cumsum(d)/max(cumsum(d)),'-sr',k,1-cumsum(dc)/max(cumsum(dc)),'-+b',k,1-cumsum(c35)/max(cumsum(c35)),'-<g')
xlabel('PAPR (dB)');ylabel('CCDF');
title('CCDF VS PAPR');grid on;hold on;
hleg=legend('ORIGINAL','clipping with CR=5 dB','clipping with CR=1 dB');
 
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victory.dj96

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how the papr is reduced when we multiplied with phase sequence in slm method.. please explain me.
 

greygoo2012

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