How to implement triple summation?

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Usha Sree Katikala
Usha Sree Katikala el 17 de Mzo. de 2021
Editada: Matt J el 9 de Abr. de 2021
  3 comentarios
Jan
Jan el 17 de Mzo. de 2021
What are h_p? What is the meaning of h_p^(q, p-q)? What is X*? Do you want a symbolical or numerical summation?
Usha Sree Katikala
Usha Sree Katikala el 9 de Abr. de 2021
Editada: Matt J el 9 de Abr. de 2021
I have made an attempt in implementing individual parts of this equation.I need help in combining them as triple summationThe code is:
.
clear all
close all
[x1,Fs] = audioread('C:\Users\admin\Downloads\OSR_us_000_0061_8k.wav'); % audio signal assumed as transmitted signal
values= x1(1:100); % array of received signal x(n)
Nsamps = 100; % number of samples taken from the input signal
t = (1/Fs)*(1:Nsamps)
plot(t, values)
title('signal');
xlabel('Time');
ylabel('Amplitude of signal');
xc=complex(values); % complex of received signal
%SI Signal model
p=3; % high non-linearity order
q=2;
M=0; % memory length
for i= 0:q
M=M+1;
first=2*(delayseq(values,M)); % x(n-m)^q
end
M=0;
for i=0:p-q
M=M+1;
xc1= 2*(delayseq(xc,M)); % x*(n-m)^p-q
end
cf=first.xc1; % x(n-m)^q * x(n-m)^p-q
g=1.012; %g-> gain imbalance
phi=1; % phi-> phase imbalance.
%%% K1 and K2 are the constants in the IQ Modulator output signal%%%
%%% model(frequency independent model)%%%
K1=1/2*(1+g*exp(i*phi)); % K1
K2=1/2*(1-g*exp(i*phi)); % K2
K1abs=abs(K1); % |K1|
K2abs=abs(K2); % |K2|
%%% h1,h2,h3,h4 are Basis function coefficients for P=3 %%%
h1=K1*K1*complex(K2);
h2=(power(K1abs,2)K1)+(2(power(K2abs,2)*K1));
h3=2*(power(K1abs,2))*K2+power(K2abs,2)*K2;
h4=K2*K2*complex(K1);
Can someone help me to solve the triple summation part?

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Respuestas (1)

Matt J
Matt J el 9 de Abr. de 2021
Editada: Matt J el 9 de Abr. de 2021
Well, for each fixed p and q, the inner sum over m is just a discrete convolution. So, perhaps the best solution is to do a double loop over p and q and add up conv() output.
r=0;
for p=1:P
for q=0:p
r=r+conv(____,shape)
end
end
One advantage to this approach is that you can use conv's 3rd argument shape to deal with array edge conditions.

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