How to reduce its execution time and why the value of e is a column vector?

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Sadiq Akbar 2023-1-2

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1887087-how-to-reduce-its-execution-time-and-why-the-value-of-e-is-a-column-vector

评论： Sadiq Akbar 2023-1-9

采纳的回答： Walter Roberson

在 MATLAB Online 中打开

I have the following piece of code.

clc;clear all;
c = 340;
f = 3400;
d = c/f/2;                          
T = 1;                             
Sig = 2;                         
M = 6;                               
N = 7;                                    
theta = linspace(-60, 60, Sig);
p = 6;                                 
SNR = 10;                               
Fc=[2*10^3:2*10^3/(Sig-1):5*10^3];  
T_Vector=1/f; 
p_N = [0:M/p:M*(N-1)/p];   
p_M = [0:N:(M-1)*N];
P = union(p_N,p_M);                    
A = zeros(length(P),Sig);               
SigVec = zeros(Sig,T);
for Q = 1:Sig        
        A(:,Q) = exp(-j*P'*2*pi*d*sin(theta(Q)*pi/180)*f/c); 
        SigVec(Q,:) = exp(1j*2*pi*Fc(Q).*T_Vector);  
end
%%%%%%%%%%%%%%%%%%
% xo calculation
%%%%%%%%%%%%%%%%%%
xo = A*SigVec; 
%%%%%%%%%%%%%%%%%%%%%%%%%%
% xe calculation
%%%%%%%%%%%%%%%%%%%%%%%%%
b=theta;
for Q = 1:Sig        
        Ae(:,Q) = exp(-j*P'*2*pi*d*sin(b(Q)*pi/180)*f/c); 
        SigVec_est(Q,:) = exp(1j*2*pi*Fc(Q).*T_Vector);  
end
xe = Ae*SigVec_est; 
%%%%%%%%%%%%%%%%%
% MSE
%%%%%%%%%%%%%%%%
e = mean(abs(xo-xe).^2,2)

I want to reduce its execution time. Further, I want a single value of e as zero but it gives me a column vector of e.

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Sadiq Akbar 2023-1-3

@Voss

Can you replace "Just the for-loops here by using concept of vectorization?". Leave the rest of the code, just replace the two for-loops here.

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采纳的回答

Walter Roberson 2023-1-2

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https://ww2.mathworks.cn/matlabcentral/answers/1887087-how-to-reduce-its-execution-time-and-why-the-value-of-e-is-a-column-vector#answer_1139482

you are taking the mean over the second dimension of a column vector (which has a second dimension of length 1)

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Sadiq Akbar 2023-1-2

Thanks a lot for your guidance dear Walter Roberson. Yes you are right. When I removed the 2 after comma, it gives single value of e now.

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Voss 2023-1-3

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1887087-how-to-reduce-its-execution-time-and-why-the-value-of-e-is-a-column-vector#answer_1140727

在 MATLAB Online 中打开

@Sadiq Akbar: OK. See below.

Note that SigVec has T columns because of its pre-allocation, but that SigVec_est has one column because it was not pre-allocated in the original code. I've kept them like that (change T to something > 1 to see the difference).

clc;clear all;
c = 340;
f = 3400;
d = c/f/2;                          
T = 1;                             
Sig = 2;                         
M = 6;                               
N = 7;                                    
theta = linspace(-60, 60, Sig);
p = 6;                                 
SNR = 10;                               
Fc=[2*10^3:2*10^3/(Sig-1):5*10^3];  
T_Vector=1/f; 
p_N = [0:M/p:M*(N-1)/p];   
p_M = [0:N:(M-1)*N];
P = union(p_N,p_M);                    
% A = zeros(length(P),Sig);               
% SigVec = zeros(Sig,T);
% 
% for Q = 1:Sig        
%         A(:,Q) = exp(-j*P'*2*pi*d*sin(theta(Q)*pi/180)*f/c); 
%         SigVec(Q,:) = exp(1j*2*pi*Fc(Q).*T_Vector);  
% end
A = exp(-1j*P(:)*2*pi*d.*sin(theta*pi/180)*f/c);
SigVec = exp(1j*2*pi*Fc(:).*T_Vector*ones(1,T));
%%%%%%%%%%%%%%%%%%
% xo calculation
%%%%%%%%%%%%%%%%%%
xo = A*SigVec; 
%%%%%%%%%%%%%%%%%%%%%%%%%%
% xe calculation
%%%%%%%%%%%%%%%%%%%%%%%%%
b=theta;
% for Q = 1:Sig        
%         Ae(:,Q) = exp(-j*P'*2*pi*d*sin(b(Q)*pi/180)*f/c); 
%         SigVec_est(Q,:) = exp(1j*2*pi*Fc(Q).*T_Vector);  
% end
Ae = exp(-1j*P(:)*2*pi*d.*sin(b*pi/180)*f/c);
SigVec_est = exp(1j*2*pi*Fc(:).*T_Vector);
xe = Ae*SigVec_est; 
%%%%%%%%%%%%%%%%%
% MSE
%%%%%%%%%%%%%%%%
e = mean(abs(xo-xe).^2,2)
e = 12×1
     0
     0
     0
     0
     0
     0
     0
     0
     0
     0

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Walter Roberson 2023-1-4

在 MATLAB Online 中打开

P = union(p_N,p_M);

p_N and p_M are built up from constants assigned inside the function, with values independent of the inputs (the inputs are b and u in this context.) So P will have a size independent of the inputs, but the details of the size could vary if you changed the constants. More duplicates could occur (or you could end up with non-integral values that might not exactly match due to round-off errors in calculations.) In this case it ends up having 12 elements.

theta=u;

You are passing in a row vector 1 x 6 so theta becomes 1 x 6

A = exp(-1j*P(:)*2*pi*d.*sin(theta*pi/180)*f/c);

With P being 1 x 12, P(:) is 12 x 1, giving A 12 rows. Meanwhile theta is the 1 x 6 passed in, so A will have 6 columns. A will therefore becomes 12 x 6.

Fc=[2*10^3:2*10^3/(Sig-1):5*10^3];

with Sig being assigned inside the function, Fc works out as being 1 x 8.

SigVec = exp(1j*2*pi*Fc(:).*T_Vector*ones(1,T));

T is 1, T_vector is a scalar. With Fc being 1 x 8, Fc(:) will be 8 x 1, and there is only a single column in the other variables in the calculation. So SigVec will be 8 x 1.

xo = A*SigVec;

So you are doing matrix multiplication between an 12 x 6 and an 8 x 1. There is no way to make that fit.

What size were you expecting as the output stored in xo ?

You run into exactly the same problems with xe = Ae*SigVec_est with Ae being the same size as A and SigVec_est being the same size as SigVec .