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removing for loop by using 3d matrix

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Parisa Salemi
Parisa Salemi on 15 Feb 2021
Commented: Jan on 19 Feb 2021
Hi. I wrote a code. I want to calculate pressure(p) of waves which produces by 1d array of elements. p is 2d variable. In the code I used 2d matrix and 2 loop. now I want to make a 3d matrix to remove one of the loops. but the problem that I have is when I creat 3d matrix p also gets a 3d matrix while it must be a 2d matrix.
clc;
close all;
clear all;
a=1;
c=1.5; % mm/us
T=5; dt=0.8; % us
fc=1;BW=3; % MHz
R=10;
xf=20;zf=20;
td=0;
dx=0.3; dz=0.3;pitch=0.3;
% x0=10; z0=0;
f=@(x) exp(-x.^2*BW^2);
g=@(x) sin(2*pi*fc*x);
H=@(x) f(x).*g(x);
Nz=120;
Nx=120;
x=[1:Nx]*dx;
z=[1:Nz]*dz;
xl=[1:Nx]*dx;
[z3,x3,xxl] = ndgrid(z,x,xl);
for t=0:dt:T
d3=sqrt((x3-xxl).^2+(z3).^2);
df=sqrt((xxl-xf).^2+(zf)^2);
td=(R-df)./c;
tt_hat=(d3./c)+td;
p = 1./sqrt(d3).*H(t-tt_hat); %
imagesc(p(:,:,100));
% colormap gray; colorbar;
% set(gca,'clim');
% title(['Time= ',num2str(t)]);
pause(0.05);
end
  3 Comments
Parisa Salemi
Parisa Salemi on 16 Feb 2021
I could to remove one of the loops. But now the problem that I have is, I do not know how can I remove the time loop in this code.
clc;
close all;
clear all;
a=1;
c=1.5; % mm/us
T=20; dt=0.8; % us
fc=1;BW=3; % MHz
R=10;
xf=20;zf=20;
td=0;
dx=0.3; dz=0.3;pitch=0.3;
% x0=10; z0=0;
f=@(x) exp(-x.^2*BW^2);
g=@(x) sin(2*pi*fc*x);
H=@(x) f(x).*g(x);
Nz=120;
Nx=120;
x=[1:Nx]*dx;
z=[1:Nz]*dz;
xl=[1:Nx]*dx;
[z3,x3,xxl] = ndgrid(z,x,xl);
xf1 = ones(size(xxl))*xf;
zf1 = ones(size(xxl))*zf;
d3=sqrt((x3-xxl).^2+(z3).^2);
df=sqrt((xxl-xf1).^2+(zf1).^2);
td=(R-df)./c;
tt_hat=(d3./c)+td;
for t=0:dt:T
p = 1./sqrt(d3).*H(t-tt_hat);
p1 = sum(p,3).^2;
imagesc(p1);
pause(0.05);
end

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Answers (1)

Jan
Jan on 16 Feb 2021
Edited: Jan on 18 Feb 2021
What is your purpose of vectorizing the code? The processing time is dominated by imagesc and pause here. I assume the loop are faster than the vectorizde code, which produces large intermediate arrays. Your original loop takes 0.86 seconds, the code in the comment above 1.33 seconds, if imagesc and pause are removed.
You can add an additional dimension:
dim = [1, size(tt_hat)];
p = H((0:dt:T).' - reshape(tt_hat, dim)) ./ reshape(sqrt(d3), dim);
p1 = sum(p, 4).^2;
But I do not see the advantage compared to clean loops.
If efficiency matters, remember that anonymous functions are expensive. So reduce them to the minimum:
% This:
f=@(x) exp(-x.^2*BW^2);
g=@(x) sin(2*pi*fc*x);
H=@(x) f(x).*g(x);
% slows down the total processing time by 10% compare to this:
H = @(x) exp(-x.^2 * BW^2) .* sin(2 * pi * fc * x);
Note: Please format your code in the forum to improve the readability. I've done this for you today.
The brute clearing header "clc;close all;clear all;" is extremly inefficient: the clear('all') removes all functions from the memory and forces Matlab to reload them from the slow disk. This wastes a lot of time and offers no advantage. Prefer using functions instead to keep your workspace clean.
  4 Comments
Jan
Jan on 19 Feb 2021
Hi. This code produces a 3D array. Of course it does, because the original code produces a set of 2D arrays. Concatenating a bunch of 2D matrices must create a 3D array.
Why do you want to get a 2D matrix? What should happen with the additional information?

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