Two variable function computation

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friet
friet 2017-1-31
回答： Walter Roberson 2017-1-31
Hello,
I am trying to reproduce the Matlab plot I found in a research paper. The function in P(t,z) (it is time and space dependent). P(t,z) have two parameters (alpha and beta) which are frequency dependent.
I have two loops for computing P(t,z). My code is below, however, I can't find a similar solution which what is mentioned. This is my code
clc
clear all
close all
f=linspace(0,20*10^6,100);
t=1./f;
x=linspace(0,0.15*10^-2,100);
alpha=(0.45*10^-11)*f.^2;
beta=(0.15*10^-4)*f;
figure(1)
subplot(2,1,1)
plot(f/10^6,alpha/100)
ylabel('alpha')
subplot(2,1,2)
plot(f/10^6,beta/100)
ylabel('beta')
p = zeros(length(x),length(t));
for it= 1:length(t)
for ix=1:length(x)
p(ix,it)= exp(-alpha(it).*x(ix)).*exp(-i.*1.*(f(it).*t(it)-beta(it).*t(it)));
end
end
figure(2)
plot3(x,t,p)
Any help will be appriciated.
Thanks
  2 个评论
Walter Roberson
Walter Roberson 2017-1-31
Note that your p is complex, so you should abs() before plotting the magnitude.
You are changing alpha and omega and beta over time, which is not part of the formula. That p formula is for some particular alpha and beta and omega.
Your definition of t is non-linear, but the plot on the right appears to use linear time.
In the right hand plot, the Distance does not appear to have a scaling factor, but you have used 10^-2 scaling factor in your definition of x.
friet
friet 2017-1-31
编辑：friet 2017-1-31
Hello,
Thanks for your comment. I have made some modification based on your kind comment. I have selected omega as 10MHz and select the corresponding values of alpha and beta. The factor 10^-2 is change from cm to m. All the units are in for x are in m and time in sec.
clc
clear all
close all
f=linspace(0,20*10^6,100);
t=linspace(0,1*10^-6,100);
x=linspace(0,0.15*10^-2,100);
alpha=(0.45*10^-11)*f.^2;
beta=(0.15*10^-4)*f;
omega=10*10^6;
a=5.5*100;
b=1.8*100;
figure(1)
subplot(2,1,1)
plot(f/10^6,alpha/100)
ylabel('alpha')
subplot(2,1,2)
plot(f/10^6,beta/100)
ylabel('beta')
p = zeros(length(x),length(t));
for it= 1:length(t)
for ix=1:length(x)
p(ix,it)= exp(-a.*x(ix)).*exp(-i.*1.*(omega.*t(it) - b.*t(it)));
end
end
tt= repmat(t,100,1);
xx= repmat(x,100,1);
figure(2)
waterfall(xx,tt,abs(p))
But still, can't find the same solution. By the way, how can I change the origins of the 3d plot, the distance and time to be start at 0 and the magnitude origin separately as it was plotted in the original plot.
Thanks

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回答（2 个）

Honglei Chen
Honglei Chen 2017-1-31
You may want to check out waterfall
https://www.mathworks.com/help/matlab/ref/waterfall.html
HTH
Walter Roberson
Walter Roberson 2017-1-31
Your expression just does not act like that, at least not over those parameter ranges.
See attached that explores a number of possibilities. (Note the vectorized implementations of the equations.) I put up several isosurface plots to explore the way the surface grows along the 4 different parameters. The last couple of plots explore the possibility that your x is too small or too large. The last one does suggest that possibly the x range should be much smaller; for example if you were to use x/200 then it would have room for only one peak.
On the isosurface plots notice that I plot the real component of p rather than the absolute magnitude. The absolute magnitude are completely boring; plotting the real component at least allows you to track through a couple of phase cycles.

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