Inverse Laplace plot error
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Guys, I have a problem with this code, I've been trying to fix it for an hour and nothing.
When I plot the inverse laplace graph, the following error appears:
Error using plot
Data must be numeric, datetime, duration or an array convertible to
double.
clear all
close all
clc
syms s t;
D=23;
QT=348000;
Tc=300;
F= (((sqrt(2)*sqrt(D^2*QT^2/s)*8*D^2*QT^2-3*Tc^2*s^2))/(8*D^2*QT^2-Tc^2*s^2));
f=ilaplace (F);
f1=symfun(f,t)
t=[0:0.1:200];
plot(t,f1(t))
Can someone help me?
3 件のコメント
Star Strider
2023 年 1 月 26 日
The function won’t plot, however 1 will, since the objective is to plot the function.
回答 (1 件)
Paul
2023 年 1 月 26 日
Hi PTK,
f1 is a symfun, so you could use fplot, or you could evaluate it at bunch of points and convert to double and the use plot, as shown below.
However, in this case ilaplace was not able to find a closed-form expression for f1, so neither approach will work. Are you sure the sqrt(1/s) is what's intended.
Also, f1(t) contains a dirac delta function. So even if f1 was in closed form otherwise, be careful using fplot because it ignores dirac delta functions, and evaluationg numerically can be tricky as well.
clear all
close all
clc
syms s t;
D=23;
QT=348000;
Tc=300;
F= (((sqrt(2)*sqrt(D^2*QT^2/s)*8*D^2*QT^2-3*Tc^2*s^2))/(8*D^2*QT^2-Tc^2*s^2))
f=ilaplace (F);
f1=symfun(f,t)
t=[0:0.1:200];
plot(t,double(f1(t)))
0 件のコメント
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