# Inverse Laplace plot error

2 vues (au cours des 30 derniers jours)
PTK le 26 Jan 2023
Commenté : Star Strider le 26 Jan 2023
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?
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Paul le 26 Jan 2023
What is the justification for substituting 1 for dirac(t)?
Star Strider le 26 Jan 2023
The function won’t plot, however 1 will, since the objective is to plot the function.

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### Réponses (1)

Paul le 26 Jan 2023
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 =
f=ilaplace (F);
f1=symfun(f,t)
f1(t) =
t=[0:0.1:200];
plot(t,double(f1(t)))
Error using symengine
Unable to convert expression containing remaining symbolic function calls into double array. Argument must be expression that evaluates to number.

Error in sym/double (line 872)
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