Inverse Laplace plot error
이전 댓글 표시
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?
댓글 수: 3
You need to use the fplot function, however the problem prior to that is that the Laplace domain function cannot be inverted —
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);
f1 = subs(f1, dirac(t), 1)
% t=[0:0.1:200];
figure
fplot(f1, [0 200])
The problem appears to be the
term. That is nonlinear in s and that is a problem for inverting Laplace transforms. (The
function can be dealt with by setting it equal to 1.)
.
Paul
2023년 1월 26일
What is the justification for substituting 1 for dirac(t)?
Star Strider
2023년 1월 26일
The
function won’t plot, however 1 will, since the objective is to plot the function.
답변 (1개)
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)))
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