f = 
Plotting inverse Laplace transform
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I want to find the inverse Laplace transform and then plot the graph. Below 👇 is the code: syms s t %defines s and t as symbolic variables.
a=0.05;
b=0.0045;
c=0.067;
f=0.0508;
g=0.2;
h=0.45;
x=2.71828;
j=232679478;
r=0.742;
k=(h+r);
F =(1-g)*b*a*j*x^(c+f))/s*(s+a*x^c)(s+b*x^f) + (r*a*j*g*b*x^(c+f))/s*(s+k)*(s+a*x^c)(s+b*x^f); %Definition of the Function F(s)
f = ilaplace(F); %calculates the inverse Laplace transform of F(s), resulting in f(t).
f_func = matlabFunction(f) %converts the symbolic function f(t) into a regular MATLAB function.
% Define the time vector for plotting
t_vec = 0:0.1:10; % time from 0 to 10
% Plot the inverse Laplace transform
plot(t_vec, f_func(t_vec))
xlabel('Time (t)')
ylabel('f(t)')
title('Inverse Laplace Transform Plot')
grid on
I got this Error message: undefined function code for input argument of type 'cher'a=0.05;b=0.0045;c=0.067;f=0.0508;g=0.2;h=0.45;x=2.71828;j=232679478;r=0.742;k=(h+r);F =(1-g)*b*a*j*x^(c+f))/s*(s+a*x^c)(s+b*x^f) + (r*a*j*g*b*x^(c+f))/s*(s+k)*(s+a*x^c)(s+b*x^f); %Definition of the Function F(s)f = ilaplace(F); %calculates the inverse Laplace transform of F(s), resulting in f(t).f_func = matlabFunction(f) %converts the symbolic function f(t) into a regular MATLAB function.% Define the time vector for plottingt_vec = 0:0.1:10; % time from 0 to 10% Plot the inverse Laplace transformplot(t_vec, f_func(t_vec))xlabel('Time (t)')ylabel('f(t)')title('Inverse Laplace Transform Plot')grid onI got this Error message: undefined function code for input argument of type 'cher'
7 个评论
Sam Chak
2025-3-21
Hi Sunday,
Could you write out the transfer function F(s) on a piece of paper and take a photo? Later, we can check if it is correctly coded. Of course, you can also verify it yourself.
You have an extra right parenthesis (that I deleted), and two missing operators (that I added as multiplication operators). The ‘f’ expression is a function of the 
function and its first and second derivative. The function itself appears to be a constant, since its integral is an ascending straight line.
function and its first and second derivative. The function itself appears to be a constant, since its integral is an ascending straight line. You most likely need to check to be certain that ‘f’ is correct, then try again.
syms s t %defines s and t as symbolic variables.
a=0.05;
b=0.0045;
c=0.067;
f=0.0508;
g=0.2;
h=0.45;
x=2.71828;
j=232679478;
r=0.742;
k=(h+r);
F =(1-g)*b*a*j*x^(c+f)/s*(s+a*x^c)*(s+b*x^f) + (r*a*j*g*b*x^(c+f))/s*(s+k)*(s+a*x^c)*(s+b*x^f); %Definition of the Function F(s)
f = ilaplace(F); %calculates the inverse Laplace transform of F(s), resulting in f(t).
f = vpa(simplify(f, 500), 5)
f_func = matlabFunction(f) %converts the symbolic function f(t) into a regular MATLAB function.
% Define the time vector for plotting
t_vec = 0:0.1:10; % time from 0 to 10
% Plot the inverse Laplace transform
plot(t_vec, f_func(t_vec))
xlabel('Time (t)')
ylabel('f(t)')
title('Inverse Laplace Transform Plot')
grid on
figure
fplot(f, [0 10])
xlabel('Time (t)')
ylabel('f(t)')
title('Inverse Laplace Transform Plot')
grid on
figure
fplot(int(f), [0 10])
xlabel('Time (t)')
ylabel('\int{f(t)}')
title('Inverse Laplace Transform Plot')
grid on
.
Walter Roberson
2025-3-21
F =(1-g)*b*a*j*x^(c+f))/s*(s+a*x^c)(s+b*x^f) + (r*a*j*g*b*x^(c+f))/s*(s+k)*(s+a*x^c)(s+b*x^f); %Definition of the Function F(s)
1 0 1 0v 0 v0 v 0 1 0v 0 v 0 v0 v
The number below each ( ) is the number of open brackets "after" the effect of the symbol above it. Here, "v" has been used to represent -1 -- the case where there has been one too many ")"
F =(1-g)*b*a*j*x^(c+f))/s*(s+a*x^c)(s+b*x^f) + (r*a*j*g*b*x^(c+f))/s*(s+k)*(s+a*x^c)(s+b*x^f); %Definition of the Function F(s)
^^ ^^
In MATLAB, the only case in which you can have two adjacent () expressions, is the case where the first one is a table variable dot index. For example,
T.(EXPRESSION)(INDEX)
is valid where EXPRESSION is either a string constant or a character vector or a positive integer.
Other than that exception for table dot indexing, adjacent () expressions are forbidden.
There is absolutely no implied multiplication in MATLAB -- not anywhere . (Including for internal MuPAD symbolic expressions.) . All multiplication must be indicated explicitly, using either the .* operator or the * operator.
Sunday Aloke
2025-3-21
Sam Chak
2025-3-21
Sunday Aloke
2025-3-21
采纳的回答
David Goodmanson
2025-3-22
编辑：David Goodmanson
2025-3-22
Hi SA
Speculative answer: I don't believe that the function F is a likely candidate for a Laplace transform. What you have is (looking at just the first term)
F = const/s*(s+a*x^c)*(s+b*x^f)
which is actually (by mistake?)
(const/s)*(s+a*x^c)*(s+b*x^f)
i.e. two factors involving s in the numerator. For the inverse transforn, that leads to stuff like the derivative of a delta function. What seems more likely is
const/(s*(s+a*x^c)*(s+b*x^f))
with all the s factors in the denominator. Similarly for the second term.
Both terms have a factor of s in the denominator. If
invLaplace(g(s)/s) = G(t)
then removing the s in the denominator effectively multplies by s and gives the time domain derivative,
invLaplace(g(s)) = dG(t)/dt.
The code below does both cases.
syms s t
a=0.05;
b=0.0045;
c=0.067;
f=0.0508;
g=0.2;
h=0.45;
x=2.71828;
j=232679478;
r=0.742;
k=(h+r);
d1 = (1-g)*b*a*j*x^(c+f)
d2 = r*a*j*g*b*x^(c+f)
F = d1/(s*(s+a*x^c)*(s+b*x^f)) + d2/(s*(s+k)*(s+a*x^c)*(s+b*x^f));
Fs = d1/( (s+a*x^c)*(s+b*x^f)) + d2/( (s+k)*(s+a*x^c)*(s+b*x^f));
f = ilaplace(F);
fs = ilaplace(Fs);
f_fun = matlabFunction(f);
fs_fun = matlabFunction(fs);
% Define the time vector for plotting
tvec = 0:2000; % extend the time
y = f_fun(tvec);
ys = fs_fun(tvec);
% Plot the inverse Laplace transform
figure(1)
plot(tvec,ys)
grid on
xlabel('Time (t)')
ylabel('df(t)/dt')
title('Inverse Laplace Transform Plot')
grid on
figure(2)
plot(tvec,y)
grid on
xlabel('Time (t)')
ylabel('f(t)')
title('Inverse Laplace Transform Plot')
grid on
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