How can I clarify an undefined function response in MATLAB?

So I've been tweaking a very simple code to plot a few functions in MATLAB using the step and tf functions coupled with the laplace and ilaplace to generate a few graphs. I first wanted to generate an ideal step response for a few inputs in the form of a transfer function, then move on to more realistic estimates using a form of the Boltzmann Sigmoid eqn, assuming several constant values. I've included the full code below for reference.
%%Ideal Unit Step Response %%
w0 = 1; %rad/s%
zeta = 0.15; %damping ratio%
wn = (w0)/((1-(zeta^2))^(1/2));
R = 1;
num = (wn^2);
den = [1 (2*zeta*wn) (wn^2)];
Ideal = tf(num,den);
t=0:0.1:30;
step(R*Ideal,t);
axis([0 30 0 2]);
stepinfo(Ideal)
%%Boltzmann Sigmoid Function (time domain method) %%
a = 0;
b = 1;
c = 6;
T = 0.1;
s = a+b/(1+exp(c*(T-1)));
% sym('Ideal');
h = ilaplace(Ideal);
y = conv(s,h);
%%Boltzmann Sigmoid Function (s-domain method) %%
S = laplace(s);
Y = S*Ideal;
y = ilaplace(Y);
The Ideal step response runs fine, but the time domain method using the approximation function stumps with
"Undefined function 'ilaplace' for input arguments of type 'tf'."
Similarly, the S domain method yields a
"Undefined function 'laplace' for input arguments of type'double'."
Is there any insight as to what may cause these issues? I thought maybe the code didn't like solving both a TF and an ilaplace at the same time, so I used the 'sym' function, but it seemed to not work. :/

 Accepted Answer

tf creates a transfer function object; even if you use tf('s'), that is a transfer function object.
laplace() and ilaplace() are symbolic toolbox, and do not take accept transfer function objects. The File Exchange routine linked to above shows how to extract information from a transfer function and use it with symbolic ilaplace

4 Comments

Sorry, I got the wrong link before; I have fixed it.
thanks bud, hopefully this will help!
Output argument "ans" (and maybe others) not assigned during call to "ilaplacetf".
Unfortunately there appears to be a small hiccup inside the custom function. Time to go diving.
ans is a special name; at some point the behaviour when you used ans as the output variable name changed. I recommend just changing the references from ans to some other otherwise unused variable name... for example output

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