Trouble with inverse laplace operation
2 views (last 30 days)
Show older comments
Deepayan Bhadra
on 18 Dec 2017
Commented: Star Strider
on 18 Dec 2017
I'm trying to run a code like this:
s = tf('s')
sys = exp(-0.1*s);
sysx = pade(sys,3);
x0 = randn(10,1);
L = D-A (A,D,L are constant matrices)
H = 1/(s+L*sysx);
K = H*x0;
ilaplace(K)
Essentially, I want to get a vector x(t) in the end as per the equation xdot = -Lx(t-0.1).
But this returns the error: Undefined function 'ilaplace' for input arguments of type 'tf'.
How do I go about this? Thanks a lot.
0 Comments
Accepted Answer
Star Strider
on 18 Dec 2017
You go about it with difficulty, because the Control System Toolbox and Symbolic Math Toolbox do not have any way of communicating with each other without your intervention.
Try this:
s = tf('s');
sys = exp(-0.1*s)
sysx = pade(sys,3)
sysxn = sysx.Numerator;
sysxd = sysx.Denominator;
x0 = randn(10,1);
syms A D L s t
L = D-A; % (A,D,L are constant matrices)
Nsysx = poly2sym(sysxn{:}, s);
Dsysx = poly2sym(sysxd{:}, s);
TFsysx = Nsysx / Dsysx;
H = 1/(s+L*TFsysx);
K = H*x0;
kh(t) = ilaplace(K, s, t);
kh(t) = vpa(kh(t), 5)
The result are a (10x1) symbolic function vector in ‘t’, ‘z’, and ‘s4’ that I will leave it to you to untangle.
Personally, I would not involve the Symbolic Math Toolbox at all, and instead evaluate the system with step, impulse, or lsim, and be happy with the result.
4 Comments
Star Strider
on 18 Dec 2017
If you do:
K = H*x0;
Num = K.Numerator
Den = K.Denominator
you will see the problem. I have no suggestions as to how to solve it.
More Answers (0)
See Also
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!