Inverse Laplace contains a complex number
9 views (last 30 days)
Show older comments
Karl Snow
on 17 Nov 2018
Edited: David Goodmanson
on 3 Dec 2018
I am using ilaplace to study circuits. Recently I drew a schematic the inverse Laplace of which (in the time domain) has a complex number. This is new to me. Is there a way to ensure that I don't get complex numbers with an inverse Laplace? One of the terms is:
0.0876468*exp(-255.018*t)*(cos(5101.72*t)*1.0i + sin(5101.72*t))
5 Comments
Walter Roberson
on 27 Nov 2018
Not having physically measurable significance is not at all the same as saying that the equations will not include the components .
Accepted Answer
David Goodmanson
on 28 Nov 2018
Edited: David Goodmanson
on 28 Nov 2018
Hi Karl
You show one of the terms, and I assume you have another one that is the complex conjugate that one so that the entire result is real. Here is a simple example of what is probably going on.
syms s a
F = 2*s/(s^2+a^2);
f = ilaplace(F)
f = 2*cos(a*t)
G = 1/(s+i*a) + 1/(s-i*a) % same as F, in partial fractions
g = ilaplace(G)
g = exp(-a*t*1i) + exp(a*t*1i)
( s^2 + a^2 ) has two complex roots, and ilaplace does not recombine the resulting complex expressions in G. However,
g1 = simplify(g)
g1 = 2*cos(a*t)
You can also try simplifying before doing ilaplace
G2 = simplify(G)
G2 = (2*s)/(a^2 + s^2) % same as F
g2 = ilaplace(G2)
g2 = 2*cos(a*t)
2 Comments
David Goodmanson
on 3 Dec 2018
Edited: David Goodmanson
on 3 Dec 2018
Hi Karl,
you're welcome, although I didn't actually provide any value added. But I do want to comment on the idea that laplace transforms can be done in the real domain only.
While the person you quoted acknowledges complex variables are incredibly useful and fundamental, he/she says that it is only is only a convention/mathematical shorthand. That makes it sound like a matter of bookkeeping, which is highly misleading at best. To begin with, the standard way to derive an inverse laplace transform involves a path of integration in the complex plane. Even for the forward laplace transform, how do you derive the simple expression
Int e^(-st) cos(at) dt = s/(s^2+a^2) ?
It can be done in real variables with integration by parts and knowledge of the differential eqn for cos(at), but it is much, much easier and more direct to split the cos expression into (1/2) exp(iat) + (1/2) exp(-iat), do integrals and recombine. More complicated expressions are done in similar fashion.
For linear circuit theory you generally get the ratio of two polynomials in s. So you can do partial fractions, pair up compex conjugare roots and create a set of all-real expressions. That processis is of course correct, but it sweeps a lot of stuff under the rug.
More Answers (0)
See Also
Categories
Find more on Number Theory in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!