# Inverse laplace in matlab gives iota

by indianaronald
Tags: control, laplace, spring mass
 P: 21 I'm applying laplace transform to a spring-mass system, the most basic one. I write this code which takes initial values x(0) and v(0) as input and I'm computing x(t) in matlab. But for some values it gives me complex roots for x(t) which doesn't seem possible. If not for laplace I can solve the same question traditional differential equation solving way and get real roots. So what's happening here?
 HW Helper Thanks PF Gold P: 7,721 Remember that a damped harmonic system may give you complex conjugate roots ##a\pm bi## with a solution pair ##\{e^{(a+bi)t},e^{(a-bi)t}\}##. But this is equivalent to the solution pair ##\{e^{at}\cos(bt),e^{at}\sin(bt)\}##. Complex roots don't necessarily cause complex solutions. In fact, if the coefficients and initial conditions are all real, you will not have complex solutions.
 P: 21 What is the meaning of solution pair? Yes, it gives me aħbi. How do I interpret it? And I don't even have damping in my system.
HW Helper
Thanks
PF Gold
P: 7,721
Inverse laplace in matlab gives iota

 Quote by LCKurtz Remember that a damped harmonic system may give you complex conjugate roots ##a\pm bi## with a solution pair ##\{e^{(a+bi)t},e^{(a-bi)t}\}##. But this is equivalent to the solution pair ##\{e^{at}\cos(bt),e^{at}\sin(bt)\}##. Complex roots don't necessarily cause complex solutions. In fact, if the coefficients and initial conditions are all real, you will not have complex solutions.
 Quote by indianaronald What is the meaning of solution pair? Yes, it gives me aħbi. How do I interpret it? And I don't even have damping in my system.
By a "solution pair" I mean the general solution is$$y = Ae^{at}\cos(bt)+e^{at}\sin(bt)$$Without actually seeing what your system is and what your work looks like, I can't be more specific about your problem.

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