Inverse laplace in matlab gives iota

[indianaronald]

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?

 

[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.

 

[indianaronald]

What is the meaning of solution pair? Yes, it gives me a±bi. How do I interpret it? And I don't even have damping in my system.

 

[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.

What is the meaning of solution pair? Yes, it gives me a±bi. 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.

 

[alphy]

It is important to note that the Laplace transform is a powerful mathematical tool for solving differential equations, but it is not foolproof. In some cases, it may give complex roots that do not seem possible, as you have observed in your spring-mass system. This could be due to the limitations of the Laplace transform in handling certain types of functions or initial conditions.

In such cases, it may be necessary to use other methods, such as traditional differential equation solving or numerical methods, to obtain more accurate and reliable solutions. It is also important to carefully check your input values and equations to ensure they are correct and appropriate for your system.

Additionally, keep in mind that complex roots can still be valid solutions in some cases, so it is important to thoroughly analyze and interpret the results of your Laplace transform in the context of your system. Overall, the inverse Laplace transform in Matlab is a useful tool, but it is important to understand its limitations and use it in conjunction with other methods for a comprehensive and accurate analysis.