Laplace transform of $Q(t)^2$?

Question

I need the transfer function of:

$$K_1 \dot Q(t) + K_2 Q(t) + K_3 Q(t)^2 + K_4 = 0 $$

Where $K_1$, $K_2$, $K_3$ and $K_4$ are constants that depend on the system parameters.

For this I need to apply the Laplace transform, but I don't know how to get the $\mathscr{L}\{Q^2(t)\}$. Any tips?

Answer 1

First of all transfer functions can only be defined for linear time invariant systems and your system is nonlinear. However, it is still possible to find an expression for the Laplace transform of your expression using the convolution theorem. Namely, the property of convolution in the time domain being equivalent to multiplication in the frequency domain also works the other way around, so multiplication in the time domain is equivalent to convolution in the frequency domain which gives $\mathcal{L}\{Q^2(t)\}(s)=(\mathcal{L}\{Q(t)\}*\mathcal{L}\{Q(t)\})(s)$ with $*$ denoting convolution.

Answer 2

The differential equation you have is a nonlinear one and cannot be solved using the Laplace transform.

The Laplace transform of the product of two functions, or a power of a time signal, is not easy to compute. For example, R. Mehrem proposes the following formula:

$$ \mathscr{L}\{Q^2(t)\}(s) = \tfrac{2}{\pi}\int_{0}^{\infty}\mathbf{im}\left(Q\left(\tfrac{s}{2}-i\tau\right)\right)^2\mathrm{d}\tau. $$

Kwin van der Veen mentioned the property $\mathscr{L}\{Q^2\}(s) = \mathscr{L}\{Q\}(s)*\mathscr{L}\{Q\}(s)$. I don't think this property is correct. For example, for $Q(t) = H_0(t)$, that is, the Heaviside function, $Q^2(t) = Q(t)$ and $\mathscr{L}\{Q^2(t)\} = \mathscr{L}\{Q(t)\} = \frac{1}{s}$. This is not equal to $\tfrac{1}{s}*\tfrac{1}{s}$.