# Laplace transform of multiplication

1. Mar 12, 2012

### Bloo_Mec

Hi everyone, I have a problem with finding the transfer function of a linear system. It happens that in some terms the input and the output are multiplying and I have no idea how to do the Laplace transform of this. The system is a car and I am only studying its speed. The input is the accelerator (0«u«1) and the output is the speed (v). Please give me a hint on how to do this. I have to isolate the input and the output. My only idea is to solve this by the integral definition of the Laplace transform, but the differential equation is big enough to scare me out of doing that.

Thank You All.

2. Mar 17, 2012

### viscousflow

This sounds like a homework problem. Also if you'd like such detailed help, it would help us if you provide the details of your work, i.e. show your working.

3. Mar 18, 2012

### X89codered89X

It's possible that the problem is more complicated than this, but it looks like the transfer function is simply $H(s) = \frac{1}{s}$. This is simply because the input is acceleration, and the output is speed, and integration of acceleration: $\int_0^t {\ddot{x}(\tau)d\tau} = \dot{x}(t)$

If the problem is more complicated than this, then you may need some heavier transfer function knowledge. Look into the derivative property of the laplace transform: ie $\int_{-\infty}^{\infty}\dot{f}(t)e^{-st}dt = s\int_{-\infty}^{\infty}f(t)e^{-st}dt - f(0^{-})$. Here the original function evaluated at time t = 0 is considered an initial condition, and is ignored in transfer functions (which are zero-state equations, if that means anything to you). This can be extended for the 2nd derivative as well. using this you can generalize something like $f(t) = 4\ddot{x}(t) + \dot{x}(t) + x(t) --> F(s) = (4s^2 + s + 1)X(s)$

Hope this helps.

Last edited: Mar 18, 2012