# Feedback Convolution

1. Jul 23, 2006

### brendan_foo

Although technically an engineering question, I think it is fitting to post it here as its general abstract.

Assuming a simple negative feedback system of feedforward transfer function $G$ and feedback transfer function $H$ it can be easily shown by cascading methods that the overall transfer function is

$$G_{cl}(s) = \frac{G(s)}{1+G(s)H(s)}$$

This is pretty standard stuff.

For arguments sake, let $G = 2$ and $H = 1$, and the input to the system is a unit step function.

Solving for $y(t)$ we get $2/3$ of course. MATLAB/simulink will generate the same output, and for the error (actuating) signal we get $1/3$...Ok thats fine.

However, at the time in which the system is initiated, there is NO transient behaviour. Of course I see this as MATLAB is simply solving these equations for fitting solutions.

Surely the initial error would be some nominal value, close to, if not zero itself. Trying to mimic this in simulink pertains to some serious divergency in the output - as is to be expected.

The Laplace transform spans the entire time domain, and is thusly used to represent the convolution of the time domain functions. I understand this, but I can't quite get it in my head as to what is actually going on in the time domain to pertain to such non transient behaviour. Not that I would expect a real system to perform to such remarkable outputs, but I wouldn't expect it to be too atypical of the MATLAB output, else the simulations are good for nothing.

Its kind of a chicken and the egg paradox.

As the impulse responses are $g(t) = 2\delta(t)$ and $h(t) = \delta(t)$ it is of course fair enough to say that:

$$y(t) = 2\cdot e(t)$$
$$e(t) = u(t) - y(t)$$
etc.. etc..

But I can see some inherent recursion right there. Wouldn't it be more appropriate to say

$$y(t) = 2\cdot e(t - \tau)$$ for some small value of $\tau$. And thusly in the S domain introducing a scalar factor of $e^{s\tau}$.

Any help or insight on this one would be greatly appreciated. I'm running circles here.

Many thanks
Brendan