Does really the dirac impulse help to find the transfer function of an electronic system?

406 Views Asked by At

Is the following statement is true:

By measuring the response of a impulse or step disturbance to a physical process we can determine the system transfer function and model or optimize the process control system. I think that's what will follow but I'm still stuck in this page (gosh I must be getting old).

And can you give an example ?

1

There are 1 best solutions below

1
On BEST ANSWER

The statement you quoted concerns LTI physical process: Linear Time Invariant physical process.

Disclaimer: since it tackles distribution theory, I should talk about the functions' spaces and so on, but I keep this "proof" simple and formal.

Let's start: for an arbitrary temporal signal $e$, the distribution theory gives you:

$$e(t)= \int_{-\infty}^{\infty} \delta(t-\tau) \; e(\tau) \mathrm{d}\tau$$

where $\delta$ is the Dirac distribution.

Let's apply the physical process to this signal $e$:

$$f(e(t)) = f\left ( \int_{-\infty}^{\infty} \delta(t-\tau)e(\tau) \mathrm{d}\tau \right )$$

$f$ being linear, we obtain:

$$f(e(t)) = \int_{-\infty}^{\infty} f(\delta(t-\tau))e(\tau) \mathrm{d}\tau$$

because $t$ is the variable.

$f$ being time invariant:

$$f(e(t)) = \int_{-\infty}^{\infty} (f \circ \delta)(t-\tau)e(\tau) \mathrm{d}\tau$$

Time invariance gives us the existence of the function $(f \circ \delta)$ because $f$ does not depends on the time when the dirac impulse is provided.

Finally, $f \circ \delta$ is the impulse response named $h$ and you have:

$$f(e(t)) = \int_{-\infty}^{\infty} h(t-\tau)e(\tau) \mathrm{d}\tau$$

which is equivalent to:

$$y(t) = f(e(t)) = (h \ast e)(t)$$

This means that the response $y$ of the LTI physical process $f$ to any entry signal $e$ is the convolution between the impulse response of $f$ and the signal $e$.

In short, all the information of $f$ is contained in $h$!

This is the reason why you study the transfer function $H$ of the physical process which is basically the Fourier transform of $h$!

$$y(t) = (h \ast e)(t) \Leftrightarrow Y(j \omega) = H(j \omega) E(j \omega)$$

(personnal opinion) This is one of the principal result regarding linear systems that I highlight a lot in my classes. This is very powerful. Next step: state-space representation (Cf. Kalman)