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I have a super basic questions. I am a not really into signal processing (more about physics), but I would like to understand an aspect of linear response function (I think the question fits for this forum).

From my basic understanding, any linear, time invariant, causal system can relate the output $S(t)$ to the input $E(t)$ under the following relations:

$$\begin{equation}S(t)=S(-\infty) + \int_{-\infty}^{+\infty} \chi(t-t') E(t') dt'\label{eq_1} \end{equation}$$

The causality holds when imposing $\chi(u<0)=0$.

Let's consider a very simple case: voltage around an inductance. I have the law: $U=L \frac{d I}{d t}$

If I express $I$ as a function of $U$, I can write down:

$$I(t)=I(-\infty) + \int_{-\infty}^{+\infty} \frac{U(t')}{L} dt'$$

My response function will simply be $\chi(u)=\frac{\Theta(u)}{L}$ (where the heaviside is for causality).

But the system in which $I$ is the input and $U$ the output is also linear. Thus I would expect to be able to express:

$$U(t)=U(-\infty) + \int_{-\infty}^{+\infty} f(t-t') I(t')$$

However because of the derivative on $I$ in the law, I don't find how it is possible.

Am I wrong in my initial statement and \ref{eq_1} ?

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Let's start with the simple system

$$y(t)=x(t)\tag{1}$$

where $y(t)$ denotes the output and $x(t)$ denotes the input. It is linear and time-invariant, and, consequently, it must be possible to represent it by the convolution integral

$$y(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau\tag{2}$$

where $h(t)$ is the system's impulse response. In this case, it turns out that $h(t)=\delta(t)$, i.e., the impulse response is a Dirac delta impulse.

Now, for a differentiator

$$y(t)=x'(t)\tag{3}$$

we "simply" (conceptually, at least) have $h(t)=\delta'(t)$, where $\delta'(t)$ is the (generalized) derivative of $\delta(t)$. So, using the concept of a generalized derivative of a distribution, we can write the input-output relationship of an ideal differentiator as the following convolution integral:

$$y(t)=\int_{-\infty}^{\infty}x(\tau)\delta'(t-\tau)d\tau\tag{4}$$

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  • $\begingroup$ Thank you for your answer. However isn't $<\delta',x> = -<\delta,x'>$ ? Thus you should have a minus in your last equation ? $\endgroup$ – StarBucK May 16 at 17:27
  • $\begingroup$ @StarBucK: Your equation is correct, but it doesn't imply a minus sign in Eq. (4). Note that $\delta'(t)$ is an odd function. $\endgroup$ – Matt L. May 16 at 17:34
  • $\begingroup$ $$y(0)=\int x(\tau)\delta'(-\tau)d\tau=-\int x(\tau)\delta'(\tau)d\tau=\int x'(\tau)\delta(\tau)d\tau=x'(0)$$ $\endgroup$ – Matt L. May 16 at 17:35

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