0

The topic of this question is the causal linear time invariant system (LTI system). I need some help to understand better how to go from this formula $$ y(t)=\int_{-\infty}^{+\infty} x(τ)h(t-τ)dτ\label{1}\tag{1}$$ to this one: $$y(t)=\int_{0}^{t} x(τ)h(t-τ)dτ.\label{2}\tag{2}$$

The system in question using a visual diagram representation is: $h(t)\longrightarrow T[x(t)]\longrightarrow y(t)$

I know that the causality criterion is: $h(t)=0$ for all $t<0$. In simple words, the output $y(t)$ is independent of all future input values.

In formula \eqref{1} do NOT swap positions of $x(t)$ and $h(t)$. That is, I do NOT want to see the analysis of the convolution integral $$\int_{-\infty}^{+\infty}h(τ)x(t-τ)dτ. $$ I already comprehend how to solve it (a.k.a. analyse it) in this manner, but i get confused when i see formula \eqref{1}.

Daniele Tampieri
  • 9,727
DontAskTheEye
  • 185

1 Answers1

0

Like you say, $h(t) = 0$ for all $t < 0$. This means that $h(t-\tau) = 0$ for all $t < \tau$ or, equivalently, for all $\tau > t$.

That is, in the range $\tau > t$, the integrand is zero, and so

$$\int_{-\infty}^\infty x(\tau) h(t-\tau) \, \mathrm{d}\tau = \int_{-\infty}^t x(\tau) h(t-\tau) \, \mathrm{d}\tau + \int_{t}^\infty 0 \, \mathrm{d}\tau = \int_{-\infty}^t x(\tau) h(t-\tau) \, \mathrm{d}\tau $$

This establishes the upper limit, so now what about the lower limit? To determine this, we must consider what defines the moment in time $t=0$? What separates it from any other arbitrary point of time? Why do we set the lower limit to 0 instead of any other arbitrary time?

$t=0$ is the time before which no input is present. That is, $x(t) = 0$ for all $t < 0$. Therefore, we can say $x(\tau) = 0$ for all $\tau < 0$, which means the integrand of the convolution goes to zero for negative $\tau$. That is

$$\int_{-\infty}^t x(\tau) h(t-\tau) \, \mathrm{d}\tau = \int_{-\infty}^0 0 \, \mathrm{d}\tau + \int_{0}^t x(\tau) h(t-\tau) \, \mathrm{d}\tau = \int_{0}^t x(\tau) h(t-\tau) \, \mathrm{d}\tau $$

So, what if there existed an input before $t=0$? What if $x(t) < 0$ for $t < -t_0$? Then the convolution integral would instead have become

$$\int_{-t_0}^t x(\tau) h(t-\tau) \, \mathrm{d}\tau $$

Involute
  • 273