# Causal LTI System Analysis

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}.

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

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