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Context: I am currently learning about convolution integrals as they apply to circuits (LTI specifically, if I understand it correctly.

(For the following, continue under the assumption that \$\theta(t-c)\$ is always used as the Heaviside step function.

Let's say we have a simple step input: \$x(t)=\theta(t-2)\$ (step input, delayed by two seconds.)

But I am provided the impulse response \$h(t)=\theta(t+3)-\theta(t-3)\$

The short version of this question is this: How should I work with an impulse response defined before \$t=0\$ (with regards to convolution)?

Some more details:

The only way I get the same answers is to do it in a way that seems wrong as I understand these concepts.

I understand that to find the output \$y(t)\$, I am interested in the convolution $$y(t)=x(t)*h(t)\equiv\int_{-\infty}^\infty [x(t)-h(t-\tau)]d\tau$$ $$\equiv\int_{-\infty}^\infty [x(t-\tau)-h(t)]d\tau$$

When I do this using the response as the translated function, I end up with a function that is defined on the boundaries of \$t<-1\$ (no overlap), \$-1\leq t<5\$ (some overlap), and \$5\leq t\$ (full overlap.)

This does not make sense to me, as I am showing output before an input is applied over the period t=\$[-1,1)\$

I also understand that I have the option of utilizing the Laplace transform to solve the convolution.

$$y(t)=x(t)*h(t)\equiv X(s)H(s)$$

(Where X(s) and H(s) are the Laplace transforms of x(t) and h(t), respectively)

This is where things start to confuse me as if I take the Laplace transform of the impulse response, it will include the following step:

$$\mathscr{L}[\theta(t+3)](s)=\int_0^\infty\theta(t+3)e^{-st}dt\equiv\int_0^\infty e^{-st}dt=\frac{1}{s}$$

As \$\theta(t+3)=1\$ over the interval \$[0,\infty)\$.

When you continue on with the product in this way, you get

$$y(t)=(t-2)\theta(t-2)-(t-5)\theta(t-5)$$

If you ignore the one-sided restriction, you get:

$$y(t)=(t+1)\theta(t+1)-(t-5)\theta(t-5)$$

The first appears unrelated, the second perfectly matches the response from earlier (when you ignore that the input starts at \$t=2\$

So my understanding of the 'impulse response' doesn't currently fit with something that is defined before \$t=0\$, and I think that is reflected in the results. I'm trying to figure out what I am missing to understand these cases where I am provided impulse that is provided with a definition such that \$h(t)\neq0\$ for \$t<0\$

SerasVita
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    An impulse response before t=0 represents an acausal system. Real analog systems obey causality, but digital systems have no such restrictions and acausal impulse responses are fairly common. You treat it like any other impulse response, and yes since it does not obey casualty, it can have an effect precede a cause. – user1850479 Jun 20 '20 at 21:27
  • But if done in a real-time system, it does require a delay. – Dave Tweed Jun 20 '20 at 22:26
  • @user1850479 That makes a bit more sense. I was going into this with the assumptions from the class I'm in (that the initial conditions are zero and the response comes from a stimulus.) That doesn't necessarily mean the problem I was given as pure math has the same constraints! – SerasVita Jun 20 '20 at 22:28
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    An acausal response still comes from a stimulus, it just happens out of order because the system allows information to be accessed out of order. As Dave Tweed said, that just means you are operating with a delay so that you can access future inputs. – user1850479 Jun 20 '20 at 23:43

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