# Impulse response of integrator

I want to get the impulse response of an LTI system where $$y(t) = \int_{t-2T}^{t-T} x(\alpha) d\alpha$$

To solve this I did: $$h(t) = \int_{t-2T}^{t-T} \delta(\alpha) d\alpha$$

Then you see that for the integrator to be around the impulse $$t > T$$ $$t < 2T$$

To me the impulse respsonse should be $$h(t)=\pi_{3T}(t-3T/2)$$ where pi is the rectangular function. Instead in the solution the answer is $$h(t)=\pi_{T}(t-3T/2)$$ According to me the subscript of the rectangular function is the width of it. How can this be T?

You showed that your impulse response is $$1$$ for
$$T < t < 2T$$
meaning that the length of its support is $$2T - T = T$$, and not $$T + 2T = 3T$$.
• So the rectangular function of length $T$ is centered at the origin. So you need to shift this to the middle of your support (i.e., $1.5T = (3/2)T$) – Metric Jan 27 '19 at 15:20
• One thing: By support I mean the interval $(T,2T)$, and the middle of this interval is $1.5T$, so we needed to shift the rectangular function that's centered at $0$ to the middle of the interval $(T,2T)$ (which is $1.5T$). The $T$ I referenced above is the length of the interval $(T,2T)$. When you see the term "support" in engineering, you can think of it as just the set of all values where the function is not zero [I should've clarified this] – Metric Jan 27 '19 at 19:18