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 $T + 2T = 3T$.

  • $\begingroup$ Okay I understand this it makes sense. But then I don't understand why it's delay is 3T/2. $\endgroup$ – Michaela H Jan 27 '19 at 15:18
  • $\begingroup$ 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., $1.5T = (3/2)T$) $\endgroup$ – Metric Jan 27 '19 at 15:20
  • $\begingroup$ Okay I understand now! So we shift it to T/2 to make it in the middle of the support. But then since we're starting from T we need to shift it to T+T/2=3T/2. Thank you so much! $\endgroup$ – Michaela H Jan 27 '19 at 19:12
  • $\begingroup$ 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] $\endgroup$ – Metric Jan 27 '19 at 19:18

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