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Dear all,

Please find attached image, I am not able to understand why is the underlined term is to be included. Shouldn't it be simply the exponential term?

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  • $\begingroup$ Paste the link to the image in plain text and someone will edit it in for you $\endgroup$ – jonsca Dec 22 '13 at 9:24
  • $\begingroup$ OK, It must be visible now. $\endgroup$ – aditya Dec 22 '13 at 9:44
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The $u(t-2)$ term needs to be included because the upper limit of the definite integral is $t-2$, not $\infty$:

\begin{align*} h(t) &= e^{-(t-2)}u(t-2)\\ y(t) &= x(t) * h(t)= \int_{-\infty}^{\infty} x(t')h(t-t') dt'\\ &= \int_{-\infty}^{\infty} x(t')e^{-(t-t'-2)}u(t-t'-2) dt'\\ u(t-t'-2)&= \begin{cases} 1 & t'<t-2 \\ 0 & \text{otherwise} \end{cases}\\ y(t) &= \int_{-\infty}^{t-2} x(t')e^{-(t-t'-2)}(1) dt' + \int_{t-2}^{\infty} x(t')e^{-(t-t'-2)} (0) dt'\\ &= \int_{-\infty}^{t-2} x(t')e^{-(t-t'-2)} dt'. \end{align*}

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  • $\begingroup$ OK Sir, now another question,what if the upper limit was ∞ ? and why is it required that upper limit should be ∞ ? Please explain.Is it because the convolution integral demands it? $\endgroup$ – aditya Dec 22 '13 at 14:56
  • $\begingroup$ The upper limit is $\infty$ because of the definition of convolution: you are summing up the effect of impulses at all times $t' \in (-\infty, \infty)$ at the current time $t$. It just so happens that in this particular case, your system only turns on for t>2, i.e. your impulse response is only non-zero for that time interval, so you can equivalently write the convolution integral with the upper limit $t-2$. $\endgroup$ – Sina Dec 22 '13 at 16:58

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