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So here in the picture \$e^{jwt}\$ is the input to the system and \$h(t)\$ is the impulse response. So, by convolution integral shouldn't the response be \$h(t-T)e^{jwT} dT\$? But here it is \$h(T)e^{jw(t-T)}dT\$. What am I missing here?

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    \$\begingroup\$ If you think of convolution as: fold; slide; multiply; add, (or fold; slide; integrate), it doesn't matter which of the signal or h(t) is folded \$\endgroup\$ – Chu Aug 31 '16 at 12:00
  • \$\begingroup\$ @Chu wow!! That just cleared everything. \$\endgroup\$ – user2626326 Aug 31 '16 at 13:24
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Nothing. Just do a variable change \$\tau'=t-\tau\$ and you'll get your integral.

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In the convolution integral we're used to seeing \$h(t - \tau)\$ when talking about impulse response \$h(t)\$, however:

$$\int_{-\infty}^\infty{f(\tau)}g(t - \tau)d\tau = \int_{-\infty}^\infty{f(t - \tau)}g(\tau)d\tau $$

In this case the left side is used with \$ f(\tau) = h(\tau)\$ and \$ g(t - \tau) = e^{j\omega_k(t - \tau)}\$ so you have:

$$\int_{-\infty}^\infty{f(\tau)}g(t - \tau)d\tau = \int_{-\infty}^\infty{h(\tau)}e^{j\omega_k(t - \tau)}d\tau = e^{j\omega_kt}\int_{-\infty}^\infty{h(\tau)}e^{-j\omega_k\tau}d\tau$$

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