# Response of an LTI system by convolution 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?

• 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 – Chu Aug 31 '16 at 12:00
• @Chu wow!! That just cleared everything. – user2626326 Aug 31 '16 at 13:24

Nothing. Just do a variable change $\tau'=t-\tau$ and you'll get your integral.
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:
$$\int_{-\infty}^\infty{f(\tau)}g(t - \tau)d\tau = \int_{-\infty}^\infty{h(\tau)}e^{j\omega_k(t - \tau)}d\tau = e^{j\omega_kt}\int_{-\infty}^\infty{h(\tau)}e^{-j\omega_k\tau}d\tau$$