# Why does inputting complex exponentials into a system give its frequency response?

Let's say I have an FIR filter with the equation:

$$y[n] = \sum_{i=0}^{N-1} h[i] x[n-i]$$

I know that to find the frequency response of this filter, I need to input a complex exponential in place of $x[n-1]$

$$x[n] = A e^{j\phi} e^{j \hat{\omega} n}$$

and the resulting frequency response will be

$$H(\hat{\omega}) = \sum_{i=0}^{N-1} h[i] e^{-j \hat{\omega} i}.$$

I understand $e^j$ to mean rotating around in a circular motion, which is a convenient way to represent complex sinusoids.

However, I don't understand how inputting this general form of a complex exponential into this system works to give the frequency response of the system. How can I understand this intuitively?

Technically, you need to input all of the complex exponentials in the angular frequency domain (where the response is not aliased by some periodicity in the system) to get the frequency response of the system. I think that's an intuitive argument for why $i$ needs to be part of the argument to the complex exponential.
• how do you define the "frequency response" ? Obviously, if the input is $x[n] = e^{i \omega n}$ then the output is $y[n] = H(\omega) e^{i \omega n}$ where $H(\omega) = \sum_n h[n]e^{-i \omega n}$ hence $H(\omega)$ is by definition the frequency response – reuns Oct 8 '16 at 19:04