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?


1 Answer 1


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.

Another argument works like this: If you want to know the pitch of a tuning fork or bell, you strike it, right? That's a physical impulse, which transforms (in Fourier) to a complex exponential whose argument is a function of the time delta. But other systems are much more complex than a tuning fork or bell, and may have more complex harmonic behavior--so I need to perturb the system more thoroughly, at different times.

Transforms make this process rigorous, and mathematically show that they "cover everything" so that you don't have to work by trial and error.

  • $\begingroup$ 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 $\endgroup$
    – reuns
    Oct 8, 2016 at 19:04
  • $\begingroup$ @user1952009 Agreed, but OP wanted to hear a more intuitive argument. $\endgroup$ Oct 8, 2016 at 19:11
  • $\begingroup$ I don't think there is anything more intuitive than what I wrote : the frequency response is the output for a pure (complex) sinusoidal input $\endgroup$
    – reuns
    Oct 8, 2016 at 19:14

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