Suppose we're given the following: $ x[n] = 2 + (-1)^n $, and are given the impulse response $ h[n] = u[n] a^n $, of an LTI system where $ |a| < 1$. We're asked to find the output $y[n]$, if $x[n]$ is input into $y[n]$.

My approach to this problem was to find the frequency response to $x[n]$ and the frequency response to $h[n]$, multiply the two, and then find the inverse fourier transform. However, this method was quite complicated, and I ended up not being able to solve it, since I essentially got back to a form where it was just $x[n]$ convolved with $h[n]$, which is difficult to compute by hand.

The answer key has a far simpler approach where $x[n]$ is written as $2 e^{j 0 n} + e^{ j \pi n}$, evaluates the frequency response $$ H = \frac{1}{1-a e^{-j \omega}} $$ at $\omega$; values of $0$ and $\pi$, multiplies and adds the result to yield:

$$ y[n] = \frac{2}{1-a} + \frac{1}{1-a} (-1)^n $$

Why can this be done? I'm not sure I follow why $y[n]$ can essentially by written as $H(\omega)$ evaluated at a value of $\omega * x[n]$.

Can this always be done for LTI systems--even continuous systems?

  • 1
    $\begingroup$ Note: in this particular case, a direct calculation by hand of the convolution is rather easy. Using Fourier analysis here is a good, but more difficult, exercise. It will be the opposite in other situations $\endgroup$
    – Damien
    Mar 19, 2019 at 14:45

1 Answer 1


The sequences $e^{j\omega_0n}$ are eigensequences of discrete-time LTI systems, i.e., the response to such a sequence is the same sequence scaled by a complex constant (the eigenvalue). This can be shown as follows:

$$\begin{align}y[n]&=(h\star x)[n]\\&=\sum_{k=-\infty}^{\infty}h[k]x[n-k]\\&=\sum_{k=-\infty}^{\infty}h[k]e^{j\omega_0(n-k)}\\&=e^{j\omega_0n}\sum_{k=-\infty}^{\infty}h[k]e^{-j\omega_0k}\\&=e^{j\omega_0n}H(\omega_0)\end{align}$$

where $h[n]$ is the system's impulse response, and $H(\omega)$ is its frequency response. A completely analogous proof can be given for continuous-time systems.


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