# Discrete Time Fourier Analysis

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?

• 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 – Damien Mar 19 at 14:45

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.