# Output of LTI (in time and frequency $\omega$ domain) : when input goes through LPF

I would like to raise a mathematical question :

Let's say we are been given : $$x(t) = \begin{cases} \cos(\pi t) & |t| \leq 0.5 \\ 0 & \textrm{otherwise} \end{cases}$$ If $$x(t)$$ goes through an LPF : $$H(\omega) = \begin{cases} \pi - 0.5|\omega| & |\omega| \leq 1.5\pi\\ 0 & \textrm{otherwise} \end{cases}$$
Compute and sketch :

1. Fourier coefficients of $$x(t)$$, as well as $$X(\omega)$$
2. $$y(t)$$ and $$Y(\omega)$$ where $$y$$ is the output of the LTI. Is y(t) periodic? If so compute it's period and its trigonmetric fourier anlysis coefficients I did the math and I've got this: $$a_n = \frac{2}{\pi}\frac{\cos(\frac{n\pi}{2})}{1-n^2}$$ which can not be defined for $$n=-1$$ or $$n=1$$
update:* We have to compute $$a_1$$ and $$a_{-1}$$ in seperate: If we do the computations we gain : $$a_1= a_{-1}=0$$ I then did : $$c_m = \frac{1}{2}(a_m + jbm) \rightarrow c_m = \frac{a_m}{2}$$ $$x(t) = \sum_{m=-\infty}^{\infty}c_me^{jm \omega_o t} \rightarrow X(\omega) =2\pi \sum_{m=-\infty}^{\infty}c_m\delta(\omega - m\omega_o)$$

What I tried:
Now we know for an LTI : $$y(t) = x(t) * h(t) \rightarrow Y(\omega) = X(\omega)H(\omega)$$
but since $$H(\omega)$$ exists only in $$\delta=[-1.5\pi, 1.5\pi]$$ and $$\omega_o = \pi$$ then we only need $$X(\omega)$$ in $$\delta$$ .
But , $$X(\omega)$$ is $$0$$ for $$|m|=1 \rightarrow |\omega| =\pi$$ and therefore $$X(\omega) \neq 0 \rightarrow m=0$$ in $$\delta$$ so :
$$X(\omega) = 2\pi a_0 \delta(\omega)$$ in $$\delta$$

we also know $$\int_{-a}^{a}f(x)\delta(x)dx=f(0)$$ so:

\begin{align} y(t) &= \frac{1}{2\pi}\int_{-\infty}^{\infty}Y(\omega)e^{j\omega t}d\omega \rightarrow \int_{-1.5\pi}^{1.5\pi}\frac{1}{2\pi}Y(\omega)e^{j\omega t}d\omega \\ &\rightarrow \int_{-1.5\pi}^{1.5\pi}\frac{1}{2\pi}(\pi^2a_0 -0.5a_0|\omega|)e^{j\omega t}\delta(\omega)d\omega = \frac{1}{2\pi}\pi^2 a_0 = \frac{1}{2\pi}\pi^2 \frac{2}{\pi} = 1 \end{align}

I assume that it should have been periodic but it's not. So where did my solution go wrong?

• Homework ? From its definition, $x(t)$ (and thus $y(t)$) is not periodic, how come you compute its CTFCs? Dec 4 '20 at 10:56
• Are you sure that $X(\omega) = 2 \pi a_0 \delta(\omega)$? If you're using $\delta(\omega)$ to denote the Dirac delta functional, then that's saying that $x(t)$ is constant -- is that right? Dec 4 '20 at 19:31
• i am not saying it's $2\pi a_0\delta(\omega)$ for all $\omega$ but only in $\delta=[-1.5\pi, 1.5\pi]$ Dec 4 '20 at 20:31

4. You can model the input as $$x(t)$$ as cosine multiplied with a rectangular function.
• does the constant output like $y(t)$ count as periodic? I searched the stack exchnage and I assume that it is considered periodic ( since it satisfies the definition) but there is no fundumental period. Dec 4 '20 at 17:47