# Impulse Response of a circuit

The following periodic function, $x(t)$, is the input to a linear, time-invariant system The impulse response of this system is $$h(t) = \frac{\sin t}{t}$$ which I converted to $$H(\mathrm{i}\omega) = \pi\,\mathrm{rect}\bigl(\frac{\omega}{2}\bigr).$$ I want to determine an explicit expression for the output $y(t)$. I have determined that the period $T_0=3$ and $\omega_0= 2\pi/3$. I know the equations for the fourier series transform is $$x(t) = \sum_{n=-\infty}^\infty C_n \mathrm{e}^{2\mathrm{i}nt}$$ and similarly I have $$y(t) = \sum_{n=-\infty}^\infty D_n \mathrm{e}^{2\mathrm{i}nt}$$ where $D_n = H(\mathrm{i}n\omega_0)C_n$ but I don't know how to proceed.

• We all had to do this when we were students, so it's homework which is OK as long as you show us what you tried and explain why you're stuck. You should find the answer or a method to find it in most textbooks about the subject. Commented Mar 16, 2016 at 8:25
• Wait, i did show my work here, why am I still being downvoted? @FakeMoustache
– user91567
Commented Mar 16, 2016 at 8:30
• It wasn't there when I saw it ! I'll remove the downvote. Commented Mar 16, 2016 at 8:39

Note that sin(t)/t is a function called a "sinc". Convolving with a sinc yields some very special properties. These are best to consider in the frequency domain. Look up something called a "dual".

You should go thru the math once, but a sinc and its dual is something you should remember outright. Again, go look this up. These are worth knowing without having to do the calculations each time.

In this case, convolution by a sinc yields a special frequency characteristic. However, due to how duals work, this other function in the time domain results in a sinc in frequency domain. Again, this is one pair really worth knowing.

• I don't think that he needs to bother with duals right now. They're very advanced signal processing category, while he (as I can see from this and other his posts) is in layman classes. I say he can solve it simply by multiplication in frequency domain. Commented Mar 16, 2016 at 12:49
• @Rok: A dual is just a concept, which he's at the threshold of now. It also seems he is doing the Fourier transform math already. Commented Mar 16, 2016 at 14:01

First, redo your math - FT of sin(t)/t is not π rect(ω/2). It's 1/sqrt(2π)*rect(ω/2π).

Okay, so your LTI system is a low pass system with cut-off frequency 2π rad/s, right? So, all of the frequencies higher than 2π rad/s of your input shall be removed at the output. Your input signal base frequency is 2π/3 rad/s, first harmonic is 4π/3 rad/s and 2nd harmonic 6π/3 rad/s. All higher frequency components shall be removed.

Now, you need to know that for frequencies 2π/3 and 4π/3 rad/s the filter attenuation is 1/sqrt(2π) while for the 6π/3 rad/s it is 0.5/sqrt(2π). What you need to do is to find the Fourier coeffs for the these three components and multiply them with these factors and ofc include phase for each of them. The output is a sum of all of them.

y = 1/sqrt(2π) * C1 sin(2π/3 t + Phi1) + 1/sqrt(2π) * C2 sin(4π/3 t + Phi2) + 0.5/sqrt(2π) * C3 sin(2π/3 t + Phi2).