$y(t)=y(t+12), y(t) = x(t) \ast h(t)$

yt The continuous time signal output $y(t)$ is a periodic square wave, 50% duty cycle pulse.

The impulse response is a box function.($A = 1, T = 2$) ht

By using Fourier series and Fourier transform, I thought I can guess the input $x(t)$, assuming the input is also periodic with the period 12 as well.

$$ w_0=\frac{2\pi}{12}\\ x(t) = x(t+12) = \sum_{k=-\infty}^{\infty} a_k e^{jkw_0t}, a_0 = \frac{1}{T} \int_{0}^{T} x(t)dt, a_k = \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}} x(t) e^{-jkw_0t}dt \\ y(t) = y(t+12) = \sum_{k=-\infty}^{\infty} b_k e^{jkw_0t}, b_0 = \frac{1}{T} \int_{0}^{T} y(t)dt, b_k = \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}} y(t) e^{-jkw_0t}dt \\ $$

By using the Fourier transform, $$ X(jw) = 2\pi \sum_{k=-\infty}^{\infty} a_k \delta (w-kw_0)\\ Y(jw) = 2\pi \sum_{k=-\infty}^{\infty} b_k \delta (w-kw_0)\\ H(jw)=2sinc(w) \\ X(jw) = Y(jw)/H(jw) $$

I was stuck at this point; I thought I can write an equation about these coefficients, but the sinc function was a hassle for me.

  • How should I write an equation which is related to $a_k, b_k$?
  • Also, in order to guess the input, do I have to use the Fourier Transform like this? Or is there a characteristic of the convolution for periodic/box functions?

I wasn't sure whether my approach was correct to find inputs with the given output and the impulse response. Thanks for your time.

  • 1
    $\begingroup$ i corrected the other post, but David, you should know that $w$ is not $\omega$. $\endgroup$ – robert bristow-johnson May 30 '19 at 1:40

The output signal of an LTI system is periodic only if its input signal is periodic as well, your reasoning was right. Now, a signal periodic with a certain period $T_p$ has a line spectrum, i.e., it only contains frequency contents at $n f_0$, $n \in \mathbb{Z}$ where $f_0 = \frac{1}{T_p}$. What it basically means is that your input only excites these frequencies so that to obtain the input output relation, it only matters how the LTI system responds to these particular frequencies. You can therefore consider the impulse response of your LTI system periodic as well, it changes nothing.

That said, you can express the input-output relation entirely using Fourier series. If your input signal $x(t)$ has Fourier coefficients $C_{x}[\mu]$, your output signal $y(t)$ has Fourier coefficients $C_{y}[\mu]$ and your system is characterized by an impulse response $g(t)$ with Fourier coefficients $C_g[\mu]$ then, the input output relation is

$$y(t) = x(t) * g(t) \quad \Leftrightarrow \quad C_y[\mu] = C_g[\mu] \cdot C_x[\mu],$$

where * represents circular convolution. Therefore, if you are given $y(t)$ and $g(t)$, you can find $x(t)$ via:

  • Obtaining the Fourier coefficients $C_y[\mu]$ of $y(t)$
  • Obtaining the Fourier coefficients $C_g[\mu]$ of the (periodified) $g(t)$
  • Obtaining the Fourier coefficients $C_x[\mu] = C_y[\mu]/C_g[\mu]$
  • Finding $x(t)$ via its Fourier series expansion using $C_x[\mu]$.

Let me elaborate shortly on the difference between Fourier transforms and Fourier series: The Fourier transform takes a continuous time signal $g(t)$ and returns its amplitude density $G(\omega)$ for every $\omega$. You find it via $$G(\omega) = \int_{-\infty}^\infty g(t) {\rm e}^{-\jmath \omega t} {\rm d}t.$$ On the other hand, the Fourier series is defined for periodic functions only. It takes a periodic time signal and returns coefficients belonging to the integer multiples of its base frequency. You find the Fourier coefficients via $$C_g[\mu] = \frac{1}{t_p} \int_{t_p} g(t) {\rm e}^{-\jmath 2\pi \mu t/t_p} {\rm d}t.$$ if $g(t)$ is $t_p$ periodic. These two definitions may look very similar. Besides normalization it may seem that $C_g[\mu]$ is the same as $G(\omega)$ at $\omega = \mu \cdot \frac{2\pi}{t_p}$. However, this is only true if $g(t)$ is either actually periodic or at least time-limited to the interval $t_p$ you are considering (which is true in your case!) since in this case, periodifying it does not change it at all. If it is not limited to $t_p$, the Fourier series integral limited to the interval $t_p$ would truncate the function which would lead to wrong results. Therefore, one would have to periodify $g(t)$ first in this case to get correct results.

  • $\begingroup$ Wow, thanks! Is the meaning of periodified 𝑔(𝑡)'s coefficient equivalent to the Fourier transform result? How can I make my non-periodic impulse response to a periodic function? Also, can I ask the reason why "impulse response of your LTI system periodic as well"? $\endgroup$ – David Badger May 29 '19 at 8:59
  • $\begingroup$ I edited my reply to answer your questions. $\endgroup$ – Florian May 29 '19 at 9:10
  • $\begingroup$ This is a great answer Florian. $\endgroup$ – David Jun 3 '19 at 4:23
  • $\begingroup$ Thanks! Good to hear it's helpful. $\endgroup$ – Florian Jun 3 '19 at 13:20

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