I am working on an exercise in binary transmission systems. The pulses are modeled using a special sinc-function in the time-domain, $f_0$ is the bitrate but just a constant in time domain:

$s(t) = f_0\operatorname{sinc} (\pi f_0(t- \frac 2 f_0))^2$

My gut feeling tells me this can be simplified, is the next deduction valid?

$s(t) = f_0\operatorname{sinc}(\pi f_0(t- \frac 2 f_0))^2 = f_0\operatorname{sinc}(\pi f_0 t -2\pi))^2 = f_0 \operatorname{sinc}(\pi f_0 t)^2$

Or does somebody see a better trick? I think this should be possible because I have to transform this using a Fourier transform and this would give me a triangular function symmetrical around the $y$ axis.

  • $\begingroup$ No, not valid - sinc is not translation invariant. $\endgroup$ – Dirk Aug 16 '15 at 16:27
  • $\begingroup$ You need to group f_0 by inclosing it in curly braces ({}); otherwise the \frac extends only up to the f and the _0 is interpreted as a subscript for the fraction as a whole (as you can see in the result). $\endgroup$ – joriki Aug 16 '15 at 17:04

What kind of trick are you looking for?

First, when you discarded the $2\pi$, you assumed that the sinc (squared) function is periodic with period $2\pi$; it is not periodic and so you have thrown away valuable information.

If you want to Fourier transform this function, it is fine as you originally wrote it. There are two important things to notice. First, it has a Fourier transform that has a magnitude that is symmetric about $f=0$ in the frequency domain; that is what you would get no matter what the time-domain delay is, which in your function is $2/f_{0}$. But that delay causes the phase to be a linear function of frequency, modulo $2\pi$ of course.

If you want a "trick," leave $s\left(t\right)$ as is, but make a kind of a temporary function, say $r\left(t\right)=f_0 \text{sinc}\left(\pi f_0 t\right)^2$ and compute or look up the Fourier transform of that (which is the triangle function as you mention), then recognize that $s\left(t\right)=r\left(t-2/f_{0}\right)$ and apply the shift theorem to your Fourier transform result for $r\left(t\right)$, that is, $S\left(f\right)=R\left(f\right)\exp\left(-j\omega2/f_{0}\right)$


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