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For a rectangular pulse that has as a fourier transform of a sinc function like so:

enter image description here

How can I estimate its sinc period in the frequency domain and compute the width of the rectangular pulse in the point-spread function?

I would like to do it in Matlab (if necessary), but I have no idea how to start off.

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Let's look at what is generally known:

Time domain is generally inversely related to frequency domain in as much as if you have a narrow width in the time domain the frequency domain spectrum will be wider. and vice versa. When you have narrower time domain pulses you need higher frequencies to represent it in the frequency domain therefore more bandwidth -> wider plot.

Looking at your sinc function to find when the function will go to zero you simply need to look at the sin aspect of it. At what values does sin = \$0\$ ? Answer \$ 0, \pi , 2\pi , 3\pi \$ ... That means the value \$ {\tau}f = 0, \pi.2\pi,3\pi ... \$

There is one exception here, as \$ {\tau}f\$ -> \$0\$ sinc ->\$1\$ becasue the sin and the \$\tau\$ cancel in the limit. so you only need to look at \$\pi,2\pi,3\pi, ...\$

Lets pick \$\tau=1\$ then the rect function will run from \$ -\frac{\tau}{2}\$ to \$ \frac{\tau}{2}\$ and the first zero in the frequency domain will be at f=1.

Lets pick \$\tau=2\$ then the rect function will run from \$ -\frac{\tau}{1}\$ to \$ \frac{\tau}{1}\$ and the first zero in the frequency domain will be at f=\$\frac{1}{2}\$.

a wider time pulse means a narrower frequency band.

Lets pick \$\tau=\frac{1}{2}\$ then the rect function will run from \$ -\frac{\tau}{4}\$ to \$ \frac{\tau}{4}\$ and the first zero in the frequency domain will be at f=\$\frac{2}{1}= 2\$.

a narrower time pulse means a wider frequency band.

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  • \$\begingroup\$ Thanks! Just one thing, how do you determine where the rect function will from and to? Like with \$\tau=2\$, how did you know that the rect function will run from time \$-\frac{\tau}{1}\$ to \$\frac{\tau}{1}\$? \$\endgroup\$ – xenon Mar 14 '13 at 15:53
  • \$\begingroup\$ From the definition ... substitute 2 into Tau in that defn. \$\endgroup\$ – placeholder Mar 14 '13 at 15:55
  • \$\begingroup\$ Pardon my slowness. I am very new to this. I have been trying to substitute but I'm confuse with how after substituting I still get Tau. Just in case I substitute into the wrong part of the definition, which part of the equation should I use to sub Tau into? \$\endgroup\$ – xenon Mar 14 '13 at 17:15

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