Week of 11/5

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{{mathematica|filename=Sincinterpolation.nb|title=Sinc function interpolation via the samping theorem}}
 
{{mathematica|filename=Sincinterpolation.nb|title=Sinc function interpolation via the samping theorem}}
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Mathematica can  find the fourier transform of a box function of width h centered on zero times  <math>sin(2 \pi x)</math>:
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<math>-\frac{i \sqrt{\frac{2}{\pi }} \left(2 \pi  \cos (h \pi ) \sin \left(\frac{h k}{2}\right)-k \cos \left(\frac{h k}{2}\right) \sin (h \pi )\right)}{4 \pi ^2-k^2}
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</math>

Revision as of 22:56, 5 November 2007

Here is the display of my oscilloscope when the input is a 1000 pulse per second output of a time-code generator. (The time-code generator is a device that locks to the 10 MHz output of an atomic clock and produces 1 Hz or 1 KHz pulse trains as well as human readable time synchronized to the atomic standard.)

Squarewave.png



I imported the data and plotted it along with its periodogram.


Sqwavexmgr.png


Notice that only the odd harmonics are present.

Here is a mathematica notebook that simulates this.

Mathematica.png Download lots of fourier transform examples




Sampling theorem. See 10/31/07 lecture notes


We end up with 
f(t) = \sum _ {n = - \infty} ^ \infty f(n/2f_s) \frac{\sin(\pi (2 f_s t -n))}{\pi (2 f_s t -n)}


This amounts to taking samples of the data every 1 / 2fs and multiplying them by a sinc function and adding up the results.

Mathematica.png Download Sinc function interpolation via the samping theorem



Mathematica can find the fourier transform of a box function of width h centered on zero times sin(2πx):

-\frac{i \sqrt{\frac{2}{\pi }} \left(2 \pi  \cos (h \pi ) \sin \left(\frac{h k}{2}\right)-k \cos \left(\frac{h k}{2}\right) \sin (h \pi )\right)}{4 \pi ^2-k^2}

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