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| − | I'll use the signal that I picked for the last homework to demonstrate the sampling rate idea. My signal was tan(t). If you sample this function at a rate of | + | I'll use the signal that I picked for the last homework to demonstrate the sampling rate idea. My signal was tan(t). If you sample this function at a rate of <math>\pi</math>, every sample will be identical, as long as it's not shifted by |
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| − | <math>pi</math>, every sample will be identical, as long as it's not shifted by | + | |
<math>\frac{\pi}{2}</math> as <math>\tan(\frac{\pi}{2}+n*\pi)</math> for any integer n is undefined. | <math>\frac{\pi}{2}</math> as <math>\tan(\frac{\pi}{2}+n*\pi)</math> for any integer n is undefined. | ||
Revision as of 14:06, 11 September 2008
Part 1
I'll use the signal that I picked for the last homework to demonstrate the sampling rate idea. My signal was tan(t). If you sample this function at a rate of $ \pi $, every sample will be identical, as long as it's not shifted by
$ \frac{\pi}{2} $ as $ \tan(\frac{\pi}{2}+n*\pi) $ for any integer n is undefined.
However, if you sample this function with a period of anything OTHER than $ \pi $ then you get random dots all over the place.
Part 2
I picked a pretty easy function for a non-periodic one for homework 1, so I'll use it again! :) I chose $ y=x $ According to the definition,
$ y(t+k*T) $
should be periodic for any k and constant T, so lets see. We get a sum that looks about like
$ (t-5) + (t-4) + (t-3) + (t-2) + (t-1) + t + (t+1) + (t+2) + (t+3) + ... $ for T=1 and k=...-5,-4,-3,-2,-1,0,1,2,3,4... When you shift this by 1 you get $ (t-4) + (t-3) + (t-2) + (t-1) + (t) + (t+1) + (t+2) + (t+3) + (t+4) $ As you can see if this sum were taken to infinity, the shift of 1 would result in the exact same signal, thus it would be periodic.
