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==Part 1== | ==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 | + | 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 |
| − | <math>pi</math> | + | <math>\frac{\pi}{2}</math> |
| − | + | since | |
| + | |||
| + | <math>\tan(\frac{\pi}{2}+n*\pi)</math> for any integer n is undefined. | ||
However, if you sample this function with a period of anything OTHER than <math>\pi</math> then you get random dots all over the place. | However, if you sample this function with a period of anything OTHER than <math>\pi</math> then you get random dots all over the place. | ||
Latest revision as of 13:07, 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} $
since
$ \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.
