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I picked a pretty easy function for a non-periodic one for homework 1, so I'll use it again! :) I chose <math>y=x</math> | I picked a pretty easy function for a non-periodic one for homework 1, so I'll use it again! :) I chose <math>y=x</math> | ||
| − | According to the definition, <math>y(t+k*T)</math> should be periodic for any k and constant T, so lets see. | + | According to the definition, |
| + | |||
| + | <math>y(t+k*T)</math> | ||
| + | |||
| + | should be periodic for any k and constant T, so lets see. | ||
We get a sum that looks about like | We get a sum that looks about like | ||
| − | <math>t + (t+1) + (t+2) + (t+3) + ... </math> for T=1 and k=0,1,2,3,4... | + | |
| + | <math>(t-5) + (t-4) + (t-3) + (t-2) + (t-1) + t + (t+1) + (t+2) + (t+3) + ... </math> for T=1 and k=...-5,-4,-3,-2,-1,0,1,2,3,4... | ||
| + | When you shift this by 1 you get | ||
| + | <math>(t-4) + (t-3) + (t-2) + (t-1) + (t) + (t+1) + (t+2) + (t+3) + (t+4) </math> | ||
| + | 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. | ||
Revision as of 14:05, 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.
