(→Part 1) |
(→Part 1) |
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| Line 24: | Line 24: | ||
| − | This can be seen in the following plot | + | This can be seen in the following plot: |
[[Image:Jkubasci dt periodic_ECE301Fall2008mboutin.jpg]] | [[Image:Jkubasci dt periodic_ECE301Fall2008mboutin.jpg]] | ||
| Line 34: | Line 34: | ||
<math>\,z[n]=x[\frac{n}{2\pi}]=2cos(n)\,</math> | <math>\,z[n]=x[\frac{n}{2\pi}]=2cos(n)\,</math> | ||
| − | + | is not periodic in DT, since there is no integer <math>N\in \mathbb{Z}, N>0</math> such that | |
<math>\,z[n]=z[n+N], \forall n\in \mathbb{Z}\,</math> | <math>\,z[n]=z[n+N], \forall n\in \mathbb{Z}\,</math> | ||
| Line 42: | Line 42: | ||
This would be true when | This would be true when | ||
| − | <math>\,N=2\pi k, k\in \mathbb{Z}\,</math> | + | <math>\,N=2\pi k, k\in \mathbb{Z}, k\not= 0\,</math> |
| − | This can be seen in the following plot | + | <math>\,\frac{N}{k}=2\pi \,</math> |
| + | |||
| + | but, <math>2\pi</math> is irrational, thus there are no values for the integers <math>\,N,k\,</math> that will satisfy this equation. | ||
| + | |||
| + | This can be seen in the following plot: | ||
[[Image:Jkubasci dt nonperiodic_ECE301Fall2008mboutin.jpg]] | [[Image:Jkubasci dt nonperiodic_ECE301Fall2008mboutin.jpg]] | ||
== Part 2 == | == Part 2 == | ||
Revision as of 16:43, 11 September 2008
Part 1
The function was chosen at random from HW1: HW1.4 Hang Zhang - Periodic vs Non-period Functions_ECE301Fall2008mboutin
$ \,x(t)=2cos(2\pi t)\, $
Periodic Signal in DT:
If $ x(t) $ is sampled at $ period=0.1 $, the function
$ \,y[n]=x[0.1n]=2cos(\frac{2\pi n}{10})\, $
is periodic, since
$ \,\exists N\in \mathbb{Z}\, $ such that $ \,y[n]=y[n+N], \forall n\in \mathbb{Z}\, $
$ \,2cos(\frac{2\pi n}{10})=2cos(\frac{2\pi n}{10}+\frac{2\pi N}{10})\, $
This is true when
$ \,\frac{2\pi N}{10}=2\pi \, $
$ \,N=10\, $
This can be seen in the following plot:
Non-Periodic Signal in DT:
However, if $ x(t) $ is sampled at $ period=1/2\pi $, the function
$ \,z[n]=x[\frac{n}{2\pi}]=2cos(n)\, $
is not periodic in DT, since there is no integer $ N\in \mathbb{Z}, N>0 $ such that
$ \,z[n]=z[n+N], \forall n\in \mathbb{Z}\, $
$ \,2cos(n)=2cos(n+N)\, $
This would be true when
$ \,N=2\pi k, k\in \mathbb{Z}, k\not= 0\, $
$ \,\frac{N}{k}=2\pi \, $
but, $ 2\pi $ is irrational, thus there are no values for the integers $ \,N,k\, $ that will satisfy this equation.
This can be seen in the following plot:
