(New page: == Part 1 == == Part 2 ==)
 
(Part 2)
 
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== Part 1 ==
 
== Part 1 ==
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The function was chosen at random from HW1: [[HW1.4 Hang Zhang - Periodic vs Non-period Functions_ECE301Fall2008mboutin]]
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<math>\,x(t)=2cos(2\pi t)\,</math>
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'''Periodic Signal in DT:'''
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If <math>x(t)</math> is sampled at <math>period=0.1</math>, the function
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<math>\,y[n]=x[0.1n]=2cos(\frac{2\pi n}{10})\,</math>
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is periodic, since
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<math>\,\exists N\in \mathbb{Z}, N\not= 0\,</math> such that <math>\,y[n]=y[n+N], \forall n\in \mathbb{Z}\,</math>
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<math>\,2cos(\frac{2\pi n}{10})=2cos(\frac{2\pi n}{10}+\frac{2\pi N}{10})\,</math>
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This is true when
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<math>\,\frac{2\pi N}{10}=2\pi \,</math>
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<math>\,N=10\,</math>
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This can be seen in the following plot:
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[[Image:Jkubasci dt periodic_ECE301Fall2008mboutin.jpg]]
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'''Non-Periodic Signal in DT:'''
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However, if <math>x(t)</math> is sampled at <math>period=1/2\pi</math>, the function
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<math>\,z[n]=x[\frac{n}{2\pi}]=2cos(n)\,</math>
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is not periodic in DT, since there is no integer <math>N\in \mathbb{Z}, N\not= 0</math> such that
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<math>\,z[n]=z[n+N], \forall n\in \mathbb{Z}\,</math>
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<math>\,2cos(n)=2cos(n+N)\,</math>
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This would be true when
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<math>\,N=2\pi k, k\in \mathbb{Z}, k\not= 0\,</math>
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<math>\,\frac{N}{k}=2\pi \,</math>
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but, <math>2\pi</math> is irrational, thus there are no values for the integers <math>\,N,k\,</math> that will satisfy this equation.
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This can be seen in the following plot:
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[[Image:Jkubasci dt nonperiodic_ECE301Fall2008mboutin.jpg]]
  
 
== Part 2 ==
 
== Part 2 ==
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The non-periodic function was chosen from HW1: [[HW1.4 Ben Horst_ECE301Fall2008mboutin]].  It appears to be possible with this function since the function decays in the positive and negative x directions.
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<math>\,x(t)=\frac{sin(t)}{t}\,</math>
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The function could be made periodic by:
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<math>\,y(t)=\sum_{k\in \mathbb{Z}}x(t+2\pi k)=\sum_{k\in \mathbb{Z}}\frac{sin(t+2\pi k)}{t+2\pi k}\,</math>
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'''Proof:'''
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Need to show <math>\,\exists T\in \mathbb{R}, T\not= 0\,</math> such that
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<math>\,y(t)=y(t+T), \forall t\in \mathbb{R}\,</math>
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<math>\,\sum_{k\in \mathbb{Z}}\frac{sin(t+2\pi k)}{t+2\pi k}=\sum_{k\in \mathbb{Z}}\frac{sin(t+T+2\pi k)}{t+T+2\pi k}\,</math>
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<math>\,\sum_{k\in \mathbb{Z}}\frac{sin(t)}{t+2\pi k}=\sum_{k\in \mathbb{Z}}\frac{sin(t+T)}{t+T+2\pi k}\,</math>
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Choose <math>\,T=2\pi \,</math>
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<math>\,\sum_{k\in \mathbb{Z}}\frac{sin(t)}{t+2\pi k}=\sum_{k\in \mathbb{Z}}\frac{sin(t+2\pi )}{t+2\pi +2\pi k}\,</math>
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<math>\,\sum_{k\in \mathbb{Z}}\frac{sin(t)}{t+2\pi k}=\sum_{k\in \mathbb{Z}}\frac{sin(t)}{t+2\pi (k+1)}\,</math>
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With the infinite sum, there will be always be matching terms for the <math>k</math> and <math>k+1</math>, since we are covering all <math>k\in \mathbb{Z}</math> with the summation.  Therefore, the two are equal and <math>\,y(t)\,</math> is periodic.

Latest revision as of 05:47, 12 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}, N\not= 0\, $ 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:

Jkubasci dt periodic ECE301Fall2008mboutin.jpg

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\not= 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:

Jkubasci dt nonperiodic ECE301Fall2008mboutin.jpg

Part 2

The non-periodic function was chosen from HW1: HW1.4 Ben Horst_ECE301Fall2008mboutin. It appears to be possible with this function since the function decays in the positive and negative x directions.

$ \,x(t)=\frac{sin(t)}{t}\, $


The function could be made periodic by:

$ \,y(t)=\sum_{k\in \mathbb{Z}}x(t+2\pi k)=\sum_{k\in \mathbb{Z}}\frac{sin(t+2\pi k)}{t+2\pi k}\, $


Proof:

Need to show $ \,\exists T\in \mathbb{R}, T\not= 0\, $ such that

$ \,y(t)=y(t+T), \forall t\in \mathbb{R}\, $

$ \,\sum_{k\in \mathbb{Z}}\frac{sin(t+2\pi k)}{t+2\pi k}=\sum_{k\in \mathbb{Z}}\frac{sin(t+T+2\pi k)}{t+T+2\pi k}\, $

$ \,\sum_{k\in \mathbb{Z}}\frac{sin(t)}{t+2\pi k}=\sum_{k\in \mathbb{Z}}\frac{sin(t+T)}{t+T+2\pi k}\, $


Choose $ \,T=2\pi \, $

$ \,\sum_{k\in \mathbb{Z}}\frac{sin(t)}{t+2\pi k}=\sum_{k\in \mathbb{Z}}\frac{sin(t+2\pi )}{t+2\pi +2\pi k}\, $

$ \,\sum_{k\in \mathbb{Z}}\frac{sin(t)}{t+2\pi k}=\sum_{k\in \mathbb{Z}}\frac{sin(t)}{t+2\pi (k+1)}\, $


With the infinite sum, there will be always be matching terms for the $ k $ and $ k+1 $, since we are covering all $ k\in \mathbb{Z} $ with the summation. Therefore, the two are equal and $ \,y(t)\, $ is periodic.

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