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Consider the function <math>{e^{-4t}}</math>,
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Consider the function <math>f(t) = e^{2\pi j}</math>  
  
ω0 = -4
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In order to determine if the function is periodic or not we need to use the formula:
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<math>\omega0/(2\pi) = k/N</math>
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where in this case the function is periodic because <math>\omega0 = 2\pi</math> so k and N can be any number such that k = N and k/N is the rational number 1.
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For the function to be non-periodic, if <math>\omega0 = \sqrt{2}\pi</math> then the function would be non-periodic because k/N would be equal to a irrational number.
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The Periodic Signal is shown below:
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[[Image:figure1_ECE301Fall2008mboutin.jpg]]
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And here is the non-periodic signal:
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[[Image:figure2_ECE301Fall2008mboutin.jpg]]

Latest revision as of 14:16, 12 September 2008

Consider the function $ f(t) = e^{2\pi j} $

In order to determine if the function is periodic or not we need to use the formula:

$ \omega0/(2\pi) = k/N $

where in this case the function is periodic because $ \omega0 = 2\pi $ so k and N can be any number such that k = N and k/N is the rational number 1.

For the function to be non-periodic, if $ \omega0 = \sqrt{2}\pi $ then the function would be non-periodic because k/N would be equal to a irrational number.

The Periodic Signal is shown below:

Figure1 ECE301Fall2008mboutin.jpg

And here is the non-periodic signal:

Figure2 ECE301Fall2008mboutin.jpg

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Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010