(Periodic Signal)
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==1. Creating two DT signals (one periodic and one non-periodic) from a periodic CT signal==
 
==1. Creating two DT signals (one periodic and one non-periodic) from a periodic CT signal==
  
Let x(t) = sin (t), which is a periodic CT signal
+
Let <math>x(t) = sin (2\pi t),</math> which is a periodic CT signal
  
'''x(t) = sin (t)'''
+
<math>x(t) = sin (2\pi t)</math>
 
[[Image:Sin1_ECE301Fall2008mboutin.jpg]]
 
[[Image:Sin1_ECE301Fall2008mboutin.jpg]]
  

Revision as of 09:26, 11 September 2008

1. Creating two DT signals (one periodic and one non-periodic) from a periodic CT signal

Let $ x(t) = sin (2\pi t), $ which is a periodic CT signal

$ x(t) = sin (2\pi t) $ Sin1 ECE301Fall2008mboutin.jpg


Sampling every t = 0.01 Samp0 ECE301Fall2008mboutin.jpg


Periodic Signal

Sampling every $ t = \pi $ Samp pi ECE301Fall2008mboutin.jpg

This discrete time signal was produced from a CT sine wave by sampling at a frequency of $ \frac{1}{\pi} $.

As can be seen from the graph, the values of x[n] are periodic because they repeat after every period of $ t = 2\pi $.

Therefore, $ x[n + 2\pi] = x[n] $

However, this still does not fulfill the requirement as $ N = 2\pi $ is not an integer. For the signal to become periodic, the CT waveform has to be modified to $ x(t) = sin(2\pi t) $ and sampled at a frequency of 1 Hz.

Non Periodic Signal

Sampling every t = 2 Samp2 ECE301Fall2008mboutin.jpg

For this discrete time signal which was produced by sampling the same sine wave at a frequency of 0.5, the values of x[n] are non-periodic because the discrete time signal is scattered all over the place with no indication of a pattern. Therefore, $ x[n + k] \neq x[n] $

2. Create a periodic signal by summing shifted copies of a non-periodic signal

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