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[[Image:Samp2_ECE301Fall2008mboutin.jpg]] | [[Image: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 <math>x[n + k] \neq x[n]</math> | + | 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, <math>x[n + k] \neq x[n]</math> | ||
==2. Create a periodic signal by summing shifted copies of a non-periodic signal== | ==2. Create a periodic signal by summing shifted copies of a non-periodic signal== | ||
Revision as of 08:03, 11 September 2008
Contents
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
x(t) = sin (t)
Sampling every t = 0.01
Periodic Signal
Sampling every $ t = \pi $
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] $
Non Periodic Signal
Sampling every t = 2
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] $
