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Therefore, <math>x[n + 2\pi] = x[n]</math> | Therefore, <math>x[n + 2\pi] = x[n]</math> | ||
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For this discrete time signal which was produced by sampling the same sine wave at a frequency of 1, 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 1, the values of x[n] are non-periodic because <math>x[n + k] \neq x[n]</math> | ||
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| + | ==2. Create a periodic signal by summing shifted copies of a non-periodic signal== | ||
Revision as of 09:00, 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 = 1
For this discrete time signal which was produced by sampling the same sine wave at a frequency of 1, the values of x[n] are non-periodic because $ x[n + k] \neq x[n] $
