(Part A)
(Part A)
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The plot on the left shows the same signal but in DT (y[n] = sin(n)).  The sampling frequency is at <math>\pi/2</math> (y[n] = sin(<math>(\pi/2)</math>*n)).  The signal repeats itself every 4 sec so that y[n] = y[n+4].  No = k * <math>\frac{2\pi}{\pi/2}</math>  ==>  No = k * 4  so that the signal repeats itself every 4 seconds.
 
The plot on the left shows the same signal but in DT (y[n] = sin(n)).  The sampling frequency is at <math>\pi/2</math> (y[n] = sin(<math>(\pi/2)</math>*n)).  The signal repeats itself every 4 sec so that y[n] = y[n+4].  No = k * <math>\frac{2\pi}{\pi/2}</math>  ==>  No = k * 4  so that the signal repeats itself every 4 seconds.
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The right picture is taken at a frequency of 10000 Hz so that y[n] = sin(10000 * n).  As you can see the signal is not periodic because the value of No is not an integer.  <math>No = k * \frac{2\pi}{10000}  =  \frac{\pi}{5000}</math> and this is not a rational integer.  Therefore the signal in DT is not periodic.
  
 
[[Image:sahw1a1_ECE301Fall2008mboutin.jpg]]  [[Image:sahw1a3_ECE301Fall2008mboutin.jpg]]
 
[[Image:sahw1a1_ECE301Fall2008mboutin.jpg]]  [[Image:sahw1a3_ECE301Fall2008mboutin.jpg]]

Revision as of 18:07, 10 September 2008

Part A

The original signal shown in the first plot is y(t) = sin(t) with a period of $ 2\pi $

Sahw1a2 ECE301Fall2008mboutin.jpg

The plot on the left shows the same signal but in DT (y[n] = sin(n)). The sampling frequency is at $ \pi/2 $ (y[n] = sin($ (\pi/2) $*n)). The signal repeats itself every 4 sec so that y[n] = y[n+4]. No = k * $ \frac{2\pi}{\pi/2} $ ==> No = k * 4 so that the signal repeats itself every 4 seconds.

The right picture is taken at a frequency of 10000 Hz so that y[n] = sin(10000 * n). As you can see the signal is not periodic because the value of No is not an integer. $ No = k * \frac{2\pi}{10000} = \frac{\pi}{5000} $ and this is not a rational integer. Therefore the signal in DT is not periodic.

Sahw1a1 ECE301Fall2008mboutin.jpg Sahw1a3 ECE301Fall2008mboutin.jpg

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