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| − | I chose Kathleen Schremser's CT signal | + | I chose Kathleen Schremser's CT signal y(t)=sin(3/4*t). |
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| + | In the given signal: | ||
| + | |||
| + | frequency=w=3/4 | ||
| + | |||
| + | In order for a CT signal to be periodic in DT the frequency divided by 2*pi must be a rational number. | ||
| + | |||
| + | k is any integer | ||
| + | |||
| + | therefore: | ||
| + | |||
| + | <math>w/(2*pi)=k/N | ||
| + | |||
| + | (3/4)/(2*pi)=0.119366207318922 | ||
| + | </math> | ||
| + | |||
| + | Since this is not a rational number the signal is not periodic in DT. | ||
| + | |||
| + | |||
| + | |||
| + | If we change the signal to y(t) = sin(2*pi*(3/4)*t) then; | ||
| + | |||
| + | frequency=w=3/4 * 2*pi | ||
| + | |||
| + | <math>w/(2*pi)=3/4(/math> | ||
| + | |||
| + | Since this is a rational number the signal is periodic in DT | ||
Revision as of 14:38, 10 September 2008
I chose Kathleen Schremser's CT signal y(t)=sin(3/4*t).
In the given signal:
frequency=w=3/4
In order for a CT signal to be periodic in DT the frequency divided by 2*pi must be a rational number.
k is any integer
therefore:
$ w/(2*pi)=k/N (3/4)/(2*pi)=0.119366207318922 $
Since this is not a rational number the signal is not periodic in DT.
If we change the signal to y(t) = sin(2*pi*(3/4)*t) then;
frequency=w=3/4 * 2*pi
$ w/(2*pi)=3/4(/math> Since this is a rational number the signal is periodic in DT $
