| (2 intermediate revisions by the same user not shown) | |||
| Line 1: | Line 1: | ||
| − | A '''time invariant''' system depends on the order of cascades to return an output with a given input. In order to check for time invariance, a time delay must be used in two cases. In one case, the time delay must | + | A '''time invariant''' system depends on the order of cascades to return an output with a given input. In order to check for time invariance, a time delay must be used in two cases. In one case, the time delay must precede the signal being sent through the system. In the other, the time delay must come after the signal being sent through the system. If the returned output is equal in both cases, the system is time invariant. |
== Example: A time invariant system == | == Example: A time invariant system == | ||
| Line 25: | Line 25: | ||
<math>z_2(t)=\pi^2 x(t-t_0)-\pi (t-t_0)</math><br> | <math>z_2(t)=\pi^2 x(t-t_0)-\pi (t-t_0)</math><br> | ||
| − | <math>z_1</math and <math>z_2</math> are not equal; therefore, the system is not time-invariant. | + | <math>z_1</math> and <math>z_2</math> are not equal; therefore, the system is not time-invariant. |
Latest revision as of 07:00, 12 September 2008
A time invariant system depends on the order of cascades to return an output with a given input. In order to check for time invariance, a time delay must be used in two cases. In one case, the time delay must precede the signal being sent through the system. In the other, the time delay must come after the signal being sent through the system. If the returned output is equal in both cases, the system is time invariant.
Example: A time invariant system
System: $ y(t)=\pi^2 x(t) $
Time-delay, then system:
$ y(t)=x(t-t_0) $
$ z_1(t)=\pi^2 x(t-t_0) $
System, then time-delay:
$ y(t)=\pi^2 x(t) $
$ z_2(t)=\pi^2 x(t-t_0) $
$ z_1 $ and $ z_2 $ are equal; therefore, the system is time-invariant.
Example: A time variant system
System: $ y(t)=\pi^2 x(t)-\pi t $
Time-delay, then system:
$ y(t)=x(t-t_0) $
$ z_1(t)=\pi^2 x(t-t_0)-\pi t $
System, then time-delay:
$ y(t)=\pi^2 x(t)-\pi t $
$ z_2(t)=\pi^2 x(t-t_0)-\pi (t-t_0) $
$ z_1 $ and $ z_2 $ are not equal; therefore, the system is not time-invariant.
