(New page: ==Definition: Time Invariance== A system is time invariant if any input that is first shifted and then put through the system yields the same result as putting the signal through the syste...) |
|||
| (7 intermediate revisions by the same user not shown) | |||
| Line 5: | Line 5: | ||
<math>y[n] = e^{x[n]}</math> | <math>y[n] = e^{x[n]}</math> | ||
| − | |||
| − | + | Let | |
| − | <math> | + | <math>z[n] = x[n+k]</math> |
| + | |||
| + | <math>a = n + k</math> | ||
| + | |||
| + | |||
| + | <math>y[z[n]]</math> =?= <math>y[a]</math> | ||
| + | |||
| + | <math>e^{z[n]}</math> =?= <math>e^{x[a]}</math> | ||
<math>e^{x[n+k]} = e^{x[n+k]}</math> | <math>e^{x[n+k]} = e^{x[n+k]}</math> | ||
| + | |||
| + | |||
| + | Therefore, the system is time-invariant. | ||
| + | |||
| + | ==Example 2: Time-Variant System== | ||
| + | <math>y[n] = sin(x[2n])</math> | ||
| + | |||
| + | |||
| + | Let | ||
| + | |||
| + | <math>z[n] = x[n+k]</math> | ||
| + | |||
| + | <math>a = n + k</math> | ||
| + | |||
| + | |||
| + | <math>y[z[n]]</math> =?= <math>y[a]</math> | ||
| + | |||
| + | <math>sin(z[2n])</math> =?= <math>sin(x[2a])</math> | ||
| + | |||
| + | <math>sin(x[2n+k])</math> =?= <math>sin(x[2(n+k)])</math> | ||
| + | |||
| + | <math>sin(x[2n+k]) \ne sin(x[2n + 2k])</math> | ||
| + | |||
| + | |||
| + | Therefore, the system is time-variant. | ||
Latest revision as of 19:53, 8 September 2008
Definition: Time Invariance
A system is time invariant if any input that is first shifted and then put through the system yields the same result as putting the signal through the system first and then shifting the output, provided the magnitude of the shift is the same in both instances.
Example 1: Time-Invariant System
$ y[n] = e^{x[n]} $
Let
$ z[n] = x[n+k] $
$ a = n + k $
$ y[z[n]] $ =?= $ y[a] $
$ e^{z[n]} $ =?= $ e^{x[a]} $
$ e^{x[n+k]} = e^{x[n+k]} $
Therefore, the system is time-invariant.
Example 2: Time-Variant System
$ y[n] = sin(x[2n]) $
Let
$ z[n] = x[n+k] $
$ a = n + k $
$ y[z[n]] $ =?= $ y[a] $
$ sin(z[2n]) $ =?= $ sin(x[2a]) $
$ sin(x[2n+k]) $ =?= $ sin(x[2(n+k)]) $
$ sin(x[2n+k]) \ne sin(x[2n + 2k]) $
Therefore, the system is time-variant.
