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Output signal y(t) can be <math>10e^t</math> by system<br> | Output signal y(t) can be <math>10e^t</math> by system<br> | ||
Prove.<br> | Prove.<br> | ||
| − | 1. <math>e^t</math> is changed to <math>e^{t-t0}</math> by time delay.<br> | + | 1. <math>e^t</math> is changed to <math>e^{(t-t0)}</math> by time delay.<br> |
| − | <math>e^(t-t0) -> 10e^(t-t0)</math> by system.<br> | + | <math>e^{(t-t0)} -> 10e^{(t-t0)}</math> by system.<br> |
2. <math>e^t -> 10e^t </math> by system.<br> | 2. <math>e^t -> 10e^t </math> by system.<br> | ||
| − | <math>10e^t -> 10e^{t-t0}</math><br> | + | <math>10e^t -> 10e^{(t-t0)}</math><br> |
The output signals are same. Then we can say that the system is time-invariant.<br> | The output signals are same. Then we can say that the system is time-invariant.<br> | ||
Revision as of 13:59, 9 September 2008
A time-invariant system
For any input signal x(t), a system yelids y(t). Now, suppose input signal shifted t0, x(t-t0). Then output signal also shifted t0, y(t-t0). Then we can say a system is time-invariant.
Example of a tume-invariant system
x(t) = $ e^t $
Output signal y(t) can be $ 10e^t $ by system
Prove.
1. $ e^t $ is changed to $ e^{(t-t0)} $ by time delay.
$ e^{(t-t0)} -> 10e^{(t-t0)} $ by system.
2. $ e^t -> 10e^t $ by system.
$ 10e^t -> 10e^{(t-t0)} $
The output signals are same. Then we can say that the system is time-invariant.
