Revision as of 12:55, 9 September 2008 by Park1 (Talk)

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 -> 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.

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Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett