(Removing all content from page) |
|||
(One intermediate revision by the same user not shown) | |||
Line 1: | Line 1: | ||
+ | A time invariant system is a system for which when a signal passes through a system and then is time shifted, it is equal to when the signal is time shifted and then passed through the system. | ||
+ | |||
+ | == Example of Time Invariant == | ||
+ | |||
+ | |||
+ | System --> sqrt[of signal] = Time invariant | ||
+ | |||
+ | In other words, | ||
+ | |||
+ | X(t) is input signal | ||
+ | |||
+ | X(t) --> system --> y(t) = sqrt[X(t)] --> delay --> z(t) = sqrt[X(t - t0)] | ||
+ | |||
+ | is equivalent to | ||
+ | |||
+ | X(t) --> delay --> Y(t) = X(t - t0) --> system --> z(t) = sqrt[X(t - t0)] | ||
+ | |||
+ | Therefore it is Time invariant. | ||
+ | |||
+ | |||
+ | |||
+ | -Note: if i am wrong about this example, let me know. Thanks. | ||
+ | |||
+ | |||
+ | == Example of not Time Invariant == | ||
+ | |||
+ | x(t) --> system --> x(2t) | ||
+ | |||
+ | X(t) --> system --> y(t) = x(2t) --> delay --> z(t) = x(2t - t0) | ||
+ | |||
+ | X(t) --> delay --> y(t) = x(t - t0) --> system --> z(t) = x(2*(t-t0)) = x(2t - 2t0) | ||
+ | |||
+ | These two outputs are not equal, so it is not a time invariant system. |
Latest revision as of 13:20, 11 September 2008
A time invariant system is a system for which when a signal passes through a system and then is time shifted, it is equal to when the signal is time shifted and then passed through the system.
Example of Time Invariant
System --> sqrt[of signal] = Time invariant
In other words,
X(t) is input signal
X(t) --> system --> y(t) = sqrt[X(t)] --> delay --> z(t) = sqrt[X(t - t0)]
is equivalent to
X(t) --> delay --> Y(t) = X(t - t0) --> system --> z(t) = sqrt[X(t - t0)]
Therefore it is Time invariant.
-Note: if i am wrong about this example, let me know. Thanks.
Example of not Time Invariant
x(t) --> system --> x(2t)
X(t) --> system --> y(t) = x(2t) --> delay --> z(t) = x(2t - t0)
X(t) --> delay --> y(t) = x(t - t0) --> system --> z(t) = x(2*(t-t0)) = x(2t - 2t0)
These two outputs are not equal, so it is not a time invariant system.