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When we shift the same function first, we get <math>(t - 5) - 12\!</math>. After we put that through the system, we are left with <math>2(t - 5) - 24\!</math>, which is, once again, the same as <math>2t - 34\!</math>. Thus, the two orders of operations give the same result, which means the system is time invariant. | When we shift the same function first, we get <math>(t - 5) - 12\!</math>. After we put that through the system, we are left with <math>2(t - 5) - 24\!</math>, which is, once again, the same as <math>2t - 34\!</math>. Thus, the two orders of operations give the same result, which means the system is time invariant. | ||
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| + | ==Time variant== | ||
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| + | A system would be considered time variant if it did not follow the above criteria. An example of such a system would be as follows: | ||
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| + | <math>y(t) = tx(t)\!</math> | ||
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| + | Let's assume that <math>x(t)\!</math> is the same as it was in the previous example, <math>t - 12\!</math>. If we put it through the system first, we get <math>t^2 - 12t\!</math>. After a time shift of <math>5\!</math>, we get <math>(t - 5)^2 - 12(t - 5)\!</math>. | ||
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| + | If we shift it first, we get <math>t - 5 - 12\!</math>. We then put it through the system, and we get <math>t^2 - 5t - 12t\!</math>. Obviously, the two answers are not the same, and thus, the system is time variant. | ||
Latest revision as of 13:57, 11 September 2008
Time Invariance
A system is considered time-invariant if the following two orders of operations performed on a function $ x(t)\! $ yield the same result:
1. The function is put through the system, and then, the function is shifted in time.
2. The function undergoes a time shift, and then, the function goes through the system.
An example of a time invariant system is as follows:
$ y(t) = 2x(t)\! $
The proof for this is rather simple. Suppose $ x(t) = t - 12\! $. After going through the system, we are left with $ 2t - 24\! $. After a time shift of, let's say $ 5\! $, we are left with $ 2(t - 5) - 24\! $, which is the same as $ 2t - 34\! $.
When we shift the same function first, we get $ (t - 5) - 12\! $. After we put that through the system, we are left with $ 2(t - 5) - 24\! $, which is, once again, the same as $ 2t - 34\! $. Thus, the two orders of operations give the same result, which means the system is time invariant.
Time variant
A system would be considered time variant if it did not follow the above criteria. An example of such a system would be as follows:
$ y(t) = tx(t)\! $
Let's assume that $ x(t)\! $ is the same as it was in the previous example, $ t - 12\! $. If we put it through the system first, we get $ t^2 - 12t\! $. After a time shift of $ 5\! $, we get $ (t - 5)^2 - 12(t - 5)\! $.
If we shift it first, we get $ t - 5 - 12\! $. We then put it through the system, and we get $ t^2 - 5t - 12t\! $. Obviously, the two answers are not the same, and thus, the system is time variant.
