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This should yield 4Y(t). Now, shift the function to the right by two units. The result is 4X(t-2). Since both of these | This should yield 4Y(t). Now, shift the function to the right by two units. The result is 4X(t-2). Since both of these | ||
procedures yielded the same end result, we can conclude that System A is time-invariant. | procedures yielded the same end result, we can conclude that System A is time-invariant. | ||
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If the input function is shifted by the same value T and put into System B we obtain the output function: 12(t)Y(t-2). Now, we will | If the input function is shifted by the same value T and put into System B we obtain the output function: 12(t)Y(t-2). Now, we will | ||
take the original input function and put it through System B. This gives us 12t*X(t). Now, we shift it 2 units to the right. | take the original input function and put it through System B. This gives us 12t*X(t). Now, we shift it 2 units to the right. | ||
This gives us 12(t-2)*X(t-2). These end results do not match. Therefore, we can say that System B is time-variant. | This gives us 12(t-2)*X(t-2). These end results do not match. Therefore, we can say that System B is time-variant. | ||
Revision as of 14:48, 11 September 2008
TIME INVARIANCE
DEFINITION
A system is defined as "time-invariant" when its output is not explicitly dependent on time (t). In other words, if one were to shift the input/output along the time axis, it would not effect the general form of the function.
METHOD
To check if a system is time-invariant, we can shift the function by a given value of T. Then, we send the function through the system and obtain an output. Now, take the same input function and put it into the system without shifting it first. Then take the output of the system and shift it by the value of T used previously. If these two processes yield the same results, then the system is called "time invariant."
SYSTEMS
A.) h1(t) = 2X1(t)
B.) h2(t) = 6t*X2(t)
Input function: P = 2Y(t)
Shift the input function on the time scale by 2 units to the right. This will give you: Pnew = 2Y(t-2). Now, put the function through System A. The system should output 4X(t-2). Now take the original function, P = 2y(t), and put it through the System A. This should yield 4Y(t). Now, shift the function to the right by two units. The result is 4X(t-2). Since both of these procedures yielded the same end result, we can conclude that System A is time-invariant.
If the input function is shifted by the same value T and put into System B we obtain the output function: 12(t)Y(t-2). Now, we will
take the original input function and put it through System B. This gives us 12t*X(t). Now, we shift it 2 units to the right.
This gives us 12(t-2)*X(t-2). These end results do not match. Therefore, we can say that System B is time-variant.
