(New page: When you see a system with LTI inside of it, it stands for Linear Time Invariant. It is easy to test to determine if a system is LTI or not. You simply put a function through a delay (z^...) |
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| − | + | ==Part A== | |
| + | Can the system be time invariant? | ||
| + | |||
| + | The amplitude of the output depends on the shift in time. | ||
| + | |||
| + | Comparing this simple statement to the definition of time invariant it is easy to identify this as a function that varies with time. The time directly influences what the output will look like, since it VARIES with time it can't be a TI system. | ||
| + | |||
| + | ==Part B== | ||
| + | The system does a phase shift to the right by 1 unit and then multiplies the amplitude of the function by the square of total shift. | ||
| + | |||
| + | if Y[n] = u[n-1] | ||
| + | |||
| + | then in the function | ||
| + | <math>\delta[n-(k+1)]</math> | ||
| + | k must be equal to 0 | ||
| + | |||
| + | we then multiply the amplitude by the square of the total shift (-1), which has no effect because it is the same as multiplying by 1. | ||
| + | |||
| + | We can finally solve and say that in order to produce Y[n], the input must be <math>x(t) = u(t)</math> | ||
Latest revision as of 13:08, 10 September 2008
Part A
Can the system be time invariant?
The amplitude of the output depends on the shift in time.
Comparing this simple statement to the definition of time invariant it is easy to identify this as a function that varies with time. The time directly influences what the output will look like, since it VARIES with time it can't be a TI system.
Part B
The system does a phase shift to the right by 1 unit and then multiplies the amplitude of the function by the square of total shift.
if Y[n] = u[n-1]
then in the function $ \delta[n-(k+1)] $ k must be equal to 0
we then multiply the amplitude by the square of the total shift (-1), which has no effect because it is the same as multiplying by 1.
We can finally solve and say that in order to produce Y[n], the input must be $ x(t) = u(t) $
