(New page: The problem on the test I most screwed up was this one: Is the system <math>y(t)=x(1-t)</math> Time Invaiant? The best way to solve this is to just go slow and if nessecary change a vari...) |
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The best way to solve this is to just go slow and if nessecary change a variable to make sure you get the expression correct | The best way to solve this is to just go slow and if nessecary change a variable to make sure you get the expression correct | ||
| − | x(t) -->|system|-->y(t)=x(1-t)-->|delay|-->y(t-t0)=x(1-t-t0) | + | x(t) -->|system|-->y(t)=x(1-t)-->|delay|-->y(t-t0)=x(1-(t-t0))=x(1-t+t0) |
| − | x(t) -->|delay|-->x(t-t0)-->|system|-->y(t-t0)=x(1- | + | x(t) -->|delay|-->x(t-t0)-->|system|-->y(t-t0)=x(1-t-t0) |
So since the outputs are not the same, the system is not Time Invariant. | So since the outputs are not the same, the system is not Time Invariant. | ||
Latest revision as of 17:51, 15 October 2008
The problem on the test I most screwed up was this one:
Is the system $ y(t)=x(1-t) $ Time Invaiant?
The best way to solve this is to just go slow and if nessecary change a variable to make sure you get the expression correct
x(t) -->|system|-->y(t)=x(1-t)-->|delay|-->y(t-t0)=x(1-(t-t0))=x(1-t+t0)
x(t) -->|delay|-->x(t-t0)-->|system|-->y(t-t0)=x(1-t-t0)
So since the outputs are not the same, the system is not Time Invariant.
