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Checking for time invariance

  • Thread starter teknodude
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  • #1
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Homework Statement



Consider the following input-output relationship:
[tex] y(t) = \int_0^\infty e^{-\sigma}x(t-\sigma) d\sigma [/tex]

A) Is the system time-invariant?
B) Find the output y(t) when the input to the system is [tex] x(t) = \mid t \mid , -\infty < t < \infty [/tex]

Homework Equations



These are the equations to check for time invariance

A system with an input-output transformation y(*) = T[x(*)] is time invariant if for any t and [tex] \tau [/tex]

[tex] y(t) = T[x(t)] [/tex]

[tex] z(t) = x(t-\tau)[/tex]

[tex] T[z(t)] = y(t-\tau)
[/tex]

The Attempt at a Solution



Edit: Ok I think it's time to get some sleep. This problem was actually pretty simple and I totally screwed it up earlier.

[tex] z(t) = T[x(t-\tau)] = \int_0^\infty e^{-\sigma}z(t - \tau - \sigma) d\sigma [/tex]

[tex] y(t-\tau) = \int_0^\infty e^{-\sigma}z(t - \tau - \sigma) d\sigma [/tex]

therefore [tex] T[x(t-\tau)] = y(t-\tau) [/tex] , so it is time-invariant

Now how do I atempt part B? I'm not seeing how the given input can be included in the given output function. My main thought is to just substitute [tex] x(t-\sigma) [/tex] with [tex] x(t) [/tex]

My other minor thought is maybe I should use the change of variable rule from calculus to the y(t) fxn.
 
Last edited:

Answers and Replies

  • #2
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Now how do I atempt part B? I'm not seeing how the given input can be included in the given output function. My main thought is to just substitute [tex] x(t-\sigma) [/tex] with [tex] x(t) [/tex]
Since x(t) = |t|, then [itex]x(t - \sigma) = |t - \sigma| [/itex]. So, all you have to do is evaluate:

[tex] y(t) = \int_0^\infty e^{-\sigma} |t - \sigma| \, d\sigma [/tex]
 

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