Is This System Time-Invariant and How Does It Respond to Absolute Time Inputs?

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SUMMARY

The system defined by the input-output relationship y(t) = ∫₀^∞ e^(-σ)x(t-σ) dσ is confirmed to be time-invariant. This conclusion is drawn from the transformation properties where T[x(t-τ)] = y(t-τ) holds true. For part B of the problem, the input x(t) = |t| leads to the evaluation of y(t) as y(t) = ∫₀^∞ e^(-σ)|t-σ| dσ, which requires substituting x(t-σ) with |t-σ|.

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  • Understanding of integral calculus, specifically improper integrals.
  • Familiarity with the concept of time invariance in systems theory.
  • Knowledge of Laplace transforms and their applications in system analysis.
  • Basic proficiency in manipulating absolute value functions within integrals.
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teknodude
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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:
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teknodude said:
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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