# Checking for time invariance

## Homework Statement

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

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

## 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 $$\tau$$

$$y(t) = T[x(t)]$$

$$z(t) = x(t-\tau)$$

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

## 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.

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

$$y(t-\tau) = \int_0^\infty e^{-\sigma}z(t - \tau - \sigma) d\sigma$$

therefore $$T[x(t-\tau)] = y(t-\tau)$$ , 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 $$x(t-\sigma)$$ with $$x(t)$$

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

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## Answers and Replies

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 $$x(t-\sigma)$$ with $$x(t)$$

Since x(t) = |t|, then $x(t - \sigma) = |t - \sigma|$. So, all you have to do is evaluate:

$$y(t) = \int_0^\infty e^{-\sigma} |t - \sigma| \, d\sigma$$