Finding LTI (Linear Time-Invariant Systems)

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SUMMARY

The discussion focuses on determining the linearity and time-invariance of three specific systems defined by integrals. All three systems, represented as \(y(t) = \int_{-\infty}^t x(\tau)d\tau\), \(y(t) = \int_{0}^t x(\tau)d\tau\), and \(y(t) = \int_{t-1}^{t+1} x(\tau)d\tau\), are confirmed to be linear due to the scaling and additive properties of integrals. Additionally, they are classified as time-invariant since a delay in the input results in an equivalent delay in the output, as demonstrated in the analysis of the first integral.

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Homework Statement



Tell whether following systems are i. linear and ii. time-invariant:

$$y(t) = \int_{-\infty}^t x(\tau)d\tau$$
$$y(t) = \int_{0}^t x(\tau)d\tau$$
$$y(t) = \int_{t-1}^{t+1} x(\tau)d\tau$$​

Homework Equations


N/A

The Attempt at a Solution



I'm a little thrown off by the integrals but here's my best explanation, all 3 of them are linear, why? Because the constants can be pulled out of the integrals thus fulfilling the scaling property that y(t) = x(t) and Cy(x) = C(x); similarly the concept of an integral inherently proves the additive property necessary for linearity.

As for time-invariance, I'm also inclined to say all of them are time-invarient, why? If we take a look at the systems ( once they've already been integrated we can see this - example of the first integral):

$$y(t) = C(x_{integrated}(t) - x_{integrated}(-\infty))$$​

It seems to me that this clearly shows that if you delay t by 1 your output will also be delayed by one.

Let me know if I have any errors in my thought process or how I went about this, LTI systems are still a little difficult for me to "reason" through, although I am very comfortable with what they stand for the analysis is just...magic.
 
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