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I have a system and I want to find which properties hold for this system

$$ y(t) = \left.\begin{cases} x(t) & t \geq 1 \\ \\ 0 & -1 < t < 1 \\ \\ -x(t) & t \leq -1 \end{cases} \right\} \\ \\ $$

is the system linear, time invariant, causal , invertible, stable?

My problem is that I don't know how to work, because the function is piecewise defined.If I try to prove for example that each piece of function is linear then the whole function is linear ? With this methodology I found that the system is linear, causal and stable.Am I right?

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  • \$\begingroup\$ Your piecewise function has conflicting limits, the first condition and the last condition both include the point t = 1. \$\endgroup\$ – KillaKem Apr 5 '14 at 18:13
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The system is linear, causal, stable, but not time-invariant.

It is linear because from the definition if \$y_1(t)\$ is the response to input \$x_1(t)\$, and \$y_2(t)\$ is the response to input \$x_2(t)\$, then the response to the input signal \$ax_1(t)+bx_2(t)\$ is given by

$$y(t)=ay_1(t)+by_2(t)$$

It is causal because in order to compute the current output signal, not future values of the input signal are necessary, i.e. to compute \$y(t_0)\$ we only need to know \$x(t_0)\$ (and not even its past values, i.e. the system has no memory).

The system is stable in the BIBO (bounded-input bounded-output sense), because any bounded input signal \$|x(t)|<K\$ produces a bounded output signal \$|y(t)|<L\$ (in our case \$K=L\$).

The system is time-variant because the response to a shifted version of the input signal is not equal to the shifted output signal, i.e. if \$y(t)\$ is the response to \$x(t)\$, then \$y(t-t_0)\$ is generally not the response to \$x(t-t_0)\$. You can see this by noting that the output is always zero for \$-1<t<1\$, no matter how the input signal is shifted. If the system were time-invariant, then also the zero portion of the output would need to be shifted with the input signal.

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  • \$\begingroup\$ A system is not causal if it is non-zero for negative values of t, so this system will only be causal in the trivial case of \$x(t) = 0\$, and if it is not causal then there will be no stability to talk of because the system will not even be realisable. \$\endgroup\$ – KillaKem Apr 5 '14 at 18:49
  • \$\begingroup\$ Please read my comment to your answer. I'm afraid your statements do not make much sense. Causality of a system does obviously not depend on its input signal, it is a property of the system. And of course there are non-causal stable systems. Please read up on some systems theory. \$\endgroup\$ – Matt L. Apr 5 '14 at 18:55
  • \$\begingroup\$ What makes you think \$x(t)\$ is an input? \$\endgroup\$ – KillaKem Apr 5 '14 at 18:58
  • \$\begingroup\$ Well, I interpret the system description as an input output relation with x(t) the input signal, and y(t) the output signal. What is your interpretation then? \$\endgroup\$ – Matt L. Apr 5 '14 at 19:01
  • \$\begingroup\$ I take \$y(t)\$ as being the impulse response and \$x(t)\$ as some unknown function of t. \$\endgroup\$ – KillaKem Apr 5 '14 at 19:07

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