# system and properties of linearity,causality, time invariance, stability, invertibility

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?

• Your piecewise function has conflicting limits, the first condition and the last condition both include the point t = 1. – KillaKem Apr 5 '14 at 18:13

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.

• 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. – KillaKem Apr 5 '14 at 18:49
• 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. – Matt L. Apr 5 '14 at 18:55
• What makes you think $x(t)$ is an input? – KillaKem Apr 5 '14 at 18:58
• 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? – Matt L. Apr 5 '14 at 19:01
• I take $y(t)$ as being the impulse response and $x(t)$ as some unknown function of t. – KillaKem Apr 5 '14 at 19:07