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Given the system: $$y(t)=x(t+1)+x(t−1)$$ is the system linear?

For a system to be a linear first it should satisfy zero input and zero output. How can we calculate output at 0 input if the system depends on future or past or both? Please explain with steps.

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    $\begingroup$ Don't use that property of linear systems, since non-linear systems can behave like that too. Rather, use the superposition principle. $\endgroup$ – MBaz Oct 16 '17 at 14:27
  • $\begingroup$ @MBaz The zero input, zero output condition (a special case of homogeneity) is one of the two things you need for superposition to hold. $\endgroup$ – Peter K. Oct 16 '17 at 14:38
  • $\begingroup$ Control books require zero state and zero input linearity. $\endgroup$ – Stanley Pawlukiewicz Oct 16 '17 at 14:48
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    $\begingroup$ @PeterK. I wanted to point out precisely that: it is a necessary, but not sufficient condition for linearity. But upon second reading, I see that I missed the "first" in "first it must satisfy...", so the OP probably already knows this. $\endgroup$ – MBaz Oct 16 '17 at 15:10
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How can we calculate output at 0 input if the system depends on future or past or both?

Well, surely zero input just means; $$ x(t) = 0~~~~\forall t $$ and the $\forall t$ means for all time: positive and negative.

Substituting that into the equation: $$ y(t) = x(t+1) + x(t-1) = 0 + 0 = 0 $$

So the system is homogeneous.

Well, as @Dilip points out, this isn't sufficient for homogeneity: we need the output to be $a y(t)$ for all inputs $a x(t)$. The case above just looks at $a=0$.

The next question is: does it satisfy additivity?

If $$ x_{\tt total}(t) = x_1(t) + x_2(t) $$ then $$ y_{\tt total}(t) = x_{\tt total}(t+1) + x_{\tt total}(t-1)\\ = x_1(t+1) + x_2(t+1) + x_1(t-1) + x_2(t-1)\\ = y_1(t) + y_2(t) $$ where $$ y_1(t) = x_1(t+1) + x_1(t-1)\\ y_2(t) = x_2(t+1) + x_2(t-1) $$ so it satisfies additivity also.

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  • $\begingroup$ How can we calculate output at 0 input if the system depends on future or past or both? Video scan or review of recorded data? $\endgroup$ – rrogers Oct 17 '17 at 17:58
  • $\begingroup$ @rrogers : That means the system is not causal (if it depends on the future). That's OK, theoretically. It's also OK practically if your independent variable isn't "time". As you suggest, in image processing the independent variable is $x$ or $y$ and in recorded data, you can play it forwards or backwards. It's obviously not OK practically for real-time systems. $\endgroup$ – Peter K. Oct 17 '17 at 18:31
  • $\begingroup$ Surely two different checks are unnecessary: just check whether the output $y_{\tt total}(t)$ corresponding to input $x_{\tt total}(t) = a x_1(t) + bx_2(t)$ equals $ay_1(t) + by_2(t)$ (which it does in this case) and you have proved linearity. $\endgroup$ – Dilip Sarwate Oct 18 '17 at 3:18
  • $\begingroup$ @DilipSarwate The OP pulled out homogeneity separately, so I answered in kind. You are, as usual, correct. :-) $\endgroup$ – Peter K. Oct 18 '17 at 12:10
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    $\begingroup$ Yes, but checking that zero input produces zero output is no guarantee of homogeneity. For example, a square-law device that maps $x(t)$ into $y(t) = x^2(t)$ is not homogeneous at all even though a zero input produces zero output. I wish you had not said $$ y(t) = x(t+1) + x(t-1) = 0 + 0 = 0 $$ So the system is homogeneous" because what has been checked does not prove homogeneity. Please consider an edit to point out to the OP that checking for zero output does not guarantee homogeneity. $\endgroup$ – Dilip Sarwate Oct 18 '17 at 20:28
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This is a linear descriptor system. Assuming that +1, and -1 refers to discrete time quantities then it is also a discrete time system. In state space you can represent it via

$$ \begin{align*} \begin{bmatrix}1 &0 &0\\0&1&0\\0&1&0\end{bmatrix} \begin{bmatrix}x(t)\\x(t+1)\\x(t+2)\end{bmatrix} &= \begin{bmatrix}0 &1 &0\\0&0&1\\0&0&1\end{bmatrix} \begin{bmatrix}x(t-1)\\x(t)\\x(t+1)\end{bmatrix} \\ y(t) &=\begin{bmatrix}1&0&1 \\\end{bmatrix}\begin{bmatrix}x(t-1)\\x(t)\\x(t+1)\end{bmatrix} \end{align*} $$

So this is basically $$ \begin{align*} Ex &= Ax\\ y&=Cx \end{align*} $$

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