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Given two inputs $\: x_1[n]\: x_2[n]\:$Is the system $\:y[n]=x_1[n]\times x_2[n]\:$ linear ?

My Approach:

$(x_1\times x_2)[n]=S_1[n]\rightarrow Y_1[n]$

$(x_3\times x_4)[n]=S_2[n]\rightarrow Y_2[n]$

$(ax_1\times x_2)[n]+(b x_3\times x_4)[n]= a S_1[n]+ b S_2 [n] \rightarrow a Y_1[n]+bY_2[n]$

If that is correct it is linear. But I am not sure about this solution. I will appreciate any help.

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closed as unclear what you're asking by Dilip Sarwate, A_A, lennon310, MBaz, AlexTP Apr 28 '18 at 14:37

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Is there a comma missing in $\: x_1[n]\: x_2[n]\:$???? What is the meaning of the symbol $\times$? Is it multiplication? If so, how is $(x_1\times x_2)[n]$ related to $\: x_1[n]\: x_2[n]\:$ $\endgroup$ – Dilip Sarwate Apr 19 '18 at 16:03
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To be linear you must have a function $f$ that satisfy:

$ f(a\overline {x_1} + b \overline{x_2}) = af(\overline{x_1})+ b f(\overline{x_2}) $

I define $ \overline{x}=\left [\begin{matrix}\overline{x}(1) \\ \overline{x}(2) \end{matrix} \right] $ the vector of your two inputs at instant $n$

For each $n$ you have $ y=f(\overline x)=\overline{x}(1) \cdot \overline{x}(2) $

Substituting to linearity definition: $ f(a\overline {x_1} + b \overline{x_2}) = [a\overline{x_1}(1)+b\overline{x_2}(1)] \cdot [a\overline{x_1}(2)+b\overline{x_2}(2)] $

That is not equal to $ af(\overline{x_1})+ b f(\overline{x_2}) = a\overline{x_1}(1)\overline{x_1}(2)+ b\overline{x_2}(1)\overline{x_2}(2)$

It means that this system is not linear

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