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How do I prove that $y(t) = x(t)x(t-1)$ is a non linear system?

I tried the following proof but it seems not to getting the desired effect. Most likely I solve it wrongly.

Superposition property:

$$y(t)=T[ x(t) ] = T[ X_1(t) + X_2(t) ] = X_1(t)X_1(t-1) + X_2(t)X_2(t-1)=y_1(t)+y_2(t)$$

Scaling property:

$$y(t)=T[ c x(t) ] = c T [x(t)]$$

Lorenzo B.
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DontAskTheEye
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1 Answers1

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Simply note that

$$y(t)=T[c\cdot x(t)]=c\cdot x(t)\cdot c \cdot x(t-1)=c^2\cdot x(t)\cdot x(t-1)=c^2\cdot T[x(t)]$$

user
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  • i do comprehend why your reply is correct :( Let me give you a similar example and how i reason it: let $y(t) = x(t) + x(t-1). $, then $cT[x(t)]=cx(t)+cx(t-1)$, and...$T[cx(t)]=cx(t)+cx(t-1)$ – DontAskTheEye Mar 01 '18 at 19:38
  • @DontAskTheEye Yes in this case scaling property holds and also superposition, how you can verify, thus it is linear. – user Mar 01 '18 at 19:43
  • i try to mimic these steps and i cannot see why you end up having $c^2$ – DontAskTheEye Mar 01 '18 at 19:45
  • @DontAskTheEye Those are completely different cases, in the OP you have multiplication and the transformation is $T[x]=x(t)\cdot x(t-1)$ then $T[c\cdot x(t)]=(c\cdot x(t))\cdot (c\cdot x(t-1))=c^2 \cdot (x(t)\cdot x(t-1))=c^2 \cdot T[x(t)]$. Note that c goes in both x functions. – user Mar 01 '18 at 19:50
  • Thus, if i have a multiplication i multiply c constant with each input x and if i have addition i do not. Does not really make sense, i have the definition of a linear system which is not helping at all to prove that $c^2$ holds true. $y(t)=T[ax_1(t)+bx_2(t)] = aT[x_1(t)]+b[x_2(t)] = ay_1(t)+by_2(t).$ confused :( – DontAskTheEye Mar 01 '18 at 20:08
  • Also for addition you put $c$ in both term exactly how you did here, but $c$ don't multiply each other, can you see the difference? with the sum the system il linear with product it is non linear. – user Mar 01 '18 at 20:09