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According to the book: signals and systems by HWEI P. HSU, chapter 1, A system is linear if it is:

1) additive, $T[x_1 + x_2] =y_1+y_2$

2) homogeneous, $T[cx] =cy$ , c = scalar

Question 1:

I will use the methodology described in the book: signals and systems edition 2 by Oppenheim, chapter 1, to prove why y(t)=x(t)x(t−1) is not linear. I would like to request a feedback , is my proof correct or wrong? If wrong, can you show me?If it is correct(!), is there a better way to prove it?

$x_1(t)->y_1(t) = x_1(t)\cdot x_1(t-1)\\ x_2(t)->y_2(t) = x_2(t)\cdot x_2(t-1)\\ Let \,\,\, x_3(t) = x_1(t)+ x_2(t)\\ x_3(t)->y_3(t) = x_3(t)\cdot x_3(t-1),below\,we\,rewriting\, the\,same...\\ y_3(t) = T[x_3(t)] = x_3(t)\cdot x_3(t-1),\,below\, we\, begin\,replacing...\\ y_3(t) =\overbrace{\big[x_1(t)\cdot x_1(t-1)+x_2(t)\cdot x_2(t-1)\big]}^{x_3(t)}\cdot \overbrace{\big[x_1(t-1)\cdot x_1(t-2)+x_2(t-1)\cdot x_2(t-2)\big]}^{x_3(t-1)} $

This result does not obey the additive property because of t-2, it will never be the same as: $y_3(t) = y_1(t)+ y_2(t) = x_1(t)\cdot x_1(t-1) + x_2(t)\cdot x_2(t-1)$

Question 2: After going through the theory content included in the above two books(chapters 1) i conclude that the following proof of homogeneous property is wrong: $$y(t)=T[c\cdot x(t)]=c\cdot x(t)\cdot c \cdot x(t-1)=c^2\cdot x(t)\cdot x(t-1)=c^2\cdot T[x(t)]$$

Reason: I was not able to discover any single piece of theory from the books which would clearly say that we can multiply constant c as shown above. The definition of a linear system does not even imply such a thing. I suspect that i can test the homogeneous property using the same proof as in question 1, i did not use a,b coefficients above there :

$$x_3(t) = ax_1(t)+b x_2(t)\\ y_3(t) =\overbrace{\big[ax_1(t)\cdot x_1(t-1)+bx_2(t)\cdot x_2(t-1)\big]}^{x_3(t)}\cdot \overbrace{\big[ax_1(t-1)\cdot x_1(t-2)+bx_2(t-1)\cdot x_2(t-2)\big]}^{x_3(t-1)}$$

i do not completely solve this, i suspect that if i do the distribution law and rearrange the terms it is not scalable. Am i right? If not can you show the theory and a solid proof i can (hopefully) understand?

DontAskTheEye
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  • I don't understand what you are asking in the second question. – copper.hat Mar 02 '18 at 15:59
  • A system is linear if 1) it's additive 2) $T[cx]=cT[x]$. Your system does not satisfy either property. But proving that 2) is not satisfied is much simpler. – Math Lover Mar 02 '18 at 16:01
  • @copper.hat question 2 i want to prove that it is NOT allowed to write: $y(t)=T[c⋅x(t)]=c⋅x(t)⋅c⋅x(t−1)$, that is wrong. Today,<>, I searched the two books i mention in the OP and i could not find any piece of info. Also, in question 2, i do not know if it is correct the method i use to prove that the system is not scalable. That is why i said 'i suspect', i would like someone to confirm that my solution for question 2 is going to produce the correct solution(system is not scalable) – DontAskTheEye Mar 02 '18 at 16:21
  • That is how the system is defined, I don't follow why you say it is not allowed. – copper.hat Mar 02 '18 at 16:33
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    Don't confuse the domain of $T$ with the domain of it's argument. – copper.hat Mar 02 '18 at 16:37
  • Looking the scaling definition, to my understanding : $T[c\cdot x_3(t)]$ , this has to be the same as $cT[ x_3(t)]$ in order for the system to be scalable. I showed in question 2 that it won't happen. That is how i end up to the conclusion that you can't write $c⋅x(t)⋅c⋅x(t−1) $ but you can write$,, c⋅x(t) ⋅x(t−1) $ – DontAskTheEye Mar 02 '18 at 16:40
  • @ copper.hat ''Don't confuse the domain of T with the domain of it's argument'' hmm, let me take an other example that has multiplication: $y(t)=x(t)cos(t)$, is this scalable? $$x_3(t)=x_1(t)+x_2(t)\ y(t)=T[c⋅x_3(t)] = [c(,x_1(t)+x_2(t)) ,]cos(t)\ c \cdot T[x_3(t)] =, same, as,T[c⋅x_3(t)],$$ that is what i understand. So, i conclude that you can't write $c\cdot x(t)\cdot c\cdot x(t-1),$ but you can write: $c\cdot x(t)\cdot x(t-1)$ – DontAskTheEye Mar 02 '18 at 16:56
  • If $(T_1 x)(t) = \cos t \cdot x(t)$, then, yes, this is a linear operator. The issue with the operator $(T_2 x)(t) = x(t) x(t-1)$ is exactly the same as the as why the function (from reals to reals) given by $x \mapsto x^2$ is not linear. The time shift is sort of irrelevant. – copper.hat Mar 02 '18 at 17:14
  • This is a notational nitpick of mine, but I would write $(T[c \cdot x_3])(t)$ rather than $T[c \cdot x_3(t)]$. Meaning compute $T[c \cdot x_3]$ and evaluate at $t$. An equivalent notation that shows the time dependence is $(T[\tau \mapsto c \cdot x_3(\tau)])(t)$, where I have changed symbols to avoid confusion. – copper.hat Mar 02 '18 at 17:18
  • @copper.hat thanks for the last 1 comment, i like your nitpicking notation $(T[c\cdot cx_3])(t)$. Can you edit your answer below('it is not clear what your domain is...') and complete my partial solution here in this comment?( homogeneous proof) :$$y(t)=x(t)x(t-1)\ x_3(t) = x_1(t) + x_2(t) \scalability, rule:c\cdot (T[x_3])(t), must,be,equal,,to:, (T[c\cdot x_3])(t). \c\cdot (T[x_3])(t) = c \cdot x_3(t)\cdot(x_3(t-1)),is,this,equal,to,(T[c\cdot x_3])(t)??$$ $$(T[c\cdot x_3])(t)=???,,the, part,i ,am,stuck/confuse:\ (T[c\cdot x_3])(t)= c\cdot(,x_3(t)\cdot x_3(t-1),)=???$$ – DontAskTheEye Mar 02 '18 at 18:00
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    Note that $(T[c \cdot x_3])(t) = (c \cdot x_3)(t) \cdot (c \cdot x_3)(t-1) = c^2 x_3(t) x_3(t-1)$. – copper.hat Mar 02 '18 at 18:02
  • ok, i believe you are right, but i was expecting to see: $c\cdot x_3(t) \cdot x_3(t-1)$!!! I'm a bit embarrassed but.... whatever, maybe i just don't get it :( – DontAskTheEye Mar 02 '18 at 18:10
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    Think of it as a purely symbolic operation. If $Tx = x(t) \cdot x(t-1)$, then $Tabc = (abc)(t)\cdot(abc)(t-1)$. Then expand using the usual rules, as in $(c \cdot x)(t) = c \cdot x(t)$, etc. Every where you see $x$, replace by $(c \cdot x)$ (in the formula for $T$, I mean). – copper.hat Mar 02 '18 at 18:18
  • $$(T[c\cdot x_3])(t)= c^2\cdot x_3(t)\cdot x_3(t-1) = \ c^2\cdot (x_1(t)+x_2(t),)\cdot (x_1(t-1)+x_2(t-1),)$$. Is this correct? If not,could you please tell the correct one? – DontAskTheEye Mar 02 '18 at 22:11

1 Answers1

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It is not clear what your domain is.

If $x$ is a constant function, or is (a non zero) constant on some interval $[t_0,t_0+1]$, then we see that $(Tx)(t_0+1) = x(t_0)^2 = T (-x)(t_0+1)$, hence this violates the $T (\lambda x) = \lambda Tx$ requirement for linearity.

copper.hat
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  • concerning the domain you say, please look at the oppenheim's book towards the end of chapter 1 second edition,it has 4 solved problems,it does not mention any domain,all these kinds of questions are theoretical. If you do not have access to the book,try google docs, or other pdf.... – DontAskTheEye Mar 02 '18 at 16:31