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Let's say a system is additive, so that inputting $x_1(n) + x_2(n) \rightarrow y_1(n) + y_2(n)$. Can we show that this system is also subtractive without assuming that it is homogeneous? (Homogeneous in this case means that if $x_1(n) \rightarrow y_1(n)$ then $c \cdot x_1(n) \rightarrow c \cdot y_1(n)$ for any real or complex $c$). So can we show that $x_1(n) - x_2(n) \rightarrow y_1(n) - y_2(n)$ ?

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  • $\begingroup$ What is x1(n) ? $\endgroup$
    – miracle173
    Oct 9, 2017 at 5:48
  • $\begingroup$ @miracle173 x1(n) is an input signal. y1(n) is the output signal. The arrow represents the "black box" system. $\endgroup$
    – ItM
    Oct 9, 2017 at 5:52
  • $\begingroup$ no questions to my answer? $\endgroup$
    – miracle173
    Oct 9, 2017 at 19:08

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Yes, we can. But it is hard to show this using your notation. So I leave it to you to transform it in your notation.

We want to show that if $f$ is addititve, so $$ f(x_1+x_2)=f(x_1)+f(x_2)$$ then $$f(x_1-x_2)=f(x_1)-f(x_2)$$ Proof:

$$0=f(0)=f(x-x)=f(x+(-x))=f(x)+f(-x)$$ Now we subtract $f(x)$ from this equation $$0=f(x)+f(-x)\;\bigg\rvert-f(x)$$ and get $$-f(x)=f(-x).$$ So we have $$f(x_1-x_2)=f(x_1+(-x_2))=f(x_1)+f(-x_2)=f(x_1)-f(x_2)$$

But we assumed here that $f(0)=0$. Is this always true?

Yes, this follows from $$f(0)=f(0+0)=f(0)+f(0).$$ We subtract $f(0)$ from this equation $$f(0)=f(0)+f(0)\;\bigg\rvert -f(0)$$

and get $$0=f(0).$$

Note

From additivity follows

$$f(cx)=cf(x),\;\forall c \in \mathbb{Q},$$ but not multiplicativity in general.

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  • $\begingroup$ Two things: I don't see why you say "Therefore f(-x)=-f(x)". Also, in what you've described, x1=x and x2=-x. But we want to show that for all x1 and x2. Also if f(0)=0 you've assumed the system is homogeneous. I'd appreciate if you could clear these questions up a bit. Thanks! $\endgroup$
    – ItM
    Oct 9, 2017 at 20:42
  • $\begingroup$ No, I didn't assume that the system is homogeneous. I proved that f(0)=0 follows from additivity in the second part of the post. I edited the post. please read it again, $\endgroup$
    – miracle173
    Oct 10, 2017 at 5:11
  • $\begingroup$ I see. The confusion rises from the fact that we can't really subtract from both sides (one side you apply the signal, one side you receive the output signal). But if we do subtract like that, then yes, makes sense. I think that's good enough for what I was looking for. Thanks! $\endgroup$
    – ItM
    Oct 10, 2017 at 5:16
  • $\begingroup$ So on your Note you mention that from additivity follows homogeneity. That's not true however. There are systems that are additive but not homogeneous. $\endgroup$
    – ItM
    Oct 10, 2017 at 5:30
  • $\begingroup$ you didn't read my note carefully. I explicitly stated that must not be homogeneous, we always have $f(cx)=cf(x)$ for all rational c but that does not mean that this holds for all real c. $\endgroup$
    – miracle173
    Oct 10, 2017 at 7:18

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