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I am trying to find an example of a continuous system that is homogeneous, but not additive. So far, the only example I could find was an example from this page, which describes system as $y(t) = \overline{x(t)}$, where (as I understand it) $\overline{a}$ is the conjugate of a complex number $a$.

This, however, will work for a scaling constant $a$ that is complex, but not real, and I am wondering if it is possible to find a system $S$ which is additive, but not homogeneous, and this holds even when we scale the input signal with a real constant?

I have been thinking a lot about it and I could not come with an example, nor could I prove it does not exist

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    $\begingroup$ Can you define additive and homogeneous systems? I'm not familiar with them. $\endgroup$ – MBaz Dec 7 '14 at 23:43
  • $\begingroup$ Of course. Additivity means that for input $x_{3}(t) = x_{1}(t) + x_{2}(t)$, we get output $y_{3}(t) = y_{1}(t) + y_{2}(t)$ where $y_{1}(t)$ is the output for the input signal $x_{1}(t)$, and $y_{2}(t)$ is the output for the input signal $x_{2}(t)$. Homogeneity means that for input $ax_{1}(t)$ we get output $ay_{1}(t)$. I hope it's clear now what I am asking. Thanks! $\endgroup$ – Momonga Dec 7 '14 at 23:46
  • $\begingroup$ According to this answer, a system that is additive is also linear, and in consequence it's also homogeneous. Assuming that answer is correct (I haven't verified it myself, but I tend to believe it is), then the answer to your question is no, there are no additive systems that are not homogeneous. $\endgroup$ – MBaz Dec 8 '14 at 0:02
  • $\begingroup$ Hm, that is quite confusing, because why would we be talking about linear systems when we could just be talking about additive systems, it kind of confuses me. I am also not quite sure how that posts shows that additive system means linear system, it's not obvious to me. Also, as I wrote, there is an example of a homogeneous, but not additive system, though the property holds only for the complex factors, and I would like to know about the real factors. Thanks for comments! $\endgroup$ – Momonga Dec 8 '14 at 0:10
  • $\begingroup$ Indeed, the answer I linked to does not prove it, just mentions it as a fact. But it says that it is true for complex scalars, which your link proves otherwise. However, it is obviously true for integer real scalars. Just like you, I've been unable to find either a proof or a counterexample with arbitrary reals. Sorry I can't be of more help! $\endgroup$ – MBaz Dec 8 '14 at 0:38
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You are misinterpreting the issues entirely.

A system with complex inputs and complex outputs that maps its input $x(t)$ to output $[x(t)]^*$, the complex conjugate of $x(t)$, is additive, is not homogeneous for complex scalars and is homogeneous for real scalars.

Consider the system with complex inputs and complex outputs that maps its input to its complex conjugate. Thus, it maps $$x_1(t) \to [x_1(t)]^*, \quad x_2(t)\to [x_2(t)]^*.$$ So, what happens if the input is the sum $x_1(t)+x_2(t)$? well, the rule is that the input gets mapped to its complex conjugate and so $$x_1(t) + x_2(t) \to [x_1(t) + x_2(t)]^* = [x_1(t)]^* + [x_2(t)]^*.$$ Since this holds for arbitrary choice of $x_1(t)$ and $x_2(t)$, we conclude that the system is indeed additive: the response to a sum of signals is the sum of the responses.

Let $\alpha$ denote a complex number. Then, this "complex conjugator" system maps $$\alpha\cdot x(t) \to [\alpha\cdot x(t)]^* = [\alpha]^*\cdot [x(t)]^* \neq \alpha\cdot [x(t)]^*$$ unless $\alpha$ happens to be a real number. So, the system is not homogeneous since the required condition does not hold for arbitrary (complex) $\alpha$. However, it is true that if $\alpha$ is restricted to be a real number, then the scaling property (a.k.a. homogeneity) does hold.


Addendum: The OP wants

a system that maps strictly real inputs to strictly real outputs, and that cannot be scaled with a complex (only with real) constants, and that is additive, but not homogeneous. That's my initial question,....

Since the input and output are required to be real, and not complex, the notion of "cannot be scaled with a complex constant" does not make sense any more because the scaling of the input makes it a complex-valued signal, and the system allows only for a real-valued input. So essentially the question boils down to

"Does there exist a system with real-valued inputs and outputs that has the additive property but not the homogeneity property?"

to which the answer is No, as pointed out in the Wikipedia article on linear systems. Most of the ideas there are reproduced in the dsp.SE answer mentioned in a comment by MBaz but the dsp.SE answer goes on to make the additional claim (not in Wikipedia) that

"not much else is needed to extend to $a \in \mathbb C$"

which is false. Indeed, as MattL notes carefully in a comment, MattL's answers are applicable to scaling by real numbers only.

An outline of the proof that for real-valued systems, additivity implies homogeneity is as follows:

Consider a system with real-valued inputs and outputs that maps any given $x$ to $A(x)$. The function $A(\cdot)$ has the property that $A(x_1+x_2) = A(x_1)+A(x_2)$ and the system is called additive since the response to the sum $x_1+x_2$ is the sum of the responses $A(x_1)$ and $A(x_2)$. Mathematical induction extends this to $$A\left(\sum_{i=1}^n x_i \right) = \sum_{i=1}^n A(x_i).\tag{1}$$

  • Choosing all the $x_i$ in $(1)$ to have value $0$, we see that $A(0) = nA(0)$ and so $A(0)=0$.

  • Choosing all the $x_i$ in $(1)$ to be $x$, we see that $A(nx)=nA(x)$ and so the scaling property holds for positive integers $n$.

  • Choosing $n=2$ and $x_2 = -x_1$ in $(1)$, we have that $A(x_1+(-x_1)) = A(x_1)+A(-x_1)$. But $x_1+(-x_1)=0$ and so we see that $A(0) = 0 = A(x_1)+A(-x_1)$, that is, $$A(-x_1)=A((-1)x_1) = -A(x_1) =(-1)A(x_1).$$ So the scaling property holds for $-1$, and by mathematical induction for all negative integers as well.

  • Scaling holds for rational numbers $\displaystyle \frac 1n$ as well. Choose $\displaystyle x_i = \frac 1n x$ in $(1)$ to get $$A(x)=nA\left(\frac 1n x\right) \Rightarrow A\left(\frac 1n x\right) = \frac 1n A(x).$$ Mathematical induction allows us to extend this to more general rational numbers $\displaystyle \frac mn$ as well.

  • Finally, to show that scaling applies to arbitrary (not necessarily rational) real numbers, Wikipedia says that we need to assume that the mapping is continuous and use the fact the rational numbers are a dense subset of the reals.

So, except for this last part which requires some mathematics beyond high school, we can show straightforwardly that additivity implies homogeneity.

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  • $\begingroup$ Thanks for the answer! However, I was wondering if we can come up with a system that maps strictly real inputs to strictly real outputs, and that cannot be scaled with a complex (only with real) constants, and that is additive, but not homogeneous. That's my initial question, let me know if that makes sense to you. Thanks. $\endgroup$ – Momonga Dec 8 '14 at 15:48
  • $\begingroup$ @user2007674 See the edited answer. $\endgroup$ – Dilip Sarwate Dec 9 '14 at 14:00
  • $\begingroup$ I think this is what was I looking for, thanks! :) $\endgroup$ – Momonga Dec 9 '14 at 17:37

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