Does scaling property imply superposition?

For a system to be linear,it follow the principles of scaling and superposition.Does scaling imply superposition?If so why are two different conditions given for linearity?If not can u specify an example for which only one of them is satisfied.

• Please have a look at this related question and its answer. I guess this should answer your question. If not, please edit your question to address the points that remain unclear. – Matt L. Jul 31 '15 at 11:47
• @MattL. I think the "related question" asked whether superposition (i.e. additivity) implied scaling (i.e. homogeneity) whereas this one is asking whether scaling implies superposition. The answer to the other question was Yes for systems with real-valued inputs and outputs (subject to continuity of the transformation and No for systems with complex-valued inputs and outputs. The answer to skt9's question is No as pointed out by Laurent Duval. – Dilip Sarwate Jul 31 '15 at 18:47

The answer is "no" in general. Let us see the details.

The notion of linearity goes beyond systems. Let us start from vector, or linear spaces. They deal with two objects: vectors and "scalars". In a vector space, one can add vectors, and multiply a vector by a scalar. More formally, a vector space over a field $F$ is a set $V$ together with two operations, vector addition and scalar multiplication. Those operations in $V$ satisfy a number of axioms. The field $F$ possesses two operations as well, satisfying different axioms, namely field axioms, ruling the addition or product of scalars. Those two operations are often assimilated to addition and multiplication. But in general the addition/multiplication in $F$ are not the same as the vector addition/vector scaling in $V$. Indeed, if $f \in F$, $v\in V$, then $f.v \in V$, but is not anymore an element of $F$. The scalar has been cast to a vector. This plays a role in the standard confusion in defining linearity.

Now take for instance the field of rationals $F=\mathbb{Q}$ with the two standard operations. Take $V=\mathbb{R}$ the vector space of reals over $F$ with the standard real addition and real product. To relate this to the introduction, for instance, $1.\sqrt{2}$ is not rational anymore, although the multiplication seems to be the same, because you mix a scalar ($1$) and a vector $(\sqrt{2})$.

Suppose you have a system $S$ that provides the following output: if $v_1\in V$ is rational, $S(v_1) = v_1$; if $v_2\in V$ is irrational, $S(v_2) = -v_2$. For any rational $q$, $q.v_1$ is rational, and $q.v_2$ is irrational. Resultingly, $S(q.v_1) = q.v_1 = q.S(v_1)$, and $S(q.v_2) = -q.v_2 = q.S(v_2)$. Hence, $s(q.v) = q.S(v)$ for any $v\in V$, so $S$ verifies the scaling property.

However, $v_1+v_2$ is irrational. Hence, $S(v_1+v_2) = -v_1 -v_2$, which is different in general from $S(v_1)+S(v_2) = v_1 - v_2$, take $v_1 = 1$ and $v_2 = \sqrt{2}$ for instance.

So scaling does not imply superposition (in your sense) in general.

But there exists somehow converse statements. In other domains, one sometimes calls the scaling "homogeneity", and with additivity we get the superposition principle for a system $S$: $S(q_1.v_1+q_2.v_2) = S(q_1.v_1)+S(q_2.v_2)$. Keeping with standard fields, it can be shown that additivity implies homogeneity with rational scalars, by playing on $S(v+v) = S(v)+S(v) = 2.S(v)$ etc. To go beyond rational scalar often requires additional assumptions like continuity, which could be troublesome depending on the sets you choose, see for instance Why doesn't superposition imply linearity? Why is homogeneity needed?

Let's try an example without division by zero problems and with some practical relevance. An RMS envelope estimator like $E\{x[n]\}=\sqrt{\sum_{k=n}^{n+N} x[n]^2}$ is an example of a system that is homogeneous for non-negative numbers while not being linear.

Clearly, we want a system that is homogeneous for all real numbers, and we can simply do that by replacing the square(-root) by a cube(-root) as in $$Q\{x[n]\}=\left( \sum_{k=n}^{n+N} x[n]^3 \right)^{1/3}$$

This is an estimator related to skewness in statistics, so it too has some practical relevance.

no. and to prove that, all you need is a counter example, one that satisfies scaling without satisfying superposition:

input $x(t)$ and output $y(t)$ and a system that satisfies scaling:

$$y(t) = \frac{x(t) \cdot x(t-\tau_1)}{x(t-\tau_2)}$$

where $\tau_1 \ne \tau_2$ and neither are zero.

pretty easy to show that scaling works with this and superposition does not.

but, if you're willing to neglect scaling by irrational numbers, you can show that superposition implies scaling. all you really need, to get to "linearity" is superposition.

scaling is really not a necessary axiom unless you're gonna be very anal about irrational numbers and cannot make any assumptions about continuity of the alleged LTI system.

• Does your system satisfy scaling only for those input signals that never take on the value $0$? Else, what if $t$ is such that $x(t-\tau_2) = 0$? What os the output $y(t)$ in this case? – Dilip Sarwate Aug 2 '15 at 21:05
• Then I don't understand the point of your answer. You suggest a system that does not satisfy the scaling law (even though you claim it does) and use this as an example of a system in which scaling holds but superposition doesn't. Duh! Then you claim that superposition can be used to prove scaling (which is a different question altogether, and one that has been asked and answered on dsp.SE already) with anatomical references about the completeness of this result. So how is your answer relevant to the question that has been asked here? – Dilip Sarwate Aug 3 '15 at 2:01
• "That's completely meaningless." indeed ... "Even if ... Following your argument, ... " i've only said two (or three) things: 1. $y(t) = \frac{x(t) \cdot x(t-\tau_1)}{x(t-\tau_2)}$ satisfies Scaling but does not satisfy Superposition (and it's only one example of many). 2. Superposition satisfies Scaling, at least for rational scalers. 3. i hadn't been worrying about $x(t-\tau2)=0$ nor about complex scalers (and i made no bones about it). – robert bristow-johnson Aug 6 '15 at 20:24
• @DilipSarwate Hi! I think you made a little unjustification over RBJ's example system. Afaik assuming that $x(t-\tau) \neq 0$ does not reduce the validity of the conclusion that the system satisfied scaling property for the restricted set of inputs. Indeed it's the same thing that in discrete-time systems we consider $x[n]=e^{j3n}$ as an aperiodic signal as you restrict the argument $n$ to be an integer, whereas if you (can) assume $n$ as a continuous variable then $x[n]$ will be periodic. So, making a restrictive assumption may not defer the validity of the conclusion... – Fat32 Oct 10 '18 at 22:37
• @DilipSarwate on the other hand, to prove that N vectors $x_k$ are linearly independent, one shows that $\sum_k c_k x_k = 0$ holds iff all $c_k$ are zero. So in this case restricting the range of $c_k$ will be wrong; i.e.; stating that $x_k$ are independent based on $\sum c_k x_k \neq 0$ whenever all $c_k \neq 0$ given that all $c_k < A$ for some positive $A$ will not be a valid conclusion as there might be a set of $c_k > A$ such that the sum $\sum_k c_k x_k = 0$ is satisfied, hence making $x_k$ a dependent set... So restrictions should be carefully used and justified. – Fat32 Oct 10 '18 at 23:00