Re: any deep thinking on why linear systems are commutative?

From: Tim Wescott (tim_at_wescottnospamdesign.com)
Date: 10/25/04 

Date: Mon, 25 Oct 2004 12:35:03 -0700

kiki wrote:

> Dear all,
>
> Could you please help me understand better by providing some deep thoughts
> on why linear systems are commutative?
>
> If T1 and T2 are linear systems, then y1=T2(T1(x)) and y2=T1(T2(x))
>
> y1 and y2 are the same.
>
> After seeing some examples of non-linear system and linear systems. I got
> convinced that for non-linear system this property does not hold.
>
> But any deeper thinking? Proof?
>
> Thanks a lot!
>
>
Kiki, David, Norm:

I think that Kiki means a linear system in the signal processing sense,
i.e. the definition in Oppenheim and Willsky, "Signals and Systems",
1983. It states that a system is "any process that results in the
transformation of signals." "Thus, a system has an input signal and an
output signal which is related to the input through the system
transformation".

So a system is one where for two scalar signals x(t) and y(t), and a
system h, h will transform x(t) into y(t) such that y(t) = h(x(t), t)
depends only on the value of x over all time, on the value of t and on
the properties of h.

Take two multipliers a_1 and a_2, and two input signals x_1 and x_2.
Generate an input signal equal to

x(t) = a_1*x_1(t) + a_2*x_2(t)

A system h is linear in this sense if and only if

y(t) = h(x(t), t) = a_1*h(x_1(t), t) + a_2*h(x_2(t), t).

Linear time invariant systems (i.e. a system h where y(t-a) = h(x(t-a),
t) iff y(t) = h(x(t), t)) are commutative -- one finds that any linear
time invariant system can be described as a convolution operation, then
one finds that convolution operations are commutative, and viola! one
has a proof. Oppenheim has, I believe, a proof. I believe that
time-invariance is not necessary, but I don't know and I'm too lazy to
look it up or prove it at the moment.

Nonlinear systems are, in general, not commutative. For instance the
(nonlinear) system h_1(x(t), t) = x(t)^2 doesn't commute with the
(linear) system h_2(x(t), t) = 42*x(t) -- applying h_1 first yields y =
42*x^2, where applying it second yields y = 1764*x^2, so they clearly
aren't equal. 

-- 
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

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