linear system linear?

Supposedly simple questions:

Is every "linear system" under all conditions linear? Of course, what is
the precice definition of "linear system"? Is it only the homogenous
solution or homogenous+particulary?

Is a system with nonzero initial conditions linear?

It is a trivial fact that all systems with zero initial conditions are
linear because they can be represented with their transfer functions and
arbitrarily interchanged.

However, consider two systems H1 and H2. Suppose that the initial
conditions for H1=0 and for H2!=0. It makes a difference if I have
H1(s)*H2(s) or H2(s)*H1(s). Why? In the first case, the non-zero i.c. is
affected by the exponentials of H2 only. In the latter case of both!

From this perspective, they do not form a linear system because H1*H2 !=
H2*H1.

But both are linear systems.

What is the truth?

Peter



PS: For each system: x' = Ax + bu , y=c^T x + du
in general:
y(t) = c^T \Phi(t)x_0 + \int_0^t c^T x(t-\tau)b u(\tau) d\tau

The non-zero initial condition (x_0) can be regarded as an additional
system input "x_01/s" etc. If x_0=0, everything works out as expected.
But if x01!=0 two *linear* systems cannot be interchanged without giving
two different results.

One remark: In my opinion, a linear system is linear is
f(ax+by)=af(x)+bf(y). This does not imply anything about commutative
property, much like matrix multiplication (which is linear too).

Does that mean that the fact that we can commute linear systems is just
a convenient result of the fact that we assume zero initial conditions?


On 2014-10-18 4:13, Peter Mairhofer wrote:
> Supposedly simple questions:
> 
> Is every "linear system" under all conditions linear? Of course, what is
> the precice definition of "linear system"? Is it only the homogenous
> solution or homogenous+particulary?
> 
> Is a system with nonzero initial conditions linear?
> 
> It is a trivial fact that all systems with zero initial conditions are
> linear because they can be represented with their transfer functions and
> arbitrarily interchanged.
> 
> However, consider two systems H1 and H2. Suppose that the initial
> conditions for H1=0 and for H2!=0. It makes a difference if I have
> H1(s)*H2(s) or H2(s)*H1(s). Why? In the first case, the non-zero i.c. is
> affected by the exponentials of H2 only. In the latter case of both!
> 
> From this perspective, they do not form a linear system because H1*H2 !=
> H2*H1.
> 
> But both are linear systems.
> 
> What is the truth?
> 
> Peter
> 
> 
> 
> PS: For each system: x' = Ax + bu , y=c^T x + du
> in general:
> y(t) = c^T \Phi(t)x_0 + \int_0^t c^T x(t-\tau)b u(\tau) d\tau
> 
> The non-zero initial condition (x_0) can be regarded as an additional
> system input "x_01/s" etc. If x_0=0, everything works out as expected.
> But if x01!=0 two *linear* systems cannot be interchanged without giving
> two different results.
> 
> 
> 

we were arguing about this at the dsp.stackexchange site not long ago.

r b-j

Am 18.10.2014 13:13, schrieb Peter Mairhofer:

> Is every "linear system" under all conditions linear?

Yes, that's what "linear" means. There is nothing like "conditioned 
linear". A system is either linear or not.

> Of course, what is
> the precice definition of "linear system"? Is it only the homogenous
> solution or homogenous+particulary?

An initial value system (I believe that is what you mean by "system") is 
linear if and only if its solution space is a linear manifold. That is, 
if f and g are solutions, so is f+g, and \lambda f for all \lambda.

> Is a system with nonzero initial conditions linear?

No. It's an affine system, not a linear system. That is, its solution 
space is a linear manifold that is shifted away from the origin (namely 
by the initial conditions). That is, the generic solution of such a 
system is given by a specific ("particular") solution of the 
inhomogeneous system (the "offset" that moves you to the manifold), and 
a generic solution of the homogeneous system.

> It is a trivial fact that all systems with zero initial conditions are
> linear because they can be represented with their transfer functions and
> arbitrarily interchanged.

It depends on what you mean by "system": Of course, linear systems are 
linear, but that is just the definition. A linear differential equation, 
plus homogeneous boundary conditions (or initial conditions) is 
trivially linear. But of course, one can also discuss non-linear 
differential equations, for example the Navier-Stokes equation. It's 
solutions are much more complicated, and in general, one does not know 
much about their solution spaces.

> However, consider two systems H1 and H2. Suppose that the initial
> conditions for H1=0 and for H2!=0. It makes a difference if I have
> H1(s)*H2(s) or H2(s)*H1(s).

I don't know what you mean by "*" or H(s).

> Why? In the first case, the non-zero i.c. is
> affected by the exponentials of H2 only. In the latter case of both!

Differential operators typically do not commute if this is what confuses 
you.


>  From this perspective, they do not form a linear system because H1*H2 !=
> H2*H1.

That is not a requirement for a system to be linear. That's at best a 
requirement for two operators to commute. But that's in general not 
true. The differential operator d/dx and the multiplication operator 
that multiplies by x are both linear operators, but they do not commute.

Greetings,
	Thomas

On Sat, 18 Oct 2014 04:13:17 -0700, Peter Mairhofer wrote:

> Supposedly simple questions:
> 
> Is every "linear system" under all conditions linear? Of course, what is
> the precice definition of "linear system"? Is it only the homogenous
> solution or homogenous+particulary?
> 
> Is a system with nonzero initial conditions linear?
> 
> It is a trivial fact that all systems with zero initial conditions are
> linear because they can be represented with their transfer functions and
> arbitrarily interchanged.
> 
> However, consider two systems H1 and H2. Suppose that the initial
> conditions for H1=0 and for H2!=0. It makes a difference if I have
> H1(s)*H2(s) or H2(s)*H1(s). Why? In the first case, the non-zero i.c. is
> affected by the exponentials of H2 only. In the latter case of both!
> 
> From this perspective, they do not form a linear system because H1*H2 !=
> H2*H1.
> 
> But both are linear systems.
> 
> What is the truth?
> 
> Peter
> 
> 
> 
> PS: For each system: x' = Ax + bu , y=c^T x + du in general:
> y(t) = c^T \Phi(t)x_0 + \int_0^t c^T x(t-\tau)b u(\tau) d\tau
> 
> The non-zero initial condition (x_0) can be regarded as an additional
> system input "x_01/s" etc. If x_0=0, everything works out as expected.
> But if x01!=0 two *linear* systems cannot be interchanged without giving
> two different results.

Oh, look!  Wikipedia has an article on linear systems RIGHT HERE: 
http://en.wikipedia.org/wiki/Linear_system

You could have found it and read most of it in the time that it took you 
to post your questions above.  All it would have taken is a little teeny 
bit of self-motivation.

-- 
www.wescottdesign.com

On Sat, 18 Oct 2014 04:13:17 -0700, Peter Mairhofer
<63832452@gmx.net> wrote:

>Supposedly simple questions:
>
>Is every "linear system" under all conditions linear? Of course, what is
>the precice definition of "linear system"? Is it only the homogenous
>solution or homogenous+particulary?
>
>Is a system with nonzero initial conditions linear?
>
>It is a trivial fact that all systems with zero initial conditions are
>linear because they can be represented with their transfer functions and
>arbitrarily interchanged.

I don't understand that.  Zero initial conditions tell you
nothing about the system.  Consider a diode with no applied
voltage:  Zero in, zero out... but the system is certainly
not linear.

Best regards,


Bob Masta
 
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On Sat, 18 Oct 2014 16:31:12 +0200
Thomas Richter <thor@math.tu-berlin.de> wrote:

> Am 18.10.2014 13:13, schrieb Peter Mairhofer:
> 
> > Is every "linear system" under all conditions linear?
> 
> Yes, that's what "linear" means. There is nothing like "conditioned 
> linear". A system is either linear or not.
> 

Disagree.  I have an op-amp and two resistors for a gain of +2; that's
a (nearly) linear element in my circuit.  If I power it from +/- 5V, and
put 5V in, I'll rail the output and take it non-linear.  That doesn't
mean that my amplifier isn't linear, it just means that it's only linear
under the designed conditions of |Vin| < 2.5V.

-- 
Rob Gaddi, Highland Technology -- www.highlandtechnology.com
Email address domain is currently out of order.  See above to fix.

On 2014-10-18 6:21, robert bristow-johnson wrote:
> we were arguing about this at the dsp.stackexchange site not long ago.

Hi Robert,

Thanks. Do you have a pointer to this by any chance?
Was it this discussion?
http://dsp.stackexchange.com/questions/18431/is-this-system-linear

Thanks,
Peter

On Saturday, October 18, 2014 8:21:01 AM UTC-5, robert bristow-johnson wrote:
> we were arguing about this at the dsp.stackexchange site not long ago.
> 
> 
> 
> r b-j

This was also discussed here earlier. Randy Yates had the correct answer, but I don't think anyone agreed with him. To be linear the system must satisfy 

f(ax+by)=af(x)+bf(y).

This means that the response must go through (0,0). To me, this does not mean the initial conditions must be (0,0). I could allow initial conditions of (-1,0), as long as the response goes through (0,0).

f(x) = 2x + 3 is not linear. f(1) = 5, but f(2) = 7, and f(2x) != 2*f(x).

f(x) = 2x + 3 is called affine, not linear. Engineering students are routinely thought that a linear system is one with a straight-line response, but not true. Engineers often admit this without knowing it. For example, an opamp operating from +5 and ground with a quiescent output of 2.5v. The amp has a gain of 2. The engineer puts in a 1v sinusoid and sees a 2v sinusoid out and says see, it's twice the input. In truth the output is the 2v sinusoid and the 2.5v quiescent. By looking at just the sinusoid, the equation f(x) = 2x + 2.5 was turned into f(x) = 2x. Which is linear.

There is a difference between a linear system and a linear equation

Maurice Givens

On Monday, October 20, 2014 3:30:02 PM UTC-4, Peter Mairhofer wrote:
> On 2014-10-18 6:21, robert bristow-johnson wrote:
> 
> > we were arguing about this at the dsp.stackexchange site not long ago.
> 
> 
> 
> Hi Robert,
> 
> 
> 
> Thanks. Do you have a pointer to this by any chance?
> 
> Was it this discussion?
> 
> http://dsp.stackexchange.com/questions/18431/is-this-system-linear
> 


yup.  and also this one:

 http://dsp.stackexchange.com/questions/18557/does-instability-make-an-otherwise-lti-system-nonlinear-or-time-variant

my position is that you should be able to apply the test of "homogeneity" directly to the difference equation and that any non-zero initial state can be, if you allow for negative time, expressed in the same convolution summation that you use for non-negative time.

r b-j