linear/non-linear system

Jerry Avins wrote:

> Oli Filth wrote:
> > Jerry Avins said the following on 26/08/2006 17:59:
> >> Andor wrote:
> >>
> >>
> >>>    ... It is not possible to verify linearity from
> >>> knowing the system's response on just two input signals (u1 and u2).
> >>> Therefore, if the problem is well posed, the system has to be
> >>> non-linear.
> >>
> >> I would say that it's not possible to know if it is linear. the Scots
> >> trial verdict of "not proven" isn't the same as "not guilty".
> >>
> >
> > I think Andor's point might have been that because the original question
> > was from a textbook or a homework assignment, it has probably been
> > designed to have a definitive answer.  As it is impossible to
> > definitively say that the system is linear, the only remaining
> > possibility is that the equations in the problem demonstrate that the
> > system must be non-linear.
>
> So this is another example of "You had better guess what I mean because
> I'm incapable of expressing it clearly?"

Oli interpreted what I said correctly. However, some more thought shows
that both options are not possible: one can neither show that the
system is linear nor that it is non-linear. To see this, one can
construct a non-linear and a linear system, respectively, which both
transform u1 to y1 and u2 to y2. I've done the first part already, the
second part is an exercise to the reader: show there exists a linear
system which transforms u1 to y1 and u2 to y2. :-)

The only thing one can say with certainty is that the system is
non-LTI.

Regards,
Andor

Andor wrote:
> Jerry Avins wrote:
> 
>> Oli Filth wrote:
>>> Jerry Avins said the following on 26/08/2006 17:59:
>>>> Andor wrote:
>>>>
>>>>
>>>>>    ... It is not possible to verify linearity from
>>>>> knowing the system's response on just two input signals (u1 and u2).
>>>>> Therefore, if the problem is well posed, the system has to be
>>>>> non-linear.
>>>> I would say that it's not possible to know if it is linear. the Scots
>>>> trial verdict of "not proven" isn't the same as "not guilty".
>>>>
>>> I think Andor's point might have been that because the original question
>>> was from a textbook or a homework assignment, it has probably been
>>> designed to have a definitive answer.  As it is impossible to
>>> definitively say that the system is linear, the only remaining
>>> possibility is that the equations in the problem demonstrate that the
>>> system must be non-linear.
>> So this is another example of "You had better guess what I mean because
>> I'm incapable of expressing it clearly?"
> 
> Oli interpreted what I said correctly. However, some more thought shows
> that both options are not possible: one can neither show that the
> system is linear nor that it is non-linear. To see this, one can
> construct a non-linear and a linear system, respectively, which both
> transform u1 to y1 and u2 to y2. I've done the first part already, the
> second part is an exercise to the reader: show there exists a linear
> system which transforms u1 to y1 and u2 to y2. :-)
> 
> The only thing one can say with certainty is that the system is
> non-LTI.

What I said is that one can't tell about nonlinearity. Did it seem 
otherwise?

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

Jerry Avins wrote:

> Andor wrote:
> > Jerry Avins wrote:
> >
> >> Oli Filth wrote:
> >>> Jerry Avins said the following on 26/08/2006 17:59:
> >>>> Andor wrote:
> >>>>
> >>>>
> >>>>>    ... It is not possible to verify linearity from
> >>>>> knowing the system's response on just two input signals (u1 and u2).
> >>>>> Therefore, if the problem is well posed, the system has to be
> >>>>> non-linear.
> >>>> I would say that it's not possible to know if it is linear. the Scots
> >>>> trial verdict of "not proven" isn't the same as "not guilty".
> >>>>
> >>> I think Andor's point might have been that because the original question
> >>> was from a textbook or a homework assignment, it has probably been
> >>> designed to have a definitive answer.  As it is impossible to
> >>> definitively say that the system is linear, the only remaining
> >>> possibility is that the equations in the problem demonstrate that the
> >>> system must be non-linear.
> >> So this is another example of "You had better guess what I mean because
> >> I'm incapable of expressing it clearly?"
> >
> > Oli interpreted what I said correctly. However, some more thought shows
> > that both options are not possible: one can neither show that the
> > system is linear nor that it is non-linear. To see this, one can
> > construct a non-linear and a linear system, respectively, which both
> > transform u1 to y1 and u2 to y2. I've done the first part already, the
> > second part is an exercise to the reader: show there exists a linear
> > system which transforms u1 to y1 and u2 to y2. :-)
> >
> > The only thing one can say with certainty is that the system is
> > non-LTI.
>
> What I said is that one can't tell about nonlinearity. Did it seem
> otherwise?

You said one cannot say if the system is linear. I agree and add: one
also cannot say if the system is non-linear.

Andor wrote:

   ...

> You said one cannot say if the system is linear. I agree and add: one
> also cannot say if the system is non-linear.

I see now. A nice distinction.

jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

VijaKhara wrote:

> hi all,
> 
> 1. For a system, if the input function is u1(t), the output is y1(t)
> and if the input u2(t) the output y2(t). They ask if this system is
> linear or not. I simply look for the transfer function:
> H1(s)=Y1(s)/U1(s) and H2(s)=Y2(s)/U2(s). If H1(s)=H2(s) the system is
> linear, if not it nonlinear. But I am confused. What if the system is
> linear but not time invariant. Does the proof above still work?
> 
You can't make a Laplace transfer function of the system if it is non 
linear or time varying.  You have to apply the definition of a linear 
system in the time domain.

> 2. For my specific case:
> u1(t)=tstep(t), y2(t)=(t+1)^3step(t+1).
> u2(t)=3(t-1)step(t-1), y2(t)=9step(t).
> 
> Step(t) is the step function/

As pointed out elsewhere, if you have two _definite_ signals you can't 
assure global linearity.  You can only find out if a system is 100% 
linear from it's mathematical description, by manipulating symbols.

If it's a real system you can just assume that it's nonlinear, because 
all real systems are nonlinear when you push them hard enough.  The only 
question you can ask is are they linear _enough_ that you can _pretend_ 
that they're linear for the purposes of your analysis.
> 
> Is there any another simpler way to verify the linearity?
> 
No.
> 
> 3. In control books, they often use Laplace transform to find the
> transfer function. Why don't they use Fourier Transform? For some
> problems, I feel FOurier Transform is more convenient because we can
> compute the transform in both sides (negative and positive). 
> 
Because the Laplace transform is more general, and more useful when 
you're interested in the time-domain results.  The Fourier transform is 
better for analyzing systems when you just care about their frequency 
response, and has some handy properties to analyze systems that are 
periodically time varying, so it's good for communications systems.

-- 

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

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"VijaKhara" <VijaKhara@gmail.com> wrote in message
news:1156604500.446483.201560@p79g2000cwp.googlegroups.com...
> hi all,
>
> 1. For a system, if the input function is u1(t), the output is y1(t)
> and if the input u2(t) the output y2(t). They ask if this system is
> linear or not. I simply look for the transfer function:
> H1(s)=Y1(s)/U1(s) and H2(s)=Y2(s)/U2(s). If H1(s)=H2(s) the system is
> linear, if not it nonlinear. But I am confused. What if the system is
> linear but not time invariant. Does the proof above still work?
>
> 2. For my specific case:
> u1(t)=tstep(t), y2(t)=(t+1)^3step(t+1).
> u2(t)=3(t-1)step(t-1), y2(t)=9step(t).
>
> Step(t) is the step function/
>
> Is there any another simpler way to verify the linearity?
>
>
> 3. In control books, they often use Laplace transform to find the
> transfer function. Why don't they use Fourier Transform? For some
> problems, I feel FOurier Transform is more convenient because we can
> compute the transform in both sides (negative and positive).
>
> Thanks
>

The simplest engineering explanation of  linearity is that if you put two
sine wave in at different frequencies then you get out two sine waves at teh
same frequencies shifted in phase and changed in amplitude. If you get
cross-product terms and other terms then it has non-linearities eg

(AcosW1t+Bcos(W2t))^2 is non -linear - a square law and it has a cross
product term and terms at 2W1 and 2W2.

This is of course superposition in action.
M.



-- 
Posted via a free Usenet account from http://www.teranews.com

VijaKhara wrote:

> hi all,
>
> 1. For a system, if the input function is u1(t), the output is y1(t)
> and if the input u2(t) the output y2(t). They ask if this system is
> linear or not.
...
> 2. For my specific case:
> u1(t)=tstep(t), y2(t)=(t+1)^3step(t+1).
> u2(t)=3(t-1)step(t-1), y2(t)=9step(t).
>
> Step(t) is the step function/

VijaKhara,

I don't know if you have been following my discussion with Jerry, but
here is the result: those two transformations are not enough to show
that the system is non-linear or linear. The reason is that there exist
both linear and non-linear transformations that map u1 to y1 and u2 to
y2.

However, it is possible to show that the system in question is not LTI
(linear and time-invariant). Do you see how?

Regards,
Andor

Major Misunderstanding said the following on 27/08/2006 04:01:
> The simplest engineering explanation of  linearity is that if you put two
> sine wave in at different frequencies then you get out two sine waves at teh
> same frequencies shifted in phase and changed in amplitude. If you get
> cross-product terms and other terms then it has non-linearities

This is actually only the case for LTI, not linear.

For instance, "multiplication by t" is a linear system, t.u1(t) + 
t.u2(t) = t.(u1(t) + u2(t)) for all u1 and u2; "superposition in 
action".  However, it certainly won't give you two sinusoids out for two 
sinusoids in.


-- 
Oli

Major Misunderstanding wrote:
> 
> The simplest engineering explanation of  linearity is that if you put two
> sine wave in at different frequencies then you get out two sine waves at teh
> same frequencies shifted in phase and changed in amplitude. If you get
> cross-product terms and other terms then it has non-linearities eg

That may be a meaningful definition of, say, a linear amplifier, but it 
is much too narrow to cover linear systems in general.
> 
> (AcosW1t+Bcos(W2t))^2 is non -linear - a square law and it has a cross
> product term and terms at 2W1 and 2W2.
> 
> This is of course superposition in action.

Regards,
Steve

The question itself is ill-defined. The definition of linearity stems
from mathematics, so you have to go to basics. A  function (or a
transformation) on a set is defined  only if the definition exists on
each of its domain elements. Otherwise the function is not defined to
begin with. Linearity is defined on a function. In your case the domain
is the set of all possible functions on time axis. You have defined
your function on just two domain elements. There are infinitely many
possible functions that satisfy your criteria. The question arises -
which unique function is the question refering to?? Some will be linear
and some non-linear.

A definite answer that the system is nonlinear would be determined if
we know what f(0) is?  '0' here means the function on time axis which
takes on zero value at all  times t. For example,  x(t) = 0 for every
't'.
The proof is shown here -

Theorem - if  f is linear then f(0) = 0.
Proof - f(a) = f(0+a) = f(0) + f(a) ...by the defintion of linearity
 Therefore f(0) = f(a) - f(a) = 0.

Hope this helps,

-rajesh