SP riddle

Friends,

since nobody picked up my "exercise to the reader" in another thread,
or challenged me to its validity, I'll repost it as a riddle:

Consider the functions

u1(t) = t step(t),
y1(t) = (t+1)^3 step(t+1),
u2(t) = 3 (t-1) step(t-1), and
y2(t) = 9 step(t).

Give an example of a linear system that transforms u1 to y1 and u2 to
y2!

Regards,
Andor

> Consider the functions
>
> u1(t) = t step(t),
> y1(t) = (t+1)^3 step(t+1),
> u2(t) = 3 (t-1) step(t-1), and
> y2(t) = 9 step(t).
>
> Give an example of a linear system that transforms u1 to y1 and u2 to
> y2!
>



H(s)= -exp(-s)*Ei(s)+1/s+1/s^2+2/s^3 

> and u2 to
> y2!
>


-3*exp(-s)*Ei(s) 

Hans S=F8rensen wrote:

> > Consider the functions
> >
> > u1(t) =3D t step(t),
> > y1(t) =3D (t+1)^3 step(t+1),
> > u2(t) =3D 3 (t-1) step(t-1), and
> > y2(t) =3D 9 step(t).
> >
> > Give an example of a linear system that transforms u1 to y1 and u2 to
> > y2!
> >
>
>
>
> H(s)=3D -exp(-s)*Ei(s)+1/s+1/s^2+2/s^3

Hello Hans,

thanks for participating! However, note that the system in question,
while linear, is clearly not time-invariant, as can be derived from the
equations. Thus it does not have a well-defined transfer function (or
impulse response)...

Regards,
Andor

>thanks for participating! However, note that the system in question,
>while linear, is clearly not time-invariant, as can be derived from the
>equations. Thus it does not have a well-defined transfer function (or
>impulse response)...
>
>Regards,
>Andor
>
>

Andor, I spent some time thinking about this problem last night.  I didn't
get an answer but I don't want to see this thread die.

My first idea was to look at analyzing the inputs and ouptputs in the
frequency domain using LaPlace transforms.  However, I found that this was
a dead end because Laplace (one sided at least?) works under the assumption
that you are analyzing an LTI system, which is not the case as you pointed
out.

So, I thought that considering a two sided LaPlace Transform, being more a
more  general approach, might mititgate this restriction.  However, I got
the impression from your post that we would not be able to express this
system in closed form, that using any analytic technique would be a dead
end.  Plus, it was bedtime so I was pretty tired :)

I don't make the habbit of following a path just because someone says it's
the correct way so can you justify your reasoning for the following
statement?

>The system in question,
>while linear, is clearly not time-invariant, as can be derived from the
>equations. Thus it does not have a well-defined transfer function (or
>impulse response)...

If this is so, then how do you expect to get an answer without any
"well-defined" way of talking about it?

My intuition says that there is no linear system that will represent
system 1 and there is a linear system that will represent system 2.

Thank you for posting this riddle.  I'm having fun thinking about it.  

bmh161 wrote:
> >thanks for participating! However, note that the system in question,
> >while linear, is clearly not time-invariant, as can be derived from the
> >equations. Thus it does not have a well-defined transfer function (or
> >impulse response)...
> >
> >Regards,
> >Andor
> >
> >
>
> Andor, I spent some time thinking about this problem last night.  I didn't
> get an answer but I don't want to see this thread die.

Ok :-).

>
> My first idea was to look at analyzing the inputs and ouptputs in the
> frequency domain using LaPlace transforms.  However, I found that this was
> a dead end because Laplace (one sided at least?) works under the assumption
> that you are analyzing an LTI system, which is not the case as you pointed
> out.

Indeed.

>
> So, I thought that considering a two sided LaPlace Transform, being more a
> more  general approach, might mititgate this restriction.  However, I got
> the impression from your post that we would not be able to express this
> system in closed form, that using any analytic technique would be a dead
> end.  Plus, it was bedtime so I was pretty tired :)
>
> I don't make the habbit of following a path just because someone says it's
> the correct way so can you justify your reasoning for the following
> statement?
>
> >The system in question,
> >while linear, is clearly not time-invariant, as can be derived from the
> >equations. Thus it does not have a well-defined transfer function (or
> >impulse response)...

Looking back at the definitions of input and output, you'll notice that
u2(t) = 3 u1(t-1).
If the system were LTI, this would mean that
y2(t) = 3 y1(t-1),
which is not true, so the system cannot be LTI. This does not imply,
however, that the system cannot be linear!

>
> If this is so, then how do you expect to get an answer without any
> "well-defined" way of talking about it?

Linearity is a well defined concept (for example here:
http://en.wikipedia.org/wiki/Linear_system). One can only use transfer
functions if the systems in question are LTI.

> My intuition says that there is no linear system that will represent
> system 1 and there is a linear system that will represent system 2.

Remember, we are looking for one single system that transforms both u1
-> y1 and u2 -> y2.

>
> Thank you for posting this riddle.  I'm having fun thinking about it.

To help you along a bit, here are some hints:

- Integration is linear, ie. transforming
u(t) -> y(t) = int_c^t u(tau) dtau,
for any starting constant c is a linear system.

- Sampling is linear, ie. transforming
u(t) -> y(t) = u(t_0),
for any sampling time t_0 is a linear system.

- Linear combinations of linear systems are also linear systems.

- Concatenating two linear systems (series connection) also results in
a linear system.

Have fun!

Regards,
Andor

Is no news good news?

Here is another hint:

We are looking for a linear system H with

y1 = H u1

and

y2 = H u2.

One way to go about this is to partition H into two systems, H = H1 +
H2, such that

y1 = H1 u1, 
0 = H1 u2,

and

y2 = H2 u2, 
0 = H2 u1.

Regards,
Andor

Andor wrote:

> Here is another hint:

Before I had no clue but that makes it grunt work ;)

> We are looking for a linear system H with
> y1 = H u1
> and
> y2 = H u2.
> 
> One way to go about this is to partition H into two systems,
> H = H1 + H2

and H1{x} = T1{x - u2}, H2{x} = T2{x - u1} where T1, T2 are LTI 
systems whose transfer functions, imposing

> y1 = H1 u1, 
> y2 = H2 u2, 

pop out as
  T1(s) = 6e^s / (s^2(1 - 3e^-s)),
  T2(s) = 9s / (3e^-s - 1).
That makes the easiest-checked form of the overall system
  Y(s) = (6e^s/s^2 (X(s) - 3e^-s/s^2) + 9(1/s - X(s)s))
         / (1 - 3e^-s).


Martin

-- 
Be the change you're trying to create.
--Mahatma Gandhi

Martin Eisenberg wrote:
> Andor wrote:
>
> > Here is another hint:
>
> Before I had no clue but that makes it grunt work ;)
>
> > We are looking for a linear system H with
> > y1 = H u1
> > and
> > y2 = H u2.
> >
> > One way to go about this is to partition H into two systems,
> > H = H1 + H2
>
> and H1{x} = T1{x - u2}, H2{x} = T2{x - u1} where T1, T2 are LTI
> systems whose transfer functions, imposing
>
> > y1 = H1 u1,
> > y2 = H2 u2,
>
> pop out as
>   T1(s) = 6e^s / (s^2(1 - 3e^-s)),
>   T2(s) = 9s / (3e^-s - 1).

Nice try :-). However, your system is not linear - do you see why?

Regards,
Andor

Andor wrote:
> Martin Eisenberg wrote:
>> Andor wrote:
>>
>> > Here is another hint:
>>
>> Before I had no clue but that makes it grunt work ;)

Some day I'm really gonna pay for this kind of confidence...

>> > We are looking for a linear system H with
>> > y1 = H u1
>> > and
>> > y2 = H u2.
>> >
>> > One way to go about this is to partition H into two systems,
>> > H = H1 + H2
>>
>> and H1{x} = T1{x - u2}, H2{x} = T2{x - u1} where T1, T2 are LTI

> Nice try :-). However, your system is not linear - do you see why?

I do, taking me back to my baseline state: The ansatz implies
H{2 u1} - 2 y1 = T1{u2} + T2{u1} != 0 in general.


Martin

-- 
Quidquid latine scriptum sit, altum viditur.