Is a voltage controlled oscillator system time invariant?

Hi,
Is the following system time invariant?
y(t)=T_VCO[x(t)]=cos[wt+a(t)]
where
a(t)=\int_{-\infty}^{t}x(tau) d tau

My answer:
No it is not because:
y1(t)=T_VCO[x(t-t1)]=cos[wt+a(t)]
where
a(t)=\int_{-\infty}^{t}x(tau-t1) d tau
Therefore
y1(t) is not equal y(t-t1)


Oliver Faust

faust_o wrote:
> Hi,
> Is the following system time invariant?
> y(t)=T_VCO[x(t)]=cos[wt+a(t)]
> where
> a(t)=\int_{-\infty}^{t}x(tau) d tau
> 
> My answer:
> No it is not because:
> y1(t)=T_VCO[x(t-t1)]=cos[wt+a(t)]
> where
> a(t)=\int_{-\infty}^{t}x(tau-t1) d tau
> Therefore
> y1(t) is not equal y(t-t1)

Good question. Homework?

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

>
> Good question. Homework?
>
> Jerry

Dear Jerry,
I quite agree that we should not do or even discuss the homework of
students in the forum. However, this is a 'real' question which came up
after we discussed the PLL system. I just want to make sure that my
answer is correct. 

Sincerely,
Oliver Faust

faust_o wrote:
>> Good question. Homework?
>>
>> Jerry
> 
> Dear Jerry,
> I quite agree that we should not do or even discuss the homework of
> students in the forum. However, this is a 'real' question which came up
> after we discussed the PLL system. I just want to make sure that my
> answer is correct. 

Faust,

We discuss homework, even giving hints; we just don't give blind 
answers. Your answer seems reasonable to me, but I won't say so 
definitely. I've too often been wrong about such distinctions.

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

faust_o wrote:
> Hi,
> Is the following system time invariant?
> y(t)=T_VCO[x(t)]=cos[wt+a(t)]
> where
> a(t)=\int_{-\infty}^{t}x(tau) d tau
> 
> My answer:
> No it is not because:
> y1(t)=T_VCO[x(t-t1)]=cos[wt+a(t)]
> where
> a(t)=\int_{-\infty}^{t}x(tau-t1) d tau
> Therefore
> y1(t) is not equal y(t-t1)
> 
> 
> Oliver Faust
> 
By your reasoning it is, indeed, time varying.  Furthermore it should be 
pretty easy to show that y(t) = T_VCO[x(t)] is also not linear.

Given any real system (not a mathematical model), you should never have 
to ask "Is this system a linear time invariant system".  Why?  Because 
all real systems are nonlinear and time varying (see the "Nonlinear 
Systems" chapter of my book for the rational).

So if you're presented with a system that you want to analyze, such as a 
VCO, you shouldn't ask the above question.  What you _can_ and _should_ 
ask, however, is "how can I pretend that this system is linear and time 
invariant, and will the model be good enough for me to proceed".

In the case of a VCO, if you can accurately measure the deviation of the 
phase from wt, then you will find that y(t) = phase(T_VCO[x(t)]) is 
pretty darn close to a linear time-invariant system.  This is, in fact, 
exactly what you do for most PLL analysis, then you treat any artifacts 
of the phase measurement as noise that's magically injected into your 
system.  From this point of view, then, you can treat the VCO as linear 
and time invariant enough (but never absolutely linear and time invariant).

-- 

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

Posting from Google?  See http://cfaj.freeshell.org/google/

"Applied Control Theory for Embedded Systems" came out in April.
See details at http://www.wescottdesign.com/actfes/actfes.html

Dear Tim Wescott,

Thank you for your reply.


> By your reasoning it is, indeed, time varying.

Thank you very much. That helped me a lot, I will update my notes.


> Given any real system (not a mathematical model), you should never have
> to ask "Is this system a linear time invariant system".  Why?  Because
> all real systems are nonlinear and time varying (see the "Nonlinear
> Systems" chapter of my book for the rational).

I don't agree with your reasoning. In my opinion the only thing each
of us has is a model of the world around. The model in our brain is
similar to a computer model and therefore similar to a mathematical
model. All models are used to make predictions about certain events
around us. Say, a 'real world' signal processing system takes input
and produces output. A model predicts the output from a given input.
Therefore, one can judge a model based on the precision of prediction,
i.e. we observe the output of the 'real system' and compare it with
the predicted output from our model. To follow your reasoning,
nonlinear and time-variant models are able to predict certain events
with a higher precision compared with linear time-invariant models.
However, I strongly disagree that 'a real system' is non-linear and
time variant, because in my opinion only models can have such
properties.


> So if you're presented with a system that you want to analyze, such as a
> VCO, you shouldn't ask the above question.  What you _can_ and _should_
> ask, however, is "how can I pretend that this system is linear and time
> invariant, and will the model be good enough for me to proceed".

In general I agree with you, however I consider this as a second
question. Before, I can pretend that a model (system) is linear and
time invariant I need to know that it is not. In case of the PLL I need
to show that a multiplication system is non-linear but time invariant
and a VCO is non-linear and time variant. Only after this step, I can
introduce the simplifications which lead to a linear time invariant
model.

Oliver Faust

faust_o wrote:


>   ...   In my opinion the only thing each
> of us has is a model of the world around. The model in our brain is
> similar to a computer model and therefore similar to a mathematical
> model. All models are used to make predictions about certain events
> around us. ...

I doesn't matter whether our model of a nail bends or not. The real 
nails that we hit with hammers tell the whole story.

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

faust_o wrote:

> Dear Tim Wescott,
> 
> Thank you for your reply.
> 
> 
> 
>>By your reasoning it is, indeed, time varying.
> 
> 
> Thank you very much. That helped me a lot, I will update my notes.
> 
> 
> 
>>Given any real system (not a mathematical model), you should never have
>>to ask "Is this system a linear time invariant system".  Why?  Because
>>all real systems are nonlinear and time varying (see the "Nonlinear
>>Systems" chapter of my book for the rational).
> 
> 
> I don't agree with your reasoning. In my opinion the only thing each
> of us has is a model of the world around. The model in our brain is
> similar to a computer model and therefore similar to a mathematical
> model. All models are used to make predictions about certain events
> around us. Say, a 'real world' signal processing system takes input
> and produces output. A model predicts the output from a given input.
> Therefore, one can judge a model based on the precision of prediction,
> i.e. we observe the output of the 'real system' and compare it with
> the predicted output from our model. To follow your reasoning,
> nonlinear and time-variant models are able to predict certain events
> with a higher precision compared with linear time-invariant models.
> However, I strongly disagree that 'a real system' is non-linear and
> time variant, because in my opinion only models can have such
> properties.

Then your original question is deeply flawed, and the answer to it is "a 
VCO can't be time invariant, only a model thereof can be".

However, I say that you can choose any physical system to define as a 
'system' in the mathematical sense, with any input/output pair relating 
to that system that you choose to choose.

You will find that _in reality_, without recourse to any math higher 
than that necessary to measure the inputs and outputs, a great enough 
input will find a nonlinearity in the output.  If nothing else you'll 
concentrate so much energy into such a small area that space will bend. 
  For most problems, long before you see relativistic effects you'll see 
some measurable nonlinear effect.

End of story.

-- 

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

Posting from Google?  See http://cfaj.freeshell.org/google/

"Applied Control Theory for Embedded Systems" came out in April.
See details at http://www.wescottdesign.com/actfes/actfes.html

Jerry Avins <jya@ieee.org> writes:

> faust_o wrote:
>
>
>>   ...   In my opinion the only thing each
>> of us has is a model of the world around. The model in our brain is
>> similar to a computer model and therefore similar to a mathematical
>> model. All models are used to make predictions about certain events
>> around us. ...
>
> I doesn't matter whether our model of a nail bends or not. The real
> nails that we hit with hammers tell the whole story.

I was just reading today in Popular Mechanics that a chief reason
for delay of the Airbus 380 is that the software modeling the
cable assemblies was faulty. 

It's important to get these models to match reality.
-- 
%  Randy Yates                  % "Though you ride on the wheels of tomorrow,
%% Fuquay-Varina, NC            %  you still wander the fields of your
%%% 919-577-9882                %  sorrow."
%%%% <yates@ieee.org>           % '21st Century Man', *Time*, ELO
http://home.earthlink.net/~yatescr

Randy Yates wrote:
> Jerry Avins <jya@ieee.org> writes:
> 
> 
>>faust_o wrote:
>>
>>
>>
>>>  ...   In my opinion the only thing each
>>>of us has is a model of the world around. The model in our brain is
>>>similar to a computer model and therefore similar to a mathematical
>>>model. All models are used to make predictions about certain events
>>>around us. ...
>>
>>I doesn't matter whether our model of a nail bends or not. The real
>>nails that we hit with hammers tell the whole story.
> 
> 
> I was just reading today in Popular Mechanics that a chief reason
> for delay of the Airbus 380 is that the software modeling the
> cable assemblies was faulty. 
> 
> It's important to get these models to match reality.

They should probably avoid the Naomi Campbell style of model. :-)

Steve