lucy wrote:
> The problem is that I could not find a good example to test them...

A unit impulse is the simplest test probe for impulse response. By 
definition.

> My feeling tells that for convolution defined from -inf to +inf,
> 
> Cov(diff(f(t)), a(t))
> =Cov(f(t), diff(a(t)))
> =diff(Cov(f(t), a(t)))
> 
> am I right?
> 
> For convolution defined from 0 to t,
> the above do not hold,
> we have to take initial conditions into consideration,
> diff(Cov(f(t), a(t)))
> =f(0)*a(t)+Cov(diff(f(t)), a(t))
> =f(t)*a(0)+Cov(f(t), diff(a(t)))
> 
> am I right?

You can think of the initial conditions at t = 0 being the result of 
convolution from -infinity to zero, if you like.

> I think I am near that point which clarifies everything...

I'm really pleased. I hope you are too.

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

The problem is that I could not find a good example to test them...

My feeling tells that for convolution defined from -inf to +inf,

Cov(diff(f(t)), a(t))
=Cov(f(t), diff(a(t)))
=diff(Cov(f(t), a(t)))

am I right?

For convolution defined from 0 to t,
the above do not hold,
we have to take initial conditions into consideration,
diff(Cov(f(t), a(t)))
=f(0)*a(t)+Cov(diff(f(t)), a(t))
=f(t)*a(0)+Cov(f(t), diff(a(t)))

am I right?

I think I am near that point which clarifies everything...

lucy wrote:
> I still don't understand it...
>
> Please tell me if you know which derivation is correct...

Although trying a known example will not tell you if a
derivation is correct, it might tell you if a derivation
is incorrect.  Do you know of any example step responses?
Have you tried them using the methods of each derivation?


-- 
rhn

lucy wrote:
> I still don't understand it...
> 
> Please tell me if you know which derivation is correct...
> 
> Thanks a lot!

Your functions are all in time. Since y(t_step) --the step response-- is

              t
           integral[y(t_impulse)]dt
              0

(where t_impulse is the impulse response and the impulse itself occurs 
at t=0), Obtain the step response by direct integration and then decide 
whether convolution or multiplication makes sense.

Did I meet you half way?

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

I still don't understand it...

Please tell me if you know which derivation is correct...

Thanks a lot!

lucy wrote:
> Hi all,
> 
> Suppose the forced input to the system is f(t), the step response of
> the system is a(t) and the output is y(t).
> 
> Now we want to find y(t),
> 
> I am confused:

   ...

I won't answer your question because I think you can answer it yourself. 
One way to discern the correct method is applying both to a simple case 
with known result and observing which method gives that result. When you 
know which is correct, I think you will easily see why it must be so.

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

Hi all,

Suppose the forced input to the system is f(t), the step response of
the system is a(t) and the output is y(t).

Now we want to find y(t),

I am confused:

Which of the following is the correct output y(t)?

(1)       y(t)=convolution(differentiate(a(t)), f(t))

and

(2)       y(t)=differentiate(convolution(a(t), f(t))

???

All "differentiate" and "convolve" operations are w.r.t. "t"...

Using infinite summation of piecewise response to the input f(t) and
then take limit as n->infinity,

I can obtain

y(t)=differentiate(convolution(a(t), f(t));


But when think about the diff(step-response of the system)=impulse
response of the system,

I obtained

y(t)=convolution(differentiate(a(t)), f(t));

Please zoom in the following picture to see the detailed derivations...

http://www.yourupload.com//uploads/losemind/dc46a-Capture9.JPG

--------------------------

Note there the convolution is defined as integration(a(t-u)*f(u), u
from 0 to t), instead of the integration(a(t-u)*f(u), u from -infinity
to +infinity)...

I think this makes a big difference... for the above two equations (1)
and (2)...

If the convolution is defined as

integration(a(t-u)*f(u), u from -infinity to +infinity)

the above two equations should be the same...

Am I right?