s the output of an analog filter real or the magnitude?(2)

Hello again and thank you for your answers
I will try to describe my scenario
1. I am working with UWB signals, which is real, s(t), and has Fourier
transform S(w).
2. I apply one filter H(w), which has gain A(w) and phase Theta(w), w =
angular frequency. Then H(w) = A(w)*exp(j*Tetha(w)*to);
3. In digital domain, I decide what I get, lets say the real part,
real(ifft(S(w)*H(w))), or the imaginary part (the signal represented by
the imaginary part) or the magnitude (abs(A(W))).
4. My question is about the analog domain. From the answers I got, it
seems that naturally I get the real part, and can extract the others (imag
and abs) by different circuits. But what about the ENERGY of the filtered
signal? is it equal to the incoming signal (assuming a passive filter) or
there is some energy left somewhere.
In digital domain I can get the energy of both, real(S(w)*H(w)) or
imag(S(w)*H(w)). However in analog domain it is not that obvious, most of
books which I have access to, deal with design, and evaluation of filters,
but so far I couldnt find the answer I want, In fact, I NEED THE ENERGY OF
THE ANALOG FILTERED SIGNAL. 

The reason why I ask this question is because of my laboratory experiments
on alternatig current, i.e. at the ouput of a resonant circuit you may
measure zero volts, however if you measure at the edges of L or C, there
is indeed voltage. And also antennas can be modeled as filters where gain
and phase functions are provided. again what is my output? the real part?

thanks

Greg
		
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"Gregwise" <gregoguti@mixmail.com> wrote in message
news:5qGdnVj94_PAecjeRVn-2A@giganews.com...
> Hello again and thank you for your answers
Hi Greg - I didn't attempt any sort of answer before as I was confused as to
what you were looking for; now I'm even more confused but I'm going to try
anyway - hope some of this helps.
> I will try to describe my scenario
> 1. I am working with UWB signals, which is real, s(t), and has Fourier
> transform S(w).
Are you assuming that this real signal is provided from a constant impedance
source having only a resistive component across the full frequency band that
your signal has any significant energy spectral density in?

> 2. I apply one filter H(w), which has gain A(w) and phase Theta(w), w =
> angular frequency. Then H(w) = A(w)*exp(j*Tetha(w)*to);
I don't know what your *to factor is.

> 3. In digital domain, I decide what I get, lets say the real part,
> real(ifft(S(w)*H(w))), or the imaginary part (the signal represented by
> the imaginary part) or the magnitude (abs(A(W))).
Why is there no IFFT or S(w) term in your magnitude ?

> 4. My question is about the analog domain. From the answers I got, it
> seems that naturally I get the real part, and can extract the others (imag
> and abs) by different circuits. But what about the ENERGY of the filtered
> signal? is it equal to the incoming signal (assuming a passive filter) or
> there is some energy left somewhere.
Generally the filter will reflect some of the source signal energy back into
the source and/or will dissipate some of the energy in resistive losses.

> In digital domain I can get the energy of both, real(S(w)*H(w)) or
> imag(S(w)*H(w)). However in analog domain it is not that obvious, most of
> books which I have access to, deal with design, and evaluation of filters,
> but so far I couldnt find the answer I want, In fact, I NEED THE ENERGY OF
> THE ANALOG FILTERED SIGNAL.
O.K.  you maybe need to calculate how much of the energy supplied by the
source is available to a load at the other side of your filter.
>
> The reason why I ask this question is because of my laboratory experiments
> on alternatig current, i.e. at the ouput of a resonant circuit you may
> measure zero volts, however if you measure at the edges of L or C, there
> is indeed voltage. And also antennas can be modeled as filters where gain
> and phase functions are provided. again what is my output? the real part?
Your output is whatever you define it to be.  You are presumably doing all
of this to cause some wanted effect somewhere while avoiding some other
unwanted effects elsewhere (e.g. have a high probablity that your
transmitted pulse is detected by a known receiving system at some remote
location in the presence of interference while making sure that you don't
trip the operator's pacemaker very often.) , your wanted receiver may be a
test probe and detector circuitry in a particular reference configuration -
until you have this defined I don't see how you can start to talk about your
output.  Is it that you want to calulate the energy dissipated in a purely
resistive load of constant known impedance at the output of your filter and
that the transfer function you have is calculated taking the effects of
source and load impedance into account? If so 
http://en.wikipedia.org/wiki/Parseval's_theorem
should help.

Otherwise I don't know where you should start.
>
> thanks
Best of Luck
Mike 

Gregwise wrote:
> Hello again and thank you for your answers
> I will try to describe my scenario
> 1. I am working with UWB signals, which is real, s(t), and has Fourier
> transform S(w).
> 2. I apply one filter H(w), which has gain A(w) and phase Theta(w), w =
> angular frequency. Then H(w) = A(w)*exp(j*Tetha(w)*to);
> 3. In digital domain, I decide what I get, lets say the real part,
> real(ifft(S(w)*H(w))), or the imaginary part (the signal represented by
> the imaginary part) or the magnitude (abs(A(W))).

If the filter is realizable in the time domain without giving it 
explicitly complex coefficients then ifft(S(w)*H(w)) will always be real.

Do you mean to say that you only know how to do filtering using Fourier 
transforms?

> 4. My question is about the analog domain. From the answers I got, it
> seems that naturally I get the real part, and can extract the others (imag
> and abs) by different circuits. But what about the ENERGY of the filtered
> signal? is it equal to the incoming signal (assuming a passive filter) or
> there is some energy left somewhere.

Energy conservation rules must apply.  Real-world passive filters either 
bounce energy back out the port it's coming in, pass it through to some 
other port, or dissipate it as heat.

> In digital domain I can get the energy of both, real(S(w)*H(w)) or
> imag(S(w)*H(w)). 

This is not the case that you cited in (3) above.  There you were taking 
the inverse transform, here you are doing things entirely in the 
frequency domain.  In general ifft(real(S(w)*H(w))) != 
real(ifft(S(w)*H(w))).

> However in analog domain it is not that obvious, most of
> books which I have access to, deal with design, and evaluation of filters,
> but so far I couldnt find the answer I want, In fact, I NEED THE ENERGY OF
> THE ANALOG FILTERED SIGNAL. 

Well, I've already told you how to do that.  And don't SHOUT, 
particularly to folks who are trying to help.  It makes them want to 
ignore you.
> 
> The reason why I ask this question is because of my laboratory experiments
> on alternatig current, i.e. at the ouput of a resonant circuit you may
> measure zero volts, however if you measure at the edges of L or C, there
> is indeed voltage.

At the output of a real-world resonant circuit with real-world 
components you will never attain complete nulling of the signal, 
although you may be able to pound it down small enough that your 
measuring instruments won't detect it.

You will, indeed, see high circulating energy in the interior of an LC 
circuit that is operating at it's resonant frequency.  They'll show up 
as high voltages in a series LC and high currents in a parallel LC.  But 
the effective impedance will be such that in an ideal circuit there 
won't be any energy being dissipated.

> And also antennas can be modeled as filters where gain
> and phase functions are provided. again what is my output? the real part?
> 
In the real world it's always the real part.

I think you have the mistaken notion that the frequency domain is 
something real.  Pull your head out of your assumptions and look around 
you.  Everything that you experience, everything that affects your 
software and your circuits happens in the time domain.  The Fourier 
transform and the frequency domain that goes with it is just a 
mathematical tool that helps you solve problems in linear system 
analysis.  Since nothing in this universe is linear the frequency domain 
is not the primary reality, it is merely a highly useful reflection.

-- 

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

Gregwise wrote:
> Hello again and thank you for your answers
> I will try to describe my scenario
> 1. I am working with UWB signals, which is real, s(t), and has Fourier
> transform S(w).
> 2. I apply one filter H(w), which has gain A(w) and phase Theta(w), w =
> angular frequency. Then H(w) = A(w)*exp(j*Tetha(w)*to);
> 3. In digital domain, I decide what I get, lets say the real part,
> real(ifft(S(w)*H(w))), or the imaginary part (the signal represented by
> the imaginary part) or the magnitude (abs(A(W))).
> 4. My question is about the analog domain. From the answers I got, it
> seems that naturally I get the real part, ...

No. You get the magnitude -- sqrt(Re^2 + Im^2) -- and the phase. In 
short, you get the filtered signal. Be aware: both Re and Im are real. 
It takes two wires (aside from ground) to represent a complex signal. 
Note "represent". The signal on each wire is real.

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