Channel design
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From: Walter Roberson on 21 Jul 2010 18:33 Marcin wrote: > Walter Roberson <roberson(a)hushmail.com> wrote in message >> What does the X axis in the graph represent? It is unlabeled. For >> example, does it represent distance? Frequency in MHz? >> >> What would the inputs to this "channel" (or filter) be? Do you need >> multiple output taps? > > Oh, I haven't even noticed it wasn't labeled. I apologize. It's > frequency of course. My input is a single digital signal and the output > should also be a single digital signal but "affected" by the channel > (the graph simply shows the channel attinuation in domain of the signal > spectrum - fyi. it was measured for the 100m piece of some kind of a > plastic fiber). The axis is frequency? And this is fibre optics? Fibre has a peak transmission frequency with increased attenuation as you move away from the peak, rather than a frequency of 0 Hz having the greatest transmittance as would be implied by that diagram if the axis is frequency (rather than, e.g., frequency _difference_ ) http://en.wikipedia.org/wiki/File:Zblan_transmit.jpg I believe one could make theoretical arguments that transmittance cannot be simply symmetrical in frequency _difference_ : if it were symmetric then the implication would be that there would have to be some kind of "negative frequencies" that would have well defined transmittance; as "negative frequency" makes no physical sense, an assumption of symmetry around the peak frequency cannot be valid. Thus it seems to me that the diagram's axes must not represent frequency nor absolute frequency _difference_ . I thus do not think the problem has been completely specified as yet.
From: Marcin on 22 Jul 2010 11:34 Walter Roberson <roberson(a)hushmail.com> wrote in message: > The axis is frequency? And this is fibre optics? Fibre has a peak transmission > frequency with increased attenuation as you move away from the peak, rather > than a frequency of 0 Hz having the greatest transmittance as would be implied > by that diagram if the axis is frequency (rather than, e.g., frequency > _difference_ ) > http://en.wikipedia.org/wiki/File:Zblan_transmit.jpg Yes, this is fiber optics but the graph represents electrical-to-electrical domain transmitance, meaning that if you modulate a laser diode or anything else with an electrical signal of a flat spectrum in range from 0 to 200 MHz, then send the modulated light beam through the fibre, you'll receive a signal characterized by the spectrum that was shown in the graph. But it doesn't matter what the medium is, my question is how to model a channel that would attenuate the input signal according to the graph I have presented. Marcin.
From: Marcin on 24 Jul 2010 11:10 Idin Motedayen-Aval <run.signature(a)MATLAB.for.email> wrote in message > As for how to design it, you can try firpm (or try the "Arbitrary > Magnitude" option in FDATool). You may have to specify a pretty high Oh thank you for pointing me on this option of the fdatool. This is exactly what I've needed. It's much easier for me to do this with the gui tool than with the command line tools you both proposed. Anyway, thank you both for you taking interest in my problem! See you around.
From: Marcin on 25 Jul 2010 15:50
Idin Motedayen-Aval <run.signature(a)MATLAB.for.email> wrote in message > You could also do this manually: > - Take your frequency domain measurements at regular frequency intervals > (interpolate if you have to), call that "shape" > - Construct a phase vector of same length: 0, pi, 0, pi, 0, ... > - Then: > mag = [shape, fliplr(shape(2:end))]; > ph = [-Phase, fliplr(Phase(2:end))]; > H = mag.*exp(1i*ph); > h = ifft(H); % This is the filter coefficients Hej Idin, Out of curiosity, why do we need a phase vector and why is it 0, pi, 0, pi and not constant. Isn't the phase response of the filter we design constant? And one another, why do you make this flipped concatenations? shape = 1 2 3 -> mad = 1 2 3 3 2. May I ask what for? Is it to fill the negative frequencies (then why 2:end not 1:end)? Sorry if I'm being too pushy with the questions ;) Kind regards, Marcin. |