Thread View: comp.dsp
2 messages
2 total messages
Started by M.Iqbal
Mon, 02 Nov 1992 00:56
PAM
Author: M.Iqbal
Date: Mon, 02 Nov 1992 00:56
Date: Mon, 02 Nov 1992 00:56
69 lines
2760 bytes
2760 bytes
Hi,
I hope this question belongs to this news group as I could not find a
group which talks only about A/D conversion specifically.
PAM is a quite simple concept, yet, no two authors seems to agree on
one thing. I have read several books and could not understand what
the authors in general mean.
I made the following two diagrams that I have not seen in any
communication book. This is what I think about the PAM natural
sampling and flat-topped sampling. Please, if someone disagrees with
me send me an email, though, I will appreciate any comments.
1) The following is natural sampling PAM diagram that I made with a LPF
and an impulse train. What I have seen in almost every communication book
is a pulse train at the mixer. However, I modeled pulse train with an
impulse train and a LPF. I think that in doing so one can understand
flat-topped PAM better which is shown in the figure of #2.
Mixer
Analog(message) ----------(X)------> PAM with natural sampling
Signal input |
|
|
-------
| LPF |
-------
|
|
Impluse train--------|
2) All I did here is moved the LPF in the output branch from the input
branch. Since LPF transfer function is similar to an integrator,
any product of the impulse and the analog-message siganal will be held.
Thus, we will have sample and hold circuitry in the output branch.
-------
Analog(message) ----------(X)------| LPF |----> Flat-topped PAM
Signal input | -------
|
|
|
|
Impluse train--------|
I have eliminated math as it is not possible to the summation
signs and other symbols in ascii.
iqbal@sioux.eel.ufl.edu
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Iqbal@sioux.eel.ufl.edu (o o) (o) (o)
University of Florida [ | ] |
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Re: PAM
Author: David Kastrup
Date: Tue, 03 Nov 1992 17:17
Date: Tue, 03 Nov 1992 17:17
71 lines
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3297 bytes
iqbal@pawnee.eel.ufl.edu (M.Iqbal) writes: >I made the following two diagrams that I have not seen in any >communication book. This is what I think about the PAM natural >sampling and flat-topped sampling. Please, if someone disagrees with >me send me an email, though, I will appreciate any comments. >1) The following is natural sampling PAM diagram that I made with a LPF >and an impulse train. What I have seen in almost every communication book >is a pulse train at the mixer. However, I modeled pulse train with an >impulse train and a LPF. I think that in doing so one can understand >flat-topped PAM better which is shown in the figure of #2. > Mixer > > Analog(message) ----------(X)------> PAM with natural sampling > Signal input | > | > | > ------- > | LPF | > ------- > | > | > Impluse train--------| >2) All I did here is moved the LPF in the output branch from the input >branch. Since LPF transfer function is similar to an integrator, >any product of the impulse and the analog-message siganal will be held. >Thus, we will have sample and hold circuitry in the output branch. > ------- >Analog(message) ----------(X)------| LPF |----> Flat-topped PAM > Signal input | ------- > | > | > | > | > Impluse train--------| >I have eliminated math as it is not possible to the summation >signs and other symbols in ascii. Boy have you eliminated math! The first circuit lowpass filters the impulse train. What remains is an assortment of frequencies being multiples of the impulse train rate. With that you multiply your signal, getting a highly modulated mishmash. If you use a short-time integrator as your LPF, the impulse train gets changed to a constant, and your circuit does NOTHING whatsoever. The second one is the possibly more correct version (depending on what you want to do). IFF your signal is already bandlimited (the possible range of frequency less than half the sampling rate), the sampling with the impulse train singles out values representing the original signal completely. By using a short time integrator, you get flat topped PAM. For analog transmission, however, a better LPF would lead better results. Your fundamental mistake in your diagram might be the assumption that multiplication with a signal is the same as using a filter. It is not.
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