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The Gabor Limit states that it's impossible to simultaneously localize a signal in both frequency and time. FM communication modulates the instantaneous frequency of the carrier in step with changes in the signal. This suggests it's impossible to perfectly determine the instantaneous signal even in a theoretical channel with zero noise, or that one has to assume that the signal is changing slowly (band-limited). How is this overcome in practice? If demodulation requires the assumption of a band-limited signal, what is the maximum spectral efficiency of frequency modulation for a given input bandwidth?

EDIT: To clarify, this is a theoretical question. The Gabor Limit seems to imply that "instantaneous frequency" isn't well-defined if I understand it correctly. I'm not sure I do, though. The question boils down to:

  1. Do I understand the Gabor Limit correctly? Is "intantaneous frequency" an unmeasurable quantity?

  2. If I do understand the Gabor Limit correctly, how does FM modulation and demodulation work in spite of it? Is there a requirement that the demodulator assume that the signal being transmitted is band-limited even on a theoretical noise-free channel?

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  • \$\begingroup\$ "FM communication modulates the instantaneous frequency of the carrier in step with changes in the signal." I am having a hard time understanding what you are saying here. \$\endgroup\$ – Kellenjb Jan 20 '12 at 19:30
  • \$\begingroup\$ Would it help analysis if the signal being modulated were free of DC bias, and the modulation were slight enough that the signal never gained or lost a full cycle as compared with an unmodulated carrier? If so, one might regard an FM signal as being phase-shift modulated by the integral of the real signal to be sent; a square wave would be sent as a triangle wave under such a scenario. \$\endgroup\$ – supercat Jan 20 '12 at 19:32
  • \$\begingroup\$ it is not so related but, why if signal is band limited it is changing slowly?! you can have BW = 100hz near 100THZ. one more thing, if the demodulator says it is 10khz instead 10.000001khz it is the same for almost every practical target :) \$\endgroup\$ – 0x90 Jan 20 '12 at 19:33
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    \$\begingroup\$ @Kellenjb: I think he's saying that if one were to e.g. try to send a square wave, the carrier frequency would instantaneously jump between values, without requiring the output waveform itself to have a jump. The difficulty I think is that the information content in the signal stems from its continuously varying phase with regard to a reference wave; to decode an FM signal, one must take the derivative of that phase changes. When taking a derivative of a noisy signal, one must use a low-pass filter or else the derivative will be swamped by high-frequency junk. \$\endgroup\$ – supercat Jan 20 '12 at 19:42
  • \$\begingroup\$ @supercat That is what I had a feeling he was saying, but the way it was worded was causing me to have to think very hard about it. \$\endgroup\$ – Kellenjb Jan 20 '12 at 20:55
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This suggests it's impossible to perfectly determine the instantaneous signal even in a theoretical channel with zero noise

I'd turn this around and say it's impossible to instantaneously determine the signal.

one has to assume that the signal is changing slowly (band-limited). How is this overcome in practice?

In practice, our message signals are bandlimited, so this is not a difficulty. In fact, our message signals generally have much less bandwidth than the carrier.

To approach your theoretical question, is a band limit strictly required, Imagine trying to modulate a 1 Hz carrier with a 1 MHz signal --- the result would be unusable. So in fact there must be some kind of limit.

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Maybe this link will help you. In short, scientists made experiments and found out that human hearing easily beats Gabor limit, so, as a conclusion, our brain does not use Fourier transform to process sound waves, it is much more complex. So no need to worry about it)

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  • \$\begingroup\$ Link is now dead. \$\endgroup\$ – pjc50 Nov 20 '13 at 15:15
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    \$\begingroup\$ What's human hearing got to do with the question? \$\endgroup\$ – immibis Sep 5 '15 at 23:35
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Kind of a while ... but ... Remember that FM takes any input signal, even perfectly bandlimited, and sends sidebands out to infinity. AM, by contrast, takes a perfectly bandlimited input and keeps it bandlimited, just shifted up to the carrier in frequency space. This illustrates the Gabor Limit as applied to the fact that it is the frequency state (not amplitude state) of the modulated signal that is being analyzed, and takes the instantaneous nature of the baseband signal and transfers it to infinity in the frequency domain. There is no practical way to observe an infinite bandwidth signal, therefore, the demodulated signal must be distorted compared to the original baseband signal. That uncertainty is arbitrarily reduced only by approaching a zero-noise infinite bandwidth analysis capability. This contrasts against AM where zero-noise is all that is needed. (non-linear distortion and channel phase distortion are left as an exercise ... :-)

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FM waveforms are very delocalized in frequency -- they have more than enough bandwidth to locate features in time with the required resolution.

The instantaneous frequency is a different thing entirely. If you have a waveform: \$v(t) = cos(\phi(t))\$, then the instantaneous frequency is by definition \$f(t) = \dot{\phi}(t)\$.

To make this clear, lets ignore the FM part, and just consider a simple waveform:

\begin{equation} x(t) = \left\{ \begin{array}{cc} 0 & \mbox{if }t < 0\\ \sin(2\pi t) & \mbox{if }0<t<1\\ 0 & \mbox{if }t > 1 \end{array}\right. \end{equation}

You can describe this waveform as having an instantaneous frequency of 1 over the interval [0,1], and an instantaneous frequency of 0 elsewhere. That isn't a statement about the frequency spectrum. It is actually a time domain description in disguise! If I Fourier transform this waveform, I find that as expected it is not perfectly localized, but has some spread, represented as a sinc function centered at 1.

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