What Nyquist Didn't Say

Thanks for the comment. The motivation for the article came about because I'd see these 15 word questions ("I need to monitor a power line, so I can sample it at 120Hz, right?"). Unfortunately, when I tried to answer them I'd get up to a few hundred words I'd give up, because there's just too dang much background that needs to be given -- so now I can just point to the article, maybe with a "see section X.x", and cover the question.

As far as pdf documents go -- I have been doing them in HTML out of fear that folks will just print out the PDF document and never come back to my web site. It's a royal PITA. Now I'm thinking that maybe I'll be better off to go ahead and print them as PDF documents, with embedded hyperlinks and all that cool stuff so that folks can print it out and _still_ get pointed back to my website.

Hmm.

must

You're correct, but there's more to it than that. Strictly greater is necessary for reconstruction, but equal is good enough to avoid aliasing. Moreover, while it's true that sampling at 2f_0 + f_e allows reconstruction, you need a significant fraction of 1/f_e seconds of signal in order to do it. It's the same as trying to distinguish f_e/2 from DC.

Jerry

I read the page a couple of times, and I like it. Apart from some very minor readability issues it gets to the meat of the subject without hashing it to death mathematically - I've seen the comment about possibly removing all the formulae, but I am not sure it could be done.

As a practical example (that I am sure most here have seen), I'll see if I can dig out a picture I took of a digital sampling scope with terrible aliasing in certain frequency / resolution bands (i.e. the problem would only show up at certain timebase settings) which can be very confusing to the beginner - certainly it's a known issue with sampling scopes.

So kudos to you for finishing it off.

Cheers

PeteS

Modulation effect? All you have are two different domains (s and z).

aliasing is

er*

Ring before the signal arrives? That sound non-causal to me.

ion

.=2E

For audio signal processing, the filters are generally active, so no inductors are used. [speaker crossovers excepted.]

In many applications, ringing cannot be tolerated. Scales for instance.

ISTR Roger Lagadec at Studer (and Sony) pointing this out in the early

martin

Ring before the main part of the signal arrives. FIR filter responses are often shown as if zero delay were the middle of the filter, effectively subtracting out the constant delay added by the filter.

Please read more carefully. The filter rings before the main part of the output step *emerges* but after the step arrives at the input. The filter's inherent delay makes that quite possible.

Jerry

-- "The rights of the best of men are secured only as the rights of the vilest and most abhorrent are protected." - Chief Justice Charles Evans Hughes, 1927 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Its only non-causal if it arrives before it was sent. :-)

There is no reason why the main signal should not arrive latter than some of the crud which accompanies it. Its perfectly normal in a dispersive medium.

Steve

May I ask what software you used to render the maths on that page? It looks clearer than the stuff I produce. MathML is getting into browsers now, but the rendering of that looks so bad with anything I have tried, that inserted images in HTML pages still seems the only practical approach.

I'd still like to see a web page I can point people to when they say a

Steve

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Nothing wrong with my reading. Now if you are somehow looking at the output to interpret where the large transition occurred, that is a different story. However, any filter where the impulse response goes negative will have such ringing, be it linear phase or not. You need to visualize the convolution.

SciLab

Well, get writing!

It's called *pre-ringing* and it appears because the chunks are processed forward and backward in a row, so a unity pulse will have an identical rising and falling edge. If the filter is of the ringing type, thus the ringing occurrs twice. You are right in saying it's impossible, but only in an analog world. Digital filters do have a latency which will always be longer than the delay of the corresponding analog filter; with linear phase it will be twice the FIR size plus twice the conversion time and more than double than the analog counterpart. Well done analog filters are of the *minimum phase* type, having just the lowest possible delay for that shape of output response. This is possible to realize digitally with IIR filters only. And do not think that even a Gauss filter has only positive FIR-coefficients. This would be only true for a filter of infinite length, which apparently isn't that desirable at all. For practicable sizes the location of the poles and zeros has to be modified and one might get even negative coefficients, depending on the ratio of sampling- and filter frequency and filter length.

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lay

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The Gaussian to which I refer is S domain. If you mapped it to Z domain, it would have to be IIR, not FIR.

Even steep linear phase analogue filters will exhibit pre-ringing. If you were to linearise the phase response of, say, a Butterworth filter, by adding one or more all-pass sections, its impulse repsonse will ring before and after the main output. Of course, the overall delay must go up for the filter to remain causal.

Jeroen Belleman

Joerg wrote: (someone wrote)

(and be distinguished on the other end).

The important point being that the math is the same even though the goal is different. I suppose, then, the sample rate should be a lemma to Nyquist's telegraph channel theorem.

By the way, Gauss published the first paper on the FFT.

-- glen

...

Minimum-phase (or nearly minimum) FIRs are possible, just not symmetric FIRs. You can make maximum-phase FIRs too. Then *all* the ringing is on the leading edge.

Jerry

(Actually, Gauss never published it. It was only published posthumously as part of his notes.)