[sdiy] Filter terminology question

[Magnus Danielson]

From: Michael Baxter <mab at cruzio.com>
Subject: Re: [sdiy] Filter terminology question
Date: Wed, 19 Nov 2003 20:14:01 -0800 (PST)
Message-ID: <Pine.LNX.4.58.0311191951020.2958 at linux.local>

Hi Tom and Michael,

Oh, a signal theory question ;O) It's been some time I saw one...

> Hi Tom,
> 
> There's relatively easy answers to your question, but I will explain some 
> of the system theory that supports it.
> 
> On Wed, 19 Nov 2003, Tom Arnold wrote:
> 
> > Okay. Here's a question that I thought I knew the answer to and when I
> > realised I didnt, I started getting bogged down in the answers I was
> > finding.
> > 
> > When talking about a filter, what is the difference between Poles and Order?
> 
> Poles and Order are different first of all. Poles are related to Zeros, 
> but not directly to Order. Every filter has a transfer function, and 
> usually this is described as a frequency-dependant ratio between input (a 
> forcing function) and output. A pole is neighborhood in the frequency 
> response where the ratio becomes very large, e.g. a peak in the frequency 
> response. A zero is similarly a neighborhood in the frequency response 
> where the ratio becomes zero -- no output.

So far I agree with Michael. There is a few details skipped over, but I think
thats OK for the general overview.

> A pole would be represented for instance by a band-pass filter, and a zero 
> by a notch filter.

This is however where I start to disagree with Michael.

> Now it's easy to see how this might be slightly confusing. Order has to do
> with the number of energy storage elements, or cascaded sections. A
> "second order filter" is somewhat like a "two pole filter," but only if
> the transfer function has poles... it might have zeros instead.

Actually, in real life (and this is actually a very generic statement I am
going to make, so you may just consider it a law if it makes life easier for
you) you always have both poles and zeros in action at the same time in a
response, infact you have an equal amount of zeros and poles (I know this is
confusing to some people, but it is true).

What you do have is that different arrangements of poles and zeros give
different responses, such as lowpass, highpass, bandpass, bandstopp, allpass,
resonant, notch etc. For all of those except the resonant filter the poles are
fairly far away and the amplitude responce is dominated by the placement of
zeros. A lowpass filter has it's zeros located up in infinity (or there abouts)
where as a highpass filter has them down at 0Hz or there abouts. Bandpass has
a few up in infinity and a few down in 0Hz while bandstopp has them located
in the frequency band you want to block out. Allpass then have them located in
a geometrically balanced fashion in relation to the poles (they mirror the
poles round the jw-axis).

> The poles or zeros might be place at the same or different frequencies. If
> they're the same frequency, this generally indicates that the filter has more
> selectivity. A 2-pole bandpass filter generally has a narrower "skirt" or 
> bandwidth (which can be defined as either the upper and lower frequencies 
> where the transfer function is -3 dB or -6 dB, depending on context) than 
> a single-pole bandpass filter. Similarly, a 2-pole low-pass filter also 
> has a greater fall-off rate as frequency increases, than does a 1-pole 
> low-pass filter.

Each zero represents a 6 dB/Oct slope, the placement decided where that slope
is heading as seen from the pass-band. The slope goes down towards the zero.

> So, sometimes "Order" can be in effect mean the same thing as the number 
> of "Poles," it doesn't always have to.

The propper use of order should indicate the full number of poles (and thus
full number of zeros) that is in play. Those number of poles may not be
visible as slopes (example: allpass-filter). The word order comes from
Analysis and the polynomial order of the two polynomials z(s) and p(s) which
forms the LaPlacian responce of H(s) = z(s) / p(s). You have a zero in s when
z(s) = 0 and you have a pole in s when p(s) = 0. The order of the system is
the maximum of the orders of z(s) and p(s), which in reality is the same
(the theory does allow for them to be different).

View order as an indication of just the computing power you might be able to
use. A order of 4 (say a Moog ladder) can potentially do more than a order of 2
filter (say a SEM VCF), but that does not mean you are deviced to actually
make use of all that power. If you team up two 2-order filters they can act as
a 4-order filter. Infact, by combining 1-order and 2-order filters you can
build any other form of filter from a theoretical standpoint (it may not be
practical thought).

> A more detailed elaboration beyond this would involve forcing functions 
> that are complex exponentials, v = exp(st), where s is a complex operator, 
> and exp(st) represents complex frequency. Also confusing, complex 
> exponentials are just a way to say sine and cosine waves at the same time. 
> This notation, and way of working, involves what is called the S-plane, 
> and is based on the computational artifice called Laplace transforms, 
> which greatly simplify manipulation of the differential equations involved 
> in R-L-C networks. Basically, Laplace transforms turn differential 
> equations involving complex quantities into simpler algebra. And, there 
> are some graphical methods with the S-plane that tell you a lot about how 
> a filter (or a system) is going to behave.

The LaPlace transform is cheating big-way - or rather - it is a very powerfull
transform in the hands of capable people.

> In the Laplace notation, poles and zeros also become much more obvious,
> because each typical type of filter has a characteristic equation in
> simple algebra form. A zero is where all or part of numerator goes to
> zero, rendering and products in the numerator as zero, thus the transfer 
> function becomes zero -- the filter has no output. A pole then is one or 
> more products in the denominator of a charactertistic equation that also 
> go to zero at some specific frequency -- 1/n where n is vanishingly small 
> becomes essentially infinite, yielding a "pole" in the transfer function.

Exactly. The forward LaPlace transform will transform the time-properties into
laplacian frequency s properties. These are then for most (but not all, pure
time delay is an exception) of the time very straightforward from that point.

It's worthwhile to learn what poles and zeros is all about, because many things
start to make a hell of a lot more sense once you can bounce these around as
you see fit. I do it all the time!

Cheers,
Magnus