FPGA Central

Gentle comp.dsp Readers,

We just had some homework on bilinear transformations in my Digital
Signal Processing class and I must admit I am confused about
something. I seek the expert council of the people attending this
group.

Some authors (e.g., [1, 2] define the bilinear transform with a 2/T
factor,

2 z - 1
s = - ------ . (1)
T z + 1

However, others (e.g., [3]) omit the 2/T factor

z - 1
s = ----- . (2)
z + 1

Proakis and Manolakis explain nicely how the transform (including the
2/T) comes from numerical integration. Mitra states that, since we
usually begin with a set of requirements in the discrete domain,
inverse map those back to the equivalent analog filter requirements,
then forward transform the result back to digital, the factor of T is
irrelevent. However, nowhere have I seen it stated explicitly that
"T" is "sampling period," even though that is what that notation
typically means in these contexts.

So basically I have two questions: 1) Is T the sample period, or is it
a parameter that is independent of sample period? 2) When is T
required and when isn't it?

--Randy



[1] John G. Proakis and Dimitris G. Manolakis. Digital Signal
Processing: Principles, Algorithms, and Applications. Prentice Hall,
third edition, 1996.

[2] Sanjit K. Mitra. Digital Signal Processing: A Computer-Based
Approach. McGraw- Hill, second edition, 2001.

[3] David J. DeFatta, Joseph G. Lucas, William S. Hodgkiss. Digital
Signal Processing: A System Design Approach. Wiley, 1988.
--
% Randy Yates % "Maybe one day I'll feel her cold embrace,
%% Fuquay-Varina, NC % and kiss her interface,
%%% 919-577-9882 % til then, I'll leave her alone."
%%%% <[email protected]> % 'Yours Truly, 2095', *Time*, ELO
http://home.earthlink.net/~yatescr

Hello Randy,

I experienced what you have discovered, and basically I handle converting
analog filter designs based on mapping a frequency point in the analog
domain to a point in the discrete domain.


So starting with the bilinear transform

z-1
s = c ----
z+1

And realizing that in terms of LaPlace transforms, real frequencies are on
the imaginary axis and with z transforms they are on the unit circle. And I
wish to match them up. One point in analog -> one point in discrete.


So s becomes j*OMEGA

and

z^-1 becomes exp(-j*omega)

After substituting both of these into the bilinear transform we get


j*OMEGA = j*c*tan(omega/2)

If we now define a digital frequency f relative to the sampling rate

omega=2*pi*f

Then the "c" (your 1/T) may be found to be

c = OMEGA*cot(pi*f)

Often one is converting an "s" equation that has its cutoff normalized to 1
radian/sec, so in this case OMEGA=1.

For your case T=tan(pi*f)

So you can sort of relate the T to the samping period.

IHTH,
Clay S. Turner






"Randy Yates" <[email protected]> wrote in message
news:[email protected]..
> Gentle comp.dsp Readers,
>
> We just had some homework on bilinear transformations in my Digital
> Signal Processing class and I must admit I am confused about
> something. I seek the expert council of the people attending this
> group.
>
> Some authors (e.g., [1, 2] define the bilinear transform with a 2/T
> factor,
>
> 2 z - 1
> s = - ------ . (1)
> T z + 1
>
> However, others (e.g., [3]) omit the 2/T factor
>
> z - 1
> s = ----- . (2)
> z + 1
>
> Proakis and Manolakis explain nicely how the transform (including the
> 2/T) comes from numerical integration. Mitra states that, since we
> usually begin with a set of requirements in the discrete domain,
> inverse map those back to the equivalent analog filter requirements,
> then forward transform the result back to digital, the factor of T is
> irrelevent. However, nowhere have I seen it stated explicitly that
> "T" is "sampling period," even though that is what that notation
> typically means in these contexts.
>
> So basically I have two questions: 1) Is T the sample period, or is it
> a parameter that is independent of sample period? 2) When is T
> required and when isn't it?
>
> --Randy
>
>
>
> [1] John G. Proakis and Dimitris G. Manolakis. Digital Signal
> Processing: Principles, Algorithms, and Applications. Prentice Hall,
> third edition, 1996.
>
> [2] Sanjit K. Mitra. Digital Signal Processing: A Computer-Based
> Approach. McGraw- Hill, second edition, 2001.
>
> [3] David J. DeFatta, Joseph G. Lucas, William S. Hodgkiss. Digital
> Signal Processing: A System Design Approach. Wiley, 1988.
> --
> % Randy Yates % "Maybe one day I'll feel her cold
embrace,
> %% Fuquay-Varina, NC % and kiss her
interface,
> %%% 919-577-9882 % til then, I'll leave her
alone."
> %%%% <[email protected]> % 'Yours Truly, 2095', *Time*, ELO
> http://home.earthlink.net/~yatescr

Randy Yates wrote:

> Gentle comp.dsp Readers,
>
> We just had some homework on bilinear transformations in my Digital
> Signal Processing class and I must admit I am confused about
> something. I seek the expert council of the people attending this
> group.
>
> Some authors (e.g., [1, 2] define the bilinear transform with a 2/T
> factor,
>
> 2 z - 1
> s = - ------ . (1)
> T z + 1
>
> However, others (e.g., [3]) omit the 2/T factor
>
> z - 1
> s = ----- . (2)
> z + 1
>
> Proakis and Manolakis explain nicely how the transform (including the
> 2/T) comes from numerical integration. Mitra states that, since we
> usually begin with a set of requirements in the discrete domain,
> inverse map those back to the equivalent analog filter requirements,
> then forward transform the result back to digital, the factor of T is
> irrelevent. However, nowhere have I seen it stated explicitly that
> "T" is "sampling period," even though that is what that notation
> typically means in these contexts.
>
> So basically I have two questions: 1) Is T the sample period, or is it
> a parameter that is independent of sample period? 2) When is T
> required and when isn't it?
>
> --Randy
>
>
>
> [1] John G. Proakis and Dimitris G. Manolakis. Digital Signal
> Processing: Principles, Algorithms, and Applications. Prentice Hall,
> third edition, 1996.
>
> [2] Sanjit K. Mitra. Digital Signal Processing: A Computer-Based
> Approach. McGraw- Hill, second edition, 2001.
>
> [3] David J. DeFatta, Joseph G. Lucas, William S. Hodgkiss. Digital
> Signal Processing: A System Design Approach. Wiley, 1988.

z - 1
I don't know the texts. I do know that ----- is dimensionless, while s
z + 1
has the dimension of 1/t. Therefore, the factor T is necessary for
consistency. Its omission might make no numerical difference in the long
run, but dimensional consistency is important in mathematics if only for
pedagogic reasons. One of my first brouhahas in comp.dsp was in defense
of R.B-J's position on a very similar construction involving impulses.

Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Randy Yates <[email protected]> writes:
> [...]

PS: My homework report is accessible at

http://www.uspsdata.org/hw.pdf
--
% Randy Yates % "Watching all the days go by...
%% Fuquay-Varina, NC % Who are you and who am I?"
%%% 919-577-9882 % 'Mission (A World Record)',
%%%% <[email protected]> % *A New World Record*, ELO
http://home.earthlink.net/~yatescr

Jerry Avins wrote:
(snip)

> z - 1
> I don't know the texts. I do know that ----- is dimensionless, while s
> z + 1
> has the dimension of 1/t. Therefore, the factor T is necessary for
> consistency. Its omission might make no numerical difference in the long
> run, but dimensional consistency is important in mathematics if only for
> pedagogic reasons. One of my first brouhahas in comp.dsp was in defense
> of R.B-J's position on a very similar construction involving impulses.

Well, in physics and engineering, at least.

Not so long ago I was thinking about the different way physics and
engineering look at units in equations. It seems to me that in
physics variables describe quantities with units, while in engineering
the units are factored out.

Physics will say F=ma, and F will have the units of m multiplied by
the units of a. (Though it might have a different name.)

Engineers might say F(in Newtons)=m(in kg)* a (in m/s**2).
One then converts the quantities supplied to the appropriate
units before using the equation.

Certainly there is a lot of overlap between physics and engineering,
and even then it may not be so convincing, but it seems to me to
be at least a little bit true.

-- glen

Randy Yates wrote:
> Gentle comp.dsp Readers,
>
> We just had some homework on bilinear transformations in my Digital
> Signal Processing class and I must admit I am confused about
> something. I seek the expert council of the people attending this
> group.
>
> Some authors (e.g., [1, 2] define the bilinear transform with a 2/T
> factor,
>
> 2 z - 1
> s = - ------ . (1)
> T z + 1
>
> However, others (e.g., [3]) omit the 2/T factor
>
> z - 1
> s = ----- . (2)
> z + 1
>
> Proakis and Manolakis explain nicely how the transform (including the
> 2/T) comes from numerical integration. Mitra states that, since we
> usually begin with a set of requirements in the discrete domain,
> inverse map those back to the equivalent analog filter requirements,
> then forward transform the result back to digital, the factor of T is
> irrelevent. However, nowhere have I seen it stated explicitly that
> "T" is "sampling period," even though that is what that notation
> typically means in these contexts.
>
> So basically I have two questions: 1) Is T the sample period, or is it
> a parameter that is independent of sample period? 2) When is T
> required and when isn't it?
>
> --Randy
>
>
>
> [1] John G. Proakis and Dimitris G. Manolakis. Digital Signal
> Processing: Principles, Algorithms, and Applications. Prentice Hall,
> third edition, 1996.
>
> [2] Sanjit K. Mitra. Digital Signal Processing: A Computer-Based
> Approach. McGraw- Hill, second edition, 2001.
>
> [3] David J. DeFatta, Joseph G. Lucas, William S. Hodgkiss. Digital
> Signal Processing: A System Design Approach. Wiley, 1988.

If you are seriously interested in approximating a transfer function in
the s domain with one in the z domain then you want to keep the 2/T.
The Laplace transform of z is z = e^sT, where T is the sampling time,
and 2(s-1)/(T(s+1)) is a fairly close approximation when s*T << 1.

Leaving out the 2/T doesn't sit well with me at all, but I don't have
the context. I _do_ prefer to construct discrete-time controllers
without explicitly factoring in the 2/T business -- it makes it harder
to adjust the sampling rate, but that's usually one of the first things
to get decided, and by the time you're actually tuning things it's
usually set in concrete.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

"glen herrmannsfeldt" schrieb
>
> Physics will say F=ma, and F will have the units of m
> multiplied by the units of a. (Though it might have a
> different name.)
>
> Engineers might say F(in Newtons)=m(in kg)* a (in m/s**2).
> One then converts the quantities supplied to the appropriate
> units before using the equation.
>
Well, physics is concerned with how the world works, while
engineering must make a gadget work.
Ultimately you'll want to build the gadget and then the
dimensions matter a lot.
For thinking about how the world works, you can work in any
unit system you like (as long as the equations are consistent),
the laws should come out identical.
In Engineering, OTOH, it is important that all players use the
same units (witness the trouble with a space shuttle once
where one group used inches and the other millimeters). Hence
the obsession with engineers with units. I use to do a unit
consistency check always first, if it fails, I consider the
equation useless.

Just my 0.02$.
Martin

Hi Randy,

The "T" is definitely the reciprocal of the
Fs sampling frequency.

I found that keeping T in the equations allows us
to mathematically describe the "frequency warping"
that occurs when we determine what digital-domain
frequency (that's always in the range of -Fs/2 to +Fs/2)
that equates to some analog frequency (that can be in
the range of -infinity to +infinity).

Not that it improves on any of your references, but
ya' might take a peek at the "Bilinear transform"
discussion in Section 6.5 of my book.
(Maybe you've already done that.) That material
might contain a little snippet of useful information.
Who knows.

Your homework solution(s) required 16 pages huh?
Whew! That must have taken you a while.

See Ya',
[-Rick-]

----------------------------------------
On Thu, 14 Oct 2004 23:15:30 GMT, Randy Yates <[email protected]> wrote:

>Gentle comp.dsp Readers,
>
>We just had some homework on bilinear transformations in my Digital
>Signal Processing class and I must admit I am confused about
>something. I seek the expert council of the people attending this
>group.
>
>Some authors (e.g., [1, 2] define the bilinear transform with a 2/T
>factor,
>
> 2 z - 1
> s = - ------ . (1)
> T z + 1
>
>However, others (e.g., [3]) omit the 2/T factor
>
> z - 1
> s = ----- . (2)
> z + 1
>
>Proakis and Manolakis explain nicely how the transform (including the
>2/T) comes from numerical integration. Mitra states that, since we
>usually begin with a set of requirements in the discrete domain,
>inverse map those back to the equivalent analog filter requirements,
>then forward transform the result back to digital, the factor of T is
>irrelevent. However, nowhere have I seen it stated explicitly that
>"T" is "sampling period," even though that is what that notation
>typically means in these contexts.
>
>So basically I have two questions: 1) Is T the sample period, or is it
>a parameter that is independent of sample period? 2) When is T
>required and when isn't it?
>
>--Randy
>
>
>
>[1] John G. Proakis and Dimitris G. Manolakis. Digital Signal
>Processing: Principles, Algorithms, and Applications. Prentice Hall,
>third edition, 1996.
>
>[2] Sanjit K. Mitra. Digital Signal Processing: A Computer-Based
>Approach. McGraw- Hill, second edition, 2001.
>
>[3] David J. DeFatta, Joseph G. Lucas, William S. Hodgkiss. Digital
>Signal Processing: A System Design Approach. Wiley, 1988.
>--
>% Randy Yates % "Maybe one day I'll feel her cold embrace,
>%% Fuquay-Varina, NC % and kiss her interface,
>%%% 919-577-9882 % til then, I'll leave her alone."
>%%%% <[email protected]> % 'Yours Truly, 2095', *Time*, ELO
>http://home.earthlink.net/~yatescr

Martin Blume wrote:

> "glen herrmannsfeldt" schrieb
>
>>Physics will say F=ma, and F will have the units of m
>>multiplied by the units of a. (Though it might have a
>>different name.)
>>
>>Engineers might say F(in Newtons)=m(in kg)* a (in m/s**2).
>>One then converts the quantities supplied to the appropriate
>>units before using the equation.
>>
>
> Well, physics is concerned with how the world works, while
> engineering must make a gadget work.
> Ultimately you'll want to build the gadget and then the
> dimensions matter a lot.
> For thinking about how the world works, you can work in any
> unit system you like (as long as the equations are consistent),
> the laws should come out identical.
> In Engineering, OTOH, it is important that all players use the
> same units (witness the trouble with a space shuttle once
> where one group used inches and the other millimeters). Hence
> the obsession with engineers with units. I use to do a unit
> consistency check always first, if it fails, I consider the
> equation useless.
>
> Just my 0.02$.
> Martin

Even in physics, if the units are inconsistent, you know the equation is
wrong. Consistent units don't guarantee correctness, but inconsistent
units are a sure sign of error. I learned that in physics class.

Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

r.lyons@_BOGUS_ieee.org (Rick Lyons) writes:

> Hi Randy,
>
> The "T" is definitely the reciprocal of the
> Fs sampling frequency.
>
> I found that keeping T in the equations allows us
> to mathematically describe the "frequency warping"
> that occurs when we determine what digital-domain
> frequency (that's always in the range of -Fs/2 to +Fs/2)
> that equates to some analog frequency (that can be in
> the range of -infinity to +infinity).
>
> Not that it improves on any of your references, but
> ya' might take a peek at the "Bilinear transform"
> discussion in Section 6.5 of my book.
> (Maybe you've already done that.) That material
> might contain a little snippet of useful information.
> Who knows.

Hi Rick,

Of course I would have referred to your book if I had
the opportunity, but your book was at the office and I
did the assignment from home. I will have a look.

How do other folks solve this problem? It seems that no
matter where I keep my books, they're always in the wrong
place when I need them!

> Your homework solution(s) required 16 pages huh?
> Whew! That must have taken you a while.

Nah, I'm just a windbag. Besides, once you take out the
title page, TOC, all those figures, and the original problem
text, there isn't much left.

--Randy


--
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
[email protected], 919-472-1124

Randy Yates <[email protected]> wrote in message news:<[email protected]>...
> Gentle comp.dsp Readers,
>
> We just had some homework on bilinear transformations in my Digital
> Signal Processing class and I must admit I am confused about
> something. I seek the expert council of the people attending this
> group.
>
> Some authors (e.g., [1, 2] define the bilinear transform with a 2/T
> factor,
>
> 2 z - 1
> s = - ------ . (1)
> T z + 1
>
> However, others (e.g., [3]) omit the 2/T factor
>
> z - 1
> s = ----- . (2)
> z + 1
>
> Proakis and Manolakis explain nicely how the transform (including the
> 2/T) comes from numerical integration. Mitra states that, since we
> usually begin with a set of requirements in the discrete domain,
> inverse map those back to the equivalent analog filter requirements,
> then forward transform the result back to digital, the factor of T is
> irrelevent. However, nowhere have I seen it stated explicitly that
> "T" is "sampling period," even though that is what that notation
> typically means in these contexts.
>
> So basically I have two questions: 1) Is T the sample period, or is it
> a parameter that is independent of sample period? 2) When is T
> required and when isn't it?
>
T is the sampling period

The bilinear transformation is created by modeling integration using
trapezoidal approximation.

starting from scratch, a difference equation of a trapezoidal
integrator sampled every T seconds is

Yn= T/2*(Xn+Xn-1) + Yn-1 = area inside a trapezoid + previous value

or
Yn-Yn-1 = T/2(Xn+Xn-1)

which is
Y (1-z-1) = T/2 X (1+z-1)

then
Y/X = T/2(1+z-1)/(1-z-1)

or
Y/X = T/2(z+1)/(z-1)

since this is an integrator, Y/X = 1/s, so

s = 1/1/s = 2/T(z-1)/(z+1)

which is the bilinear transformation. I mean you have to have the T
for it to work, perhaps removing the 2/T is a normilized version.

Digital Signal Processing, A Practical Approach
Ifeachor & Jarvis
Addison-Wesley 1993
ISBN 0 201 54413 X

Chapter 7
Section 7.4.8(1) Page 393.....

......suggests that if prewarping is used then the factor of 1 may
be used because the factor of 2/T cancels out.

Implying that if the transform is derived from first principles and then
used without pre-warping that the factor of 2/T is required.

In an article by Jack Crenshaw, "On Getting It, Part Deux" (sorry, URL not
available and it's just as easy for you to find it as it is for me), Under
the chapter
headed, "Back To Work" on page 4, the bi-linear transformation is shown as
being derived as the first term in a power series expansion for ln(x).
(Although
there appears to be a typo in the equation (12) in that the factor is
presented
as "1/2" instead of, "2".

Randy Yates <[email protected]> wrote in message
news:<[email protected]>...
> Some authors (e.g., [1, 2] define the bilinear transform with a 2/T
> factor,
> 2 z - 1
> s = - ------ . (1)
> T z + 1
> However, others (e.g., [3]) omit the 2/T factor
> z - 1
> s = ----- . (2)
> z + 1
> 2) When is T required and when isn't it?

The derivation for the BLT (not to be confused with the
appetising sandwich with the same TLA) is to take
the equation....

(1) z = e^(sT)

and to recast it as

(2) s = (1/T) * ln(z)

and then to seek a power series expansion for ln(x)

(3) ln(x) = 2[ (z-1)/(z+1) + {1/3}(z-1)^3/(z+1)^3 + {1/5}(z-1)^5/(z+1)^5
...... ]

and ignoring all but the first term in the expansion to give

(4) ln(x) = 2(z-1)/(z+1)

-----ooooo-----

So, although there may be a similar equation deriving from numerical
integration, I do not believe it to be relevant because our purpose is
parameter substitution and not integration per se. YMMV.

"steve" <[email protected]> wrote in message
news:[email protected] m...

> The bilinear transformation is created by modeling integration using
> trapezoidal approximation.
>
> starting from scratch, a difference equation of a trapezoidal
> integrator sampled every T seconds is
>
> Yn= T/2*(Xn+Xn-1) + Yn-1 = area inside a trapezoid + previous value

I forgot the final derivation.....

Substituting (4) back into (2) to give.....

(5) s = (2/T)[ (z-1)/(z+1)]

"Airy R. Bean" <[email protected]> wrote in message
news:[email protected]..
> The derivation for the BLT (not to be confused with the
> appetising sandwich with the same TLA) is to take
> the equation....
>
> (1) z = e^(sT)
>
> and to recast it as
>
> (2) s = (1/T) * ln(z)
>
> and then to seek a power series expansion for ln(x)
>
> (3) ln(x) = 2[ (z-1)/(z+1) + {1/3}(z-1)^3/(z+1)^3 + {1/5}(z-1)^5/(z+1)^5
> ..... ]
>
> and ignoring all but the first term in the expansion to give
>
> (4) ln(x) = 2(z-1)/(z+1)

"Airy R. Bean" <[email protected]> wrote in message news:<[email protected]>...

> So, although there may be a similar equation deriving from numerical
> integration, I do not believe it to be relevant because our purpose is
> parameter substitution and not integration per se. YMMV.
>
I suppose, but I was always taught (or maybe somehow incorrectly
inferred) that all transformations from H(s) to H(z) can simply be
boiled down to how the continuous integration was approximated. That
idea made alot of sense to me, especially since I could derive the
transformations without any expansion table and the characteristics of
various numerical integration techniques have been well know for a
long time.

OK - but that is new to me - in your article you started off
with a trapezoidal approximation to an integral already
presumed - can you point me in the direction of a URL,
or otherwise explain how this integral came about, because,
as I said, I thought it was a matter of parameter substitution
only to go from H(s) to H(z)? Ta.

"steve" <[email protected]> wrote in message
news:[email protected] om...
> "Airy R. Bean" <[email protected]> wrote in message
news:<[email protected]>...
>
> > So, although there may be a similar equation deriving from numerical
> > integration, I do not believe it to be relevant because our purpose is
> > parameter substitution and not integration per se. YMMV.
> >
> I suppose, but I was always taught (or maybe somehow incorrectly
> inferred) that all transformations from H(s) to H(z) can simply be
> boiled down to how the continuous integration was approximated. That
> idea made alot of sense to me, especially since I could derive the
> transformations without any expansion table and the characteristics of
> various numerical integration techniques have been well know for a
> long time.

NOTE: This message is in a much higher quality PDF format at
http://home.earthlink.net/~yatescr/lyonsr.pdf

Hi Rick et al.,

I'm responding to you directly, Rick, but I actually intend a wider
audience. I owe the group collectively a response on this
topic. Sorry for the delay, but in the intervening time I have been
able to size up my thoughts on the situation and come to some
significant conclusions.

Ho-kay, where do I start? How about with you, Rick. I did read your
([1]) section around p.259 on the topic. You explain the mapping
properties (jw axis to unit circle, etc.) very nicely. You define
the transformation as


2 z - 1
z = -------- (1)
t z + 1
s

(equation 6-88 in your 2ed), asserting that t is sample period, and in
a footnote state that it comes from the theory of s complex variables.

The problem is, in the theory of complex variables, at least in [2],
the bilinear transformation is defined as

az + b
w = ------ (2)
cz + d

where a,b,c, and d are complex constants. (Churchill also calls this
a linear fractional transformation. No 2=t_s , and the coefficients
are generalized. In order to distinguish between the "engineering" BT
and "mathematical" BT, let me denote them "EBT" and "MBT,"
respectively. So I don't really get an intuitive understanding of why
t_s is required to represent sample rate in the EBT, and as you will
see if you keep reading, I actually disagree on this point.

Consider this. The EBT goes from the analog domain, in which frequency
is absolute and sample rate doesn't matter, to the digital domain, in
which frequency is normalized and sample rate doesn't matter. So why
should sample rate come into a transformation between two domains in
which sample rate doesn't matter?

The concept that numerical integration is the basis of the bilinear
transformation doesn't really hold water with me either. In its most
general sense, the BT is a mapping from one domain to another. It is,
as far as I can tell, a coincidence that the numerical integration
process yields a bilinear transformation. I'd like to think it isn't,
and perhaps it really isn't and I just can't see the connection.

Finally, to Jerry, Martin, et al., unit consistencies are nice things.
I agree that to maintain unit consistency there must be a factor of
[1/seconds] in the units. I just don't think it necessarily has to be
related to the sample rate.

So to summarize, I make the following conclusions:

1. The bilinear transformation is a mathematical transformation
between two domains and, taken as two domains in the abstract sense
(i.e., taken as simply R^2 ) does not need to have a factor
involving the sample period. Further, even when taken as a mapping
between the Laplacian s-plane and the z-transform z-plane, the use
of sample period does not make sense.

2. I do not think the BT arises from numerical integration, and I
think that the fact that this process corresponds to a BT is a
coincidence.

3. The 2=t_s factor is useful for matching up analog frequency
points in the analog domain with those in the digital domain, but a
simple factor a, as in

z - 1
s = a ----- (3)
z + 1

will do the same and need not be related to
any t_s , sample period or otherwise.

References

[1] Richard G. Lyons. Understanding Digital Signal Processing.
Prentice Hall, second edition, 2004.

[2] Ruel V. Churchill, James Ward Brown. Complex Variables and
Applications. McGraw-Hill, fifth edition, 1990.


--
% Randy Yates % "Maybe one day I'll feel her cold embrace,
%% Fuquay-Varina, NC % and kiss her interface,
%%% 919-577-9882 % til then, I'll leave her alone."
%%%% <[email protected]> % 'Yours Truly, 2095', *Time*, ELO
http://home.earthlink.net/~yatescr

Randy Yates wrote:

> NOTE: This message is in a much higher quality PDF format at
> http://home.earthlink.net/~yatescr/lyonsr.pdf
>
> Hi Rick et al.,
>
> I'm responding to you directly, Rick, but I actually intend a wider
> audience. I owe the group collectively a response on this
> topic. Sorry for the delay, but in the intervening time I have been
> able to size up my thoughts on the situation and come to some
> significant conclusions.
>
> Ho-kay, where do I start? How about with you, Rick. I did read your
> ([1]) section around p.259 on the topic. You explain the mapping
> properties (jw axis to unit circle, etc.) very nicely. You define
> the transformation as
>
>
> 2 z - 1
> z = -------- (1)
> t z + 1
> s
>
> (equation 6-88 in your 2ed), asserting that t is sample period, and in
> a footnote state that it comes from the theory of s complex variables.
>
> The problem is, in the theory of complex variables, at least in [2],
> the bilinear transformation is defined as
>
> az + b
> w = ------ (2)
> cz + d
>
> where a,b,c, and d are complex constants. (Churchill also calls this
> a linear fractional transformation. No 2=t_s , and the coefficients
> are generalized. In order to distinguish between the "engineering" BT
> and "mathematical" BT, let me denote them "EBT" and "MBT,"
> respectively. So I don't really get an intuitive understanding of why
> t_s is required to represent sample rate in the EBT, and as you will
> see if you keep reading, I actually disagree on this point.
>
> Consider this. The EBT goes from the analog domain, in which frequency
> is absolute and sample rate doesn't matter, to the digital domain, in
> which frequency is normalized and sample rate doesn't matter. So why
> should sample rate come into a transformation between two domains in
> which sample rate doesn't matter?

Some rate is needed for consistency of units. Do you have an alternate
proposal? The choice of 2/T matches up the frequencies at the low end.

> The concept that numerical integration is the basis of the bilinear
> transformation doesn't really hold water with me either. In its most
> general sense, the BT is a mapping from one domain to another. It is,
> as far as I can tell, a coincidence that the numerical integration
> process yields a bilinear transformation. I'd like to think it isn't,
> and perhaps it really isn't and I just can't see the connection.

It has been said that there are no coincidences in mathematics. Putting
that aside, I agree with your analysis.

> Finally, to Jerry, Martin, et al., unit consistencies are nice things.

More than nice. Imperative.

> I agree that to maintain unit consistency there must be a factor of
> [1/seconds] in the units. I just don't think it necessarily has to be
> related to the sample rate.

To maximize its engineering utility, a domain mapping should preserve as
much of the original as possible. The MBT takes us from absolute
frequency to a dimensionless frequency ratio. Consistency demands the
inclusion of a rate. The EBT choses the rate to match the original and
transformed absolute frequencies.

> So to summarize, I make the following conclusions:
>
> 1. The bilinear transformation is a mathematical transformation
> between two domains and, taken as two domains in the abstract sense
> (i.e., taken as simply R^2 ) does not need to have a factor
> involving the sample period. Further, even when taken as a mapping
> between the Laplacian s-plane and the z-transform z-plane, the use
> of sample period does not make sense.

Hmm. At least if there is sense, we haven't found it yet. Or maybe it
just makes the numbers work out. We need newtons because we insist that
f = ma. If we defined f = gma, there's be no need. There are three
systems in use with English units. One defines mass in slugs, another
defines force in poundals, and the third, the engineering system,
defines f = gma. Life is simpler that way, even if not as pure.

> 2. I do not think the BT arises from numerical integration, and I
> think that the fact that this process corresponds to a BT is a
> coincidence.
>
> 3. The 2=t_s factor is useful for matching up analog frequency
> points in the analog domain with those in the digital domain, but a
> simple factor a, as in
>
> z - 1
> s = a ----- (3)
> z + 1
>
> will do the same and need not be related to
> any t_s , sample period or otherwise.

Perhaps not, but the constant must be a real frequency.

...

Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Jerry Avins <[email protected]> writes:

> Some rate is needed for consistency of units. Do you have an alternate
> proposal? The choice of 2/T matches up the frequencies at the low end.

I would say that we simply use a "units transformation" of [1/second]. That
could be interpreted as a rate of 1 Hz.

> It has been said that there are no coincidences in mathematics.

Yeah, that's bugging me.

> > Finally, to Jerry, Martin, et al., unit consistencies are nice things.
>
> More than nice. Imperative.
>
> > I agree that to maintain unit consistency there must be a factor of
> > [1/seconds] in the units. I just don't think it necessarily has to be
> > related to the sample rate.
>
> To maximize its engineering utility, a domain mapping should preserve as
> much of the original as possible. The MBT takes us from absolute
> frequency to a dimensionless frequency ratio. Consistency demands the
> inclusion of a rate. The EBT choses the rate to match the original and
> transformed absolute frequencies.

It uses sample rate - I don't think that makes sense.

> > So to summarize, I make the following conclusions:
> >
> > 1. The bilinear transformation is a mathematical transformation
> > between two domains and, taken as two domains in the abstract sense
> > (i.e., taken as simply R^2 ) does not need to have a factor
> > involving the sample period. Further, even when taken as a mapping
> > between the Laplacian s-plane and the z-transform z-plane, the use
> > of sample period does not make sense.
>
> Hmm. At least if there is sense, we haven't found it yet.

I would say "put up or shutup." It's kinda a copout to say "you might
be wrong but I don't know how." Sorry, Jerry - not intending to be
mean - just not mincing words.

> Or maybe it
> just makes the numbers work out.

No, it makes the units work out. The mapping that it results in
is arbitrary.

> We need newtons because we insist that
> f = ma. If we defined f = gma, there's be no need. There are three
> systems in use with English units. One defines mass in slugs, another
> defines force in poundals, and the third, the engineering system,
> defines f = gma. Life is simpler that way, even if not as pure.

I've already acknowledged that making the units work out is the right
thing to do.

> > 2. I do not think the BT arises from numerical integration, and I
> > think that the fact that this process corresponds to a BT is a
> > coincidence.
> >
> > 3. The 2=t_s factor is useful for matching up analog frequency
> > points in the analog domain with those in the digital domain, but a
> > simple factor a, as in
> >
> > z - 1
> > s = a ----- (3)
> > z + 1
> >
> > will do the same and need not be related to
> > any t_s , sample period or otherwise.
>
> Perhaps not,

And if it isn't, then that's an extremely significant discrepancy in
understanding between what really "is" and the view put forward in
several texts.

> but the constant must be a real frequency.

Fine. Let a = 1 [1/second] for the obvious practical reason. This
means that an analog frequency of 1 Hz corresponds to a normalized
digital frequency of approximately 0.45.
--
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
[email protected], 919-472-1124

Randy Yates wrote:

> NOTE: This message is in a much higher quality PDF format at
> http://home.earthlink.net/~yatescr/lyonsr.pdf
>
> Hi Rick et al.,
>
> I'm responding to you directly, Rick, but I actually intend a wider
> audience. I owe the group collectively a response on this
> topic. Sorry for the delay, but in the intervening time I have been
> able to size up my thoughts on the situation and come to some
> significant conclusions.
>
> Ho-kay, where do I start? How about with you, Rick. I did read your
> ([1]) section around p.259 on the topic. You explain the mapping
> properties (jw axis to unit circle, etc.) very nicely. You define
> the transformation as
>
>
> 2 z - 1
> z = -------- (1)
> t z + 1
> s
>
> (equation 6-88 in your 2ed), asserting that t is sample period, and in
> a footnote state that it comes from the theory of s complex variables.
>
> The problem is, in the theory of complex variables, at least in [2],
> the bilinear transformation is defined as
>
> az + b
> w = ------ (2)
> cz + d
>
> where a,b,c, and d are complex constants. (Churchill also calls this
> a linear fractional transformation. No 2=t_s , and the coefficients
> are generalized. In order to distinguish between the "engineering" BT
> and "mathematical" BT, let me denote them "EBT" and "MBT,"
> respectively. So I don't really get an intuitive understanding of why
> t_s is required to represent sample rate in the EBT, and as you will
> see if you keep reading, I actually disagree on this point.
>
The simple answer is that for engineering problems, when you set a = 2,
c = T_s, and b = d = 1, then you have a fairly good approximation for s
for the purposes of finding a z-domain transfer function in s.

> Consider this. The EBT goes from the analog domain, in which frequency
> is absolute and sample rate doesn't matter, to the digital domain, in
> which frequency is normalized and sample rate doesn't matter. So why
> should sample rate come into a transformation between two domains in
> which sample rate doesn't matter?
>
Because the sample rate makes a critical difference if you're actually
sampling a real signal.

> The concept that numerical integration is the basis of the bilinear
> transformation doesn't really hold water with me either. In its most
> general sense, the BT is a mapping from one domain to another. It is,
> as far as I can tell, a coincidence that the numerical integration
> process yields a bilinear transformation. I'd like to think it isn't,
> and perhaps it really isn't and I just can't see the connection.
>
It's a very happy coincidence. It's what makes your "EBT" useful to to
real work. If it didn't hold your "MBT" would be a mere curiosity.

> Finally, to Jerry, Martin, et al., unit consistencies are nice things.
> I agree that to maintain unit consistency there must be a factor of
> [1/seconds] in the units. I just don't think it necessarily has to be
> related to the sample rate.
>
With a real sampler you have to relate to the sample rate. _Everything_
changes with sample rate in the real world. I have never had a
mathematical solution hold water if it could not first hold unit
consistency.

> So to summarize, -- snip --
>

To do an exact conversion of an s-domain transfer function you must take
the sample rate into account. In fact, you need to also take the
characteristics of the sampler and the output device (usually a fairly
good approximation of a zero-order-hold) into account. This exact
conversion is _not_ a bilinear transform, and involves evaluating
partial fractions or e^{some matrix}, so many people want a decent
approximation for it. The bilinear transformation _is_ a fairly good
approximation, as long as the poles of your transfer function are
relatively slow compared to the sample rate.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Randy Yates wrote:

> Consider this. The EBT goes from the analog domain, in which
> frequency is absolute and sample rate doesn't matter, to the
> digital domain, in which frequency is normalized and sample rate
> doesn't matter. So why should sample rate come into a
> transformation between two domains in which sample rate doesn't
> matter?

Put that way, I'd guess it determines the relation between absolute
and normalized frequency, i.e., it defines what "normalization" is.
What do you think of this?

> The concept that numerical integration is the basis of the
> bilinear transformation doesn't really hold water with me
> either. In its most general sense, the BT is a mapping from one
> domain to another. It is, as far as I can tell, a coincidence
> that the numerical integration process yields a bilinear
> transformation. I'd like to think it isn't, and perhaps it
> really isn't and I just can't see the connection.

Steve showed in his message
<[email protected]> how the bilinear
transform can be derived from trapezoidal integral approximation. But
there is nothing enforcing that particular rule. If I connect the
samples with quadratics rather than straight lines by:

1) fitting a parabola through the second
last, previous, and current samples,
2) symbolically integrating that parabola
between the previous and current sample time,
3) and adding the result of 2) to the approximate
integral up to the previous sample time,

then I get the substitution for s:

s = 12/T * (z^2 - z) / (5*z^2 + 8*z - 1),

modulo goof-ups and without trying to see what this actually does to
the complex plane. So it's not, in a sense, a coincidence that an
integration rule employing linear functions gives a bilinear
substitution. I hope that's pertinent to your doubts.


Martin

--
Quidquid latine dictum sit, altum viditur.

Randy Yates <[email protected]> writes:

> Jerry Avins <[email protected]> writes:
>
> > Some rate is needed for consistency of units. Do you have an alternate
> > proposal? The choice of 2/T matches up the frequencies at the low end.
>
> I would say that we simply use a "units transformation" of [1/second]. That
> could be interpreted as a rate of 1 Hz.

I have a slightly different and hopefully more simplistic take on
this based on D. Schlichtharle's book on Digital Filters.

We know that the mapping from discrete to/from continuous time is defined by:

z = e^(jwT)

or

jw = ln z / T

We want ln z to be a rational function since those are the kind we
know how to implement in real systems.

One series expansion for ln z is:

ln z = (z-1)/z + (z-1)^2/2z^2 + (z-1)^3/3z^3 + ...

real(z)>1/2

by truncating this at the first term, we get:

ln z = (z-1)/z (approx.)

This is the "diffence method" for finding the z transform.

Another series expansion for ln z is:

ln z = 2( (z-1)/(z+1) + 1/3*((z-1)/(z+1))^3 + ...)

real(z) > 0

by truncating this at the first term we get the bilinear transform
complete with with the factor of 2. This is also the secant
approximation for the derivative through two points; in this case the
constant 2 arises from the "average" of slopes at the two points. Same
math, two interpretations.

By going back to our original

jw = ln z / T

substituting our approximation for ln z gives us:

jw = (2(z-1)/(z+1)) / T (approx)

Just my $0.02.

[email protected] wrote:

...

> Just my $0.02.

There's that factor of 2 again!

Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Randy Yates <[email protected]> wrote in message news:<[email protected]>...

> Consider this. The EBT goes from the analog domain, in which frequency
> is absolute and sample rate doesn't matter, to the digital domain, in
> which frequency is normalized and sample rate doesn't matter. So why
> should sample rate come into a transformation between two domains in
> which sample rate doesn't matter?

Ehum... eh... how do you reach the conclusion that "frequency is
normalized and sample rate doesn't matter" in discrete-time domain?
I don't understand your line of reasoning, mainly because of this
subtle point.

The normalizations in frequency domain I am aware of, are of the type

f' = f/f_s

where f' is normalized frequency, f is physical frequency and f_s is
sampling frequency. As we know, scanning the circumfrerence of the unit
circle once, from z=(1,0) in the counterclockwise direction, relates
to scanning the physical frequency band [0,f_s]. A mapping between the
two domains needs, as far as I can see, to preserve this connection
or it is useless.

How can you get from physical frequency in s domain to normalized
frequency in z domain without a reference to the sampling frequency?
More importantly, how do you get back, assuming you run a digital
filter on some signal that eventually is played back using a DAC?

Rune

[email protected] (Rune Allnor) writes:

> Randy Yates <[email protected]> wrote in message news:<[email protected]>...
>
>> Consider this. The EBT goes from the analog domain, in which frequency
>> is absolute and sample rate doesn't matter, to the digital domain, in
>> which frequency is normalized and sample rate doesn't matter. So why
>> should sample rate come into a transformation between two domains in
>> which sample rate doesn't matter?
>
> Ehum... eh... how do you reach the conclusion that "frequency is
> normalized and sample rate doesn't matter" in discrete-time domain?

If I give you a point on the unit circle in the z-domain at an angle of
pi/4, what is the frequency? Answer: pi/4. Where is the sample rate in
this answer? You do NOT get to go into the analog domain from the z-domain
since the BT is NOT a mapping to the analog domain but the z-domain!!!

> I don't understand your line of reasoning, mainly because of this
> subtle point.
>
> The normalizations in frequency domain I am aware of, are of the type
>
> f' = f/f_s
>
> where f' is normalized frequency, f is physical frequency and f_s is
> sampling frequency. As we know, scanning the circumfrerence of the unit
> circle once, from z=(1,0) in the counterclockwise direction, relates
> to scanning the physical frequency band [0,f_s]. A mapping between the
> two domains needs, as far as I can see, to preserve this connection
> or it is useless.
>
> How can you get from physical frequency in s domain to normalized
> frequency in z domain without a reference to the sampling frequency?
> More importantly, how do you get back, assuming you run a digital
> filter on some signal that eventually is played back using a DAC?

Rune, rather than address your questions and points individually, let
me jump to some conclusions. The z-domain is the z-domain is the
z-domain. There is no sample rate in the z-domain. I know you know
this. What you're doing is implicitly mapping the sample-rateless z-domain to the
analog domain, and to do THAT you need sample rate, but then you're
implying that the bilinear transform is a mapping from the Laplacian
s-plane to something other than the z-plane, and it ain't.

There's even more to this argument that no one has even touched on
yet. I am trying to decide how to express it. I hope to post about
it within the next few days.
--
% Randy Yates % "With time with what you've learned,
%% Fuquay-Varina, NC % they'll kiss the ground you walk
%%% 919-577-9882 % upon."
%%%% <[email protected]> % '21st Century Man', *Time*, ELO
http://home.earthlink.net/~yatescr