> Hi again. You said that the discretized filter becomes unstable when using
> the bilinear transform on the new zero (by the differentiation). What I
> then find confusing is that there are many analog prototypes which have a
> zero like this. And the BLT of a stable filter is also a stable filter, so
> what's wrong here?

Your last statement is wrong. You have the counter-example.
To be more specific: the BLT maps the left-half complex plane
(excluding the imaginary line) into the unit circle. The pole of the
continuous time differntiator at infinity gets mapped onto the Nyquist
frequency. For analog designs, the pole at infinity is usually
countered by some kind of (possibly system inherent) lowpass.
Regards,
Andor

Reply by gkn●May 15, 20072007-05-15

>On May 8, 3:42 am, "gkn" <gei...@ii.uib.no> wrote:
>> As differentiation in the time domain corresponds to multiplication by

s =

>in
>> the Laplace domain, it seems reasonable (to me) in terms of the

Bilinear

>> transform that multiplying the same digitized transfer function by
>>
>> 1-z^=AF=B9
>> ------
>> 1+z^=AF=B9
>>
>> should give the same desired effect in discrete case. But why doesn't

this

>> work? (If I, however, just multiply by 1-z=AF=B9 it ofcourse

corresponds =

>to a
>> finite difference in the time domain.)
>>
>> Thanks
>>
>> _____________________________________
>> Do you know a company who employs DSP engineers?
>> Is it already listed athttp://dsprelated.com/employers.php?
>
>I wasn't sure exactly what you were after by you after by your post,
>but just in case you are looking for a method to perform
>differentiation in real time, there are other means to perform this
>function. Specifically, there are methods that calculate a real time
>derivative based on the difference between successive samples and the
>accuracy depends on the 'order' of your differences. As an example of
>what I mean by order, if you take the difference between two samples,
>this first order and the difference of these (first order) differences
>is second order and so on. If you are interested, Jack Crenshaw's
>book on Math Programming for real time systems goes into the subject
>in great detail.
>
>
>

Hi again. You said that the discretized filter becomes unstable when using
the bilinear transform on the new zero (by the differentiation). What I
then find confusing is that there are many analog prototypes which have a
zero like this. And the BLT of a stable filter is also a stable filter, so
what's wrong here?
_____________________________________
Do you know a company who employs DSP engineers?
Is it already listed at http://dsprelated.com/employers.php ?

Reply by Noway2●May 9, 20072007-05-09

On May 8, 3:42 am, "gkn" <gei...@ii.uib.no> wrote:

> As differentiation in the time domain corresponds to multiplication by s =

in

> the Laplace domain, it seems reasonable (to me) in terms of the Bilinear
> transform that multiplying the same digitized transfer function by
>
> 1-z^=AF=B9
> ------
> 1+z^=AF=B9
>
> should give the same desired effect in discrete case. But why doesn't this
> work? (If I, however, just multiply by 1-z=AF=B9 it ofcourse corresponds =

to a

> finite difference in the time domain.)
>
> Thanks
>
> _____________________________________
> Do you know a company who employs DSP engineers?
> Is it already listed athttp://dsprelated.com/employers.php?

I wasn't sure exactly what you were after by you after by your post,
but just in case you are looking for a method to perform
differentiation in real time, there are other means to perform this
function. Specifically, there are methods that calculate a real time
derivative based on the difference between successive samples and the
accuracy depends on the 'order' of your differences. As an example of
what I mean by order, if you take the difference between two samples,
this first order and the difference of these (first order) differences
is second order and so on. If you are interested, Jack Crenshaw's
book on Math Programming for real time systems goes into the subject
in great detail.

Reply by Andor●May 9, 20072007-05-09

On 9 Mai, 11:59, "gkn" <gei...@ii.uib.no> wrote:

> >On 9 Mai, 09:53, "gkn" <gei...@ii.uib.no> wrote:
> >> >On 8 Mai, 11:24, "gkn" <gei...@ii.uib.no> wrote:
> >> >> >gkn wrote:
> >> >> >> As differentiation in the time domain corresponds to
> multiplication
> >> by
> >> >> s =3D3D
> >> >> >in
> >> >> >> the Laplace domain, it seems reasonable (to me) in terms of the
> >> >> Bilinear
> >> >> >> transform that multiplying the same digitized transfer function
> by
>
> >> >> >> 1-z^=3D3DAF=3D3DB9
> >> >> >> ------
> >> >> >> 1+z^=3D3DAF=3D3DB9
>
> >> >> >> should give the same desired effect in discrete case. But why
> >> doesn't
> >> >> this
> >> >> >> work? (If I, however, just multiply by 1-z=3D3DAF=3D3DB9 it ofco=

urse

> >> >> corresponds =3D3D
> >> >> >to a
> >> >> >> finite difference in the time domain.)
>
> >> >> >Multiplying with the above term makes a stable transfer function
> >> >> >unstable (due to the pole at Nyquist).
>
> >> >> >Differentiators are usually approximated with FIR filters. The
> >> >> >approximation that works for you depends strongly on the band
> which
> >> >> >you want to differentiate. First difference, second difference,
> >> >> >antisymmetric difference etc. are all simple approximations to the
> >> >> >ideal differentiator.
>
> >> >> >Regards,
> >> >> >Andor
>
> >> >> Thank you, Andor. Suppose the filter is just a simple one-pole
> >> resonator,
> >> >> is it possible to make a better differentiator (in this narrow
> >> passband)
> >> >> than just by multiplication with 1-z=3DAF=3DB9?
>
> >> >Is a one-pole resonator a filter of the form
>
> >> >H(z) =3D3D b0 + b1 z^-1 + b2 z^-2 / (1 - 2 cos(w0) z^-1 + z^-2)
>
> >> >?
>
> >> >If so, you can apply a simple two-coefficient FIR filter that has a
> >> >gain of w0 at w0 and a phase of 90=3DB0 at w0 (these two values
> completely
> >> >specify the FIR). If I didn't screw it up, the filter
>
> >> >h=3D3D[w0 cot(w0) -w0 csc(w0)]
>
> >> >should have the required frequency response.
>
> >> >Regards,
> >> >Andor
>
> >> Do you have any references? I need to understand this in detail..
>
> >Is this filter what you were looking for? Do we agree on what a "one
> >pole resonator" is?
>
> >What exactly don't you understand? I don't have any references, I
> >calculated the filter coefficients while replying to your post.
>
> >Regards,
> >Andor
>
> Well, I am not experienced with this, so I guess I should read some
> introductory material about differentiators. Do you know where to look?
> Anyway, my filter is actually a two (repeated) pole complex resonator with
> the digital transferfunction
>
> b0=B2
> H(z) =3D ----------
> (1-a1*z=AF=B9)=B2
>
> a1 can be chosen to be
>
> a1 =3D R*y =3D exp(-pi*df*T)*exp(2*pi*i*fc*T)
>
> where df is the bandwidth in Hz, and fc is the center frequency in Hz, and
> T is the sampling period.
>
> So yes, the filter is of the form you assumed (by choosing some of the
> coefficients to be zero). Now, is it easy to explain how you came up with
> your coefficients using this filter instead?

The exact filter doesn't matter.
What matters is that you want an approximation of a differentiator
over a narrow band centered at w0. If you decide for a two-tap FIR,
then you have two degrees of freedom to define the frequency response
of the FIR filter. In general, the frequency response of a real two-
tap FIR is
H(w) =3D h1 + h2 exp(-j w).
If you want H(w0) =3D w0 exp(j pi/2) (which is the response of the
differentiator at w0), then you have
w0 exp(j pi/2) =3D h1 + h2 exp(-j w0)
from which you can determine h1 and h2 using Euler's identites:
w0 cos(pi/2) =3D 0 =3D h1 + h2 cos(-w0)
w0 sin(pi/2) =3D w0 =3D h2 sin(-w0)
This gives you the coefficient values that I posted.
Regards,
Andor

Reply by gkn●May 9, 20072007-05-09

>On 9 Mai, 09:53, "gkn" <gei...@ii.uib.no> wrote:
>> >On 8 Mai, 11:24, "gkn" <gei...@ii.uib.no> wrote:
>> >> >gkn wrote:
>> >> >> As differentiation in the time domain corresponds to

multiplication

>> by
>> >> s =3D
>> >> >in
>> >> >> the Laplace domain, it seems reasonable (to me) in terms of the
>> >> Bilinear
>> >> >> transform that multiplying the same digitized transfer function

by

>>
>> >> >> 1-z^=3DAF=3DB9
>> >> >> ------
>> >> >> 1+z^=3DAF=3DB9
>>
>> >> >> should give the same desired effect in discrete case. But why
>> doesn't
>> >> this
>> >> >> work? (If I, however, just multiply by 1-z=3DAF=3DB9 it ofcourse
>> >> corresponds =3D
>> >> >to a
>> >> >> finite difference in the time domain.)
>>
>> >> >Multiplying with the above term makes a stable transfer function
>> >> >unstable (due to the pole at Nyquist).
>>
>> >> >Differentiators are usually approximated with FIR filters. The
>> >> >approximation that works for you depends strongly on the band

which

>> >> >you want to differentiate. First difference, second difference,
>> >> >antisymmetric difference etc. are all simple approximations to the
>> >> >ideal differentiator.
>>
>> >> >Regards,
>> >> >Andor
>>
>> >> Thank you, Andor. Suppose the filter is just a simple one-pole
>> resonator,
>> >> is it possible to make a better differentiator (in this narrow
>> passband)
>> >> than just by multiplication with 1-z=AF=B9?
>>
>> >Is a one-pole resonator a filter of the form
>>
>> >H(z) =3D b0 + b1 z^-1 + b2 z^-2 / (1 - 2 cos(w0) z^-1 + z^-2)
>>
>> >?
>>
>> >If so, you can apply a simple two-coefficient FIR filter that has a
>> >gain of w0 at w0 and a phase of 90=B0 at w0 (these two values

completely

>> >specify the FIR). If I didn't screw it up, the filter
>>
>> >h=3D[w0 cot(w0) -w0 csc(w0)]
>>
>> >should have the required frequency response.
>>
>> >Regards,
>> >Andor
>>
>> Do you have any references? I need to understand this in detail..
>
>Is this filter what you were looking for? Do we agree on what a "one
>pole resonator" is?
>
>What exactly don't you understand? I don't have any references, I
>calculated the filter coefficients while replying to your post.
>
>Regards,
>Andor
>
>

Well, I am not experienced with this, so I guess I should read some
introductory material about differentiators. Do you know where to look?
Anyway, my filter is actually a two (repeated) pole complex resonator with
the digital transferfunction
b0�
H(z) = ----------
(1-a1*z��)�
a1 can be chosen to be
a1 = R*y = exp(-pi*df*T)*exp(2*pi*i*fc*T)
where df is the bandwidth in Hz, and fc is the center frequency in Hz, and
T is the sampling period.
So yes, the filter is of the form you assumed (by choosing some of the
coefficients to be zero). Now, is it easy to explain how you came up with
your coefficients using this filter instead?
Regards,
GK
_____________________________________
Do you know a company who employs DSP engineers?
Is it already listed at http://dsprelated.com/employers.php ?

Reply by Andor●May 9, 20072007-05-09

On 9 Mai, 09:53, "gkn" <gei...@ii.uib.no> wrote:

> >On 8 Mai, 11:24, "gkn" <gei...@ii.uib.no> wrote:
> >> >gkn wrote:
> >> >> As differentiation in the time domain corresponds to multiplication
> by
> >> s =3D
> >> >in
> >> >> the Laplace domain, it seems reasonable (to me) in terms of the
> >> Bilinear
> >> >> transform that multiplying the same digitized transfer function by
>
> >> >> 1-z^=3DAF=3DB9
> >> >> ------
> >> >> 1+z^=3DAF=3DB9
>
> >> >> should give the same desired effect in discrete case. But why
> doesn't
> >> this
> >> >> work? (If I, however, just multiply by 1-z=3DAF=3DB9 it ofcourse
> >> corresponds =3D
> >> >to a
> >> >> finite difference in the time domain.)
>
> >> >Multiplying with the above term makes a stable transfer function
> >> >unstable (due to the pole at Nyquist).
>
> >> >Differentiators are usually approximated with FIR filters. The
> >> >approximation that works for you depends strongly on the band which
> >> >you want to differentiate. First difference, second difference,
> >> >antisymmetric difference etc. are all simple approximations to the
> >> >ideal differentiator.
>
> >> >Regards,
> >> >Andor
>
> >> Thank you, Andor. Suppose the filter is just a simple one-pole
> resonator,
> >> is it possible to make a better differentiator (in this narrow
> passband)
> >> than just by multiplication with 1-z=AF=B9?
>
> >Is a one-pole resonator a filter of the form
>
> >H(z) =3D b0 + b1 z^-1 + b2 z^-2 / (1 - 2 cos(w0) z^-1 + z^-2)
>
> >?
>
> >If so, you can apply a simple two-coefficient FIR filter that has a
> >gain of w0 at w0 and a phase of 90=B0 at w0 (these two values completely
> >specify the FIR). If I didn't screw it up, the filter
>
> >h=3D[w0 cot(w0) -w0 csc(w0)]
>
> >should have the required frequency response.
>
> >Regards,
> >Andor
>
> Do you have any references? I need to understand this in detail..

Is this filter what you were looking for? Do we agree on what a "one
pole resonator" is?
What exactly don't you understand? I don't have any references, I
calculated the filter coefficients while replying to your post.
Regards,
Andor

Reply by gkn●May 9, 20072007-05-09

>On 8 Mai, 11:24, "gkn" <gei...@ii.uib.no> wrote:
>> >gkn wrote:
>> >> As differentiation in the time domain corresponds to multiplication

by

>> s =3D
>> >in
>> >> the Laplace domain, it seems reasonable (to me) in terms of the
>> Bilinear
>> >> transform that multiplying the same digitized transfer function by
>>
>> >> 1-z^=3DAF=3DB9
>> >> ------
>> >> 1+z^=3DAF=3DB9
>>
>> >> should give the same desired effect in discrete case. But why

doesn't

>> this
>> >> work? (If I, however, just multiply by 1-z=3DAF=3DB9 it ofcourse
>> corresponds =3D
>> >to a
>> >> finite difference in the time domain.)
>>
>> >Multiplying with the above term makes a stable transfer function
>> >unstable (due to the pole at Nyquist).
>>
>> >Differentiators are usually approximated with FIR filters. The
>> >approximation that works for you depends strongly on the band which
>> >you want to differentiate. First difference, second difference,
>> >antisymmetric difference etc. are all simple approximations to the
>> >ideal differentiator.
>>
>> >Regards,
>> >Andor
>>
>> Thank you, Andor. Suppose the filter is just a simple one-pole

resonator,

>> is it possible to make a better differentiator (in this narrow

passband)

>> than just by multiplication with 1-z=AF=B9?
>
>Is a one-pole resonator a filter of the form
>
>H(z) =3D b0 + b1 z^-1 + b2 z^-2 / (1 - 2 cos(w0) z^-1 + z^-2)
>
>?
>
>If so, you can apply a simple two-coefficient FIR filter that has a
>gain of w0 at w0 and a phase of 90=B0 at w0 (these two values completely
>specify the FIR). If I didn't screw it up, the filter
>
>h=3D[w0 cot(w0) -w0 csc(w0)]
>
>should have the required frequency response.
>
>Regards,
>Andor
>
>

Do you have any references? I need to understand this in detail..
thank you!
_____________________________________
Do you know a company who employs DSP engineers?
Is it already listed at http://dsprelated.com/employers.php ?

Reply by Andor●May 8, 20072007-05-08

On 8 Mai, 11:24, "gkn" <gei...@ii.uib.no> wrote:

> >gkn wrote:
> >> As differentiation in the time domain corresponds to multiplication by
> s =3D
> >in
> >> the Laplace domain, it seems reasonable (to me) in terms of the
> Bilinear
> >> transform that multiplying the same digitized transfer function by
>
> >> 1-z^=3DAF=3DB9
> >> ------
> >> 1+z^=3DAF=3DB9
>
> >> should give the same desired effect in discrete case. But why doesn't
> this
> >> work? (If I, however, just multiply by 1-z=3DAF=3DB9 it ofcourse
> corresponds =3D
> >to a
> >> finite difference in the time domain.)
>
> >Multiplying with the above term makes a stable transfer function
> >unstable (due to the pole at Nyquist).
>
> >Differentiators are usually approximated with FIR filters. The
> >approximation that works for you depends strongly on the band which
> >you want to differentiate. First difference, second difference,
> >antisymmetric difference etc. are all simple approximations to the
> >ideal differentiator.
>
> >Regards,
> >Andor
>
> Thank you, Andor. Suppose the filter is just a simple one-pole resonator,
> is it possible to make a better differentiator (in this narrow passband)
> than just by multiplication with 1-z=AF=B9?

Is a one-pole resonator a filter of the form
H(z) =3D b0 + b1 z^-1 + b2 z^-2 / (1 - 2 cos(w0) z^-1 + z^-2)
?
If so, you can apply a simple two-coefficient FIR filter that has a
gain of w0 at w0 and a phase of 90=B0 at w0 (these two values completely
specify the FIR). If I didn't screw it up, the filter
h=3D[w0 cot(w0) -w0 csc(w0)]
should have the required frequency response.
Regards,
Andor

Reply by gkn●May 8, 20072007-05-08

>gkn wrote:
>> As differentiation in the time domain corresponds to multiplication by

s =

>in
>> the Laplace domain, it seems reasonable (to me) in terms of the

Bilinear

>> transform that multiplying the same digitized transfer function by
>>
>> 1-z^=AF=B9
>> ------
>> 1+z^=AF=B9
>>
>> should give the same desired effect in discrete case. But why doesn't

this

>> work? (If I, however, just multiply by 1-z=AF=B9 it ofcourse

corresponds =

>to a
>> finite difference in the time domain.)
>
>Multiplying with the above term makes a stable transfer function
>unstable (due to the pole at Nyquist).
>
>Differentiators are usually approximated with FIR filters. The
>approximation that works for you depends strongly on the band which
>you want to differentiate. First difference, second difference,
>antisymmetric difference etc. are all simple approximations to the
>ideal differentiator.
>
>Regards,
>Andor
>
>

Thank you, Andor. Suppose the filter is just a simple one-pole resonator,
is it possible to make a better differentiator (in this narrow passband)
than just by multiplication with 1-z��?
Thanks again
_____________________________________
Do you know a company who employs DSP engineers?
Is it already listed at http://dsprelated.com/employers.php ?

Reply by Andor●May 8, 20072007-05-08

gkn wrote:

> As differentiation in the time domain corresponds to multiplication by s =

in

> the Laplace domain, it seems reasonable (to me) in terms of the Bilinear
> transform that multiplying the same digitized transfer function by
>
> 1-z^=AF=B9
> ------
> 1+z^=AF=B9
>
> should give the same desired effect in discrete case. But why doesn't this
> work? (If I, however, just multiply by 1-z=AF=B9 it ofcourse corresponds =

to a

> finite difference in the time domain.)

Multiplying with the above term makes a stable transfer function
unstable (due to the pole at Nyquist).
Differentiators are usually approximated with FIR filters. The
approximation that works for you depends strongly on the band which
you want to differentiate. First difference, second difference,
antisymmetric difference etc. are all simple approximations to the
ideal differentiator.
Regards,
Andor