hi all
  i would like to know the technical description or derivation about
the slope of a filter
ie
why we say that 1st order filters have a 20 dB/decade (or 6 dB/octave)
slope.

with regards 
rammya

On 28 Jan, 13:57, "rammya.tv" <rammya...@ymail.com> wrote:
> hi all
> =A0 i would like to know the technical description or derivation about
> the slope of a filter
> ie
> why we say that 1st order filters have a 20 dB/decade (or 6 dB/octave)
> slope.

I suppose the general derivation would be something like

slope/bandwidth =3D (|H(w_0)|/|H(w_1)|)/(w_1 - w_0)

Fill in a specific filter transfer function and add
the missing logarithms, and you should have your
explanation.

Rune

rammya.tv wrote:
> hi all
>   i would like to know the technical description or derivation about
> the slope of a filter
> ie
> why we say that 1st order filters have a 20 dB/decade (or 6 dB/octave)
> slope.

Consider the voltage divider that consists of a resistor connected to 
the signal source at one end and a capacitor to ground at the other. 
This is a low-pass filter and the junction is the filter's output. With 
constant excitation: At very low frequencies, the capacitor has 
negligible effect and the output is constant. At frequencies where the 
capacitor's impedance (1/jwC) is comparable to the resistor's (R) the 
output is in transition. At frequencies where the capacitor's impedance 
  is much less than the resistor's, the output is inversely proportional 
to frequency. A response proportional to 1/f falls off at 20 dB/decade.

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

On Jan 28, 7:57=A0am, "rammya.tv" <rammya...@ymail.com> wrote:
> hi all
> =A0 i would like to know the technical description or derivation about
> the slope of a filter
> ie
> why we say that 1st order filters have a 20 dB/decade (or 6 dB/octave)
> slope.
>
> with regards
> rammya

For frequencies way higher than the "knee" frequency, the frequency
response of a 1st order lowpass filter behaves like c/f where "c" is a
gain constant and "f" is the frequency.

So find the ratio in this limiting case of the filter at "f" and at
"2f" and then find 20*log() of that ratio. You will find you get
-6.020599913... dB/octave.

IHTH,
Clay


p.s. The magnitude response of a nth butterworth lowpass filter is
simply

A(f) =3D 1/sqrt(1+(f/fc)^2n)

On Jan 29, 1:57=A0am, "rammya.tv" <rammya...@ymail.com> wrote:
> hi all
> =A0 i would like to know the technical description or derivation about
> the slope of a filter
> ie
> why we say that 1st order filters have a 20 dB/decade (or 6 dB/octave)
> slope.
>
> with regards
> rammya

In radians/s we have that the transfer function is

H(jw)=3D1/(1+jwT) where T is the time-constant. We can write this as 1/
(1+jw/wc) where wc=3D1/T, the cut-off freq.

Now convert to dB magnitude by taking 20Log10

dBMag=3D -20log10sqrt((1+(w/wc)^2)). Now asymptotically is when w--.>inf
and hence we can ignore the 1 giving

dBMag=3D-20Log10(w/wc)

or a slope of -20dB/decade. test by putting w=3D1 and w=3D10/wc (a decade
for normalised freq).The dBMag drops by 20dB This is the exact same as
-6dB/octave. Check by putting w=3D1 and w=3D2/wc and see the dBMag drop by
=3D6dB.

Hardy

On Jan 29, 8:59=A0am, Clay <c...@claysturner.com> wrote:
> On Jan 28, 7:57=A0am, "rammya.tv" <rammya...@ymail.com> wrote:
>
> > hi all
> > =A0 i would like to know the technical description or derivation about
> > the slope of a filter
> > ie
> > why we say that 1st order filters have a 20 dB/decade (or 6 dB/octave)
> > slope.
>
> > with regards
> > rammya
>
> For frequencies way higher than the "knee" frequency, the frequency
> response of a 1st order lowpass filter behaves like c/f where "c" is a
> gain constant and "f" is the frequency.
>
> So find the ratio in this limiting case of the filter at "f" and at
> "2f" and then find 20*log() of that ratio. You will find you get
> -6.020599913... dB/octave.
>
> IHTH,
> Clay
>
> p.s. The magnitude response of a nth butterworth lowpass filter is
> simply
>
> A(f) =3D 1/sqrt(1+(f/fc)^2n)

Only with 3dB passband ripple..


Hardy

HardySpicer wrote:
> On Jan 29, 8:59 am, Clay <c...@claysturner.com> wrote:

   ...

>> p.s. The magnitude response of a nth butterworth lowpass filter is
>> simply
>>
>> A(f) = 1/sqrt(1+(f/fc)^2n)
> 
> Only with 3dB passband ripple..

Hunh? Show us. (Hint: show that a derivative goes to zero somewhere 
other than f=0.)

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

On Jan 31, 6:20=A0am, Jerry Avins <j...@ieee.org> wrote:
> HardySpicer wrote:
> > On Jan 29, 8:59 am, Clay <c...@claysturner.com> wrote:
>
> =A0 =A0...
>
> >> p.s. The magnitude response of a nth butterworth lowpass filter is
> >> simply
>
> >> A(f) =3D 1/sqrt(1+(f/fc)^2n)
>
> > Only with 3dB passband ripple..
>
> Hunh? Show us. (Hint: show that a derivative goes to zero somewhere
> other than f=3D0.)
>
> Jerry
> --
> Engineering is the art of making what you want from things you can get.
> =AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF

eh? This is standard theory for Butterworth filters but many people
only get taught the 3dB version.
You specify the attenuation in the passband and find the ripple factor
eps for that attenuation. Turns out the the ripple factor is unity
(nearly) when the passband attenuation is 3dB. The poles lie in a
circle radius unity for 3dB passband ripple but on a circle radius (1/
eps)^1/n if I remember right for a ripple factor eps and order n. The
equation  A(f) =3D 1/sqrt(1+(f/fc)^2n) also changes. Look it up.


Hardy

HardySpicer wrote:
> On Jan 31, 6:20 am, Jerry Avins <j...@ieee.org> wrote:
>> HardySpicer wrote:
>>> On Jan 29, 8:59 am, Clay <c...@claysturner.com> wrote:
>>    ...
>>
>>>> p.s. The magnitude response of a nth butterworth lowpass filter is
>>>> simply
>>>> A(f) = 1/sqrt(1+(f/fc)^2n)
>>> Only with 3dB passband ripple..
>> Hunh? Show us. (Hint: show that a derivative goes to zero somewhere
>> other than f=0.)
>>
>> Jerry
>> --
>> Engineering is the art of making what you want from things you can get.
>> �����������������������������������������������������������������������
> 
> eh? This is standard theory for Butterworth filters but many people
> only get taught the 3dB version.
> You specify the attenuation in the passband and find the ripple factor
> eps for that attenuation. Turns out the the ripple factor is unity
> (nearly) when the passband attenuation is 3dB. The poles lie in a
> circle radius unity for 3dB passband ripple but on a circle radius (1/
> eps)^1/n if I remember right for a ripple factor eps and order n. The
> equation  A(f) = 1/sqrt(1+(f/fc)^2n) also changes. Look it up.

I can only guess what you're driving at. "Butterworth" is the same as 
"maximally flat". A Butterworth filter exhibits no ripple at all. True, 
the edge of the passband is 3 dB down from the flat top, but that's not 
ripple. Are you by chance thinking of Butterworth as the flat limit of a 
Chebychev filter?

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

On Jan 31, 3:22=A0pm, Jerry Avins <j...@ieee.org> wrote:
> HardySpicer wrote:
> > On Jan 31, 6:20 am, Jerry Avins <j...@ieee.org> wrote:
> >> HardySpicer wrote:
> >>> On Jan 29, 8:59 am, Clay <c...@claysturner.com> wrote:
> >> =A0 =A0...
>
> >>>> p.s. The magnitude response of a nth butterworth lowpass filter is
> >>>> simply
> >>>> A(f) =3D 1/sqrt(1+(f/fc)^2n)
> >>> Only with 3dB passband ripple..
> >> Hunh? Show us. (Hint: show that a derivative goes to zero somewhere
> >> other than f=3D0.)
>
> >> Jerry
> >> --
> >> Engineering is the art of making what you want from things you can get=
..
> >> =AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF
>
> > eh? This is standard theory for Butterworth filters but many people
> > only get taught the 3dB version.
> > You specify the attenuation in the passband and find the ripple factor
> > eps for that attenuation. Turns out the the ripple factor is unity
> > (nearly) when the passband attenuation is 3dB. The poles lie in a
> > circle radius unity for 3dB passband ripple but on a circle radius (1/
> > eps)^1/n if I remember right for a ripple factor eps and order n. The
> > equation =A0A(f) =3D 1/sqrt(1+(f/fc)^2n) also changes. Look it up.
>
> I can only guess what you're driving at. "Butterworth" is the same as
> "maximally flat". A Butterworth filter exhibits no ripple at all. True,
> the edge of the passband is 3 dB down from the flat top, but that's not
> ripple. Are you by chance thinking of Butterworth as the flat limit of a
> Chebychev filter?
>
> Jerry
> --
> Engineering is the art of making what you want from things you can get.
> =AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF

It's a bit of a missnoma... it is called ripple and ripple factor in
the literature even though there is no ripple!
This is due to the generalised form of polynomials that define such
filters. For Butterworth a better term is just the dB attenuation in
the passband.
This (with your equations) is defined as 3dB whereas it could be say
1dB. It's just the droop you get at the end of the passband - no
ripple.
Of course in the case of Chebychev there is a real ripple but the
polynomial is different.
The general form is H(v)^2=3D1/1+eps^2.Ln(v)^2 where Ln(v) is the
polynomial in normalised freq v of order n. Substitute L, the
polynomial of your choice,Butterworth,Chebychev etc. The "square-root"
of this gives the filter. (actually the spectral factorization)

Hardy

HardySpicer wrote:
> On Jan 31, 3:22 pm, Jerry Avins <j...@ieee.org> wrote:
>> HardySpicer wrote:
>>> On Jan 31, 6:20 am, Jerry Avins <j...@ieee.org> wrote:
>>>> HardySpicer wrote:
>>>>> On Jan 29, 8:59 am, Clay <c...@claysturner.com> wrote:
>>>>    ...
>>>>>> p.s. The magnitude response of a nth butterworth lowpass filter is
>>>>>> simply
>>>>>> A(f) = 1/sqrt(1+(f/fc)^2n)
>>>>> Only with 3dB passband ripple..
>>>> Hunh? Show us. (Hint: show that a derivative goes to zero somewhere
>>>> other than f=0.)
>>>> Jerry
>>>> --
>>>> Engineering is the art of making what you want from things you can get.
>>>> �����������������������������������������������������������������������
>>> eh? This is standard theory for Butterworth filters but many people
>>> only get taught the 3dB version.
>>> You specify the attenuation in the passband and find the ripple factor
>>> eps for that attenuation. Turns out the the ripple factor is unity
>>> (nearly) when the passband attenuation is 3dB. The poles lie in a
>>> circle radius unity for 3dB passband ripple but on a circle radius (1/
>>> eps)^1/n if I remember right for a ripple factor eps and order n. The
>>> equation  A(f) = 1/sqrt(1+(f/fc)^2n) also changes. Look it up.
>> I can only guess what you're driving at. "Butterworth" is the same as
>> "maximally flat". A Butterworth filter exhibits no ripple at all. True,
>> the edge of the passband is 3 dB down from the flat top, but that's not
>> ripple. Are you by chance thinking of Butterworth as the flat limit of a
>> Chebychev filter?
>>
>> Jerry
>> --
>> Engineering is the art of making what you want from things you can get.
>> �����������������������������������������������������������������������
> 
> It's a bit of a missnoma... it is called ripple and ripple factor in
> the literature even though there is no ripple!
> This is due to the generalised form of polynomials that define such
> filters. For Butterworth a better term is just the dB attenuation in
> the passband.
> This (with your equations) is defined as 3dB whereas it could be say
> 1dB. It's just the droop you get at the end of the passband - no
> ripple.
> Of course in the case of Chebychev there is a real ripple but the
> polynomial is different.
> The general form is H(v)^2=1/1+eps^2.Ln(v)^2 where Ln(v) is the
> polynomial in normalised freq v of order n. Substitute L, the
> polynomial of your choice,Butterworth,Chebychev etc. The "square-root"
> of this gives the filter. (actually the spectral factorization)

We're on the same page, then. What did you mean when contradicting Clay 
by writing "Only with 3dB passband ripple.."?

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

On 31 Jan, 03:22, Jerry Avins <j...@ieee.org> wrote:
> HardySpicer wrote:
> > On Jan 31, 6:20 am, Jerry Avins <j...@ieee.org> wrote:
> >> HardySpicer wrote:
> >>> On Jan 29, 8:59 am, Clay <c...@claysturner.com> wrote:
> >> =A0 =A0...
>
> >>>> p.s. The magnitude response of a nth butterworth lowpass filter is
> >>>> simply
> >>>> A(f) =3D 1/sqrt(1+(f/fc)^2n)
> >>> Only with 3dB passband ripple..
> >> Hunh? Show us. (Hint: show that a derivative goes to zero somewhere
> >> other than f=3D0.)
>
> >> Jerry
> >> --
> >> Engineering is the art of making what you want from things you can get=
..
> >> =AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF
>
> > eh? This is standard theory for Butterworth filters but many people
> > only get taught the 3dB version.
> > You specify the attenuation in the passband and find the ripple factor
> > eps for that attenuation. Turns out the the ripple factor is unity
> > (nearly) when the passband attenuation is 3dB. The poles lie in a
> > circle radius unity for 3dB passband ripple but on a circle radius (1/
> > eps)^1/n if I remember right for a ripple factor eps and order n. The
> > equation =A0A(f) =3D 1/sqrt(1+(f/fc)^2n) also changes. Look it up.
>
> I can only guess what you're driving at. "Butterworth" is the same as
> "maximally flat". A Butterworth filter exhibits no ripple at all. True,
> the edge of the passband is 3 dB down from the flat top, but that's not
> ripple. Are you by chance thinking of Butterworth as the flat limit of a
> Chebychev filter?

Clay is right in the formula for the Butterworth frequency
response, and in that formula a characteristic frequency
appears that indicate the frequency of the 3dB point.

However, that characteristic frequency needs not be the
cut-off frequency of the filter spec. Given any ripple
and corner frequency, one can dedude the corresponding
3dB frequency of the filter.

So you discussion hinges on exactly what meaning is assigned
to the factor 'fc' in the fomula: The characteristic 3dB
frequency, or the passband corner frequency?

Rune

On Jan 31, 6:45=A0pm, Rune Allnor <all...@tele.ntnu.no> wrote:
> On 31 Jan, 03:22, Jerry Avins <j...@ieee.org> wrote:
>
>
>
> > HardySpicer wrote:
> > > On Jan 31, 6:20 am, Jerry Avins <j...@ieee.org> wrote:
> > >> HardySpicer wrote:
> > >>> On Jan 29, 8:59 am, Clay <c...@claysturner.com> wrote:
> > >> =A0 =A0...
>
> > >>>> p.s. The magnitude response of a nth butterworth lowpass filter is
> > >>>> simply
> > >>>> A(f) =3D 1/sqrt(1+(f/fc)^2n)
> > >>> Only with 3dB passband ripple..
> > >> Hunh? Show us. (Hint: show that a derivative goes to zero somewhere
> > >> other than f=3D0.)
>
> > >> Jerry
> > >> --
> > >> Engineering is the art of making what you want from things you can g=
et.
> > >> =AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF
>
> > > eh? This is standard theory for Butterworth filters but many people
> > > only get taught the 3dB version.
> > > You specify the attenuation in the passband and find the ripple facto=
r
> > > eps for that attenuation. Turns out the the ripple factor is unity
> > > (nearly) when the passband attenuation is 3dB. The poles lie in a
> > > circle radius unity for 3dB passband ripple but on a circle radius (1=
/
> > > eps)^1/n if I remember right for a ripple factor eps and order n. The
> > > equation =A0A(f) =3D 1/sqrt(1+(f/fc)^2n) also changes. Look it up.
>
> > I can only guess what you're driving at. "Butterworth" is the same as
> > "maximally flat". A Butterworth filter exhibits no ripple at all. True,
> > the edge of the passband is 3 dB down from the flat top, but that's not
> > ripple. Are you by chance thinking of Butterworth as the flat limit of =
a
> > Chebychev filter?
>
> Clay is right in the formula for the Butterworth frequency
> response, and in that formula a characteristic frequency
> appears that indicate the frequency of the 3dB point.
>
> However, that characteristic frequency needs not be the
> cut-off frequency of the filter spec. Given any ripple
> and corner frequency, one can dedude the corresponding
> 3dB frequency of the filter.
>
> So you discussion hinges on exactly what meaning is assigned
> to the factor 'fc' in the fomula: The characteristic 3dB
> frequency, or the passband corner frequency?
>
> Rune

The convention is to define the dB attenuation in the passband. So we
say for example (for a lowpass filter design)  you need 1dB
attenuation in the passband at 1kHz. The 3dB freq is not normally
defined here though of course it exists in all LTI systems. So 1kHz is
the passband 'edge" but it is nearly always defined as unity for
normalised frequency. My point is that you cannot use the 3dB
Butterworth solution for this. For the 1dB case the polynomial changes
since the poles no longer lie in a circle radius unity (though the
pole  angles will be the same). So for a 1dB passband attenuation of
1kHz and a stopband attenuation of 40dB at 10kHz (as a simple example)
we use unity as the passband freq and 10 (normalised) as the stopband
freq. The poles lie on a circle radius (1/eps)^(1/n) if I remember
right. eps is found from Ap (passband attenuation)=3D10Log10.(1+eps^2)
and will be 1/sqrt(2) for the 3dB case. Here we can find eps for the
1dB case from eps^2=3D[10^0.1Ap]-1 =3D0.25 so that eps=3D0.5088. Giving the
pole radius as 1.402 and not unity.
Anyway we then find the pole polynomial (since we know the angles - we
also need the order which I have  taken as 2 for simplicity here
though we can easily work this out from the attenuation in the
stopband). Once we have the pole polynomial we get the NORMALISED
transfer function and then apply the low-pass to low-pass
transformation viz s-->s/Wx where Wx=3D2pi*1000 to get the real transfer
function.

We then end up with a 2nd order Butterworth with a passband freq of
1kHz at 1dB.


Hardy

On 31 Jan, 08:15, HardySpicer <gyansor...@gmail.com> wrote:
> On Jan 31, 6:45=A0pm, Rune Allnor <all...@tele.ntnu.no> wrote:
>
>
>
>
>
> > On 31 Jan, 03:22, Jerry Avins <j...@ieee.org> wrote:
>
> > > HardySpicer wrote:
> > > > On Jan 31, 6:20 am, Jerry Avins <j...@ieee.org> wrote:
> > > >> HardySpicer wrote:
> > > >>> On Jan 29, 8:59 am, Clay <c...@claysturner.com> wrote:
> > > >> =A0 =A0...
>
> > > >>>> p.s. The magnitude response of a nth butterworth lowpass filter =
is
> > > >>>> simply
> > > >>>> A(f) =3D 1/sqrt(1+(f/fc)^2n)
> > > >>> Only with 3dB passband ripple..
> > > >> Hunh? Show us. (Hint: show that a derivative goes to zero somewher=
e
> > > >> other than f=3D0.)
>
> > > >> Jerry
> > > >> --
> > > >> Engineering is the art of making what you want from things you can=
 get.
> > > >> =AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF
>
> > > > eh? This is standard theory for Butterworth filters but many people
> > > > only get taught the 3dB version.
> > > > You specify the attenuation in the passband and find the ripple fac=
tor
> > > > eps for that attenuation. Turns out the the ripple factor is unity
> > > > (nearly) when the passband attenuation is 3dB. The poles lie in a
> > > > circle radius unity for 3dB passband ripple but on a circle radius =
(1/
> > > > eps)^1/n if I remember right for a ripple factor eps and order n. T=
he
> > > > equation =A0A(f) =3D 1/sqrt(1+(f/fc)^2n) also changes. Look it up.
>
> > > I can only guess what you're driving at. "Butterworth" is the same as
> > > "maximally flat". A Butterworth filter exhibits no ripple at all. Tru=
e,
> > > the edge of the passband is 3 dB down from the flat top, but that's n=
ot
> > > ripple. Are you by chance thinking of Butterworth as the flat limit o=
f a
> > > Chebychev filter?
>
> > Clay is right in the formula for the Butterworth frequency
> > response, and in that formula a characteristic frequency
> > appears that indicate the frequency of the 3dB point.
>
> > However, that characteristic frequency needs not be the
> > cut-off frequency of the filter spec. Given any ripple
> > and corner frequency, one can dedude the corresponding
> > 3dB frequency of the filter.
>
> > So you discussion hinges on exactly what meaning is assigned
> > to the factor 'fc' in the fomula: The characteristic 3dB
> > frequency, or the passband corner frequency?
>
> > Rune
>
> The convention is to define the dB attenuation in the passband. So we
> say for example (for a lowpass filter design) =A0you need 1dB
> attenuation in the passband at 1kHz. The 3dB freq is not normally
> defined here though of course it exists in all LTI systems. So 1kHz is
> the passband 'edge" but it is nearly always defined as unity for
> normalised frequency. My point is that you cannot use the 3dB
> Butterworth solution for this.

Well, given the exact example you state, then no, the equation
does not work. But your example only states a partial spec.
A complete filter spec gives *two* constraints: The passband and
stopband frequencies, with associated attenuations.

Given passband parameters (Wp,Dp) and stopband parameters (Ws,Ds)
where W means s-domain omega and D means linear-scale ripple,
we get

Ds^2 =3D 1 / (1+ (Ws/Wc)^2n )
Dp^2 =3D 1 / (1+ (Wp/Wc)^2n )

A little algebraic massage, and the two are merged into

( Ws / Wp )^2n =3D ( Dp^2 -1 )/( Ds^2 -1 )

where the 3dB frequency Wc is gone. Apply some logarithms
and you find the formula for estimating the order of the
Butterworth filter, given the spec.

Getting all this right is intricate, though, particularly
if the goal is to end up with a digital filter: You need
to pre-warp the spec from discrete-time (DT) domain into
contionuous-time (CT) domain, come up with a normalized
CT filter, BLT it back to DT domain to a normalized DT filter,
which then has to be denormalized.

Rune

On Jan 31, 9:10=A0pm, Rune Allnor <all...@tele.ntnu.no> wrote:
> On 31 Jan, 08:15, HardySpicer <gyansor...@gmail.com> wrote:
>
>
>
> > On Jan 31, 6:45=A0pm, Rune Allnor <all...@tele.ntnu.no> wrote:
>
> > > On 31 Jan, 03:22, Jerry Avins <j...@ieee.org> wrote:
>
> > > > HardySpicer wrote:
> > > > > On Jan 31, 6:20 am, Jerry Avins <j...@ieee.org> wrote:
> > > > >> HardySpicer wrote:
> > > > >>> On Jan 29, 8:59 am, Clay <c...@claysturner.com> wrote:
> > > > >> =A0 =A0...
>
> > > > >>>> p.s. The magnitude response of a nth butterworth lowpass filte=
r is
> > > > >>>> simply
> > > > >>>> A(f) =3D 1/sqrt(1+(f/fc)^2n)
> > > > >>> Only with 3dB passband ripple..
> > > > >> Hunh? Show us. (Hint: show that a derivative goes to zero somewh=
ere
> > > > >> other than f=3D0.)
>
> > > > >> Jerry
> > > > >> --
> > > > >> Engineering is the art of making what you want from things you c=
an get.
> > > > >> =AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF
>
> > > > > eh? This is standard theory for Butterworth filters but many peop=
le
> > > > > only get taught the 3dB version.
> > > > > You specify the attenuation in the passband and find the ripple f=
actor
> > > > > eps for that attenuation. Turns out the the ripple factor is unit=
y
> > > > > (nearly) when the passband attenuation is 3dB. The poles lie in a
> > > > > circle radius unity for 3dB passband ripple but on a circle radiu=
s (1/
> > > > > eps)^1/n if I remember right for a ripple factor eps and order n.=
 The
> > > > > equation =A0A(f) =3D 1/sqrt(1+(f/fc)^2n) also changes. Look it up=
..
>
> > > > I can only guess what you're driving at. "Butterworth" is the same =
as
> > > > "maximally flat". A Butterworth filter exhibits no ripple at all. T=
rue,
> > > > the edge of the passband is 3 dB down from the flat top, but that's=
 not
> > > > ripple. Are you by chance thinking of Butterworth as the flat limit=
 of a
> > > > Chebychev filter?
>
> > > Clay is right in the formula for the Butterworth frequency
> > > response, and in that formula a characteristic frequency
> > > appears that indicate the frequency of the 3dB point.
>
> > > However, that characteristic frequency needs not be the
> > > cut-off frequency of the filter spec. Given any ripple
> > > and corner frequency, one can dedude the corresponding
> > > 3dB frequency of the filter.
>
> > > So you discussion hinges on exactly what meaning is assigned
> > > to the factor 'fc' in the fomula: The characteristic 3dB
> > > frequency, or the passband corner frequency?
>
> > > Rune
>
> > The convention is to define the dB attenuation in the passband. So we
> > say for example (for a lowpass filter design) =A0you need 1dB
> > attenuation in the passband at 1kHz. The 3dB freq is not normally
> > defined here though of course it exists in all LTI systems. So 1kHz is
> > the passband 'edge" but it is nearly always defined as unity for
> > normalised frequency. My point is that you cannot use the 3dB
> > Butterworth solution for this.
>
> Well, given the exact example you state, then no, the equation
> does not work. But your example only states a partial spec.
> A complete filter spec gives *two* constraints: The passband and
> stopband frequencies, with associated attenuations.
>
I missed that out for simplicity. Yes there is a formula for order of
the filter based on stopband attenuation.
It's a bit messy for ascii and doesn't change the results in any way.
The formula comes from

As=3D10Log10(1+eps^2.vs^2n) where v is  normalised freq, n is order and
As is stopband attenuation in dB. eps is the ripple factor previously
calculated and vs is the normalise stopband freq..
So you re-arrange the above to find vs^2n=3D(10^0.1As-1)/eps^2 and by
taking logs we find n=3D0.5Log[(10^0.1As-1)/eps^2]


 The Butterworth polynomial for say a 2nd order is

s^2+sqrt(2)s+1  but this is for 3dB passband attenuation only. This
will change for the (say) 1dB case and hence the filter with it.

so the basic formula quoted is normally for the 3dB case.

Your comments about the digital versions are all valid of course
aswith any analogue to discrete conversion.



Hardy

On Feb 1, 9:33=A0am, HardySpicer <gyansor...@gmail.com> wrote:
> On Jan 31, 9:10=A0pm, Rune Allnor <all...@tele.ntnu.no> wrote:
>
> > On 31 Jan, 08:15, HardySpicer <gyansor...@gmail.com> wrote:
>
> > > On Jan 31, 6:45=A0pm, Rune Allnor <all...@tele.ntnu.no> wrote:
>
> > > > On 31 Jan, 03:22, Jerry Avins <j...@ieee.org> wrote:
>
> > > > > HardySpicer wrote:
> > > > > > On Jan 31, 6:20 am, Jerry Avins <j...@ieee.org> wrote:
> > > > > >> HardySpicer wrote:
> > > > > >>> On Jan 29, 8:59 am, Clay <c...@claysturner.com> wrote:
> > > > > >> =A0 =A0...
>
> > > > > >>>> p.s. The magnitude response of a nth butterworth lowpass fil=
ter is
> > > > > >>>> simply
> > > > > >>>> A(f) =3D 1/sqrt(1+(f/fc)^2n)
> > > > > >>> Only with 3dB passband ripple..
> > > > > >> Hunh? Show us. (Hint: show that a derivative goes to zero some=
where
> > > > > >> other than f=3D0.)
>
> > > > > >> Jerry
> > > > > >> --
> > > > > >> Engineering is the art of making what you want from things you=
 can get.
> > > > > >> =AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF
>
> > > > > > eh? This is standard theory for Butterworth filters but many pe=
ople
> > > > > > only get taught the 3dB version.
> > > > > > You specify the attenuation in the passband and find the ripple=
 factor
> > > > > > eps for that attenuation. Turns out the the ripple factor is un=
ity
> > > > > > (nearly) when the passband attenuation is 3dB. The poles lie in=
 a
> > > > > > circle radius unity for 3dB passband ripple but on a circle rad=
ius (1/
> > > > > > eps)^1/n if I remember right for a ripple factor eps and order =
n. The
> > > > > > equation =A0A(f) =3D 1/sqrt(1+(f/fc)^2n) also changes. Look it =
up.
>
> > > > > I can only guess what you're driving at. "Butterworth" is the sam=
e as
> > > > > "maximally flat". A Butterworth filter exhibits no ripple at all.=
 True,
> > > > > the edge of the passband is 3 dB down from the flat top, but that=
's not
> > > > > ripple. Are you by chance thinking of Butterworth as the flat lim=
it of a
> > > > > Chebychev filter?
>
> > > > Clay is right in the formula for the Butterworth frequency
> > > > response, and in that formula a characteristic frequency
> > > > appears that indicate the frequency of the 3dB point.
>
> > > > However, that characteristic frequency needs not be the
> > > > cut-off frequency of the filter spec. Given any ripple
> > > > and corner frequency, one can dedude the corresponding
> > > > 3dB frequency of the filter.
>
> > > > So you discussion hinges on exactly what meaning is assigned
> > > > to the factor 'fc' in the fomula: The characteristic 3dB
> > > > frequency, or the passband corner frequency?
>
> > > > Rune
>
> > > The convention is to define the dB attenuation in the passband. So we
> > > say for example (for a lowpass filter design) =A0you need 1dB
> > > attenuation in the passband at 1kHz. The 3dB freq is not normally
> > > defined here though of course it exists in all LTI systems. So 1kHz i=
s
> > > the passband 'edge" but it is nearly always defined as unity for
> > > normalised frequency. My point is that you cannot use the 3dB
> > > Butterworth solution for this.
>
> > Well, given the exact example you state, then no, the equation
> > does not work. But your example only states a partial spec.
> > A complete filter spec gives *two* constraints: The passband and
> > stopband frequencies, with associated attenuations.
>
> I missed that out for simplicity. Yes there is a formula for order of
> the filter based on stopband attenuation.
> It's a bit messy for ascii and doesn't change the results in any way.
> The formula comes from
>
> As=3D10Log10(1+eps^2.vs^2n) where v is =A0normalised freq, n is order and
> As is stopband attenuation in dB. eps is the ripple factor previously
> calculated and vs is the normalise stopband freq..
> So you re-arrange the above to find vs^2n=3D(10^0.1As-1)/eps^2 and by
> taking logs we find n=3D0.5Log[(10^0.1As-1)/eps^2]
>
> =A0The Butterworth polynomial for say a 2nd order is
>
> s^2+sqrt(2)s+1 =A0but this is for 3dB passband attenuation only. This
> will change for the (say) 1dB case and hence the filter with it.
>
> so the basic formula quoted is normally for the 3dB case.
>
> Your comments about the digital versions are all valid of course
> aswith any analogue to discrete conversion.
>
> Hardy

That should read
 n=3D0.5Log[(10^0.1As-1)/eps^2]/log(vs)

On Feb 1, 9:33=A0am, HardySpicer <gyansor...@gmail.com> wrote:
> On Jan 31, 9:10=A0pm, Rune Allnor <all...@tele.ntnu.no> wrote:
>
> > On 31 Jan, 08:15, HardySpicer <gyansor...@gmail.com> wrote:
>
> > > On Jan 31, 6:45=A0pm, Rune Allnor <all...@tele.ntnu.no> wrote:
>
> > > > On 31 Jan, 03:22, Jerry Avins <j...@ieee.org> wrote:
>
> > > > > HardySpicer wrote:
> > > > > > On Jan 31, 6:20 am, Jerry Avins <j...@ieee.org> wrote:
> > > > > >> HardySpicer wrote:
> > > > > >>> On Jan 29, 8:59 am, Clay <c...@claysturner.com> wrote:
> > > > > >> =A0 =A0...
>
> > > > > >>>> p.s. The magnitude response of a nth butterworth lowpass fil=
ter is
> > > > > >>>> simply
> > > > > >>>> A(f) =3D 1/sqrt(1+(f/fc)^2n)
> > > > > >>> Only with 3dB passband ripple..
> > > > > >> Hunh? Show us. (Hint: show that a derivative goes to zero some=
where
> > > > > >> other than f=3D0.)
>
> > > > > >> Jerry
> > > > > >> --
> > > > > >> Engineering is the art of making what you want from things you=
 can get.
> > > > > >> =AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF
>
> > > > > > eh? This is standard theory for Butterworth filters but many pe=
ople
> > > > > > only get taught the 3dB version.
> > > > > > You specify the attenuation in the passband and find the ripple=
 factor
> > > > > > eps for that attenuation. Turns out the the ripple factor is un=
ity
> > > > > > (nearly) when the passband attenuation is 3dB. The poles lie in=
 a
> > > > > > circle radius unity for 3dB passband ripple but on a circle rad=
ius (1/
> > > > > > eps)^1/n if I remember right for a ripple factor eps and order =
n. The
> > > > > > equation =A0A(f) =3D 1/sqrt(1+(f/fc)^2n) also changes. Look it =
up.
>
> > > > > I can only guess what you're driving at. "Butterworth" is the sam=
e as
> > > > > "maximally flat". A Butterworth filter exhibits no ripple at all.=
 True,
> > > > > the edge of the passband is 3 dB down from the flat top, but that=
's not
> > > > > ripple. Are you by chance thinking of Butterworth as the flat lim=
it of a
> > > > > Chebychev filter?
>
> > > > Clay is right in the formula for the Butterworth frequency
> > > > response, and in that formula a characteristic frequency
> > > > appears that indicate the frequency of the 3dB point.
>
> > > > However, that characteristic frequency needs not be the
> > > > cut-off frequency of the filter spec. Given any ripple
> > > > and corner frequency, one can dedude the corresponding
> > > > 3dB frequency of the filter.
>
> > > > So you discussion hinges on exactly what meaning is assigned
> > > > to the factor 'fc' in the fomula: The characteristic 3dB
> > > > frequency, or the passband corner frequency?
>
> > > > Rune
>
> > > The convention is to define the dB attenuation in the passband. So we
> > > say for example (for a lowpass filter design) =A0you need 1dB
> > > attenuation in the passband at 1kHz. The 3dB freq is not normally
> > > defined here though of course it exists in all LTI systems. So 1kHz i=
s
> > > the passband 'edge" but it is nearly always defined as unity for
> > > normalised frequency. My point is that you cannot use the 3dB
> > > Butterworth solution for this.
>
> > Well, given the exact example you state, then no, the equation
> > does not work. But your example only states a partial spec.
> > A complete filter spec gives *two* constraints: The passband and
> > stopband frequencies, with associated attenuations.
>
> I missed that out for simplicity. Yes there is a formula for order of
> the filter based on stopband attenuation.
> It's a bit messy for ascii and doesn't change the results in any way.
> The formula comes from
>
> As=3D10Log10(1+eps^2.vs^2n) where v is =A0normalised freq, n is order and
> As is stopband attenuation in dB. eps is the ripple factor previously
> calculated and vs is the normalise stopband freq..
> So you re-arrange the above to find vs^2n=3D(10^0.1As-1)/eps^2 and by
> taking logs we find n=3D0.5Log[(10^0.1As-1)/eps^2]
>
> =A0The Butterworth polynomial for say a 2nd order is
>
> s^2+sqrt(2)s+1 =A0but this is for 3dB passband attenuation only. This
> will change for the (say) 1dB case and hence the filter with it.
>
> so the basic formula quoted is normally for the 3dB case.
>
> Your comments about the digital versions are all valid of course
> aswith any analogue to discrete conversion.
>
> Hardy

Just to summarise the procedure without all the equations:

Decide on type of filter reuqired - eg Lowpass,Highpass bandpass

Specify a lowpass prototype using the following (for all filter types
ie Lowpass,Highpass,Bandpass etc)

For a particular Passband freq wp and ws find the normailsed
equivalents vp=3D1 and wp/ws =3Dvs

For a particular dB  passband attenuation Ap  find the ripple factor
eps

For a particular dB  stopband attenuation As find the order n. (see
previous post)

Find the radius of the poles using eps above and n

Find the angle of the poles

Construct the pole polynomial P(s)  from the above two steps

This gives us our low-pass prototype from which we can transform to
find the real filter.

Find the real filter by first forming the prototype Hp(s)=3D1/P(s) where
P(s) is the pole polynomial found previosley.

Then transform Lowpass to Lowpass s--> s/wp or Lowpass to highpass s--
>ws/s and there are other transforms for lowpass to bandpass,lowpass
to bandsstop etc.
Lowpass to bandpass is s--->(s^2+w0^2)/Bs where B is bandwidth of
passband and w0 is the centre freq - normally the geometric root of wl
and wh which are at the lower and higher ends of the passband
respectively. Note that for this case when transforming a 2nd order
prototype only yields a first order "skirt" ie slope.

Once you have all this you can realise the filter analogue style or do
the pre-warping and Bilinear transform for digital. You won't get
linear phase of course.


Hardy

HardySpicer wrote:
> On Feb 1, 9:33 am, HardySpicer <gyansor...@gmail.com> wrote:
>> On Jan 31, 9:10 pm, Rune Allnor <all...@tele.ntnu.no> wrote:
>>
>>> On 31 Jan, 08:15, HardySpicer <gyansor...@gmail.com> wrote:
>>>> On Jan 31, 6:45 pm, Rune Allnor <all...@tele.ntnu.no> wrote:
>>>>> On 31 Jan, 03:22, Jerry Avins <j...@ieee.org> wrote:
>>>>>> HardySpicer wrote:
>>>>>>> On Jan 31, 6:20 am, Jerry Avins <j...@ieee.org> wrote:
>>>>>>>> HardySpicer wrote:
>>>>>>>>> On Jan 29, 8:59 am, Clay <c...@claysturner.com> wrote:
>>>>>>>>    ...
>>>>>>>>>> p.s. The magnitude response of a nth butterworth lowpass filter is
>>>>>>>>>> simply
>>>>>>>>>> A(f) = 1/sqrt(1+(f/fc)^2n)
>>>>>>>>> Only with 3dB passband ripple..
>>>>>>>> Hunh? Show us. (Hint: show that a derivative goes to zero somewhere
>>>>>>>> other than f=0.)
>>>>>>>> Jerry
>>>>>>>> --
>>>>>>>> Engineering is the art of making what you want from things you can get.
>>>>>>>> �����������������������������������������������������������������������
>>>>>>> eh? This is standard theory for Butterworth filters but many people
>>>>>>> only get taught the 3dB version.
>>>>>>> You specify the attenuation in the passband and find the ripple factor
>>>>>>> eps for that attenuation. Turns out the the ripple factor is unity
>>>>>>> (nearly) when the passband attenuation is 3dB. The poles lie in a
>>>>>>> circle radius unity for 3dB passband ripple but on a circle radius (1/
>>>>>>> eps)^1/n if I remember right for a ripple factor eps and order n. The
>>>>>>> equation  A(f) = 1/sqrt(1+(f/fc)^2n) also changes. Look it up.
>>>>>> I can only guess what you're driving at. "Butterworth" is the same as
>>>>>> "maximally flat". A Butterworth filter exhibits no ripple at all. True,
>>>>>> the edge of the passband is 3 dB down from the flat top, but that's not
>>>>>> ripple. Are you by chance thinking of Butterworth as the flat limit of a
>>>>>> Chebychev filter?
>>>>> Clay is right in the formula for the Butterworth frequency
>>>>> response, and in that formula a characteristic frequency
>>>>> appears that indicate the frequency of the 3dB point.
>>>>> However, that characteristic frequency needs not be the
>>>>> cut-off frequency of the filter spec. Given any ripple
>>>>> and corner frequency, one can dedude the corresponding
>>>>> 3dB frequency of the filter.
>>>>> So you discussion hinges on exactly what meaning is assigned
>>>>> to the factor 'fc' in the fomula: The characteristic 3dB
>>>>> frequency, or the passband corner frequency?
>>>>> Rune
>>>> The convention is to define the dB attenuation in the passband. So we
>>>> say for example (for a lowpass filter design)  you need 1dB
>>>> attenuation in the passband at 1kHz. The 3dB freq is not normally
>>>> defined here though of course it exists in all LTI systems. So 1kHz is
>>>> the passband 'edge" but it is nearly always defined as unity for
>>>> normalised frequency. My point is that you cannot use the 3dB
>>>> Butterworth solution for this.
>>> Well, given the exact example you state, then no, the equation
>>> does not work. But your example only states a partial spec.
>>> A complete filter spec gives *two* constraints: The passband and
>>> stopband frequencies, with associated attenuations.
>> I missed that out for simplicity. Yes there is a formula for order of
>> the filter based on stopband attenuation.
>> It's a bit messy for ascii and doesn't change the results in any way.
>> The formula comes from
>>
>> As=10Log10(1+eps^2.vs^2n) where v is  normalised freq, n is order and
>> As is stopband attenuation in dB. eps is the ripple factor previously
>> calculated and vs is the normalise stopband freq..
>> So you re-arrange the above to find vs^2n=(10^0.1As-1)/eps^2 and by
>> taking logs we find n=0.5Log[(10^0.1As-1)/eps^2]
>>
>>  The Butterworth polynomial for say a 2nd order is
>>
>> s^2+sqrt(2)s+1  but this is for 3dB passband attenuation only. This
>> will change for the (say) 1dB case and hence the filter with it.
>>
>> so the basic formula quoted is normally for the 3dB case.
>>
>> Your comments about the digital versions are all valid of course
>> aswith any analogue to discrete conversion.
>>
>> Hardy
> 
> Just to summarise the procedure without all the equations:
> 
> Decide on type of filter reuqired - eg Lowpass,Highpass bandpass
> 
> Specify a lowpass prototype using the following (for all filter types
> ie Lowpass,Highpass,Bandpass etc)
> 
> For a particular Passband freq wp and ws find the normailsed
> equivalents vp=1 and wp/ws =vs
> 
> For a particular dB  passband attenuation Ap  find the ripple factor
> eps
> 
> For a particular dB  stopband attenuation As find the order n. (see
> previous post)
> 
> Find the radius of the poles using eps above and n
> 
> Find the angle of the poles
> 
> Construct the pole polynomial P(s)  from the above two steps
> 
> This gives us our low-pass prototype from which we can transform to
> find the real filter.
> 
> Find the real filter by first forming the prototype Hp(s)=1/P(s) where
> P(s) is the pole polynomial found previosley.
> 
> Then transform Lowpass to Lowpass s--> s/wp or Lowpass to highpass s--
>> ws/s and there are other transforms for lowpass to bandpass,lowpass
> to bandsstop etc.
> Lowpass to bandpass is s--->(s^2+w0^2)/Bs where B is bandwidth of
> passband and w0 is the centre freq - normally the geometric root of wl
> and wh which are at the lower and higher ends of the passband
> respectively. Note that for this case when transforming a 2nd order
> prototype only yields a first order "skirt" ie slope.
> 
> Once you have all this you can realise the filter analogue style or do
> the pre-warping and Bilinear transform for digital. You won't get
> linear phase of course.

We speak different languages. On this side of the Atlantic, Butterworth 
means maximally flat, a condition achieved in the low-pass prototype by 
setting all the derivatives of A(f) to zero at f=0. There is no eps. The 
formula for A(f) is as Clay gave it, 1/sqrt(1+(f/fc)^2n), where n is the 
order of the filter. For a first-order filter, fc is 3 dB down and the 
rolloff is 6 dB/octve. For a second-order filter fc is 6 dB down and the 
rolloff is 12 dB/octve, and so on. Because A(f) falls monotonically as f 
increases, there is no ripple. There are many filters with no ripple, 
but only the Butterworth is maximally flat in the sense I defined above.

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

On 1 Feb, 00:52, Jerry Avins <j...@ieee.org> wrote:

> We speak different languages. On this side of the Atlantic, Butterworth
> means maximally flat,
....
> Because A(f) falls monotonically as f
> increases, there is no ripple. There are many filters with no ripple,
> but only the Butterworth is maximally flat in the sense I defined above.

I think the problem is the definition of the term 'ripple'.

If one interprets it literally, as something like 'fluctuating
back and forth', then yes, the Butterworth has no ripple.

If, on the other hand, one interprets the term as 'deviation
from desired value', then it is immediately clear that the
Butterworth lowpass filter has a non-zero ripple everywhere
except at DC.

I wasn't there to see what actually happened, but I wouldn't be
surprised if the term 'ripple' was preferred over 'deviations
from desired value' for rather pragmatic reasons...

Rune

On 31 Jan, 23:51, HardySpicer <gyansor...@gmail.com> wrote:

> Just to summarise the procedure without all the equations:

The design of IIR filters from analog prototypes is an
excellent excercise for several reasons:

- Provides hands-on experience with a number of key
  concepts in DSP (ZT vs LT, BLT, pre-warping, frequency
  transforms, numerical root solvers, the various filter
  types, normalized vs desired filters, ...)

- Produces a very useful tool at the end

- Can be used as a case study for learning how to program
  efficiently, no matter what language. There are sufficiently
  many variables that the design becomes complicated,

     - Filter type: Butterworth, Cheb 1/2, Elliptic
     - Spec type: Lowpass, Highpass, Bandpass, Bandstop

  while at the same time few enough that one is able to keep
  track of what is going on.

Try to implement *one* design procedure that can do all
of the combinations above, *without* using a case/switch
type construct (including cascaded if-tests) anywhere.

It can be done (at least in C++), but one needs to understand
both modular programming and the programming language, to
achieve it.

Rune

Rune Allnor wrote:
> On 1 Feb, 00:52, Jerry Avins <j...@ieee.org> wrote:
> 
>> We speak different languages. On this side of the Atlantic, Butterworth
>> means maximally flat,
> ...
>> Because A(f) falls monotonically as f
>> increases, there is no ripple. There are many filters with no ripple,
>> but only the Butterworth is maximally flat in the sense I defined above.
> 
> I think the problem is the definition of the term 'ripple'.
> 
> If one interprets it literally, as something like 'fluctuating
> back and forth', then yes, the Butterworth has no ripple.
> 
> If, on the other hand, one interprets the term as 'deviation
> from desired value', then it is immediately clear that the
> Butterworth lowpass filter has a non-zero ripple everywhere
> except at DC.

It does the art no good to debase clear terms. if "ripple" is 
interpreted to mean "deviation from desired value", then nearly every 
filter, calculated transfer function, and approximation will exhibit 
ripple. We will need to invent a new term. "Actual ripple" perhaps?

> I wasn't there to see what actually happened, but I wouldn't be
> surprised if the term 'ripple' was preferred over 'deviations
> from desired value' for rather pragmatic reasons...

A lot of sloppy language gets used, most of it informally. We will lose 
the ability to communicate if we adopt every illogical neologism.

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

On 1 Feb, 12:47, Jerry Avins <j...@ieee.org> wrote:
> Rune Allnor wrote:
> > On 1 Feb, 00:52, Jerry Avins <j...@ieee.org> wrote:
>
> >> We speak different languages. On this side of the Atlantic, Butterworth
> >> means maximally flat,
> > ...
> >> Because A(f) falls monotonically as f
> >> increases, there is no ripple. There are many filters with no ripple,
> >> but only the Butterworth is maximally flat in the sense I defined above.
>
> > I think the problem is the definition of the term 'ripple'.
>
> > If one interprets it literally, as something like 'fluctuating
> > back and forth', then yes, the Butterworth has no ripple.
>
> > If, on the other hand, one interprets the term as 'deviation
> > from desired value', then it is immediately clear that the
> > Butterworth lowpass filter has a non-zero ripple everywhere
> > except at DC.
>
> It does the art no good to debase clear terms. if "ripple" is
> interpreted to mean "deviation from desired value", then nearly every
> filter, calculated transfer function, and approximation will exhibit
> ripple. We will need to invent a new term. "Actual ripple" perhaps?

That's the kind of thing I have said on numerous occations.

In this particular instance, I think we have been beaten by
~80 years of tradition. I wouldn't be surprised of the term
dates back to the likes of Bode, Wold and Nyquist. Pre WWII.

> > I wasn't there to see what actually happened, but I wouldn't be
> > surprised if the term 'ripple' was preferred over 'deviations
> > from desired value' for rather pragmatic reasons...
>
> A lot of sloppy language gets used, most of it informally. We will lose
> the ability to communicate if we adopt every illogical neologism.

Again, I might agree with your point. But then, any similarity
between Homo Sapiens and a logically adept - even minimally
so - creature is superficial and coincidential.

Rune

On Jan 28, 7:57=A0am, "rammya.tv" <rammya...@ymail.com> wrote:
> hi all
> =A0 i would like to know the technical description or derivation about
> the slope of a filter
> ie
> why we say that 1st order filters have a 20 dB/decade (or 6 dB/octave)
> slope.
>
> with regards
> rammya

I see you got a lot of replies, but not an answer.  I have always had
to "understand" things like this on a physical basis rather than a
purely mathematical basis, so I think I can answer your question so
you will understand it too.

A 1st order filter would use a single reactive component such as a low
pass filter with one capacitor and one resistor.

Vin       R
>-------\/\/\------+---------> output
                   |
                   |
                  ---
                  --- C
                   |
                   |
                  ---
                   V

The output voltage is just Vout =3D Vin * Xc/(R+Xc).  Xc is 1/2pi*f*C.

When f is large (well above the corner frequency where Xc =3D R), Xc is
small and R+Xc is approximately R.  So Vout/Vin =3D Xc/R or

Vout/Vin =3D 1/(2pi*R*C*f)

So as F changes by a factor of two, the output voltage changes by a
factor of two.  A voltage change of a factor of two results in 3 dB
voltage, but dB is conventionally expressed as a power ratio, since P
=3D V**2/R the result is 6 dB power per octave.

Does that help?

I think a lot of newbies have trouble with decibels more so than the
applications of them.  For example, it is common to talk about a -3 dB
point in filters and other applications, but that does not mean the
voltage is half of the 0 dB point.  The *power* is half and R =3D Xc in
the example above, but the reactive component is not in phase with the
resistive component so that the voltages do not add up.  Each voltage
is sqrt(2) times the 0dB value.

Maybe this was just a problem for me when I was learning this stuff.
But I find a lot of people don't really get decibels.

Rick

Rune Allnor wrote:
> On 1 Feb, 12:47, Jerry Avins <j...@ieee.org> wrote:
>> Rune Allnor wrote:
>>> On 1 Feb, 00:52, Jerry Avins <j...@ieee.org> wrote:
>>>> We speak different languages. On this side of the Atlantic, Butterworth
>>>> means maximally flat,
>>> ...
>>>> Because A(f) falls monotonically as f
>>>> increases, there is no ripple. There are many filters with no ripple,
>>>> but only the Butterworth is maximally flat in the sense I defined above.
>>> I think the problem is the definition of the term 'ripple'.
>>> If one interprets it literally, as something like 'fluctuating
>>> back and forth', then yes, the Butterworth has no ripple.
>>> If, on the other hand, one interprets the term as 'deviation
>>> from desired value', then it is immediately clear that the
>>> Butterworth lowpass filter has a non-zero ripple everywhere
>>> except at DC.
>> It does the art no good to debase clear terms. if "ripple" is
>> interpreted to mean "deviation from desired value", then nearly every
>> filter, calculated transfer function, and approximation will exhibit
>> ripple. We will need to invent a new term. "Actual ripple" perhaps?
> 
> That's the kind of thing I have said on numerous occations.
> 
> In this particular instance, I think we have been beaten by
> ~80 years of tradition. I wouldn't be surprised of the term
> dates back to the likes of Bode, Wold and Nyquist. Pre WWII.
> 
>>> I wasn't there to see what actually happened, but I wouldn't be
>>> surprised if the term 'ripple' was preferred over 'deviations
>>> from desired value' for rather pragmatic reasons...
>> A lot of sloppy language gets used, most of it informally. We will lose
>> the ability to communicate if we adopt every illogical neologism.
> 
> Again, I might agree with your point. But then, any similarity
> between Homo Sapiens and a logically adept - even minimally
> so - creature is superficial and coincidential.

So if the shelf I had intended to be level is a bit low on one end, it 
ripples? Come on, now. As verbs, "ripple" and "undulate" are synonyms.

I recognize that there are certain equations with variables labeled 
"ripple" (or computer programs with data-entry boxes similarly labeled) 
where the term can be used to specify droop, but that doesn't make droop 
become ripple. Abraham Lincoln once asked his cabinet, "If we call a 
tail a leg, how many legs does a sheep have?" After getting the expected 
answer, he said, "Wrong. Calling it a leg doesn't make it one."

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

On 1 Feb, 17:50, Jerry Avins <j...@ieee.org> wrote:
> Rune Allnor wrote:
> > On 1 Feb, 12:47, Jerry Avins <j...@ieee.org> wrote:
> >> Rune Allnor wrote:
> >>> On 1 Feb, 00:52, Jerry Avins <j...@ieee.org> wrote:
> >>>> We speak different languages. On this side of the Atlantic, Butterworth
> >>>> means maximally flat,
> >>> ...
> >>>> Because A(f) falls monotonically as f
> >>>> increases, there is no ripple. There are many filters with no ripple,
> >>>> but only the Butterworth is maximally flat in the sense I defined above.
> >>> I think the problem is the definition of the term 'ripple'.
> >>> If one interprets it literally, as something like 'fluctuating
> >>> back and forth', then yes, the Butterworth has no ripple.
> >>> If, on the other hand, one interprets the term as 'deviation
> >>> from desired value', then it is immediately clear that the
> >>> Butterworth lowpass filter has a non-zero ripple everywhere
> >>> except at DC.
> >> It does the art no good to debase clear terms. if "ripple" is
> >> interpreted to mean "deviation from desired value", then nearly every
> >> filter, calculated transfer function, and approximation will exhibit
> >> ripple. We will need to invent a new term. "Actual ripple" perhaps?
>
> > That's the kind of thing I have said on numerous occations.
>
> > In this particular instance, I think we have been beaten by
> > ~80 years of tradition. I wouldn't be surprised of the term
> > dates back to the likes of Bode, Wold and Nyquist. Pre WWII.
>
> >>> I wasn't there to see what actually happened, but I wouldn't be
> >>> surprised if the term 'ripple' was preferred over 'deviations
> >>> from desired value' for rather pragmatic reasons...
> >> A lot of sloppy language gets used, most of it informally. We will lose
> >> the ability to communicate if we adopt every illogical neologism.
>
> > Again, I might agree with your point. But then, any similarity
> > between Homo Sapiens and a logically adept - even minimally
> > so - creature is superficial and coincidential.
>
> So if the shelf I had intended to be level is a bit low on one end, it
> ripples? Come on, now. As verbs, "ripple" and "undulate" are synonyms.

Again, *I* agree with you, but that's not enough to undo
3/4 century's worth of damage.

> I recognize that there are certain equations with variables labeled
> "ripple" (or computer programs with data-entry boxes similarly labeled)
> where the term can be used to specify droop, but that doesn't make droop
> become ripple. Abraham Lincoln once asked his cabinet, "If we call a
> tail a leg, how many legs does a sheep have?" After getting the expected
> answer, he said, "Wrong. Calling it a leg doesn't make it one."

Brilliant. I wish I knew that anecdote earlier.

Rune

On Feb 1, 12:52=A0pm, Jerry Avins <j...@ieee.org> wrote:
> HardySpicer wrote:
> > On Feb 1, 9:33 am, HardySpicer <gyansor...@gmail.com> wrote:
> >> On Jan 31, 9:10 pm, Rune Allnor <all...@tele.ntnu.no> wrote:
>
> >>> On 31 Jan, 08:15, HardySpicer <gyansor...@gmail.com> wrote:
> >>>> On Jan 31, 6:45 pm, Rune Allnor <all...@tele.ntnu.no> wrote:
> >>>>> On 31 Jan, 03:22, Jerry Avins <j...@ieee.org> wrote:
> >>>>>> HardySpicer wrote:
> >>>>>>> On Jan 31, 6:20 am, Jerry Avins <j...@ieee.org> wrote:
> >>>>>>>> HardySpicer wrote:
> >>>>>>>>> On Jan 29, 8:59 am, Clay <c...@claysturner.com> wrote:
> >>>>>>>> =A0 =A0...
> >>>>>>>>>> p.s. The magnitude response of a nth butterworth lowpass filte=
r is
> >>>>>>>>>> simply
> >>>>>>>>>> A(f) =3D 1/sqrt(1+(f/fc)^2n)
> >>>>>>>>> Only with 3dB passband ripple..
> >>>>>>>> Hunh? Show us. (Hint: show that a derivative goes to zero somewh=
ere
> >>>>>>>> other than f=3D0.)
> >>>>>>>> Jerry
> >>>>>>>> --
> >>>>>>>> Engineering is the art of making what you want from things you c=
an get.
> >>>>>>>> =AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF
> >>>>>>> eh? This is standard theory for Butterworth filters but many peop=
le
> >>>>>>> only get taught the 3dB version.
> >>>>>>> You specify the attenuation in the passband and find the ripple f=
actor
> >>>>>>> eps for that attenuation. Turns out the the ripple factor is unit=
y
> >>>>>>> (nearly) when the passband attenuation is 3dB. The poles lie in a
> >>>>>>> circle radius unity for 3dB passband ripple but on a circle radiu=
s (1/
> >>>>>>> eps)^1/n if I remember right for a ripple factor eps and order n.=
 The
> >>>>>>> equation =A0A(f) =3D 1/sqrt(1+(f/fc)^2n) also changes. Look it up=
..
> >>>>>> I can only guess what you're driving at. "Butterworth" is the same=
 as
> >>>>>> "maximally flat". A Butterworth filter exhibits no ripple at all. =
True,
> >>>>>> the edge of the passband is 3 dB down from the flat top, but that'=
s not
> >>>>>> ripple. Are you by chance thinking of Butterworth as the flat limi=
t of a
> >>>>>> Chebychev filter?
> >>>>> Clay is right in the formula for the Butterworth frequency
> >>>>> response, and in that formula a characteristic frequency
> >>>>> appears that indicate the frequency of the 3dB point.
> >>>>> However, that characteristic frequency needs not be the
> >>>>> cut-off frequency of the filter spec. Given any ripple
> >>>>> and corner frequency, one can dedude the corresponding
> >>>>> 3dB frequency of the filter.
> >>>>> So you discussion hinges on exactly what meaning is assigned
> >>>>> to the factor 'fc' in the fomula: The characteristic 3dB
> >>>>> frequency, or the passband corner frequency?
> >>>>> Rune
> >>>> The convention is to define the dB attenuation in the passband. So w=
e
> >>>> say for example (for a lowpass filter design) =A0you need 1dB
> >>>> attenuation in the passband at 1kHz. The 3dB freq is not normally
> >>>> defined here though of course it exists in all LTI systems. So 1kHz =
is
> >>>> the passband 'edge" but it is nearly always defined as unity for
> >>>> normalised frequency. My point is that you cannot use the 3dB
> >>>> Butterworth solution for this.
> >>> Well, given the exact example you state, then no, the equation
> >>> does not work. But your example only states a partial spec.
> >>> A complete filter spec gives *two* constraints: The passband and
> >>> stopband frequencies, with associated attenuations.
> >> I missed that out for simplicity. Yes there is a formula for order of
> >> the filter based on stopband attenuation.
> >> It's a bit messy for ascii and doesn't change the results in any way.
> >> The formula comes from
>
> >> As=3D10Log10(1+eps^2.vs^2n) where v is =A0normalised freq, n is order =
and
> >> As is stopband attenuation in dB. eps is the ripple factor previously
> >> calculated and vs is the normalise stopband freq..
> >> So you re-arrange the above to find vs^2n=3D(10^0.1As-1)/eps^2 and by
> >> taking logs we find n=3D0.5Log[(10^0.1As-1)/eps^2]
>
> >> =A0The Butterworth polynomial for say a 2nd order is
>
> >> s^2+sqrt(2)s+1 =A0but this is for 3dB passband attenuation only. This
> >> will change for the (say) 1dB case and hence the filter with it.
>
> >> so the basic formula quoted is normally for the 3dB case.
>
> >> Your comments about the digital versions are all valid of course
> >> aswith any analogue to discrete conversion.
>
> >> Hardy
>
> > Just to summarise the procedure without all the equations:
>
> > Decide on type of filter reuqired - eg Lowpass,Highpass bandpass
>
> > Specify a lowpass prototype using the following (for all filter types
> > ie Lowpass,Highpass,Bandpass etc)
>
> > For a particular Passband freq wp and ws find the normailsed
> > equivalents vp=3D1 and wp/ws =3Dvs
>
> > For a particular dB =A0passband attenuation Ap =A0find the ripple facto=
r
> > eps
>
> > For a particular dB =A0stopband attenuation As find the order n. (see
> > previous post)
>
> > Find the radius of the poles using eps above and n
>
> > Find the angle of the poles
>
> > Construct the pole polynomial P(s) =A0from the above two steps
>
> > This gives us our low-pass prototype from which we can transform to
> > find the real filter.
>
> > Find the real filter by first forming the prototype Hp(s)=3D1/P(s) wher=
e
> > P(s) is the pole polynomial found previosley.
>
> > Then transform Lowpass to Lowpass s--> s/wp or Lowpass to highpass s--
> >> ws/s and there are other transforms for lowpass to bandpass,lowpass
> > to bandsstop etc.
> > Lowpass to bandpass is s--->(s^2+w0^2)/Bs where B is bandwidth of
> > passband and w0 is the centre freq - normally the geometric root of wl
> > and wh which are at the lower and higher ends of the passband
> > respectively. Note that for this case when transforming a 2nd order
> > prototype only yields a first order "skirt" ie slope.
>
> > Once you have all this you can realise the filter analogue style or do
> > the pre-warping and Bilinear transform for digital. You won't get
> > linear phase of course.
>
> We speak different languages. On this side of the Atlantic, Butterworth
> means maximally flat, a condition achieved in the low-pass prototype by
> setting all the derivatives of A(f) to zero at f=3D0. There is no eps. Th=
e
> formula for A(f) is as Clay gave it, 1/sqrt(1+(f/fc)^2n), where n is the
> order of the filter. For a first-order filter, fc is 3 dB down and the
> rolloff is 6 dB/octve. For a second-order filter fc is 6 dB down and the
> rolloff is 12 dB/octve, and so on. Because A(f) falls monotonically as f
> increases, there is no ripple. There are many filters with no ripple,
> but only the Butterworth is maximally flat in the sense I defined above.
>
> Jerry
> --
> Engineering is the art of making what you want from things you can get.
> =AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF

As I said it's just because the "ripple factor" is a generalisation
that it is called that. The ripple effect does not apply to
Butterworth - only the attenuation in the passband which need not be
limited to 3dB or 6dB etc..

Hardy

On Feb 2, 5:50=A0am, Jerry Avins <j...@ieee.org> wrote:
> Rune Allnor wrote:
> > On 1 Feb, 12:47, Jerry Avins <j...@ieee.org> wrote:
> >> Rune Allnor wrote:
> >>> On 1 Feb, 00:52, Jerry Avins <j...@ieee.org> wrote:
> >>>> We speak different languages. On this side of the Atlantic, Butterwo=
rth
> >>>> means maximally flat,
> >>> ...
> >>>> Because A(f) falls monotonically as f
> >>>> increases, there is no ripple. There are many filters with no ripple=
,
> >>>> but only the Butterworth is maximally flat in the sense I defined ab=
ove.
> >>> I think the problem is the definition of the term 'ripple'.
> >>> If one interprets it literally, as something like 'fluctuating
> >>> back and forth', then yes, the Butterworth has no ripple.
> >>> If, on the other hand, one interprets the term as 'deviation
> >>> from desired value', then it is immediately clear that the
> >>> Butterworth lowpass filter has a non-zero ripple everywhere
> >>> except at DC.
> >> It does the art no good to debase clear terms. if "ripple" is
> >> interpreted to mean "deviation from desired value", then nearly every
> >> filter, calculated transfer function, and approximation will exhibit
> >> ripple. We will need to invent a new term. "Actual ripple" perhaps?
>
> > That's the kind of thing I have said on numerous occations.
>
> > In this particular instance, I think we have been beaten by
> > ~80 years of tradition. I wouldn't be surprised of the term
> > dates back to the likes of Bode, Wold and Nyquist. Pre WWII.
>
> >>> I wasn't there to see what actually happened, but I wouldn't be
> >>> surprised if the term 'ripple' was preferred over 'deviations
> >>> from desired value' for rather pragmatic reasons...
> >> A lot of sloppy language gets used, most of it informally. We will los=
e
> >> the ability to communicate if we adopt every illogical neologism.
>
> > Again, I might agree with your point. But then, any similarity
> > between Homo Sapiens and a logically adept - even minimally
> > so - creature is superficial and coincidential.
>
> So if the shelf I had intended to be level is a bit low on one end, it
> ripples? Come on, now. As verbs, "ripple" and "undulate" are synonyms.
>

I don't think you are right there. although the use of the word ripple
factor is clearly a miss-noma I still don't understand how you go
about designing a Butterworth filter with say a 1dB attenuation in the
passband at 1kHz and a 40 dB attenuation in the stopband at 20kHz? Do
you define only 3dB attenuations and 'work back"?


Hardy

On 1 Feb, 19:18, HardySpicer <gyansor...@gmail.com> wrote:
>  I still don't understand how you go
> about designing a Butterworth filter with say a 1dB attenuation in the
> passband at 1kHz and a 40 dB attenuation in the stopband at 20kHz? Do
> you define only 3dB attenuations and 'work back"?

Start with (view with fixed-width font)

              1
H(w) = --------------                                   [1]
        1+(Wp'/Wc)^2N

where N has been determined from the spec, as
indicated earlier. Define Wc == 1, and get

        1
Dp = ----------                                         [2]
      1+Wp'^2N

Here, Wp' is *a* frequency Wp' < 1 where the attenuation is Dp.
Solve [2] for Wp' to find

Wp' = (Dp^(-2)-1)^(1/2N)                                [3]

Proceed to find the coefficients of the N'th order filter.

Last, apply a LP -> LP frequency transform that maps the
design frequency Wp' to the target frequency Wp.

Job done. In s domain. All the DT <-> CT domain mappings
need to be added, if you want a DT filter.

Rune

HardySpicer wrote:

   ...

>> So if the shelf I had intended to be level is a bit low on one end, it
>> ripples? Come on, now. As verbs, "ripple" and "undulate" are synonyms.
>>
> 
> I don't think you are right there. although the use of the word ripple
> factor is clearly a miss-noma I still don't understand how you go
> about designing a Butterworth filter with say a 1dB attenuation in the
> passband at 1kHz and a 40 dB attenuation in the stopband at 20kHz? Do
> you define only 3dB attenuations and 'work back"?

All low-pass Butterworth filters of the same order have the same shape 
when plotted in normalized co-ordinates, f/fc and log gain. What is 
usually specifies is maximum attenuation in the passband and rate of 
attenuation in the stopband. The stopband rate in dB/decade is always a 
multiple of 20, and the multiple is the order. (A third-order filter 
rolls off at 60 dB/decade.)

Given the maximum attenuation allowed in the passband, i.e., the minimum 
A(f) in the passband, and the order n, it is easy to solve A(f) = 
1/sqrt(1+(f/fc)^2n) for fc.

For your filter (1dB attenuation in the passband at 1kHz and a 40 dB 
attenuation in the stopband at 20kHz) the rules of thumb indicate that a 
first-order filter with fc=2 KHz meets the spec. (The response will be 1 
dB down at fc/2, 3 dB down at fc, 5 dB down at 2fc, and 20 dB down at 
10fc. You picked the kind of easy numbers I would use on a quiz.)

An alternate solution that I would rarely use associates 2 values of 
A(f) with assigned values of f and solves the simultaneous equations for 
n and fc. Of course, one rounds n up to an integer.

Jerry

P.S. A first-order Butterworth (and Chebychev!) is a simple R-C.
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

On Feb 2, 9:01=A0am, Jerry Avins <j...@ieee.org> wrote:
> HardySpicer wrote:
>
> =A0 =A0...
>
> >> So if the shelf I had intended to be level is a bit low on one end, it
> >> ripples? Come on, now. As verbs, "ripple" and "undulate" are synonyms.
>
> > I don't think you are right there. although the use of the word ripple
> > factor is clearly a miss-noma I still don't understand how you go
> > about designing a Butterworth filter with say a 1dB attenuation in the
> > passband at 1kHz and a 40 dB attenuation in the stopband at 20kHz? Do
> > you define only 3dB attenuations and 'work back"?
>
> All low-pass Butterworth filters of the same order have the same shape
> when plotted in normalized co-ordinates, f/fc and log gain. What is
> usually specifies is maximum attenuation in the passband and rate of
> attenuation in the stopband. The stopband rate in dB/decade is always a
> multiple of 20, and the multiple is the order. (A third-order filter
> rolls off at 60 dB/decade.)
>
> Given the maximum attenuation allowed in the passband, i.e., the minimum
> A(f) in the passband, and the order n, it is easy to solve A(f) =3D
> 1/sqrt(1+(f/fc)^2n) for fc.
>
> For your filter (1dB attenuation in the passband at 1kHz and a 40 dB
> attenuation in the stopband at 20kHz) the rules of thumb indicate that a
> first-order filter with fc=3D2 KHz meets the spec. (The response will be =
1
> dB down at fc/2, 3 dB down at fc, 5 dB down at 2fc, and 20 dB down at
> 10fc. You picked the kind of easy numbers I would use on a quiz.)
>
> An alternate solution that I would rarely use associates 2 values of
> A(f) with assigned values of f and solves the simultaneous equations for
> n and fc. Of course, one rounds n up to an integer.
>
> Jerry
>
> P.S. A first-order Butterworth (and Chebychev!) is a simple R-C.
> --
> Engineering is the art of making what you want from things you can get.
> =AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF

Ok so you are applying the 3dB rules and working back in freq. The
other method gives you the direct solution.
I am not saying that the slopes won't be multiples of +/- -20dB/decade
- for LTI systems they have to be of course.

ok so here is what is happening. For a nomalised freq v the magnitude-
squared Butetrworth is given by

|H(v)|^2=3D1/(1+eps^2.v^2n)

what you have is

|H(v)|^2=3D1/(1+v^2n)

so you are absorbing the constand eps^2 into the term v^2n. The
problem with that is that I don't think the calculations are as simple
as you need to 'work back" so to speak.

Hardy

rickman wrote:
> On Jan 28, 7:57 am, "rammya.tv" <rammya...@ymail.com> wrote:
>> hi all
>>   i would like to know the technical description or derivation about
>> the slope of a filter
>> ie
>> why we say that 1st order filters have a 20 dB/decade (or 6 dB/octave)
>> slope.
>>
>> with regards
>> rammya
> 
> I see you got a lot of replies, but not an answer.  I have always had
> to "understand" things like this on a physical basis rather than a
> purely mathematical basis, so I think I can answer your question so
> you will understand it too.
> 
> A 1st order filter would use a single reactive component such as a low
> pass filter with one capacitor and one resistor.
> 
> Vin       R
>> -------\/\/\------+---------> output
>                    |
>                    |
>                   ---
>                   --- C
>                    |
>                    |
>                   ---
>                    V
> 
> The output voltage is just Vout = Vin * Xc/(R+Xc).  Xc is 1/2pi*f*C.
> 
> When f is large (well above the corner frequency where Xc = R), Xc is
> small and R+Xc is approximately R.  So Vout/Vin = Xc/R or
> 
> Vout/Vin = 1/(2pi*R*C*f)
> 
> So as F changes by a factor of two, the output voltage changes by a
> factor of two.  A voltage change of a factor of two results in 3 dB
> voltage, but dB is conventionally expressed as a power ratio, since P
> = V**2/R the result is 6 dB power per octave.
> 
> Does that help?
> 
> I think a lot of newbies have trouble with decibels more so than the
> applications of them.  For example, it is common to talk about a -3 dB
> point in filters and other applications, but that does not mean the
> voltage is half of the 0 dB point.  The *power* is half and R = Xc in
> the example above, but the reactive component is not in phase with the
> resistive component so that the voltages do not add up.  Each voltage
> is sqrt(2) times the 0dB value.
> 
> Maybe this was just a problem for me when I was learning this stuff.
> But I find a lot of people don't really get decibels.
> 
> Rick

Rick,
The expression you give, Vout = Vin * Xc/(R+Xc) is fine in 
the region you are talking about where Xc << R, but it is 
not very accurate otherwise.

In particular, at a frequency where R = Xc the above 
expression becomes: Vout = Vin * 1/2
This gives a gain (Vout/Vin) of 0.5 ( -6.0 dB), which is 
incorrect.


If instead of Xc, the reactance of C, you use jXc, the 
impedance of C,  you get the correect expression:
   Vout = Vin * jXc/(R+jXc)
At the frequency where R = Xc the expression becomes:
   Vout = Vin * 1/sqrt(2)
This gives a gain of 0.707 (-3.0 dB) which is correct.

Regards,
John

>Rune Allnor wrote:
>> On 1 Feb, 00:52, Jerry Avins <j...@ieee.org> wrote:
>> 
>>> We speak different languages. On this side of the Atlantic,
Butterworth
>>> means maximally flat,
>> ...
>>> Because A(f) falls monotonically as f
>>> increases, there is no ripple. There are many filters with no ripple,
>>> but only the Butterworth is maximally flat in the sense I defined
above.
>> 
>> I think the problem is the definition of the term 'ripple'.
>> 
>> If one interprets it literally, as something like 'fluctuating
>> back and forth', then yes, the Butterworth has no ripple.
>> 
>> If, on the other hand, one interprets the term as 'deviation
>> from desired value', then it is immediately clear that the
>> Butterworth lowpass filter has a non-zero ripple everywhere
>> except at DC.
>
>It does the art no good to debase clear terms. if "ripple" is 
>interpreted to mean "deviation from desired value", then nearly every 
>filter, calculated transfer function, and approximation will exhibit 
>ripple. We will need to invent a new term. "Actual ripple" perhaps?
>
>> I wasn't there to see what actually happened, but I wouldn't be
>> surprised if the term 'ripple' was preferred over 'deviations
>> from desired value' for rather pragmatic reasons...
>
>A lot of sloppy language gets used, most of it informally. We will lose 
>the ability to communicate if we adopt every illogical neologism.

Just as much sloppy language gets used quite formally. The two of us can't
communicate unless the comms channel is operating above capacity. That's
completely backwards from proper English. It makes ripple for droop seem a
minor distortion of the English language.

Names are generally chosen to obscure rather than illuminate. Professional
jargon isn't much different.

Steve

On Feb 2, 12:58=A0am, John Monro <johnmo...@optusnet.com.au> wrote:
> rickman wrote:
> > On Jan 28, 7:57 am, "rammya.tv" <rammya...@ymail.com> wrote:
> >> hi all
> >> =A0 i would like to know the technical description or derivation about
> >> the slope of a filter
> >> ie
> >> why we say that 1st order filters have a 20 dB/decade (or 6 dB/octave)
> >> slope.
>
> >> with regards
> >> rammya
>
> > I see you got a lot of replies, but not an answer. =A0I have always had
> > to "understand" things like this on a physical basis rather than a
> > purely mathematical basis, so I think I can answer your question so
> > you will understand it too.
>
> > A 1st order filter would use a single reactive component such as a low
> > pass filter with one capacitor and one resistor.
>
> > Vin =A0 =A0 =A0 R
> >> -------\/\/\------+---------> output
> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0|
> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0|
> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 ---
> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 --- C
> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0|
> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0|
> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 ---
> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0V
>
> > The output voltage is just Vout =3D Vin * Xc/(R+Xc). =A0Xc is 1/2pi*f*C=
..
>
> > When f is large (well above the corner frequency where Xc =3D R), Xc is
> > small and R+Xc is approximately R. =A0So Vout/Vin =3D Xc/R or
>
> > Vout/Vin =3D 1/(2pi*R*C*f)
>
> > So as F changes by a factor of two, the output voltage changes by a
> > factor of two. =A0A voltage change of a factor of two results in 3 dB
> > voltage, but dB is conventionally expressed as a power ratio, since P
> > =3D V**2/R the result is 6 dB power per octave.
>
> > Does that help?
>
> > I think a lot of newbies have trouble with decibels more so than the
> > applications of them. =A0For example, it is common to talk about a -3 d=
B
> > point in filters and other applications, but that does not mean the
> > voltage is half of the 0 dB point. =A0The *power* is half and R =3D Xc =
in
> > the example above, but the reactive component is not in phase with the
> > resistive component so that the voltages do not add up. =A0Each voltage
> > is sqrt(2) times the 0dB value.
>
> > Maybe this was just a problem for me when I was learning this stuff.
> > But I find a lot of people don't really get decibels.
>
> > Rick
>
> Rick,
> The expression you give, Vout =3D Vin * Xc/(R+Xc) is fine in
> the region you are talking about where Xc << R, but it is
> not very accurate otherwise.
>
> In particular, at a frequency where R =3D Xc the above
> expression becomes: Vout =3D Vin * 1/2
> This gives a gain (Vout/Vin) of 0.5 ( -6.0 dB), which is
> incorrect.
>
> If instead of Xc, the reactance of C, you use jXc, the
> impedance of C, =A0you get the correect expression:
> =A0 =A0Vout =3D Vin * jXc/(R+jXc)
> At the frequency where R =3D Xc the expression becomes:
> =A0 =A0Vout =3D Vin * 1/sqrt(2)
> This gives a gain of 0.707 (-3.0 dB) which is correct.


Did you read my post???  I clearly stated the region where the
approximation was reasonable and in the later part I discuss why the
gain is not as it would appear in the region where the reactive
component is out of phase with the resistive component.  You merely
repeated what I posted with complex math which may well be beyond the
grasp of the OP.  As I stated, I was trying to explain it in simpler
terms which someone can understand without resorting to complex
arithmetic.

Rick

rickman wrote:
> On Feb 2, 12:58 am, John Monro <johnmo...@optusnet.com.au> wrote:
>> rickman wrote:
>>> On Jan 28, 7:57 am, "rammya.tv" <rammya...@ymail.com> wrote:
>>>> hi all
>>>>   i would like to know the technical description or derivation about
>>>> the slope of a filter
>>>> ie
>>>> why we say that 1st order filters have a 20 dB/decade (or 6 dB/octave)
>>>> slope.
>>>> with regards
>>>> rammya
>>> I see you got a lot of replies, but not an answer.  I have always had
>>> to "understand" things like this on a physical basis rather than a
>>> purely mathematical basis, so I think I can answer your question so
>>> you will understand it too.
>>> A 1st order filter would use a single reactive component such as a low
>>> pass filter with one capacitor and one resistor.
>>> Vin       R
>>>> -------\/\/\------+---------> output
>>>                    |
>>>                    |
>>>                   ---
>>>                   --- C
>>>                    |
>>>                    |
>>>                   ---
>>>                    V
>>> The output voltage is just Vout = Vin * Xc/(R+Xc).  Xc is 1/2pi*f*C.
>>> When f is large (well above the corner frequency where Xc = R), Xc is
>>> small and R+Xc is approximately R.  So Vout/Vin = Xc/R or
>>> Vout/Vin = 1/(2pi*R*C*f)
>>> So as F changes by a factor of two, the output voltage changes by a
>>> factor of two.  A voltage change of a factor of two results in 3 dB
>>> voltage, but dB is conventionally expressed as a power ratio, since P
>>> = V**2/R the result is 6 dB power per octave.
>>> Does that help?
>>> I think a lot of newbies have trouble with decibels more so than the
>>> applications of them.  For example, it is common to talk about a -3 dB
>>> point in filters and other applications, but that does not mean the
>>> voltage is half of the 0 dB point.  The *power* is half and R = Xc in
>>> the example above, but the reactive component is not in phase with the
>>> resistive component so that the voltages do not add up.  Each voltage
>>> is sqrt(2) times the 0dB value.
>>> Maybe this was just a problem for me when I was learning this stuff.
>>> But I find a lot of people don't really get decibels.
>>> Rick
>> Rick,
>> The expression you give, Vout = Vin * Xc/(R+Xc) is fine in
>> the region you are talking about where Xc << R, but it is
>> not very accurate otherwise.
>>
>> In particular, at a frequency where R = Xc the above
>> expression becomes: Vout = Vin * 1/2
>> This gives a gain (Vout/Vin) of 0.5 ( -6.0 dB), which is
>> incorrect.
>>
>> If instead of Xc, the reactance of C, you use jXc, the
>> impedance of C,  you get the correect expression:
>>    Vout = Vin * jXc/(R+jXc)
>> At the frequency where R = Xc the expression becomes:
>>    Vout = Vin * 1/sqrt(2)
>> This gives a gain of 0.707 (-3.0 dB) which is correct.
> 
> 
> Did you read my post???  I clearly stated the region where the
> approximation was reasonable and in the later part I discuss why the
> gain is not as it would appear in the region where the reactive
> component is out of phase with the resistive component.  You merely
> repeated what I posted with complex math which may well be beyond the
> grasp of the OP.  As I stated, I was trying to explain it in simpler
> terms which someone can understand without resorting to complex
> arithmetic.
> 
> Rick

I did not 'merely' repeat your post; I corrected the 
expression you used.  I am sorry that you have taken 
offence.  I hope my post was useful to the OP.

Regards,
John