> 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
Reply by rickman●February 2, 20102010-02-02
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
Reply by steveu●February 2, 20102010-02-02
>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
Reply by John Monro●February 2, 20102010-02-02
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
Reply by HardySpicer●February 1, 20102010-02-01
On Feb 2, 9:01�am, Jerry Avins <j...@ieee.org> wrote:
> 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.
> �����������������������������������������������������������������������
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=1/(1+eps^2.v^2n)
what you have is
|H(v)|^2=1/(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
Reply by Jerry Avins●February 1, 20102010-02-01
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.
�����������������������������������������������������������������������
Reply by Rune Allnor●February 1, 20102010-02-01
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
Reply by HardySpicer●February 1, 20102010-02-01
On Feb 2, 5:50�am, 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.
>
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
Reply by HardySpicer●February 1, 20102010-02-01
On Feb 1, 12:52�pm, 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:
> >>>>>>>> � �...
> >>>>>>>>>> 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.
> �����������������������������������������������������������������������
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
Reply by Rune Allnor●February 1, 20102010-02-01
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