why first order filter have slope 6db/octave
in [DSP]
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From: HardySpicer on 30 Jan 2010 05:11 On Jan 29, 8:59 am, Clay <c...(a)claysturner.com> wrote: > On Jan 28, 7:57 am, "rammya.tv" <rammya...(a)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 > > 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) = 1/sqrt(1+(f/fc)^2n) Only with 3dB passband ripple.. Hardy
From: Jerry Avins on 30 Jan 2010 12:20 HardySpicer wrote: > On Jan 29, 8:59 am, Clay <c...(a)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. �����������������������������������������������������������������������
From: HardySpicer on 30 Jan 2010 19:11 On Jan 31, 6:20 am, Jerry Avins <j...(a)ieee.org> wrote: > HardySpicer wrote: > > On Jan 29, 8:59 am, Clay <c...(a)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. Hardy
From: Jerry Avins on 30 Jan 2010 21:22 HardySpicer wrote: > On Jan 31, 6:20 am, Jerry Avins <j...(a)ieee.org> wrote: >> HardySpicer wrote: >>> On Jan 29, 8:59 am, Clay <c...(a)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. �����������������������������������������������������������������������
From: HardySpicer on 30 Jan 2010 21:54
On Jan 31, 3:22 pm, Jerry Avins <j...(a)ieee.org> wrote: > HardySpicer wrote: > > On Jan 31, 6:20 am, Jerry Avins <j...(a)ieee.org> wrote: > >> HardySpicer wrote: > >>> On Jan 29, 8:59 am, Clay <c...(a)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) Hardy |