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
why first order filter have slope 6db/octave
Started by ●January 28, 2010
Reply by ●January 28, 20102010-01-28
On 28 Jan, 13:57, "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.I suppose the general derivation would be something like slope/bandwidth = (|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
Reply by ●January 28, 20102010-01-28
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. �����������������������������������������������������������������������
Reply by ●January 28, 20102010-01-28
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 > rammyaFor 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)
Reply by ●January 29, 20102010-01-29
On Jan 29, 1:57�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 > rammyaIn radians/s we have that the transfer function is H(jw)=1/(1+jwT) where T is the time-constant. We can write this as 1/ (1+jw/wc) where wc=1/T, the cut-off freq. Now convert to dB magnitude by taking 20Log10 dBMag= -20log10sqrt((1+(w/wc)^2)). Now asymptotically is when w--.>inf and hence we can ignore the 1 giving dBMag=-20Log10(w/wc) or a slope of -20dB/decade. test by putting w=1 and w=10/wc (a decade for normalised freq).The dBMag drops by 20dB This is the exact same as -6dB/octave. Check by putting w=1 and w=2/wc and see the dBMag drop by =6dB. Hardy
Reply by ●January 30, 20102010-01-30
On Jan 29, 8:59�am, Clay <c...@claysturner.com> 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 > > 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
Reply by ●January 30, 20102010-01-30
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. �����������������������������������������������������������������������
Reply by ●January 30, 20102010-01-30
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. Hardy
Reply by ●January 30, 20102010-01-30
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. �����������������������������������������������������������������������
Reply by ●January 30, 20102010-01-30
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) Hardy
