From: rammya.tv on 28 Jan 2010 07:57 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
From: Rune Allnor on 28 Jan 2010 08:10 On 28 Jan, 13:57, "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. 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
From: Jerry Avins on 28 Jan 2010 09:50 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 lowpass 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. �����������������������������������������������������������������������
From: Clay on 28 Jan 2010 14:59 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)
From: HardySpicer on 29 Jan 2010 15:44 On Jan 29, 1:57 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 In radians/s we have that the transfer function is H(jw)=1/(1+jwT) where T is the timeconstant. We can write this as 1/ (1+jw/wc) where wc=1/T, the cutoff 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

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