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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 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.
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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 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
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