I have been doing some preliminary lab calculations since I have an intense session coming up next week. I have to solve many filters, most of which are just basic, passive RC filters; others form part of Operational Amplifier circuits (which I have been told are called active filters). My lab instructor told us that the roll-off of filters after the cutoff frequency is -20 dB/decade (he was talking about a Low Pass RC filter). However, sometimes, when I am solving circuits involving OpAmps, I get that for instance, the gain at the cutoff frequency is 17 dB, and at \$10f_c\$, I get 0 dB, which is not a -20 dB roll-off. Other times I have gotten \$G(f_c)= -3dB\$ and \$G(10f_c)= -20dB\$, which is yet again a 17 dB difference. Is this normal? How can I expect these results?

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    \$\begingroup\$ The -20dB/decade is an asymptotic value; the further you go from the cutoff frequency the more accurate it gets. Also, -20dB/dec is only true for a first-order filter! \$\endgroup\$
    – Hearth
    Commented May 14, 2018 at 20:24
  • 1
    \$\begingroup\$ You have to look at all the poles and zeros of the filter. There is an academic technique to draw an approximate Bode plot based on the poles and zeros. In general, the slope of the simple low-pass filter response is 20dB per decade if there is only one reactive component. 40dB per decade if there are two reactive components, 60 if three, etc. Google techniques for drawing the Bode Plot. \$\endgroup\$
    – user57037
    Commented May 14, 2018 at 20:37
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    \$\begingroup\$ Also, the roll-off you will get with a butterworth filter compared to a chebychev filter is not the same near the cut-off frequency. And if you want to push even farther the analysis, elliptical filters don't even follow that dB/decade asymptote rule! \$\endgroup\$
    – lucas92
    Commented May 14, 2018 at 20:41
  • \$\begingroup\$ @lucas92 They do, but the transition width would have to be unrealistically wide. Inverse Chebyshev and inverse Pascal follow close, too. In rest, odd orders will follow the asymptotic -20dB/dec behaviour for the slopes beyond the last zero (for all three). \$\endgroup\$ Commented May 15, 2018 at 10:25

4 Answers 4


For an ideal first order filter, 20dB/decade is the asymptote in the stop band.

A real filter might do other things far into the stopband, due to non-idealities in the components. For instance with an RC filter, stray C across the R, or ESR in the capacitor, will both limit the ultimate attenuation.

A second order filter will go off at 40dB/decade. An Nth order filter will drop at 20N dB/decade.


Why 20dB/decade?

-20dB is actually 1/10th voltage per decade.

Why: Because your basic RC filter element is just a voltage divider.

At 10x F, reactance (X) of C will be 1/10th, so signal will be divided to 1/10th = -20dB

Why is it not exactly true close-in?

The C phase shifts the signal, so that the divider voltages don't divide "inline", but at 90 degrees. so when R=X the signal is not 1/2 but 1/sqrt(2) (the long side of a triangle), ie not 6dB but 3dB.

This is true for each single RC pole, without feedback.


A real filter will have a different cuttoff then an ideal pole. The differences in slope are because you are selecting frequencies close to the pole, which is rounded off. The further away you get away from the pole, the closer you get to a 20db rolloff.

enter image description here
Source: Wikipedia Low pass filter


A brown noise filter of white noise is staggered RC filters to give 10dB / decade in the useful range otherwise all typical filters are 20dB/dec per n order at 1 decade away from breakpoint and phase shift approach of almost 90 deg per n order takes 2 decades above or below break point. For LPF and HPF respectively.

extra point

Here I compare 4th order Butterworth and Bessel (maximally flat group delay) Both are 80dB/decade. Can you see the subtle differences just from a small difference in values.

enter image description here You differentiate or take the slope of phase to get group delay .

This group delay can have a poor effect on data where the jitter now increases due to the random data and random frequencies now getting extra delay time at the breakpoint.

If you zoom in, you can see the Bessel filter has different Q values and staggers each 2nd order stage around 1,4,1.5kHz yet has the same net -3dB 1kHz BW cutoff and the same -80dB/decade rolloff.

  • \$\begingroup\$ Roberge explains in this video how a -10dB/decade approximation is useful in opamp compensation : m.youtube.com/watch?v=fn2UGyk5DP4 \$\endgroup\$
    – HKOB
    Commented May 15, 2018 at 7:42
  • \$\begingroup\$ also useful in PLL filters \$\endgroup\$ Commented May 15, 2018 at 10:01

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