3
\$\begingroup\$

So as we know that the cutoff frequency of a filter can be determined from graph by dropping 3dB. But, how do we take the 3dB drop? From the highest point or from 0? In this case, this graph highest point is 1.2dB, so the 3dB drop should be -1.8dB or just -3dB? Thanks.

graph

\$\endgroup\$
  • 1
    \$\begingroup\$ Usually 3dB below the low frequency asymptote. \$\endgroup\$ – Chu Dec 10 '17 at 17:26
  • \$\begingroup\$ No - not "usually". (see my answer). \$\endgroup\$ – LvW Dec 11 '17 at 9:02
  • 1
    \$\begingroup\$ Does it really matter? The rolloff (there is not such thing as a true cut off) frequency is just a simple single number to quickly convey some information about the filter. For a single pole filter, that's all you need. Yours is more complicated, so you need more than a single parameter when looking at it closely anyway. \$\endgroup\$ – Olin Lathrop Dec 11 '17 at 11:54
1
\$\begingroup\$

From the graph, first you have to find the DC gain. i.e, the gain at w = 0. It's approximately 0 here. Now drop 3 DBs from there. Find the meeting point of that 3-DB level with the graph. That is the cut-off frequency.

\$\endgroup\$
  • 1
    \$\begingroup\$ This applies to magnitude responses without peaking only. For example, it does NOT apply to the shown example. \$\endgroup\$ – LvW Dec 11 '17 at 8:59
1
\$\begingroup\$

The correct way to actually find the cutoff frequency for a low pass filter would be to find the frequency at which the gain of the filter is 1/sqrt(2) times the gain at dc frequency.

$$Gain (dB)=20log(\frac{A}{\sqrt2})$$ where A is the gain in V/V at dc frequency (not dB). Substituiting respective values from the graph you have, you should be able to get fc.

\$\endgroup\$
  • 2
    \$\begingroup\$ This only applies to low-pass filters, not "any gain vs frequency plot". \$\endgroup\$ – The Photon Dec 11 '17 at 6:28
  • \$\begingroup\$ ....and it applies to Butterworth and Bessel responses only! (see my detailed answer). \$\endgroup\$ – LvW Dec 11 '17 at 9:01
0
\$\begingroup\$

There are TWO basic definitions for a low pass cut-off frequency. This definition depends on the respective low pass approximation:

1.) Magnitude response without peaking (without "ripple") in the passband (Examples: Butterworth and Bessel response): In this case it is common practice to use the -3dB definition. That means: Frequency where the magnitude is 3db down with respect to the magnitude at w=0 (DC).

2.) Magnitude with peaking in the passband (called "ripple") - as shown with the blue curve in the original question (Chebyshev and elliptic Cauer response): In this case, the cut-off frequency usually is defined at a point where the magnitude crosses the value for w=0 (DC) again. In the shown example, this gives a cut-off frequency below 38.6 Hz.

3.) The above definitions are also used to compute the various tabulated filter parameters (pole frequency and pole-Q), which can be found in relevant textbooks.

\$\endgroup\$
  • \$\begingroup\$ This is the first time I've heard your definition for Chebyshev and elliptic filters. It seems impractical, considering that any real analog filters will have unwanted but tolerated ripple and peaking due to component tolerances and such - How would you define whether to use the -3 dB point or the 0 dB point? \$\endgroup\$ – pipe Dec 11 '17 at 9:20
  • \$\begingroup\$ No - it is not MY definition. This definition can be found in each relevant text book and - as mentioned - it was used for calculating the well-known filter tables. Such a definition is, of course, independent on possible tolerance effects. It is a definition - nothing else! As an example - lets take a chebyshev response with a 3dB ripple. In this case, a -3dB cut-off definition would give a 6-dB uncertainty within the passband. This makes no sense. \$\endgroup\$ – LvW Dec 11 '17 at 10:32
  • \$\begingroup\$ References for the various low pass approximations (with passband definitions): Claude S. Lindquist (Active Network Design with Signal Filtering Applications); Carson Chen (Active Filter Design). \$\endgroup\$ – LvW Dec 11 '17 at 10:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.