how the Q of a filter in SigmaStudio works?
Hi forum,
In SigmaStudio I am using a second order general parametric EQ filter.
I am setting a f0 = 500 Hz, a dBboost of -12 and a Q of 5.
I then plug in the probe and the stimulus on the filter and check the frequency response.
The result does not match my Matlab implementation of the filter. I also compare the frequency response of the filter with the same coming from another DSP tool and it also does not match the curve I see in SigmaStudio. But it does match my matlab implementation.
I know SigmaStudio is probably defining the Q differently than other tools, according to this:
General Filters [Analog Devices Wiki]
However, none of the the Q = 5 or the Q' = 5 seem to give me what the other tools give me for a Q = 5.
I can more or less match the response of the filter in SigmaStudio if I set the Q = 2.5 and Q' = 1.25.
Why is this and how can I match the Q in a SigmaStudio filter to those that other tools give me?
thanks
Dimitris
The wiki page you linked actually remarks on this near the bottom of the "General 2nd Order" subsection:
By default, the Q is shown with the value adjusted (from the classical EE definition) so that a boost of N dB followed by a cut of N dB for identical Q and f0/Fs results in a precisely flat unity gain filter or “wire”. This is equal to A*Q, where A = 10^(dBgain/40).
The “Classic EQ” version of Q can optionally be shown by clicking the small circular button next to the Q control.
Since you apparently got the right result if you use a Q of 5*10^(-12/40) ≈ 2.5 it would seem you're using the classic Q instead of the adjusted Q.
For an alternative view on how this Q-adjustment works see attachment, which is how I managed to formulate the classic biquads after fiddling a bit with them. In particular ibq.m contains a comment on how to obtain the classic-Q versions. It also shows quite clearly that with the adjusted version, flipping the sign of the gain (in dB) is equivalent to swapping the numerator and denominator of the transfer function, hence yields the inverse filter. (In this formulation however it's not obvious that the same can be accomplished merely by adjusting the Q.)
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