Third order butterworth bandpass filter

I am designing a third order butterworth bandpass filter, I derived the transfer function using matlab which was composed of a 6th order which would give 3 quadratic equations, the resulting transfer function of the 3 quadratic equations are attached, I just have a quick question in regards to some basics, in the attachment of the transfer function, why the Wo and q differs from one stage to the other? and how can the wo/q term which is the coefficient of S in the denominator is not equal to the nominator which also should be wo/q based on the general formula of the bandpass filters? I understand this would result in some gain in the bode plot of the quadratic equation alone,however if i plot the second and third stage they both have a gain bigger than 0db while the first stage has a gain of 0db, the resulting gain of all stages is 0db which is what I wanted actually but I cannot understand how those building blocks results in my desired design when 2 of 3 stages have gain bigger than 0 dB?

Hi,

you can not simply add "gain".
See every "gain" as a vector with it´s phase shift. Then add the vectors.

Klaus

Oh yeah it makes sense now, another question why do cascaded stages have different wo and different q ?

Hi,

I´m not an expert for cascaded filters.

Afaik this is urgent to get "butterworth" behaviour.
I assume it is not possible to get butterworth behaviour with equal stages cascaded.

Klaus

why do cascaded stages have different wo and different q

Click to expand...

Because they are specifically dimensioned to implement Butterworth polynomials. See https://en.wikipedia.org/wiki/Butterworth_filter, n= 6

sherif96 said:

Oh yeah it makes sense now, another question why do cascaded stages have different wo and different q ?

Click to expand...

Example: Two identical Butterworth stages (n=2, 3db frequency at w=wo) in cascade will have at wo a damping of 6 db (each stage contributes 3 dB).
Hence, the two identical wo values must be lower as before in order to achieve 3db attenuation at the desired frequency.
However, the composite function will not have a Butterworth characteristic anymore. Hence, both Q-values must be modified accordingly (see filter tables).