The 2nd filter acts as a LOAD for the first filter. So, this load has an impedance equal to R2 + 1/jwC. Basically, at the first filters output we connect an impedance equal to R2 + 1/jwc. Right?
Correct.Okay, but now i am using an op-amp as a buffer to isolate the 2 filters, so i don`t need to increase the impedance of the 2nd stage 10x.
You calculus is off.Calculus: 33Hz - simulation: 22Hz - Reality: 23Hz
And how do i do that? What`s the formula?You calculus is off.
Remember that the standard 1/2piRC calculation is the -3db point of one RC filter.
For two filters in series that gives the -6dB point, with the -3dB point at about 22Hz as the simulation and measurements show.
So to get the -3dB point of the combined filter, you must calculate the -1.5dB point of each filter.
The -3dB point for the combined filter is where each filter has a rolloff of -1.5dB or 0.8414.And how do i do that? What`s the formula?
Normally the cutoff frequency is defined as the -3dB point.PS2: so.. for a 2nd order filter, i should look for the -6dB or the -3dB to find the cut-off frequency?
What's to explain?PS3: in the simulation the phase of the first filter at -3dB is -45*, and for both filters at -6dB is -90*...
Can you please explain me?
Well, I'm rather allergic to doing math so you're on you own here, or perhaps someone else will help.I don`t know if it`s okay and i should calculate for -3dB (Vout/Vin = 0.707). I`m stuck at calculating the absolute value of the denominator..
Because you are adding together the settling time of two time-constants in series.why does it take longer to stabilize at the output of the whole system?