Hi, this ain't homework though but i suspect something similar for a test. Last week i had a similar practical about an active filter... All i want to know is how to calculate the cut off freq of this 2stage filter with an Rload resistor. circuit - http://i49.tinypic.com/rbwsoz.jpg 3db cut off - http://i49.tinypic.com/2em0qb4.jpg Above are the practical results. So the default equation goes Fc=1/(2∏RC) I understand that Rs and R1 is in series but my equation and practical doesn't match up. Fc=1/[(2∏(R1+Rs)//R2//R3)*(C1+C2)] = 212 Hz With orcad I get 122Hz. Can someone please help me with the theory cut off freq? Thank you
I obtain a transfer function as follows where and The -3dB point would be found for ω when For which I obtain ω=762.691 radians/sec or f=121.386Hz
The second RC filter on the right overloads the first RC filter on the left so they make a very poor filter. There should be a buffer amplifier between the RC filters.
If the DC gain is simply At frequency ω [rads/sec] From which we obtain the magnitude The -3dB point on the DC gain has a gain value The value of ω which satisfies this condition is found by equating or which after multiplying through by ac gives
OK I understand but I still have a question. Why don't you use a classical form of a transfer function?
I wasn't particularly concerned with standard forms. Variable a just happened to be the term I'd factored out in the denominator. No doubt I could have made one the inverse of the other to agree with convention. I must pose the obvious question - Would it have made any difference to the final relationship?
I think that the answer look like this So if we multiplying through by c and divide by a the answer will look the same. So this equation holds for any second-order filter ?
I guess it's definitely true of a low pass type without any pass-band ripple - such as the Butterworth filter. What about the Chebyshev LPF with a 3dB passband ripple say...? I'd have to think about it some more.