# second order low pass filter simulation and configuration

Can someone explain what the sets of resistors and capacitors do? im a student in my second year in EE we were asked to simulate this circuit for various sinusoidal frequency inputs and record the amplitudes and i have done it and as i increase the frequency the amplitude of the output voltage wave decreases yet i still dont understand why theres a third opamp if the measurment is taken from the output of the second and what does each set of resistors and capacitors do individually i know that a capacitor and resistor in parallel will have a Zeq=R/1+rsc in laplace domain but how does this affect the cutoff frequency in this circuit

• Didn't you already ask the almost same question a day or two ago? Did you delete it? – Justme May 5 at 16:30

The shown circuit is a so-called "Biquad". In the literature you can find a desription under the name "Tow-Thomas-Filter".

It realizes the general biquadratic transfer function - that means: At the first output (node 14) we have a bandpass function and at the second output (node 7) there is a second-order lowpass function. The same function - however inverted - is available at the third output (node 8).

The last inverting amplifier is necessary because the overall feedback path always must contain an odd number of inversions due to DC stability reasons.

Explanation: The first opamp realizes a first-order inverting lowpass and the second one is a simple inverting integrator (another 1st-order circuit). So we have two first-order functions in series with overall negative feedback - thereby producing a complex pole pair.

How do we arrive at a bandpass function at the first output? As seen from this output we have a lowpass function (1st opamp) with another lowpass in the feedback path (an integrator is a lowpass with a very small cut-off frequency). We know that negative feedback inverts the feedback function - hence, the 2nd opamp acts as a highpass function - as far as the output of the 1st opamp is concerned.

Thus, the first opamp together with the 2nd opamp one form a lowpass-highpass combination: A bandpass.

In your circuit we have C1=C2=C and R3=R4=R. In this case, the pole frequency (of course, for both lowpass and bandpass function) is simply wp=1/RC. (This is also the midfrequency for the bandpass, wp=wo). The quality factor Qp (selectivity) is given by Qp=R2/R.

• thanks this really cleared up many things in my mind – lolls222 May 5 at 16:00