# Band-Pass Gain of a filter (exercise)

could you help me to find the band-pass gain of the following filter please?

The correct expression of pass-band gain hasn't the term $C_{B}R_{B}$ to the numerator (said my professor), instead in my expression there is:

$$G_{PB}=\left(1+\frac {R_{4}}{R_{3}} \right)\left( 1+\frac {R_{2}}{R_{1}} \right)R_{B}C_{B}$$

• That photo hurts my eyes. Please find a way to upload an easier to read schematic. Or redraw it with the in build schematic tool. You might get more help that way. Sep 25, 2016 at 15:24
• Hi @crowie, please tell me what parts of photo aren't clear. Sep 25, 2016 at 15:26
• Its all just very hard to read I'm just suggesting if you want to get help on this site make it clear and easy for people. Sep 25, 2016 at 15:33
• For starters, what is the symbol in parentheses after $V_{IN}$? Sep 26, 2016 at 9:56
• Hi @Dmitry Grigoryev, I wrote $$V_{IN}\left(s\right)$$ where s is the Laplace's variable. Sep 26, 2016 at 12:20

• Gnd symbols are too close to parts.
• contrast of grids is same as pencil so hard to read.
• adding brightness and contrast to photo could help us.

I only see one cap, Ca, not two Ca, Cb

The "unstated" assumption in this circuit is that the pulse width expected is much shorter than the time constant T=RaCa so for this condition, assume V(Ca) is a drifting DC voltage dependant on the duty cycle of the pulse and will always be between the pulse max~min levels, so the DC reference level Vlf will change with the signal and its gain is 1+|Av-| which for high gain 1+ can be neglected. Taking the difference between a signal and it's LPF signal yields the HPF response when both gains are the same (assuming Av>>1 of 2nd stage. This is an intuitive answer not mathematical such as yours which may be correct for Tpw>RaCa (low frequencies). When we want to simplify gain, we make these assumptions and state them e.g. for f>> 1/(2piRaCa) Av2= +R2/R1

You have two opamp buffers with gain totalling Av = (1 + R2/R1) * (1 + R4/R3) and two first order networks a lowpass RA/CA and a highpass RB/CB.

A first order LPF is G(w) = 1 / (1+jRaCaw)and a highpass is G(w) = jRbCbw/(1+jRbCbw)

Putting it all together I make the overall response (1/(1+jRaCaw)) * jRbCbw/(1+jRbCbw) * (1 + R2/R1) * (1 + R4/R3)

It might be worthwhile multiplying thru by a few complex conjugates to see if you can massage the thing into whatever form the prof wants.

Been a few years since I took this crap back in circuits and systems but it looks right to me.