# Narrow active band pass filter design

I'm trying to design a filter that allows through a 1kHz sine wave, based on my university lecture notes I have the following transfer function for a multiple feedback band pass filter:

$$A(s) = \frac{-H\omega_0s}{s^2+(1/Q)\omega_0 s+\omega_0^2}$$ where $\omega_0$ is the center frequency and $Q$ is the quality factor.

I have calculated $Q$ to be 16.6667 (bandwidth of 60Hz) and $\omega_0 = 2\times \pi \times 1000$.

My lecturer has informed me that I can treat $H$ in the above transfer function to be a specification for the passband gain, I wish a gain of 0dB at the center frequency so I set $H = 1$. The problem is when I calculate my capacitor and resistor values using the provided formulas in my lecture slides, my frequency response is centered at 1000Hz, however it has a gain of approx 25dB (my chosen cap values are 100nF and R1 = 1.59k, R2 = 41, R5 = 64k).

How do I appropriately choose $H$ so that I have a gain of 0dB at the passband (aka 1kHz)?

I have attached the relevant info from the lectures below.

• You typically get gain in a band pass filter. (usually equal to the Q.) You could attenuate the signal ahead or behind the filter. (depending on dynamic range.) – George Herold Nov 3 '15 at 14:28

the op-amp is correctly wired up with in an inverting circuit configuration. because of the negative feedback through passive components, the "-" terminal is a virtual ground. the node equations ($V_2$ is the voltage at the node where are $R_1$, $R_2$, $C_4$, and $C_3$ are connected) are:

$$\left(\frac{1}{R_1} + \frac{1}{R_2} + sC_4 + sC_3 \right)V_2 - sC_4 V_\text{out} = \frac{1}{R_1} V_\text{in}$$

$$sC_3 V_2 + \frac{1}{R_5} V_\text{out} = 0$$

from that, i get

\begin{align} A(s) \triangleq \frac{V_\text{out}}{V_\text{in}} & = \frac{-\frac{1}{R_1} s C_3}{\frac{1}{R_5}\left(\frac{1}{R_1} + \frac{1}{R_2} + sC_4 + sC_3 \right) + (sC_4)(sC_3) } \\ \\ & = \frac{-\frac{1}{R_1 C_4} s}{\frac{1}{R_5 C_3 C_4} \left( \frac{1}{R_1} + \frac{1}{R_2} \right) + \frac{C_4 + C_3}{R_5 C_3 C_4} s + s^2 } \\ \\ & = \frac{-H \omega_0 s}{s^2 + \frac{\omega_0}{Q} s + \omega_0^2} \\ \end{align}

equating the corresponding coefficients...

$$\omega_0^2 = \frac{1}{R_5 C_3 C_4} \left( \frac{1}{R_1} + \frac{1}{R_2} \right)$$

$$\frac{\omega_0}{Q} = \frac{C_4 + C_3}{R_5 C_3 C_4}$$

$$H \omega_0 = \frac{1}{R_1 C_4}$$

i think the intent, in the lecture notes posted in the question is that $\omega_0 \triangleq 2 \pi f_\text{m}$

so let $C_3 = C_4 \triangleq C$ and let $k \triangleq \omega_0 C$.

then $$\frac{1}{R_1} = H \omega_0 C$$ $$\frac{1}{R_2} = (2Q - H) \omega_0 C$$ $$\frac{1}{R_5} = \frac{1}{2Q} \omega_0 C$$.

so plug this in for $R_1$, $R_2$, and $R_5$ and see if equality in the three "corresponding coefficients" equations above is met. if so, the transfer function, as given in the question, is correct.

If you set $$s = \omega _0$$ $$A(s) = \frac{-H \omega^2}{(2+\frac{1}{Q})\omega^2}$$ and if A(s) = -1, then $$H = 2 + \frac{1}{Q}$$ Note that the - sign implies that the output will have a phase shift of 180 degrees.

• uhm, down arrow. $s = j \omega_0$ at the resonant frequency. $H$ should simply be equal to $\frac{1}{Q}$ to get 0 dB (with inverting gain) "in the passband". – robert bristow-johnson Nov 3 '15 at 5:27
• I redid my calculations and I'm still getting a gain of approx 30dB (more then previously), for reference my new values are C3 = C4 = 100nF, R1 = 773ohms, R2 = 51ohms & R5 = 53k. E: I think robert is correct. – brok9n Nov 3 '15 at 6:12

Your transfer function is NOT correct.

The numerator must be $N(s)=-H \omega_0^2 s$.

Therefore, the midband gain is $A_m=H \omega_0 Q$.

For $A_m=1$ we get $H=\frac 1 {\omega_0 Q}$ (note that H is given in seconds.)

Using your components we have $A_m = \frac {R_5} {R_1 + R_1C_4/C_3}$.

EDIT: I have to modify my statement that the function as given by you wouldn't be "correct". For my opinion, it is a bit uncommon to use this form - nevertheless several forms are always possible. Based on your formula the midband gain is $A_m = HQ$ and for $A_m=1$ we simply require $H=\frac 1 Q$.

• uhm, LV, dimensionally, how can there be frequency^3 in the numerator and only frequency^2 in the denominator and have $A(s)$ come out dimensionless? and it must come out dimensionless since the species of animal coming out (voltage) is the same as the species going in. i believe the transfer function is correct. – robert bristow-johnson Nov 3 '15 at 19:15
• robert, please note that in line 4 of my answer I wrote that H is given in seconds (unit "s"). Hence, both numerator and denominator are quadratic expressions. – LvW Nov 4 '15 at 8:11
• Lv, you're simply not correct. you cannot just ascribe to $H$ whatever dimension of quantity you want. $H$ is dimensionless. and to be dimensionally consistent in numerator and denominator, it's $-H \omega_0 s$ in the numerator. – robert bristow-johnson Nov 4 '15 at 20:51
• robert, sorry - but you are wrong. May I remind you to the general biquadratic transfer function with numerator N(s)=wp²*(Ao+A1s+A2s²)and the denominator D(s)=(s²+swp/Qp+wp²). – LvW Nov 5 '15 at 9:12
• continue: Here, Ao has no unit, A1 is given in seconds and A2 in (seconds)². Please note, that in our case A2=H. – LvW Nov 5 '15 at 9:18