Narrow active band pass filter design

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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\$.