3
\$\begingroup\$

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. enter image description here enter image description here

\$\endgroup\$
  • \$\begingroup\$ 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.) \$\endgroup\$ – George Herold Nov 3 '15 at 14:28
1
\$\begingroup\$

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.

\$\endgroup\$
0
\$\begingroup\$

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.

\$\endgroup\$
  • \$\begingroup\$ 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". \$\endgroup\$ – robert bristow-johnson Nov 3 '15 at 5:27
  • \$\begingroup\$ 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. \$\endgroup\$ – brok9n Nov 3 '15 at 6:12
0
\$\begingroup\$

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\$.

\$\endgroup\$
  • \$\begingroup\$ 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. \$\endgroup\$ – robert bristow-johnson Nov 3 '15 at 19:15
  • \$\begingroup\$ 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. \$\endgroup\$ – LvW Nov 4 '15 at 8:11
  • \$\begingroup\$ 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. \$\endgroup\$ – robert bristow-johnson Nov 4 '15 at 20:51
  • \$\begingroup\$ 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). \$\endgroup\$ – LvW Nov 5 '15 at 9:12
  • \$\begingroup\$ continue: Here, Ao has no unit, A1 is given in seconds and A2 in (seconds)². Please note, that in our case A2=H. \$\endgroup\$ – LvW Nov 5 '15 at 9:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.