0
\$\begingroup\$

I am trying to figure out the transfer function for this bandpass filter (images below). I am sorry for badly drawn circuit. The operational amplifier is an ideal op amp.

I'd like to know if I correctly analyzed the circuit. I don't think so because when I try to simplify the transfer function into bandpass filter transfer function, something seems to be off.

enter image description here

enter image description here

|improve this question
\$\endgroup\$
1
\$\begingroup\$

By nodal analysis, using \$\large\Sigma\$(currents away from node) = 0:

Let \$\small V\$ be the voltage at the \$\small R_1,\: R_2,\: C_3,\: C_4\$ node, then:

\$\large\frac{V-V_E}{R_1}+\frac{V}{R_2}+\frac{V}{Z_3}+\frac{V-V_S}{Z_4}=0\$

And, at the summing junction node:

\$\large\frac{-V}{Z_3}+\frac{-V_S}{R_5}=0\$

Eliminating \$\small V\$:

\$\large\frac{V_S}{V_E}=\frac{-R_2Z_4R_5}{R_1R_2(Z_3+Z_4+R_5)+(R_1+R_2)Z_3Z_4}\$

|improve this answer
\$\endgroup\$
  • \$\begingroup\$ Thanks Chu for the answer. On your nodal analysis, You used +V/R2. Is it because R2 is directly branched by the ground? (the current runs from low voltage to high voltage) \$\endgroup\$ – user167987 Nov 25 '17 at 14:45
  • \$\begingroup\$ ALWAYS use the rule: sum of currents AWAY from each node =0. This helps to avoid errors due to getting the signs wrong. In each numerator term, the node voltage under consideration will appear first, once again minimising sources of error. \$\endgroup\$ – Chu Nov 25 '17 at 18:22
  • \$\begingroup\$ You have an $R_3$ in your answer but there's no $R_3$ in the schematic \$\endgroup\$ – Andrew Aug 20 at 4:08
  • 1
    \$\begingroup\$ @Andrew Typo ... \$Z_3\$. Thanks \$\endgroup\$ – Chu Aug 22 at 21:54
  • \$\begingroup\$ user167987...after inserting Z3=1/sC3 and Z4=1/sC4 into the equation you should rewrite/simplify the formula with the goal to have the form: sT1/(1+sT2+s²T²). with T=R*C. This is the classical bandpass form for identifying bandwidth and midfrequency. \$\endgroup\$ – LvW Aug 23 at 9:37

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.