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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

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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}\$

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  • \$\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
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    \$\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

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