I have an RC network as follows:


simulate this circuit – Schematic created using CircuitLab

As far as I can tell, this has the transfer characteristic:

$$ \begin{align} \frac{V_{out}}{V_{in}}=\frac{Z_{lower}}{Z_{lower} + Z_{upper}} &=\frac{ \frac{1}{j\omega C_1 + \frac{1}{R_1}} }{ \left(\frac{1}{j\omega C_1 + \frac{1}{R_1}}\right) + \left(\frac{1}{j\omega C_2} + R_2 \right) } \\ &=\frac{ 1 }{ 1 + \left(\frac{1}{j\omega C_2} + R_2 \right) \left(j\omega C_1 + \frac{1}{R_1}\right) } \\ &=\frac{ j\omega R_1 C_2 }{ j\omega R_1 C_2 + \left(1 + j\omega R_2C_2 \right) \left(j\omega R_1 C_1 + 1\right) } \end{align} $$

Am I right in calling this a bandpass filter? How do I find the cutoff frequencies?

Or phrased alternatively, how can I choose my resistor and capacitor values given desired cutoffs?

  • 1
    \$\begingroup\$ The two cutoff frequencies are found by setting the magnitude of the transfer function to \$\frac{1}{\sqrt 2}\$ and solving the resulting polynomial equation. \$\endgroup\$ – Captainj2001 Jun 30 '16 at 21:00
  • \$\begingroup\$ Please check the complete low-entropy answer here electronics.stackexchange.com/questions/299231/… The key is to factor the expression in such a way that gain/pole/zero appear clearly ordered. As this is a second-order polynomial form in the denominator, you can re-organize it in a canonical form as shown in the proposed answer. \$\endgroup\$ – Verbal Kint May 19 '17 at 13:53

If the two cut-off frequencies are quite a distance apart then R2 (or C1) largely determines the point at which the high frequencies start to become attenuated and C2 (or R1) determines the low frequency below which frequencies are progressively attenuated. Yes, it's a band pass filter.

If the two frequencies are close then it becomes a poor band pass filter and is best visualized using something like LTSpice.

For this to be an effective BPF you have to consider mid-band attenuation - you would want to design it so that mid-band gain attenuation is not significant. To this end it is clear that R2 should be significantly smaller than R1 or you will get mid-band attenuation that is significant and generally detrimental. For those same reasons you would want C2 to be significantly larger than C1. This basically informs you that having the upper and lower frequencies quite close gives you mid-band attenuation up to 6 dB.

Based on that you clearly want upper and lower frequencies to be at least a decade apart and the upshot of this is you can analyse the two cut-offs almost without any interaction effects.

  • \$\begingroup\$ What's considered "quite a distance"? A decade? Half a decade? \$\endgroup\$ – Eric Jun 30 '16 at 20:58
  • 2
    \$\begingroup\$ A decade certainly but just try simulating it and you'll see. Hint you will want R1 >> R2 and C2 >> C1. If you do that then midband attenuation is very low however, if R1 is similar to R2 then midband gain is down about 6 dB as you would expect from a simple potential divider using two equal resistors. Ditto with the caps. This is why it is fairly pointless treating it as a mathematical exercise becuase when Fupper and lower are close the circuit performs badly. \$\endgroup\$ – Andy aka Jun 30 '16 at 20:58
  • \$\begingroup\$ R2/C1 is a pretty meaningless ratio to compare to frequency, when looked at dimensionally \$\endgroup\$ – Eric Jun 30 '16 at 23:32
  • \$\begingroup\$ @Eric - I meant the "optional" use of "/" - I'll fix the answer for clarification. \$\endgroup\$ – Andy aka Jul 1 '16 at 9:10

If \$\small R_1C_2\ll R_1C_1+R_2C_2\$ then the cut-off frequencies will be approximately \$\frac{1}{R_1C_1}\$ and \$\frac{1}{R_2C_2}\$

This is because the denominator of the TF (in Laplace form, since this makes the polynomial more amenable to inspection than the \$j\omega\$ form) may be written: $$\small s^2+\frac{R_1C_1+R_2C_2+R_1C_2}{R_1C_1R_2C_2}s+\frac{1}{R_1C_1R_2C_2}$$ which factorises to: $$\small \left (s+\frac{1}{R_1C_1}\right)\left(s+\frac{1}{R_2C_2}\right)$$

if \$\small R_1C_2\$ is small compared with the sum of the other two terms in the numerator of the s-coefficient


Here is another - rather simple - method for finding the 3db-cutoff frequencies (see the s polynominal as given by Chu in his answer):

1.) From the denominator of the bandpass transfer function you easily can derive the expressions for the bandpass center frequency wo=SQRT(1/T1T2) with T1=R1C1 and T2=R2C2.

2.) The same applies to the quality factor Q=SQRT(T1T2)/(T1+T2+R1C2)

3.) Because of the definition Q=wo/(w2-w1) and wo=SQRT(w1w2) we can find two equations for the two unknown cutoff frequencies w1 and w2 (Q and wo are known values).

4.) Note that the selectivity of this bandpass is rather poor because the maximum quality factor is Q=0.5 (bandwidth is twice the center frequency). The most common set of values (R1=R2, C1=C2) gives a quality factor of only Q=1/3. Note that for all RC bandpass configurations the quality factor never exceeds Q=0.5.


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