Determining cut offs frequencies of band-pass filter

Question

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?

Answer 1

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.

Answer 2

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

Answer 3

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.