## Analysis

The cascaded bandpass filter uses a first-order lowpass filter, followed by a first-order highpass filter, followed by an inverting amplifier to raise or lower the total passband gain to some desired level. The equation given in Alexander & Sadiku is:

\begin{align} H=\frac{V_o}{V_i}&= \left(-\frac{1}{1+j\omega C_1R} \right) \left(-\frac{j\omega C_2R}{1+j\omega C_2R} \right) \left(-\frac{R_f}{R_i}\right)\\ &=-\frac{R_f}{R_i}\frac{j\omega C_2R}{(1+j\omega C_1R)(1+j\omega C_2R)} \end{align}

and as noted in Alexander & Sadiku, defining constants:

\begin{align} \omega_1&=\frac{1}{RC_2} & \omega_2&=\frac{1}{RC_1} \end{align}

the transfer function can be written as:

\begin{align} H&=-\frac{R_f}{R_i}\frac{j\omega \omega_2}{(j\omega+\omega_1)(j\omega+\omega_2)}\\ &=-\frac{\frac{R_f}{R_i}j\omega \omega_2}{(j\omega)^2+(\omega_1+\omega_2)(j\omega)+\omega_1\omega_2} \end{align}

or, as re-cast in class,

\begin{align} H=(-1)\frac{K2\zeta\omega_n j\omega}{(j\omega)^2+2\zeta\omega_n j\omega+\omega_n^2}=(-1) \frac{\left(\frac{R_f}{R_i}\frac{\omega_2}{\omega_1+\omega_2}\right)(\omega_1+\omega_2)j\omega}{(j\omega)^2+\left(\omega_1+\omega_2\right)j\omega+\omega_1\omega_2} \end{align}

This means the passband gain (which is an absolute value), natural frequency, and damping ratio are, respectively,

\begin{align} K&=\frac{R_f}{R_i}\frac{\omega_2}{\omega_1+\omega_2}\\ \omega_n&=\sqrt{\omega_1\omega_2}\\ \zeta&=\frac{\omega_1+\omega_2}{2\sqrt{\omega_1\omega_2}} \end{align}

From this, you can determine the quality and the bandwidth:

\begin{align} Q&=\frac{1}{2\zeta}=\frac{\sqrt{\omega_1\omega_2}}{\omega_1+\omega_2}\\ BW&=\frac{\omega_n}{Q}=2\zeta\omega_n=\omega_1+\omega_2 \end{align}

This is where disagreement with the book comes in; the book assumes that the bandwidth is the difference between the highpass stage's cutoff and the lowpass stage's cutoff. This fails to take into account, however, the fact that these are not ideal filters. The highpass filter's gain is not just 1 for frequencies higher than $\omega_1$ nor is the lowpass filter's gain just 1 for frequencies lower than $\omega_2$. Because of that, the actual half-power frequencies are different from the corners of the first-order filters. Specifically,

\begin{align} \omega_{cen,lin}&=\omega_n\frac{\sqrt{1+4Q^2}}{2Q}=\frac{1}{2}\sqrt{\omega^2_1+6\omega_1\omega_2+\omega^2_2}\\ \omega_{hp}&=\omega_{cen,lin}\pm\frac{BW}{2}\\ &=\frac{1}{2}\left(\sqrt{\omega^2_1+6\omega_1\omega_2+\omega^2_2} \pm (\omega_1+\omega_2)\right) \end{align}

## Design Limitations

This particular circuit has a design limitation: the damping cannot be less than 1. As proof, given the above, we can substitute the bandwidth equation into the natural frequency equation:

\begin{align} \omega_2&=BW-\omega_1\mbox{, so}\\ \omega_n^2&=\omega_1\omega_2=\omega_1(BW-\omega_1)=BW\cdot\omega_1-\omega_1^2 \end{align}

We can then re-write this to find $\omega_1$:

\begin{align} \omega_1^2-BW\cdot\omega_1+\omega_n^2 &= 0\\ \omega_1&=\frac{BW+\sqrt{BW^2-4\omega_n^2}}{2}\\ \omega_1&=\frac{BW+\sqrt{(BW)^2-(2\omega_n)^2}}{2} \end{align}

In order for this to be a real number - and it must be for this circuit, because we want to use real capacitors and resistors - the bandwidth must be at least twice the natural frequency or else the discriminant is negative. And since the damping ratio is

\begin{align} \zeta&=\frac{1}{2Q}=\frac{BW}{2\omega_n} \end{align}

the damping ratio must be greater than 1 and the quality factor must be less than 0.5.

## Design Methodology

If given a bandpass filter to design, first check to see if the damping ratio or quality has a value attainable by this cascading circuit. If not, you will need to find some other paradigm for your circuit. If you are given two cutoff frequencies and a maximum passband gain for a filter with a quality less than 0.5, you can use the circuit in Figure 14.45 but you will need to solve for $\omega_1$ and $\omega_2$ in terms of the given cutoff values and the passband gain. That turns out to be made a bit easier by the fact that

\begin{align} BW&=\omega_{hp,h}-\omega_{hp,l}=\omega_1+\omega_2\\ \omega_n&=\sqrt{\omega_{hp,h}\omega_{hp,l}}=\sqrt{\omega_{1}\omega_{2}}\\ K&=\frac{R_f}{R_i}\frac{\omega_2}{\omega_1+\omega_2} \end{align}

Note that though the $R$ values in the first two stages are listed as being the same, the mathematics work out the same if you replace the two $R$ values in stage 1 with some $R_1$ and replace the two $R$ values in stage 2 with some $R_2$. That is to say, in general,

\begin{align} \omega_1&=\frac{1}{R_2C_2}\\ \omega_2&=\frac{1}{R_1C_1} \end{align}

for more design flexibility. The important quantity is the product of the resistance and the capacitance for the stage.

## Design Example

Assume you want a filter that has cutoff frequencies of 1 rad/s and 100 rad/s and a maximum gain of 15. First thing to check is whether this circuit can be built in cascade or not. In this case, the bandwidth is 99 rad/s and the natural frequency is 10 rad/s so the quality is well under 0.5 -- this filter can be built. Next, solve the simultaneous equations:

\begin{align} \omega_1+\omega_2&=99\\ \sqrt{\omega_1\omega_2}&=10 \end{align}

to get

\begin{align} \omega_1&=1.0206\\ \omega_2&=97.9794 \end{align}

Note that in the equations above, numerically, the two values can be swapped. For the system, however, $\omega_1$ should be the smaller of the two values. From that you can determine the resistor and capacitor values in the filter stages. For example, assuming $R$ is a 4.7 k$\Omega$ resistor,

\begin{align} C_1&=\frac{1}{\omega_2 R}=2.1715~\mu\mbox{F}\\ C_2&=\frac{1}{\omega_1 R}=208.47~\mu\mbox{F}\\ \end{align}

To get the gain, note that

\begin{align} \frac{R_f}{R_i}&=K\frac{\omega_1+\omega_2}{\omega_2}\\ \frac{R_f}{R_i}&=15.156 \end{align}
so use any reasonable resistors with a ratio of 15.156.
1. Alexander, Charles and Sadkiu, Matthew. Fundamentals of Electric Circuits. 4th ed. McGraw-Hill, 2009