Active Bandpass Limitations

This page is to demonstrate why a homework problem originally assigned in BME 153.1 in Spring 2009 was impossible.

Original Problem

Based on various problems. Assuming you only have common-value resistors at your disposal (see Table 2.2 on p. 45), along with any capacitors you need (alas, the inductor supply has disappeared), design an active band-pass filter that (a) does not require any power be delivered by the source, (b) has a maximum gain of ≈ +6dB, (c) has a linear center frequency of 660 Hz, (d) has a bandwidth of 440 Hz, and (e) uses resistances that allow the ideal op-amp assumptions to hold. You may assume that the impedance of the load is sufficiently high. Note that this particular filter’s job will be to isolate notes played on the octave (plus one note) of a piano keyboard starting at middle A. You will likely need multiple op-amps.

Analysis

Without inductors, the most likely candidate for such a filter would be the filter on p. 439 of the Rizzoni text which uses a series combination of a resistor and capacitor as \(Z_N\) and a parallel combination as \(Z_F\). This leads to an overall transfer function of:

\(\begin{align} H=-\frac{j\omega C_NR_F}{(1+j\omega C_FR_F)(1+j\omega C_NR_N)} \end{align}\)

or, as re-cast in class,

\(\begin{align} H=\frac{K2\zeta\omega_n j\omega}{(j\omega)^2+2\zeta\omega_n j\omega+\omega_n^2}= -\frac{j\omega\left(\frac{C_NR_F}{C_NR_N+C_FR_F}\right)\left(\frac{1}{C_FR_F}+\frac{1}{C_NR_N}\right)}{(j\omega)^2+\left(\frac{1}{C_FR_F}+\frac{1}{C_NR_N}\right)j\omega+\frac{1}{C_FR_FC_NR_N}} \end{align}\)

To make life a little easier, let's call

\(\begin{align} \omega_1&=\frac{1}{C_FR_F} & \omega_2&=\frac{1}{C_NR_N} \end{align}\)

which means

\(\begin{align} H=-\frac{j\omega\left(\frac{C_NR_F}{C_NR_N+C_FR_F}\right)\left(\omega_1+\omega_2\right)}{(j\omega)^2+\left(\omega_1+\omega_2\right)j\omega+\omega_1\omega_2} \end{align}\)

This means the bandwidth and natural frequency squared are, respectively,

\(\begin{align} BW&=\omega_1+\omega_2 \\ \omega_n^2&=\omega_1\omega_2 \end{align}\)

Good so far...

The Trouble Begins...

Given the above, we can substitute the bandwidth equation into the natural frequency equation:

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

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.

...And Ends

Given the problem statement, the bandwidth is 440 Hz (880\(\pi\) rad/s) and the linear center is 660 Hz (1320\(\pi\) rad/s); this gives half-power frequencies (which are different from the \(\omega_1\) and \(\omega_2\) above, as discussed in class) of

\(\begin{align} \omega_{hp}&=\omega_{c,lin}\pm\frac{1}{2}BW=1320\pi\pm440\pi\\ \omega_{hp}&=880\pi,1760\pi \end{align}\)

which gives a logarithmic center frequency - which is the same as the natural frequency - of

\(\begin{align} \omega_{n}&=\sqrt{\omega_{hp,1}\omega_{hp,2}}=\sqrt{880\pi\cdot1760\pi} \omega_n&=1244.51\pi \end{align}\)

which, in the number system most commonly used in the class, is not less than half of \(880\pi\). As an aside, it is 622.23 Hz, which is the D# above middle A.

Lesson Learned

Since the single-op-amp-with-four-components bandpass filter has a requirement that the bandwidth be at least twice the natural frequency, there is a limit on the quality of filter produced:

\(\begin{align} Q&=\frac{\omega_n}{BW}<\frac{1}{2} \end{align}\)

Now - turns out I should have figured this out very quickly for the following reason - this circuit cannot possibly be underdamped - underdamped circuits require energy storage units that are out of phase with each other (something inductive and something capacitive). Which means this must have a damping ratio \(\zeta\) of 1 or greater. And since

\(\begin{align} Q&=\frac{1}{2\zeta} \end{align}\)

this circuit can only have a quality of 0.5 or less.

Note that just having a capacitor and an inductor is insufficient to give a circuit the potential to be underdamped. For example, cascading a first-order low-pass active filter:

\(\begin{align} H_L&=\frac{K_L\omega_1}{j\omega+\omega_1} \end{align}\)

with a first-order high-pass active filter:

\(\begin{align} H_H&=\frac{K_Hj\omega}{j\omega+\omega_2} \end{align}\)

will yield a filter:

\(\begin{align} H=H_L*H_H&=\frac{K_LK_H\omega_1j\omega}{(j\omega + \omega_1)(j\omega+\omega_2)} \end{align}\)

that cannot be underdamped even if the first filter is made by placing an inductor and resistor in series for \(Z_N\) and a resistor for \(Z_F\):

\(\begin{align} H_L&=-\frac{R_{FL}}{j\omega L_{NL}+R_{NL}}= -\frac{ \frac{R_{FL}}{L_{NL}} }{j\omega+\frac{R_{NL}}{L_{NL}}} \end{align}\)

and the second is made by placing a capacitor and a resistor in series for \(Z_N\) and a resistor for \(Z_F\):

\(\begin{align} H_H&=-\frac{R_{FH}}{R_{NH}+\frac{1}{j\omega C_{NH}}}= -\frac{ j\omega\frac{R_{FH}}{R_{NH}} }{j\omega+\frac{1}{R_{NH}C_{NH}}} \end{align}\)

The Solution

Assuming inductors are no longer outlawed, one way to actually build the specific filter above would be to have a series RLC connection for \(Z_n\) and a resistor in the feedback loop. Which sounds like a great problem for Homework 11...hmmm....

Questions

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