This circuit is based on the article Percussion Instrument Synthesizer by Forrest M. Mims, Popular Electronics, March, 1976, pgs. 90 - 102. which describes an audio amplifier with a twin-T notch filter in the feedback loop, The R-C network gives a sharp peak at a single audio frequency at which feedback occurs. Disturbing the touch plate introduces a transient which starts a damped oscillation "ringing". I've added a Hall effect sensor on the input instead of a touch plate to inject the transient, and an audio power amplifer on the output.

- $i_1 - i_3 - i_6 = 0$
- $i_4 - i_2 + i_7 = 0$
- $i_3 - i_4 - i_5 = 0$
- $i_6 - i_7 - i_8 = 0$

Recall the convention that currents flowing into the node are taken as positive.

First an aside, recall the Laplace transform of a function $f$ is defined to be $\mathscr{L}\{ f(t) \} = \int_0^{\infty} f(t) e^{-st} dt$ and has the property $\mathscr{L}\{ f'(t) \} = s \mathscr{L} \{ f(t) \} - f(0)$ and that the transform is linear.

Recall that the voltage across a resistor is $V = iR$. A capacitor has the equation $\frac{d V}{d t} = {1 \over C} i$ since capacitance is defined by $CV = q$. Then we have $\mathscr{L} \{ V(t) \} = {1 \over s C} \mathscr{L} \{ i(t) \}$ if we assume the initial voltage $V = 0$ at $t = 0$

Now we apply Kirchhoff's voltage law, which is equivalent to convervation of energy (think of following the work W = q dV done by a charge going around a closed loop), on the meshes 1-4. But first we apply the Laplace transform to all voltage terms in the loop using linearity and assuming initial currents and voltages are zero. For simplicity of notation, let's use $i(t)$ to denote $\mathscr{L} \{ i(t) \}$, and $V(t)$ to denote $\mathscr{L} \{ V(t) \}$.

- $R_1 i_3 + \frac{1}{s C_1} i_5 - V_1 = 0$
- $\frac{1}{s C_2} i_6 + R_2 i_8 - V_1 = 0$
- $\frac{1}{s C_1} i_5 - V_2 - i_4 R_1 = 0$
- $R_2 i_8 - V_2 - \frac{1}{s C_2} i_7 = 0$

Now we'll solve the equations by substitution. Mesh equation 1 gives $$i_5 = (V_1 - R_1 i_3) s C_1$$ Substituting $i_5$ into node equation 3 gives $$\text{ * } i_3 - i_4 - (V_1 - R_1 i_3) s C_1 = 0$$ Substituting $i_5$ into mesh equation 3 gives $$(V_1 - R_1 i_3) - V_2 - i_4 R_1 = 0$$ then $$i_4 = \frac{V_1 - R_1 i_3 - V_2}{R_1}$$ Substituting $i_4$ into equation * above gives, after rearrangement, $$i_3 = \frac{ \frac{ V_1 - V_2}{R_1} + s C_1 V_1 }{ 2 + s C_1 R_1}$$ Now solve mesh equation 4 for $i_8$ to get $$i_8 = \frac{V_2 + \frac{i_7}{s C_2} }{R_2}$$ Substitute $i_8$ into mesh equation 2 to get after rearrangement, $$i_7 = (V_1 - V_2 - \frac{i_6}{s C_2}) s C_2$$ Substitute $i_8$ into node equation 2 to get $$\text{ ** } i_6 - i_7 - (\frac{V_2 + \frac{i_7}{s C_2}}{R_2}) = 0$$ Substitute $i_7$ into equation ** above to get after some rearrangement, $$i_6 = \frac{ (V_1 - V_2) s C_2 + \frac{V_1}{R_2}}{2 + \frac{1}{s C_2 R_2}}$$ Finally, use node equation 1 to give $$i_1 = \frac{ \frac{ V_1 - V_2}{R_1} + s C_1 V_1 }{ 2 + s C_1 R_1} + \frac{ (V_1 - V_2) s C_2 + \frac{V_1}{R_2}}{2 + \frac{1}{s C_2 R_2}}$$ This is almost what we need because it relates input voltage $V_1$ to output voltage $V_2$ given only input current $i_1$.

Returning to our complete circuit, we draw it using the twin-T notch filter block just analyzed, an ideal operational amplifer, and a voltage source. The ideal OpAmp has $\infty$ input impedance, $0$ output impedance, perfect linearity and gain $\mu \rightarrow \infty$ It's equation is $V_{o} = -\mu V_{i}$. The voltage source has internal resistance $r$. By Kirchhoff's voltage law, recalling we are using the Laplace transforms of voltage and current here, the mesh equation for the input is $$V_1 = -e + i_1 r = 0$$ Now lets eliminate $i_1$ and $V_1$ in the Twin-T equation using $$i_1 = \frac{e-V_1}{r}$$ $$V_1 = -\frac{V_2}{\mu}$$ giving $$ \frac{e + \frac{V_2}{\mu} }{r} = \frac{ \frac{ -\frac{V_2}{\mu} - V_2 }{R_1} + s C_1(-\frac{V_2}{\mu}) }{ 2 + s C_1 R_1} + \frac{ (-\frac{V_2}{\mu} - V_2) s C_2 + \frac{ -\frac{V_2}{\mu} }{R_2} }{2 + \frac{1}{s C_2 R_2}}$$ But the ideal op amp open loop gain $\mu$ is very high, so let $\mu \rightarrow \infty$ giving $$ \frac{e}{r} = \frac{ \frac{ -V_2 }{R_1} }{ 2 + s C_1 R_1 } + \frac{ (- V_2) s C_2 }{2 + \frac{1}{s C_2 R_2}}$$ Now let's put over a common denomiator, and invert to get $$ \frac{ V_2 }{ e } = -r \frac{ (2 + s C_1 R_1)(2 + \frac{1}{s C_2 R_2}) } { \frac{1}{R_1} (2 + \frac{1}{s C_2 R_2}) + s C_2 (2 + s C_1 R_1) } $$ After more rearranging we get the transfer function of our circuit, $$ H(s) = -r \frac{ (2 + s C_1 R_1) (2 s C_2 R_2 + 1)(\frac{ R_1 }{R_1 C_1 R_2 C_2 R_1 C_2}) } { s^3 + (\frac{2}{R_1 C_1})s^2 + (\frac{2}{R_1 C_1 R_1 C_2})s + (\frac{1}{R_1 C_1 R_2 C_2 R_1 C_2}) } $$ where $V_2(s) = H(s) e(s)$

To ring the bell in operation, one applies a sharp impulse to the input. To approximate this, set $e(t) = \delta(t)$, the Dirac delta function. Then $e(s) = \mathscr{L}\{ \delta(t) \} = 1$ The solution will be $V_2(t) = \mathscr{L}^{-1}\{ H(s) \}$

From complex analysis, for $t \ge 0$, the inverse Laplace transform for $Re(s) \gt \sigma$ is given by $$h(t) = \mathscr{L}^{-1}\{ H(s) \} = \sum \{ \text{ Residues of } e^{st} H(s) \text{ at each of its poles } \}$$ since $H(s)$ is analytic on $\mathbb{C}$ except at a finite number of poles, in particular, analytic on the half plane $\{ s | Re(s) \gt \sigma \}$ and $| H(s) | \le \frac{ M }{|z|} \forall |z| \ge R$ for constants $M \gt 0$ and $R \gt 0$ Recall for a simple pole $s_0$ the residue is $$Res( H(s), s = s_0 ) = lim_{s \rightarrow s_0} (s - s_0) H(s)$$ If we write $H(s) = \frac{N(s)}{D(s)}$, the roots of the cubic $D(s) = s^3 + a_2 s^2 + a_1 s + a_0 = 0$ are the poles of $H(s)$ where $$a_2 = \frac{2}{R_1 C_1}, \quad a_1 = \frac{2}{R_1 C_1 R_1 C_2}, \quad a_0 = \frac{1}{R_1 C_1 R_1 C_2 R_2 C_2}$$ and let the roots be $s_1, s_2, s_3$ Assuming the roots are all distinct, we have the solution $$h(t) = lim_{s \rightarrow s_1} {N(s) (s - s_1) e^{st} \over (s - s_1)(s - s_1)(s - s_1)} + lim_{s \rightarrow s_2} {N(s) (s - s_2) e^{st} \over (s - s_2)(s - s_2)(s - s_2)} + lim_{s \rightarrow s_3} {N(s) (s - s_3) e^{st} \over (s - s_3)(s - s_3)(s - s_3)}$$ or $$h(t) = {N(s_1) e^{s_1 t} \over (s_1 - s_2)(s_1 - s_3)} + {N(s_2) e^{s_2 t} \over (s_2 - s_1)(s_2 - s_3)} + {N(s_3) e^{s_3 t} \over (s_3 - s_1)(s_3 - s_2)}$$ Only the general form of the solution will concern us, $$V_2(t) = k_1 e^{s_1 t} + k_2 e^{s_2 t}+ k_3 e^{s_3 t}$$ That is we don't need to know the exact amplitude of the voltage, but only the frequencies of its sinusoidal components, or the relative amplitiude of its exponential solutions, if any. Or problem is then to find the general roots of the cubic equation.

From classical algebra, the roots of the cubic $s^3 + a_2 s^2 + a_1 s + a_0 = 0$ are given by the following formula, Let $Q \equiv \frac{ 3 a_1 - a_2^2 }{ 9 }$ and $R \equiv \frac{ 9 a_2 a_1 - 27 a_0 - 2 a_2^3 }{ 54 }$ Let the discriminant be $D = Q^3 + R^2$ Let $S \equiv \sqrt[3]{R + \sqrt{ D }}$ and $T \equiv \sqrt[3]{R - \sqrt{ D }}$ Then the roots are $$s_1 = -\frac{1}{3} a_2 + (S + T)$$ $$s_2 = -\frac{1}{3} a_2 - \frac{1}{2} (S + T) + \frac{1}{2} i \sqrt{3} (S - T)$$ $$s_3 = -\frac{1}{3} a_2 - \frac{1}{2} (S + T) - \frac{1}{2} i \sqrt{3} (S - T)$$

If $D = 0$ all roots are real and at least two are equal, and if $D \lt 0$ all roots are real and unequal. The solutions in this case would be of the form $k e^{ \pm u t}$, either growing exponential solutions (output increasing until limited by the circuit?) or decaying exponential solutions (maybe shows up as a click?). I'm not sure about the existence of these.

If $D \gt 0$ we have one real root and a complex conjugate pair of roots $s_1 = \gamma$, $s_2 = u + v i$, $s_3 = u - v i$ which would give us one solution of the form $k e^{ \gamma t}$ as above and two solutions $\pm k e^{ ut } sin( vt )$.

Since our cubic has real coefficients, if $Q \gt 0$ then $D = Q^3 + R^2 \gt 0$. But $Q = $$3 \left( \frac{2}{R_1^2 C_1 C_2} \right) - {\left( \frac{2}{R_1 C_1}\right)}^2 = \frac{ 6 C_1 - 4 C_2}{9 R_1^2 C_1^2 C_2} \gt 0$ iff $3 C_1 \gt 2 C_2$ since the denominator > 0.

Recall Newton's formulas for the coefficients in terms of the roots, $$s_1 + s_2 + s_3 = -a_2$$ $$s_1 s_2 + s_1 s_3 + s_2 s_3 = a_1$$ $$s_1 s_2 s_1 s_3 = -a_0$$

Then multiplying out the roots gives $$-a_2 = \gamma + u + i v + u - i v = \gamma + 2u$$ $$a_1 = \gamma (u + i v) + \gamma (u - iv) + (u + iv)(u - iv) = 2 \gamma u + (u^2 + v^2)$$ $$-a_0 = \gamma (u+iv)(u-iv) = \gamma (u^2 + v^2)$$

On the cusp of steady oscillation, $u = 0$ and we can simplfy the equations to $$-a_2 = \gamma$$ $$a_1 = v^2$$ $$-a_0 = \gamma v^2$$

Combining the equations gives $$a_0 = a_2 a_1$$ or $$\frac{1}{R_2 C_2} = \frac{4}{R_1 C_1}$$

The frequency of the oscillating solution is $$f = \frac{v}{2 \pi} = \frac{ \sqrt{a_1} }{2 \pi} = \frac{1}{2 \pi R_1} \sqrt{ \frac{2}{C_1 C_2} }$$ and the transient term decays as $$k e^{ -\frac{2}{R_1 C_1} t}$$

For this design, pick the capacitances to satisfy $$3 C_1 \gt 2 C_2$$ pick a small value for $$R_1 C_1$$ pick a frequency $f$ which gives $R_1$ as $$R_1 = \frac{1}{2 \pi f} \sqrt{ \frac{2}{C_1 C_2}}$$ finally $R_2$ is $$R_2 = R_1 \left( \frac{C_1}{4 C_2} \right)$$

Here's the circuit design schematics: Page 18 Page 19 Page 20 Page 21 Page 22 Page 23 Page 24 Page 25 Page 26 Page 27 Page 28 Page 29 Page 30 Page 31 Page 32

- TBD

Copyright © 1986-2017 by Sean Erik O'Connor. All Rights Reserved. last updated 10 Nov 17.