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 $V_2(t) = \mathscr{L}^{-1}\{ H(s) \}$

- TBD

Copyright © 1986-2017 by Sean Erik O'Connor. All Rights Reserved. last updated 01 Jan 17.