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This post will derive the transfer function for a bridged t-network, expressed using impedances, and will be the first of a series of posts I’ll be making as I’m learning about, and working on, a design for an analog drum machine similar to that of the famous Roland machines.

The post will begin with a short background for why the bridged t-network in particular will be examined. Then, the derivation of the transfer function will be given, followed by a simulation of the circuit to validate that the derived expression is correct.

I. Background

In the service manual for the tr-808 drum machine, the following principal diagram for the basic pulse-triggered sound generator can be found:

tr808_sound_generator — Fig 1, representative bridged t-network schematic from tr-808 service manual

While the circuitry in the tr-808 is not as straight forward as in Fig 1, it is an excellent starting point for a study. By understanding the principal schematic, it creates the foundation for learning about the more complex and nuanced circuits in the actual hardware.

The sound generator consists of an operational amplifier (op-amp) and a feedback circuit consisting of the bridged t-network, which forms a bandpass filter with a high resonance. The filter acts as a damped oscillator (see Fig 2), and will self-oscillate at the resonance-frequency for a short time after being stimulated by an input-pulse. This is makes up the basic drum sound.

The circuit above (Fig 1) will be able to create a tom-like sound or a kick-like sound depending on the resonant frequency’s value, which is determined by the valeus of the circuit components.

II. Transfer function for the bridged t-network

The transfer function describes the relation between the components and the circuits’ behavior, and will be derived from the following diagram:

schematic — Fig 3, bridged t-network with currents labeled I1 and I2

The transfer function is an expression for the ratio between the output voltage $V_o$ and a given input voltage $V_i$ :

$\displaystyle \textrm{transfer function} = \frac{V_o}{V_i}$

Deriving the transfer function from the schematic becomes a process of finding an expression for the output voltage as as a function of the input voltage, and factoring that expression.

The output voltage node and input voltage node are separated by the impedance $Z_1$ . The current $I_1$ via $Z_1$ causes a voltage drop, so that the output voltage can be expressed by:

$\displaystyle V_o=V_i-I_1 Z_1$

$I_1$ can be expressed by introducing a common-voltage $V_{com}$ at the junction between $Z_2$ , $Z_3$ and $Z_4$ :

$\displaystyle I_1=\frac{V_i-V_{com}}{Z_1+Z_3}$

To express the common-voltage we notice that $Z_1$ , $Z_2$ and $Z_3$ form an equivalent resistor $Z_{(1+3)//2}$ that together with $Z_4$ form a voltage divider with $V_{com}$ as its output:

$\displaystyle V_{com}=V_i \frac{Z_4}{Z_{tot}}$

where the total impedance $Z_{tot}$ is:

$\displaystyle Z_{tot}=Z_{(1+3)//2}+Z_4$

The current $I_1$ can now be rewritten as:

$\begin{aligned} \displaystyle I_1 &= V_i \frac{1}{Z_1+Z_3}-V_{com}\frac{1}{Z_1+Z_3} \\[0.5em] \displaystyle &= V_i \frac{1}{Z_1+Z_3}-V_i\frac{Z_4}{Z_{tot}}\frac{1}{Z_1+Z_3} \\[0.5em] \end{aligned}$

and then factored:

$\begin{aligned} \displaystyle &= \frac{V_i}{Z_1+Z_3} (1-\frac{Z_4}{Z_{tot}}) \\[0.5em] \displaystyle &= \frac{V_i}{Z_1+Z_3} (\frac{Z_{tot}-Z_4}{Z_{tot}}) \\[0.5em] \displaystyle &= \frac{V_i}{Z_1+Z_3} (\frac{Z_{(1+3)//2}}{Z_{tot}}) \\[0.5em] \end{aligned}$

By calculating the equivalent impedance $Z_{(1+3)//2}$ and then $Z_{tot}$ , the expression for the current can be made explicit:

$\begin{aligned} \displaystyle Z_{(1+3)//2} &= \frac{(Z_1+Z_3)Z_2}{(Z_1+Z_3)+Z_2} \\[1em] \displaystyle Z_{tot} &= \frac{(Z_1+Z_3)Z_2}{(Z_1+Z_3)+Z_2}+Z4 \\[0.5em] &= \frac{(Z_1+Z_3)Z_2}{Z_1+Z_2+Z_3}+Z4\frac{(Z_1+Z_2+Z_3)}{(Z_1+Z_2+Z_3)} \\[0.5em] &= \frac{(Z_1+Z_3)Z_2+Z_4(Z_1+Z_2+Z_3)}{Z_1+Z_2+Z_3} \\[0.5em] \end{aligned}$

The impedance-quotient then becomes:

$\displaystyle \begin{aligned} \frac{Z_{(1+3)//2}}{Z_{tot}} &= \frac {\displaystyle \frac{(Z_1+Z_3)Z_2}{Z_1+Z_2+Z_3}} {\displaystyle \frac{(Z_1+Z_3)Z_2+Z_4(Z_1+Z_2+Z_3)}{Z_1+Z_2+Z_3}} \\[0.5em] &= \frac{(Z_1+Z_3)Z_2}{(Z_1+Z_3)Z_2+Z_4(Z_1+Z_2+Z_3)} \end{aligned}$

The current $I_1$ can now be expressed with the known impedances as:

$\begin{aligned} \displaystyle I_1 &= \frac{V_i}{Z_1+Z_3} (\frac{Z_{(1+3)//2}}{Z_{tot}}) \\[0.7em] \displaystyle &= \frac{V_i}{(Z_1+Z_3)} \frac{(Z_1+Z_3)Z_2}{(Z_1+Z_3)Z_2+Z_4(Z_1+Z_2+Z_3)} \\[0.7em] \displaystyle &= V_i \frac{Z_2}{(Z_1+Z_3)Z_2+Z_4(Z_1+Z_2+Z_3)} \\[0.7em] \end{aligned}$

Finally, we can return to the output voltage expression, and factor out $V_i$ :

$\begin{aligned} \displaystyle V_o &= V_i-I_1 Z_1 \\[0.5em] \displaystyle &= V_i-V_i \frac{Z_2}{(Z_1+Z_3)Z_2+Z_4(Z_1+Z_2+Z_3)} Z_1 \\[0.5em] \displaystyle &= V_i \left(1-\frac{Z_1 Z_2}{(Z_1+Z_3)Z_2+Z_4(Z_1+Z_2+Z_3)}\right) \\[0.5em] \end{aligned}$

and then calculate the difference inside the parentheses by expanding the leading 1-term:

$\begin{aligned} \displaystyle &= V_i \frac{(Z_1+Z_3)Z_2+Z_4(Z_1+Z_2+Z_3)-Z_1 Z_2}{(Z_1+Z_3)Z_2+Z_4(Z_1+Z_2+Z_3)} \\[0.5em] \displaystyle &= V_i \frac{Z_1 Z_2+Z_2 Z_3 +Z_4(Z_1+Z_2+Z_3)-Z_1 Z_2}{(Z_1+Z_3)Z_2+Z_4(Z_1+Z_2+Z_3)} \\[0.5em] \displaystyle &= V_i \frac{Z_2 Z_3 +Z_4(Z_1+Z_2+Z_3)}{(Z_1+Z_3)Z_2+Z_4(Z_1+Z_2+Z_3)} \\[0.5em] \end{aligned}$

With the original expression fully factorized, the transfer function can now be expressed:

$\begin{aligned} \displaystyle \frac{V_o}{V_i} &= \frac{Z_2 Z_3 +Z_4(Z_1+Z_2+Z_3)}{(Z_1+Z_3)Z_2+Z_4(Z_1+Z_2+Z_3)} \\[0.5em] \end{aligned}$

And the parenthesises expanded:

$\begin{aligned} \displaystyle \frac{V_o}{V_i} &= \frac{Z_1 Z_4+Z_2 Z_3 +Z_2 Z_4+Z_3 Z_4}{Z_1 Z_2+Z_1 Z_4+Z_2 Z_3+Z_2 Z_4+Z_3 Z_4} \\[0.5em] \end{aligned}$

The transfer function for the schematic has now been derived.

III. Validating the transfer function with a simulation

In order to validate that the transfer function is correct, it will be tested using resistive values and compared to the output from the simulation software PSpice.

A simple straight forward test is to simply define all the impedances to 1kΩ:

$\begin{aligned} \displaystyle \frac{V_o}{V_i}(Z_i=1k) &= \frac{(1+1+1+1)k}{(1+1+1+1+1)k} = \frac{4k}{5k} = 0.8\\[0.5em] \end{aligned}$

sim1 — Fig 4, simulation result from PSpice for test 1

With the input voltage $V_i$ set to 1V, the output voltage $V_o$ is 0.8V as expected.

In order to make sure this isn’t just a fluke result, another set of test-values can be defined:

$\begin{aligned} \displaystyle \frac{V_o}{V_i}(V_i=i k) &= \frac{(1\cdot4+2\cdot3 +2\cdot4+3\cdot4)k}{(1\cdot2+1\cdot4+2\cdot3+2\cdot4+3\cdot4)k}\\&=\frac{4+6+8+12}{2+4+6+8+12}=\frac{30}{32}=0.9375\end{aligned}$

sim2 — Fig 5, simulation result from PSpice for test 2

For this second test the result is also the expected one, with an input voltage $V_i$ of 1 giving 937.5 mV output.

IV. Conclusion

A valid transfer function for the bridged t-network has been derived for impedances. In order to analyze its oscillation properties and bode diagram, those impedances need to be defined using resistors $R$ and capacitors $C$ .