Electronic – What happens when I connect an LC circuit to only the positive terminal of a voltage source

If the dependent voltage source was a resistor there would be no current through it since one end of it is connected to an open circuit (at node a). A resistor with no current through it has no voltage through it (since \$V = IR\$), so a resistor would not affect \$V_{\text{TH}}\$.

But the dependent voltage source has a non-zero voltage because its value is \$30\times 10^3 i_0\$ (where \$i_0\$ is the current through the upper resistor), and \$i_0\$ is non-zero.

There is still no current through the dependent source because it is connected to an open circuit at node a. That means the current through both resistors (\$i_0\$) is the same: $$i_0 = -\frac{100\text{V}}{20\text{k}\Omega + 80\text{k}\Omega} = -1\text{mA}$$

\$i_0\$ is negative based on the direction indicated in the circuit. The dependent voltage source's voltage is therefore $$V_D = 30\times 10^3 i_0 = -30\text{V}$$

This needs to be added to the voltage at the node between the resistors (which is a simple voltage divider) to calculate \$V_{\text{TH}}\$.

Since the Sonifex can be driven by 5 volt logic, you should be able to connect your logic output directly to the input lines, without the transistor. You can add the transistor if you need a logic inversion, and/or if it makes you "feel safer".

In any case, your logic ground must be connected to the Sonifex ground.

If there is any problem connecting the grounds, you should use a reed relay or opto-isolater to provide galvanic isolation.