Electronic – Ultra-basic question: where is the voltage in this simple circuit diagram

CircuitLab allows you to simulate this:

^{simulate this circuit – Schematic created using CircuitLab}

Open the circuit, click on "simulate," and figure out how it works yourself! Note that I had to simulate the inductance of the speaker (important in this circuit) by inserting a series inductor, as CircuitLab didn't include this in the speaker model.

Separately, the reason the capacitor is connected to the speaker's "top" terminal, rather than ground, is that it needs to see the voltage drop across the speaker, which fluctuates with both how the speaker conducts, and how Q2 conducts; this is what keeps the oscillation going. If the capacitor was connected directly to ground, there would be a single click, and then it would be silent.

Specifically, when Q2 is open, the voltage across C1 will drop to 0 (it will discharge.) But, as C1 discharges, it turns from an isolator to a conductor (a discharged capacitor works a bit like a wire; a charged capacitor works a bit like an isolator.)

First thing to note is that the step function is defined as

$$ u(t) = \left.\begin{cases} 0 & t<0\\ 1 & t>0\\ \end{cases}\right\} $$

so

$$ V_c = \frac{1}{C}\int_{-\infty}^{t_f} i(t) dt $$

$$ V_c = \frac{1}{C}\int_{-\infty}^{0} 0.4\sin (2t - \frac{\pi}{4} )u(t) dt + \frac{1}{C}\int_0^{t_f} 0.4\sin (2t - \frac{\pi}{4} )u(t) dt $$

$$ V_c = \frac{0.4}{C}\int_0^{t_f} \sin (2t - \frac{\pi}{4} ) dt $$

So all the step function does is set all values of t that are less than zero to zero.

You then continue to compute the integral in the normal way, so:

$$ V_c = \frac{0.4}{C} \left[ \frac{1}{2} \cos(2t - \frac{\pi}{4})\right]_0^{t_f} $$