Electronic – What are the capacitors between 5V and Gnd for

Short Answer:

Inductor: at t=0 is like an open circuit at 't=infinite' is like an closed circuit (act as a conductor)

Capacitor: at t=0 is like a closed circuit (short circuit) at 't=infinite' is like open circuit (no current through the capacitor)

Long Answer:

A capacitors charge is given by \$Vt=V(1-e^{(-t/RC)})\$ where V is the applied voltage to the circuit, R is the series resistance and C is the parallel capacitance.

At the exact instant power is applied, the capacitor has 0v of stored voltage and so consumes a theoretically infinite current limited by the series resistance. (A short circuit) As time continues and the charge accumulates, the capacitors voltage rises and it's current consumption drops until the capacitor voltage and the applied voltage are equal and no current flows into the capacitor (open circuit). This effect may not be immediately recognizable with smaller capacitors.

A nice page with graphs and some math explaining this is http://webphysics.davidson.edu/physlet_resources/bu_semester2/c11_rc.html

For an inductor, the opposite is true, at the moment of power-on, when voltage is first applied, it has a very high resistance to the changed voltage and carries little current (open circuit), as time continues, it will have a low resistance to the steady voltage and carry lots of current (short circuit).

The capacitors just act as relatively high impedances to sum the two oscillator signals at D4.

D4 then provides the non-linearity to cause mixing (multiplication) of the two signals and create sum and difference signals in the frequency domain.

In this case the two oscillators are running at approximately 250kHz. So the sum frequency will be >500kHz and the difference will be in the audio range. The audio frequency is filtered by C23 and associated resistors and then amplifier to create the audio output.