in this circuit (link to article), what is the 47k resistor to ground (R2) for?
I simulated the circuit and noticed that without the resistor, there are negative spikes across the transistor's base-emitter at every negative-going edge of the input signal. Why?

R1-C1 is basically an edge detector, correct? But it also removes DC bias such that the 12V signal becomes centered around 0V (between 6...-6 V) The reverse-biased B-E junction looks like an open circuit . Is the purpose of the 47k to simply conduct the reverse current of the diode away from the transistor?

 

and another thing, if I wanted to process a very slow signal (after adjusting the monostable period accordingly), the 555 would need a filter with a huge RC constant at the output, which also means the output would settle very slowly, correct? If so, what can be done?

 

The 47k resistor to ground ensures the transistor switches off in the absence of an input signal.
With a 1.5kHz input signal, the main use of C1 is to remove the DC component of the input signal. It would not be required if the input signal was at 0V when low.
With the input capacitor in use, the diode prevents the base from going negative. However, the diode has an internal capacitance so if the input signal has a fast falling edge a short negative spike will appear on the base. This is what you are seeing on your simulation.
When the input signal goes high, the transistor switches on and discharges the capacitor C2. If the input is above a certain frequency, C2 never reaches the threshold of the 555 before being discharged again.
For sensing very low frequencies, the values of C2 and C3 will need to be increased and may get larger than practical.
What frequency are you wanting to detect?

 

What frequency are you wanting to detect?

0.5 ...10 Hz?

For sensing very low frequencies, the values of C2 and C3 will need to be increased and may get larger than practical.

do you mean C2 and C5?

 

0.5 ...10 Hz?

You probably want to calculate the capacitance required for this circuit to take a 0.5hz signal, before you attempt to build it.

The circuit is meant to be used with a square wave. So for sinous signals, you may need a conditioning circuitry. Or the output is dependent on the duty cycle, or DC offset.

 

and another thing, if I wanted to process a very slow signal (after adjusting the monostable period accordingly), the 555 would need a filter with a huge RC constant at the output, which also means the output would settle very slowly, correct? If so, what can be done?

Is the input signal always the same duty cycle or does it vary with frequency?
In other words; a 10Hz signal could be 50ms high 50ms low or could it be 10ms high 90 ms low?