I have tried with some op-amp set-up as a precision full-wave
rectifier and AC to DC converter, but the results doesn't match the
Yes, the clue is in the term RMS - you take the square of the voltage waveform, then average that modified waveform (mean) then finally take the square root to get true RMS. Starting with your suggestion of precision full-wave rectification makes life easier and doesn't affect accuracy AND, it allows a simpler multiplication algorithm or circuit.
What components can be used to achieve that and what type of circuit
can be built?
You can still buy analogue multipliers such as the AD636. It is a "Low Level, True RMS-to-DC Converter": -
And, importantly, it's still available from digikey for example. According to their page they still have over 3,000 in stock and, it is quite expensive (circa £18). ADI no longer make the device either.
There are other options that can be pursued of course. For instance, it's quite easy to make a single-quadrant, analogue squaring circuit using op-amps, analogue-switches and pulse width modulation. This can be shown if requested but, here's a link to what I've said previously. Summary picture: -
But, RMS-to-DC conversion also requires taking the square root of a number and this may sound more challenging until you realize that using a squaring circuit in the feedback loop of an op-amp does the job (albeit inverting): -
So, -5 volts is the input and \$\sqrt5\$ (2.236068 volts) is the output.
This would be the general form of an analogue circuit using PWM to calculate power: -
On the final graph above I've plotted a red horizontal line corresponding to 0.7071 volts. As you can see, the RMS output hits the target pretty accurately. In case you want to inspect the PWM on the half-wave: -
Hopefully, you should be able to recognize that when the squared waveform is low in value, the PWM duty cycle is nearly zero. When the squared waveform is near the peak of 1 volts, the PWM duty cycle is nearly unity.
When filtered lightly you would see a nice sine-shaped waveform of twice the frequency of the input (50 Hz) and centred at 0.5 volts. Why 0.5 volts; because, if the root-mean-square output is 0.7071 volts (as we would expect) then the mean-square value would be 0.5 volts.