question on diode in the Energy storing circuit

Question

Please consider the following circuit.

For this case, my following notion is correct?

1. When v(t) is greater than diode turning-on voltage (v_d), the forwarding current (i_d) occurrs, and then energy will be harvested.
2. When v(t) is less than v_d, i_d is zero, and then energy will be not harvested.
3. Thus, the amount of harvested energy is depending on the i_d only when v(t) is greater than v_d.

For example:

1. 

2. 

First signal's averagy power (E[|v(t)|^2])is less than second one's. However, the first one turns on diode for a longer time than second one's. (that is, the first signal's sum of red-brakets time is longer.)

In this case, although second signal has more average power, the amount of harvested energy by first signal is much, right??

Thank you for reading my question.

Answer 1

The answer to your question is more or less yes, but I'm not sure you're asking the right question.

1) Only when there's a voltage across and a current through the load is energy being delivered to the load, which you've drawn as a capacitor.

2) I_d is zero when the source voltage is too low.

3) The amount of energy delivered to the load depends on i_d and v(t)-v_d

The input signal does not 'have a power', as the power it delivers depends on the load, and the load the signal sees changes with the load's voltage, due to your diode.

The input signal has a voltage, and an output impedance, the latter complicating what happens to the voltage that appears across a load when a current flows.