I try to explain everything as clear as possible, but Falstad simulations will help in understand each piece. The links are in the headers
Let's start from the base, ignoring the diode and the capacitor: you obtain a non-inverting amplifier. Skipping the 100k potentiometer to make the signals more readable, the Op-Amp forces the two inputs to be equal, and the voltage divider makes the inverting input to be half of the output:
$$
V_{-} = V_{S} \frac{R_{1}}{R_{1} + R_{2}} = \frac {V_{S}}{2} = V_{+}
$$
$$ V_{S} = \frac {R_{1} + R_{2} + R_{3}} {R_{1}} V_{e} $$
Result: the Output is 2x the Input.
Now the diode comes into play: in this new circuit, the input follows the output when it's positive, and:
$$
V_{-} = (V_{S} - V_{D}) \frac{R_{1}}{R_{1} + R_{2}} = \frac {V_{S} - V_{D}}{2} = V_{+}
$$
$$ V_{S} = \frac {R_{1} + R_{2} + R_{3}} {R_{1}} V_{e} + V_{D} $$
When the input voltage goes negative, the Op-Amp tries to push the output more negative to balance its inputs, but the diode is reverse biased and acts like an open circuit; so, the Op-Amp goes in negative saturation, the inputs are not anymore equal, and the only conductive path to the output is to the ground through the resistors.
What you obtain is a half wave rectifier.
Now let's put in the capacitor. This acts like a low-pass filter, storing the charge when the output is positive and releasing it when it goes to 0. So the charging will be fast, because the current will be supplied by the Op-Amp through the diode, but the discharge will be much slower, because it will happen through a resistance that varies between 20k and 120k.
$$
V_{S} = V_{sat}^{+} e^{- \frac{t}{RC}}
$$
$$
RC = 20 \cdot 10^{3} \cdot 4.7 \cdot 10^{-6} = 94 ms
$$
So the capacitor will discharge with a time constant (\$ \tau \$) of 94 ms (the time to discharge to 27% of the starting value), but the time between two charges is only half a period (in this case 0.5 ms), so it will go down to
$$
V_{S} = V_{sat}^{+} e^{- \frac{0.5}{94}} = 0.994 \cdot V_{sat}^{+}
$$
So with the smallest feedback resistance it will discharge to the 99.4% of the peak. The result is a pretty much continuous output voltage, with a small ripple due to the charge/discharge process.
The potentiometer has two effects: in the charging phase, it increases the gain of the amplifier, because R(feedback) becomes bigger and the Op-Amp has to push the output higher to balance the inputs; in the discharge phase, makes the discharge path of the capacitor more resistive, increasing the time constant and so reducing the ripple.
Transfer function and Output resistance
To describe these parameters, we have to consider separately the charge and the **discharge phases, because the circuit behaves differently between both and it's not possible to define them uniquely.
Charge phase
As we said, ideally the op-amp and the diode have 0 output resistance, so the capacitor is charged instantly and
V_{S} = \frac { R_{1} + R_{2} + R_{3} }{ R_{1} } V_{e} + V_{D}
Vs = ( R1 + R2 + R3 / R1 ) Ve + Vd
(LaTeX is giving problems)
The output impedance will be the parallel of the capacitor with the series of the resistance of the diode and the Op-Amp. Since both these resistances are 0, so will be the total output impedance.
Discharge phase
Here it's not possible to define the transfer function, as the output will no longer be dependent on the input because of the diode and the saturated Op-Amp.
The output impedance is shown in the figure:
And it will be given by:
Z_{out} = R_{eq} || \frac {1}{sC}
Zout = Req || (1 / sC)
(LaTeX is giving problems)
