Op-amps circuit with diodes
You can analyse it in two conditions of input ui.


When ui > 0


The op-amp output is positive and hence D1 will be forward biased and D2 will be reverse biased. Assuming ideal diodes and op-amps, we can draw the circuit like this.



Op-amps are ideal and have negative feed back. So \$V_B = V_A = ui \$

At the node B, apply KCL/nodal analysis for three branches. 

Input currents to op-amps are zero. So:

$$\frac{(V_B - uo)}{R} + 0 + 0 = 0$$
$$\implies V_B = uo$$
$$\therefore uo = ui$$


When ui &lt;0


The op-amp output is negative and hence D1 will be reverse biased and D2 will be forward biased. Assuming ideal diodes and op-amps, we can draw the circuit like this.


A little more neatly.



It is simply a buffer driving an inverting amplifier. The input to the inverting amplifier is ui itself. I can therefore write the traditional input output relation of inverting amplifier like:

$$uo = -ui(\frac{R}{R})$$
$$\therefore uo = -ui$$

Now you can plot a sine waveform for ui as \$V_{in}\$ and u0 as \$V_{out} \$ to see that it would just act as a full wave rectifier.


--EDIT--

Even if diode drops are considered, the same output is obtained at the end. This is an example of Full Wave Precision Rectifier. It acts as a rectifier with ideal diodes.