Calculating voltage at the output of a circuit
Let's just consider the first period of the sawtooth waveform, the output pattern will be similar for the later states.

$$U_e = 6t - 30, \quad 0 &lt; t &lt; 10\;ms $$

Current \$I\$ flows only when diode is forward biased i.e \$U_{QR} &gt; U_d\$

Considering \$U_p\$ as ground reference potential, while the diode is off 
\$U_p = U_r = 0\$

For the current to flow, diode must turn on.

The diode turns on when \$U_s &gt; -10 + U_d\$  (The potential at node \$Q\$ should be \$U_d\$, the voltage source behind it increases the potential of node \$T\$ (or \$S\$) by 10 V , so \$U_s\$ can be 10 V lower than \$U_d\$ for diode to turn on).

Now there are two cases to be considered:

Case I

\$-30 &lt; U_e &lt; -10 + U_d \$


Diode is off
\$I\$ does not flow


So, \$U_e = U_a\$, meaning output tracks the input waveform.

Case II

\$-10 + U_d &lt; U_e &lt; 30\$


Diode is on, hence \$I\$ flows.


In loop PSTR, applying KVL:

$$ U_e - 1.8i + 10 - U_d - 3.2i = 0 $$

$$ U_e - U_d + 10 = 5i $$

$$ i = \frac{U_e - U_d + 10}{5} $$

where \$U_e = 6t - 30\$ (you need to find exact time instant where diode turns on, I cannot do this as \$U_d\$ is not given).

In this case,

$$ U_a = U_e - 1.8i + 10 - U_d $$

In this equation you can replace \$i\$ with equation given above, and you will have all the known values except for \$U_d\$ (which is typically \$0.7V\$ for diodes), from here on it is easy to plot.

Hope it helps.