Confusion regarding diode rectification circuit graph
The question is a little difficult because neither side of the AC is ground referenced. With the bridge rectifier in place the ground keeps switching from side to side of the AC source as each is "grounded" by D3 or D4 on alternate half-cycles. That means that your nice blue sine graph has to be on a separate axis to the diode waveforms.







simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. The original circuit for simulation.



Figure 2. The results of the simulation (using real diode models).

In Figure 2 you can see the blue and orange side of the bridge get alternately "grounded" or clamped at zero volts.







simulate this circuit

Figure 3. (a) Positive half-cycle equivalent circuit. (b) Negative half-cycle equivalent circuit.

Splitting the circuit as shown in Figure 3 may help with your analysis.


On the V1 positive half-cycle D4 (not-shown) will be reverse biased so it will have the positive half-cycle voltage applied to it's cathode. But, since the question has a '+' on the anode it shows that voltage is to be measured anode to cathode so the sign is reversed.
On the V1 negative half-cycle D4 is forward biased so the voltage is zero.


This means that your large red and blue graph is correct.