Assuming the ideal diode,draw the output waveform for the circuit given in the figure. Explain the waveform.

(N/A) Here,$v_{i} = 20 \sin (\omega t)$ and so the maximum voltage is $V_{m} = 20 \text{ V}$. It means that the input voltage changes from $+20 \text{ V}$ to $-20 \text{ V}$.
$(i)$ For time intervals from $0$ to $t_{1}$ and from $t_{2}$ to $\frac{T}{2}$ when $v_{i} < 5 \text{ V}$,the potential at the anode is less than the potential at the cathode $(V_{A} < V_{K})$,so the diode is reverse biased and no current passes through it. Hence,the input signal voltage appears directly across the load resistance $R_{L}$ in the output. Thus,the waveform of the output voltage $v_{0}$ across $R_{L}$ is similar to the input voltage $v_{i}$,as shown in Figure $(3)$.
$(ii)$ At instants $t = t_{1}$ and $t = t_{2}$,when $v_{i} = 5 \text{ V}$,no current passes through the resistance $R$ and diode $D$,so $v_{0} = v_{i} = 5 \text{ V}$,which is shown in Figure $(3)$.
$(iii)$ For the time interval from $t_{1}$ to $t_{2}$,$v_{i} > 5 \text{ V}$,so the diode is forward biased. Its resistance becomes zero,and the potential difference across it is zero. Hence,in the time interval from $t_{1}$ to $t_{2}$,$v_{0} = 5 \text{ V}$ (constant,which is the battery voltage).
$(iv)$ For the negative half cycle from $\frac{T}{2}$ to $T$,the diode $D$ is reverse biased,so no current passes through it due to infinite resistance. The input signal passes directly through $R_{L}$,and the waveform of $v_{0}$ across $R_{L}$ is similar to the waveform of $v_{i}$,as shown in Figure $(3)$.