(N/A) The circuit consists of an input voltage source $v_{s}$,a resistor $R$,and a diode connected in series. The output voltage $v_{o}$ is measured across the resistor $R$.
The current $I$ in the circuit is given by:
$I = \frac{v_{i}}{R + R_{D}}$
where $R_{D}$ is the resistance of the diode.
$1$. During the positive half cycle of the input voltage $(v_{i} = +1 \text{ V})$,the diode is forward-biased. Assuming an ideal diode,$R_{D} = 0$. Therefore,the current $I = \frac{v_{i}}{R}$. The output voltage across the resistor is $v_{o} = I \cdot R = (\frac{v_{i}}{R}) \cdot R = v_{i} = +1 \text{ V}$.
$2$. During the negative half cycle of the input voltage $(v_{i} = -1 \text{ V})$,the diode is reverse-biased. For an ideal diode,$R_{D} = \infty$. Therefore,the current $I = 0$. The output voltage across the resistor is $v_{o} = I \cdot R = 0 \cdot R = 0 \text{ V}$.
Thus,the output waveform is a positive-going pulse train where the voltage is $+1 \text{ V}$ during the positive half cycle of the input and $0 \text{ V}$ during the negative half cycle.