The output of the following is:

Assume ideal diode.

Concept

A diode conducts when the positive terminal of the battery is connected to the anode and the negative terminal of the battery is connected to the cathode. Under this condition, the diode is said to be in a forward-biased condition.

In forward-biased conditions, the diode is replaced by short-circuit.

A diode does not conduct when the positive terminal of the battery is connected to the cathode and the negative terminal of the battery is connected to the anode. Under this condition, the diode is said to be in a reverse-biased condition.

In reverse-biased conditions, the diode is replaced by an open circuit.

Note: When the battery of the same polarity is present on both sides of the diode, then the battery with greater voltage magnitude will decide the biasing of the diode.

Calculation

Case 1: During +ve half cycle:

Diode is forward-biased, hence the diode is replaced by short-circuit.

\(V_o=-V_s\)

Case 2: During -ve half cycle:

(i) 0< V_in < V

Diode is forward-biased, hence the diode is replaced by short-circuit.

\(V_o=-V_s\)

(ii) V < Vin < V_m

Diode is reverse-biased, hence the diode is replaced by an open-circuit.

\(V_o=0\)

So, the output waveform is:

So, the correct answer is option 2.