In the given circuit the input voltage $$V_{in}$$ is shown in figure. The cut-in voltage of $$p-n$$ junction diode ($$D_1$$ or $$D_2$$) is 0.6 V. Which of the following output voltage ($$V_0$$) waveform across the diode is correct?

A parallel back-to-back diode clipper caps the output voltage at the diodes' forward cut-in threshold ($$\pm 0.6\text{ V}$$) during each respective half-cycle.

Positive half-cycle ($$V_{\text{in}} > 0$$):

$$V_{\text{in}} \le 0.6\text{ V} \implies \text{Both diodes off} \implies V_o = V_{\text{in}}$$

$$V_{\text{in}} > 0.6\text{ V} \implies D_1 \text{ is forward biased} \implies V_o = +0.6\text{ V}$$

Negative half-cycle ($$V_{\text{in}} < 0$$):

$$V_{\text{in}} \ge -0.6\text{ V} \implies \text{Both diodes off} \implies V_o = V_{\text{in}}$$

$$V_{\text{in}} < -0.6\text{ V} \implies D_2 \text{ is forward biased} \implies V_o = -0.6\text{ V}$$

Answer: Option (D): Symmetrical clipping occurs at $$+0.6\text{ V}$$ and $$-0.6\text{ V}$$.