The figure below shows a circuit and its inpu | vedclass

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(A) The circuit consists of two diodes $D_1$ and $D_2$ in parallel with the output,each in series with a $DC$ voltage source.$1$. For the positive hal

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(A) The circuit consists of two diodes $D_1$ and $D_2$ in parallel with the output,each in series with a $DC$ voltage source.$1$. For the positive half-cycle of $V_i$:Diode $D_1$ is forward-biased when $V_i > 1 \, V$. Once $V_i$ exceeds $1 \, V$,$D_1$ conducts and clamps the output voltage $V_o$ to $1 \, V$. Thus,for $V_i > 1 \, V$,$V_o = 1 \, V$.$2$. For the negative half-cycle of $V_i$:Diode $D_2$ is forward-biased when $V_i $3$. For the range $-3 \, V \leq V_i \leq 1 \, V$:Both diodes are reverse-biased and do not conduct. Therefore,the output voltage follows the input voltage,i.e.,$V_o = V_i$.Comparing this behavior with the given options,the graph that shows the output voltage clamped at $1 \, V$ for positive inputs and $-3 \, V$ for negative inputs is the correct representation.

The figure below shows a circuit and its input voltage $V_i$ as a function of time $t$. Assuming the diodes to be ideal,which of the following graphs depicts the output voltage $V_o$ as a function of time $t$?

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