Consider the ideal diodes \(D_1\) and \(D_2\) as shown in the figure with cut-in voltage \(V_\gamma=0\) Volt and \(v_i(t)\) is in Volt.
The maximum voltage (Volt) of the output \(v_o(t)\) is (rounded off to two decimal places).
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D1 (the 2 V branch) clamps node N before D2 ever gets a chance, so only D1 is active; find vo as a voltage divider once D1 conducts.
Step 1: Model the on diode as a plain wire.
Once $v_i(t)$ pushes past $2$ V, $D_1$ conducts and behaves like a plain wire since its cut-in voltage is $0$, so node $N$ is tied straight to the $2$ V battery. $D_2$ stays reverse biased the whole time, since its own battery sits at a higher $4$ V, so that branch can be dropped while checking the output.
Step 2: Redraw as a two-source resistor network.
The circuit for $v_i(t)>2$ V now looks like $v_i(t)$ feeding node $A$ through $R_1$, and the $2$ V battery feeding the same node $A$ through $R_2$, since $D_1$ is now a wire and node $N=2$ V exactly. The output $v_o(t)$ is just the voltage at node $A$.
Step 3: Apply superposition.
With the $2$ V source turned off, that is replaced by a short to ground for this step, $v_i(t)$ sees a divider of $R_1$ and $R_2$ to ground, both $10$ k$\Omega$:
\[ v_{o,1}=v_i(t)\cdot\frac{R_2}{R_1+R_2}=v_i(t)\cdot\frac{1}{2} \]
With $v_i(t)$ turned off instead, that is replaced by a short to ground, the $2$ V source sees the same divider from the other side:
\[ v_{o,2}=2\cdot\frac{R_1}{R_1+R_2}=2\cdot\frac{1}{2}=1\ \text{V} \]
Step 4: Add the two contributions.
\[ v_o(t)=v_{o,1}+v_{o,2}=\frac{1}{2}v_i(t)+1 \]
This matches the divider formula found the direct way, now built from first principles using superposition instead of a current substitution.
Step 5: Plug in the peak value.
The input peaks at $v_i(t)=6$ V, giving
\[ v_{o,\max}=\frac{1}{2}(6)+1=4\ \text{V} \]
Step 6: Confirm the diode assumption.
At this peak, node $N$ sits at $2$ V, fixed by $D_1$, which is still below the $4$ V needed to turn $D_2$ on, so the assumption holds.
\[ \boxed{4.00\ \text{V}} \]
