The initial state of \(Q_3Q_2Q_1Q_0\) in the digital circuit shown below is 1011. The new value of \(Q_3Q_2Q_1Q_0\) after two rising edges of the clock (CLK) will be __________.
Show Hint
Trace one clock edge at a time: every flip-flop copies the value to its left, and the leftmost flip-flop is fed by the XOR of the two rightmost bits, \(Q_1\) and \(Q_0\), taken before that edge.
A tabular (timing-diagram) approach.
Instead of substituting one edge at a time inside a formula, list every bit in a table and update the whole row together, since all four flip-flops share the same clock and change simultaneously.
Rule recap.
Each flip-flop copies its neighbor on the left, except the leftmost one, $Q_3$, which copies the XOR of the two rightmost bits, $Q_1$ and $Q_0$, read before that edge:
\[
Q_3 \leftarrow Q_1 \oplus Q_0, \quad Q_2 \leftarrow Q_3, \quad Q_1 \leftarrow Q_2, \quad Q_0 \leftarrow Q_1
\]
Table of states.
$$
\begin{array}{c|cccc}
\text{Time} & Q_3 & Q_2 & Q_1 & Q_0 \\\hline
t_0\ (\text{start}) & 1 & 0 & 1 & 1 \\
t_1\ (\text{after edge 1}) & 0 & 1 & 0 & 1 \\
t_2\ (\text{after edge 2}) & 1 & 0 & 1 & 0
\end{array}
$$
Row $t_1$ is filled by shifting the $t_0$ row one place right ($Q_2,Q_1,Q_0$ columns just copy $Q_3,Q_2,Q_1$ of $t_0$) and computing the new $Q_3 = 1\oplus 1 = 0$.
Row $t_2$ is filled the same way from row $t_1$: shift right, and compute the new $Q_3 = Q_1(t_1)\oplus Q_0(t_1) = 0 \oplus 1 = 1$.
Reading the answer.
The bottom row, $t_2$, is the state after two rising edges: $Q_3Q_2Q_1Q_0 = 1010$. Building the full table this way is a useful check against option B (0101), which is easy to pick by mistake if you stop after only the $t_1$ row.
\[ \boxed{Q_3Q_2Q_1Q_0 = 1010} \]
