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.
Step 1: Read the connections in the circuit.
There are four D flip-flops in a row, holding \(Q_3, Q_2, Q_1, Q_0\) from left to right. Each flip-flop's output feeds the D input of the flip-flop to its right, so on every clock edge the register shifts: the new \(Q_2\) becomes the old \(Q_3\), the new \(Q_1\) becomes the old \(Q_2\), and the new \(Q_0\) becomes the old \(Q_1\). The leftmost flip-flop, \(Q_3\), does not copy anyone directly, its D input is instead driven by an XOR gate. Tracing the two wires that feed the XOR gate shows they are tapped from \(Q_1\) and \(Q_0\), so the new \(Q_3\) is \(Q_1 \oplus Q_0\), read just before that clock edge. All four flip-flops share the same clock, so they all update together on each rising edge, this is a 4-bit Fibonacci-style linear feedback shift register.
Step 2: Write the update rule.
On every rising edge, using the "old" values to compute the "new" values:
\[ Q_3^{new} = Q_1^{old} \oplus Q_0^{old} \]
\[ Q_2^{new} = Q_3^{old},\qquad Q_1^{new} = Q_2^{old},\qquad Q_0^{new} = Q_1^{old} \]
Step 3: Apply the rule for the first rising edge.
The initial state is \(Q_3Q_2Q_1Q_0 = 1011\), that is \(Q_3{=}1, Q_2{=}0, Q_1{=}1, Q_0{=}1\).
\[ Q_3^{new} = Q_1 \oplus Q_0 = 1 \oplus 1 = 0 \]
\[ Q_2^{new} = Q_3 = 1,\qquad Q_1^{new} = Q_2 = 0,\qquad Q_0^{new} = Q_1 = 1 \]
So after the first rising edge, \(Q_3Q_2Q_1Q_0 = 0101\).
Step 4: Apply the rule for the second rising edge.
Now the current state is \(Q_3{=}0, Q_2{=}1, Q_1{=}0, Q_0{=}1\).
\[ Q_3^{new} = Q_1 \oplus Q_0 = 0 \oplus 1 = 1 \]
\[ Q_2^{new} = Q_3 = 0,\qquad Q_1^{new} = Q_2 = 1,\qquad Q_0^{new} = Q_1 = 0 \]
So after the second rising edge, \(Q_3Q_2Q_1Q_0 = 1010\).
Step 5: Why the other options are wrong.
0101 (option B) is the state after only ONE clock edge, stopping one edge too early is a common mistake. 1101 (option A) and 1011 (option D, the original state repeated) do not follow from applying the shift-plus-XOR-feedback rule twice and do not appear anywhere in the correct two-step trace above.
Final Answer:
After two rising clock edges, \(Q_3Q_2Q_1Q_0 = 1010\).
\[ \boxed{Q_3Q_2Q_1Q_0 = 1010} \]
Find dV/dt