سوال ۷۷ کنکور ارشد مهندسی کامپیوتر ۱۴۰۳

This is a Mealy sequential circuit. The edge labels are $x/z$, so we first extract the flip-flop input equations and the output equation, then build the next-state table.

Read the gates off the schematic. The AND gate forms the output from $x$ and the $\bar{Q}$ line of the JK flip-flop; the $J$ and $K$ inputs are tied to $x$ (one node on the right bus); the inverter together with the feedback drives $D$:

$J = K = x, \qquad z = x \cdot y_2', \qquad D = x \oplus y_2.$

Flip-flop next states. A JK flip-flop with $J=K=x$ is a toggle (T) flip-flop controlled by $x$:

$y_2^{+} = J\,y_2' + K'\,y_2 = x\,y_2' + x'\,y_2 = y_2 \oplus x.$

The D flip-flop simply latches its input:

$y_1^{+} = D = x \oplus y_2.$

State / output table (state $= y_1 y_2$):

Match to a diagram. The table is not a simple linear chain: from $00$ the machine jumps directly to $11$ (a crossing/skip arc), and $10$ has a self-loop on $x=0$ ($0/0$) while $11$ has the $-/0$ behaviour shown. Only the diagram with the crossing transitions plus self-loops on $10$ ($0/0$) and $11$ ($-/0$) and the single $1/1$ output edge reproduces every row — that is option 3.

Options that draw a plain chain $00\!-\!10\!-\!01\!-\!11$ (option 1) or that drop a reachable state down to a 3-state diagram (option 2) or place the self-loop on $01$ (option 4) all violate at least one row of the table.

Exam trick: For these "which state diagram" items, never try to eyeball the arrows. Write $J,K,D$ in terms of $x$ and the state, convert JK to $y^{+}=y\oplus x$ (toggle) or $y^{+}=Jy'+K'y$, fill the 8-row table, and then find the matching picture. The output column ($z=x\,y_2'$) is what separates near-identical diagrams.