Question 2 of 6: Synchronous JK up/down counter with enable

Question 2: Synchronous JK up/down counter with enable [3+3+3+3 = 12]

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Two JK flip-flops $A,B$; inputs $E$ (enable) and $X$ (direction); $E{=}0\Rightarrow$ hold; $E{=}1,X{=}1\Rightarrow$ count up $00{\to}01{\to}10{\to}11{\to}00$; $E{=}1,X{=}0\Rightarrow$ count down $00{\to}11{\to}10{\to}01{\to}00$. JK excitation table (from the reference sheet): $Q{=}0{\to}Q^+{=}0\!:J{=}0,K{=}X$; $Q{=}0{\to}Q^+{=}1\!:J{=}1,K{=}X$; $Q{=}1{\to}Q^+{=}0\!:J{=}X,K{=}1$; $Q{=}1{\to}Q^+{=}1\!:J{=}X,K{=}0$.

Find. The state diagram, the state transition table, minimised $J_A,K_A,J_B,K_B$ as functions of $A,B,E,X$, and the resulting circuit.

Approach. Notice the down sequence is exactly the up sequence traversed backwards ($00{\to}11{\to}10{\to}01{\to}00$ reverses every edge of $00{\to}01{\to}10{\to}11{\to}00$), so both directions live on a single 4-state ring with $X$ choosing the direction of travel. From that, tabulate present/next state for all 16 combinations of $A,B,E,X$, read $J,K$ per bit off the excitation table, then K-map each of $J_A,K_A,J_B,K_B$ over $A,B,E,X$ (16-cell maps) to minimise.

1. (b) State transition table. When $E{=}0$, every row of the excitation table reduces to $J{=}0,K{=}0$ for both flip-flops regardless of $A,B,X$ (a JK flip-flop with $J{=}K{=}0$ holds), so those 8 trivial rows are summarised in one line rather than repeated. The 8 rows with $E{=}1$:

   $AB$	$X$	$A^+B^+$	$J_AK_A$	$J_BK_B$
   00	0	11	1,1	1,1
   00	1	01	0,0	1,1
   01	0	00	0,0	1,1
   01	1	10	1,1	1,1
   10	0	01	1,1	1,1
   10	1	11	0,0	1,1
   11	0	10	0,0	1,1
   11	1	00	1,1	1,1

   When $E{=}0$: $J_A{=}K_A{=}J_B{=}K_B{=}0$ for every $A,B,X$ (hold), 8 further rows omitted as trivial.
2. (c) K-map simplification. $J_B$ and $K_B$ are $1$ on exactly the same 8 rows ($E{=}1$, any $A,B,X$) and $0$ whenever $E{=}0$ — the K-map collapses to a single variable: $$\boxed{J_B=K_B=E}.$$ For $J_A$ ($=K_A$, by the same "always both 1 or both 0" pattern visible in the table), the K-map over $A,B,E,X$ shows $A$ never appears (both $A{=}0$ and $A{=}1$ rows give identical values for the same $B,E,X$), and the two 1-rows per $E{=}1$ block are $(B,X){=}(0,0)$ and $(B,X){=}(1,1)$ — adjacent in neither the $B$ nor $X$ direction, so no further grouping is possible beyond the two minterms $E\bar B\bar X$ and $EBX$: $$\boxed{J_A=K_A=EB X+E\bar B\bar X=E(B\odot X)}$$ (using $B\odot X$ for the XNOR of $B$ and $X$, i.e. $A$ toggles when enabled and $B$ agrees with the direction bit).
3. (d) Resulting circuit. One XNOR gate combines $B$ and $X$; its output ANDed with $E$ drives $J_A$ and $K_A$ (tied together); $E$ drives $J_B$ and $K_B$ directly (also tied together). Both flip-flops share the same clock edge.
   Fig. 2(d) — combinational logic: XNOR(B,X) AND E feeds $J_A{=}K_A$; E feeds $J_B{=}K_B$ directly.