SME
Digital Logic Design · Problem type
JK flip-flop next state
Use this for “A JK flip-flop holds Q = 0 with J = K = 1. What is Q after the clock edge?”
Diagram
f = Σm(1, 4, 5, 6) = JQ′ + K′Q.
3 inputs: 2³ = 8 rows of the truth table, one cell of the map each.
Minterm m6 is row 110: JKQ′.
This is an example. Choose your own values to start a new problem.
J
Input J
example
K
Input K
example
Q
Present state
example
Q⁺
Next state
worked out (not typed)
A
Action
worked out (not typed)
m
Truth-table row
worked out (not typed)
Formulas
Q
⁺
=
J
×
(1
−
Q
)
+
(1
−
K
)
×
Q
1 = 1 × (1 − 0) + (1 − 1) × 0
m
=
4
×
J
+
2
×
K
+
Q
6 = 4 × 1 + 2 × 1 + 0
A
:
compare
J
with
K
J
=
1,
K
=
1,
so
toggle
Assumptions
•
The flip-flop changes only on the clock edge; Q⁺ is the state just after it.
•
The characteristic equation is Q⁺ = JQ′ + K′Q; for a bit, the complement Q′ is 1 − Q.
•
J = K = 0 holds, K = 1 alone resets, J = 1 alone sets, and J = K = 1 toggles.
Step-by-step
WE KNOW
J = 1, K = 1, Q = 0
FIND
next state (Q⁺), truth-table row (m), action (A)
Step 1 · Find next state (Q⁺)
Use Q⁺ = J × (1 − Q) + (1 − K) × Q
JQ′
turns
0
into
1
when
J
is
1;
K
′
Q
keeps
1
unless
K
is
1.
At
most
one
term
is
1.
Q
⁺
=
1
×
(1
−
0)
+
(1
−
1)
×
0
Q
⁺
=
1
+
0
×
0
Q
⁺
=
1
Step 2 · Find truth-table row (m)
Use m = 4 × J + 2 × K + Q
Read
J
,
K
,
Q
as
a
3-bit
number:
the
row
of
the
truth
table
and
the
cell
of
the
K
-map.
m
=
4
×
1
+
2
×
1
+
0
m
=
6
Step 3 · Find action (A)
Use A: compare J with K
Both
inputs
1:
Q
flips
to
its
complement.
J
=
1,
K
=
1:
toggle
A
=
compare
1
with
1
A
=
toggle
Check
Put the numbers back in. Both sides should match.
1 = 1 × (1 − 0) + (1 − 1) × 0 ✓
6 = 4 × 1 + 2 × 1 + 0 ✓
J
=
1,
K
=
1,
so
toggle
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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