11/15 & 11/17
CS150
Section week 12
This week: Karnaugh maps, State minimization
1.
K-maps
Making K-maps.
AB
ABCD OUT Maxterm
Minterm
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
(A+B+C+D)
(A+B+C+D)
(A+B+C+D)
(A+B+C+D)
(A+B+C+D)
(A+B+C+D)
(A+B+C+D)
(A+B+C+D)
(A+B+C+D)
(A+B+C+D)
(A+B+C+D)
(A+B+C+D)
(A+B+C+D)
(A+B+C+D)
(A+B+C+D)
(A+B+C+D)
CD
1
0
1
1
0
0
1
1
1
1
1
1
0
1
0
1
ABCD
ABCD
ABCD
ABCD
ABCD
ABCD
ABCD
ABCD
ABCD
ABCD
ABCD
ABCD
ABCD
ABCD
ABCD
ABCD
00
01
11
10
00
01
11
10
00
1
0
0
1
1
0
0
1
01
0
0
1
1
0
0
1
1
11
1
1
1
1
1
1
1
1
10
1
1
0
1
1
1
0
1
00
1
0
0
1
1
0
01
0
0
1
1
0
0
1
1
11
1
1
1
1
1
1
1
1
10
1
1
0
1
1
1
0
1
The four corners are
0 this1 circle
actually
here…
OUT=BD+AC+AD
Terms:
Maxterm:
Product of Sums ( POS ). Product of terms whose results are zero:
OUT=(A+B+C+D)(A+B+C+D)(A+B+C+D)(A+B+C+D)(A+B+C+D)
Implicant:
Anything that you circle to be part of your equation. 1s if you’re looking for
minterms, 0 if you’re looking for maxterms…
Prime implicant:
If you can draw a bigger circle around an group, in other words, if you can
draw a circle that completely envelopes a group + some, then that group is not a prime implicant. A
prime implicant is a circled group that is the biggest it can possibly be.
Essential prime implicant: If there is only one circle that includes a particular implicant, then that
circle is necessary, uh, essential. If all of one group’s members are also inside other circles, then that
circle is redundant and not necessary ( essential ).
Minterm…
What you’ve been doing up to now. Sum of Products ( SOP ). Sum of terms whose
results are one:
OUT=ABCD+ABCD+ABCD+ABCD+ABCD+ABCD+ABCD+ABCD+ABCD+ABCD+ABCD
Simplify:
_ __
__
_
__
_
_
OUT = XYZ + XYZ + XYZ + XYZ + XYZ = ___Y+XZ_____________________
using K-maps.
STT
XYZ
000
001
010
011
100
101
110
111
OUT
1
1
0
0
1
1
1
0
XY
Z OUT
0
1
00
1
1
01
0
0
11
1
0
10
1
1
2.
Glitches and K-maps:
OUT
AB
C
00
01
11
10
0
0
1
1
0
1
0
0
1
1
K-map simplified equation: OUT=BC+AC
Equation drawn in gates: You can do this…
Can a glitch occur on the 110  111 transition? Yes.
It happens when the new 0 propagates through the
circuit faster than the new 1 does. ( This will
make sense if you draw the circuit ).
How can you fix it using a K-map? Add a circle that
covers that transition. ( The red circle ).
OUT
AB
C
00
01
11
10
0
0
1
1
0
1
0
0
1
1
3. State minimization
Given the STD below, use a state implication table to find and draw the reduced state transition diagram.
Things that make two states equal:
1.
2.
Same outputs for same inputs
Same next states
0
A
[0]
0
0
G
[0]
B
[0]
1
1
C
[0]
1
0
1
0
1
F
[0]
0
E
[0]
1
1
D
[1]
0
Present
State
A
B
C
D
E
F
G
Initial STT
Next State
X=0 X=1
B
A
D
D
A
B
A
C
C
C
E
F
G
E
Output
0
0
0
1
0
0
0
Step 1: Write up implication chart and cross out squares where the output of the two states are
different. ( Cross out things that violate first rule ).
Step 2: Write in next states for each pair of states in implication table.
Step 3: Use information in table ( crossed out squares ) to determine whether next states of each
state pair are equivalent or not. Cross out those that are not equivalent because they violate the
second rule of equivalence.
Step 4: Repeat step 3 until you have a pass where you don’t cross anything out.
Step 5: The states that haven’t been crossed out are equivalent. ( This isn’t really a step… )
State Implication Chart
Step 1. D’s output is different than all other states so we can cross out
D’s rows and columns.
B
C
D
E
F
G
A
B
C
D
E
F
Step 2: Use initial STT to fill in implication table’s next states.
B
C
B
A
C
B
C
D A
D
C
C
C
C
B
A A
A D
A
C
B
F C
B A
F C
B D
F
B
C
B
G C
A A
G C
A D
G
A
F
A
G
A B
A
C
E C
E C
E
F
E G
E
D
E
F
G
A
B
C
A
D
B
E
F
Blue Xs are first pass.
Green Xs are second pass.
Nothing crossed out on third pass.
So the equivalent states are:
A=B
and
E=F=G
Reduced STD
0
1
Filling in the reduced STT will make drawing
the reduced STD easier:
Present
State
1
Reduced STT
Next State
Output
X=0 X=1
AB
AB
C
0
C
D
C
0
D
D
EFG
1
EFG
AB
EFG 0
AB
C
[0]
[0]
0
0
EFG
[0]
1
D
[1]
0
1