Homework # 14 KEY
Name ___________________
Finish the problem we started in class by adding the Karnaugh maps and determining the necessary equations for each flip-flop input. Note that I have added SR flip flops to the table below.
current |
state |
input |
next |
state |
SA |
RA |
SB |
RB |
DA |
DB |
TA |
TB |
JA |
KA |
JB |
KB |
A |
B |
x |
A |
B |
|
|
|
|
|
|
|
|
|
|
|
|
0 |
0 |
0 |
0 |
1 |
0 |
X |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
X |
1 |
X |
0 |
0 |
1 |
0 |
0 |
0 |
X |
0 |
X |
0 |
0 |
0 |
0 |
0 |
X |
0 |
X |
0 |
1 |
0 |
1 |
1 |
1 |
0 |
X |
0 |
1 |
1 |
1 |
0 |
1 |
X |
X |
0 |
0 |
1 |
1 |
0 |
0 |
0 |
X |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
X |
X |
1 |
1 |
0 |
0 |
0 |
1 |
0 |
1 |
1 |
0 |
0 |
1 |
1 |
1 |
X |
1 |
1 |
X |
1 |
0 |
1 |
1 |
0 |
X |
0 |
0 |
X |
1 |
0 |
0 |
0 |
X |
0 |
0 |
X |
1 |
1 |
0 |
1 |
0 |
X |
0 |
0 |
1 |
1 |
0 |
0 |
1 |
X |
0 |
X |
1 |
1 |
1 |
1 |
0 |
1 |
0 |
1 |
X |
0 |
0 |
1 |
1 |
0 |
X |
1 |
X |
0 |
|
B’x’ |
B’x |
Bx |
Bx’ |
A' |
000 |
001 |
011 |
010 |
A |
100 |
101 |
111 |
110 |
SA = Bx’
|
B’x’ |
B’x |
Bx |
Bx’ |
A' |
0 |
0 |
0 |
1 |
A |
0 |
X |
0 |
X |
RA = Bx + B’x’
|
B’x’ |
B’x |
Bx |
Bx’ |
A’ |
X |
X |
X |
0 |
A |
1 |
0 |
1 |
0 |
DA = Bx’ + AB’x
|
B’x’ |
B’x |
Bx |
Bx’ |
A’ |
0 |
0 |
0 |
1 |
A |
0 |
1 |
0 |
1 |
DB = A’x’ + B’x’ +
ABx
|
B’x’ |
B’x |
Bx |
Bx’ |
A’ |
1 |
0 |
0 |
1 |
A |
1 |
0 |
1 |
0 |
TA = A’Bx’ + ABx +
AB’x’
|
B’x’ |
B’x |
Bx |
Bx’ |
A’ |
0 |
0 |
0 |
1 |
A |
1 |
0 |
1 |
0 |
TB = B’x’ + Ax’ + A’Bx
|
B’x’ |
B’x |
Bx |
Bx’ |
A’ |
1 |
0 |
1 |
0 |
A |
1 |
0 |
0 |
1 |
SB = B’x’
|
B’x’ |
B’x |
Bx |
Bx’ |
A’ |
1 |
0 |
0 |
X |
A |
1 |
0 |
X |
0 |
RB = A’x + ABx’
|
B’x’ |
B’x |
Bx |
Bx’ |
A’ |
0 |
X |
1 |
0 |
A |
0 |
X |
0 |
1 |
JB = x’
|
B’x’ |
B’x |
Bx |
Bx’ |
A’ |
1 |
0 |
X |
X |
A |
1 |
0 |
X |
X |
JA = Bx’
|
B’x’ |
B’x |
Bx |
Bx’ |
A’ |
0 |
0 |
0 |
1 |
A |
X |
X |
X |
X |
KB = A’x + Ax’
|
B’x’ |
B’x |
Bx |
Bx’ |
A’ |
X |
X |
1 |
0 |
A |
X |
X |
0 |
1 |
KA = B’x’ + Bx
|
B’x’ |
B’x |
Bx |
Bx’ |
A’ |
X |
X |
X |
X |
A |
1 |
0 |
1 |
0 |
SA =
Bx’
RA =
B’x’ + Bx
DA = Bx’ + AB’x
DB = A’x’ + B’x’ +
ABx
TA = A’Bx’ + ABx +
AB’x’
TB = B’x’ + Ax’ + A’Bx
JA = Bx’
KA = B’x’ + Bx
JB = x’
KB = A’x + Ax’
SB = B’x’
RB = A’x + ABx’