ECE 362 Spring 2001 - EXAM 2 Solutions
PROBLEM 1.
S(1,2,6) OR S(0,3,5,6) = S(0,1,2,3,5,6)
S(1,2,6) AND S(0,3,5,6) = S(6)
S(1,2,6) XOR S(0,3,5,6) = S(0,1,2,3,5)
PROBLEM 2.
(1) F(c,d,g) = c'd'g + c'd + cdg' + cd'
g(a,b) = ab' + a'b
(2) Decomposition does not exist of the form F(a,b, g(c,d)).
PROBELM 3.
Here CSPF for only Primary Inputs from Top to Bottom are given:
Input1
Input 2 Input 3
0000 1111 0011 0011
0101 0101
00-- 11-- 0011 0-11
--01 -101
PROBLEM 4.
Since it is not known that a solution exist with all positive weights,
we cannot restrict the inequalities to only Prime Implicants of f
and f'. We must generate all ineqaulities from the truth table..
| x y z |
f |
inequalities |
| 0 0 0 |
0 |
0 < T |
| 0 0 1 |
0 |
az < T |
| 0 1 0 |
1 |
ay >= T |
| 0 1 1 |
0 |
ay + az < T |
| 1 0 0 |
1 |
ax >= T |
| 1 0 1 |
0 |
ax + az < T |
| 1 1 0 |
1 |
ax + ay >= T |
| 1 1 1 |
1 |
ax + ay + az <=
T |
PROBLEM 5.
Ciritical Race: (I) Stable State 00 with current input
00 or 11, next input 10, intended final state 11.
Since state 00 changes to 11, it may go through intemmediate state
10 or 01.
In either case the machine ends up in state 01. Hence critical race
exists.
(II) Stable State 10 with current input 11, next input 10, intended
final state 01.
Since state 10 changes to 01, it may go through intermmediate state
00 or 11.
In either case it ends up in the state 11. Hence critical race exists.
Essential Hazards: Stable State 00 with current input
11, next input 10, intended final state 11.
If the final state sees the old input-11 then it moves to state 01.
Finally it sees the new input-10 and moves to state 01.
Another possible Hazard: Stable State 01 with current input 10, next
input 00, intended final state 00.
PROBLEM 6.
Create Five state with the following meaning
ST = Starting State
S0 = Last inputs were 0 or 00
S1 = Last inputs were 1 or 11
S01 = Last two inputs were 01
S10 = Last two inputs were 10
Present
State
|
Input
0
|
Input
1
|
|
ST
|
S0, 0
|
S1, 0
|
|
S0
|
S0, 0
|
S01, 0
|
|
S1
|
S10, 0
|
S1, 0
|
|
S01
|
S10, 0
|
S1, 1
|
|
S10
|
S0, 0
|
S01, 1
|
PROBLEM 7.
P0 = (AGH)(BCDEF)
P1 = (A)(GH)(BCE)(DF)
P2 = (A)(GH)(B)(CE)(DF)
P3 = P2
Reduced Machine has 5 states so needs a minimum of 3 flip-flops to
encode 5 states.
Standard Form Reduced table
|
Present
State
|
00
|
01
|
10
|
11
|
Z
|
|
A: a
|
a
|
b
|
c
|
d
|
0
|
|
GH: b
|
b
|
d
|
a
|
c
|
0
|
|
DF: c
|
c
|
d
|
e
|
a
|
1
|
|
CE: d
|
d
|
c
|
b
|
a
|
1
|
|
B: e
|
c
|
c
|
b
|
a
|
1
|
PROBLEM 8.
All Compatible Pairs: (A,B) (A,E) (B,E) (C,D) (G, H)
All Maximal Compatibility Classes: (A,B,E) (C,D) (G,H) (F)