ECE255A – HW3 solutions
P. 1/2
4.2 (Random Test Generation)
Recall that the detection probability is
n
f
f
T
d
2
=
, where
f
T
is the set
of vectors that can detect fault f, and n is the number of inputs in the
circuit. In this circuit, we have 3 inputs, so there is a total of 8
possible vectors.
(b) For e/1, one vector, abc = 010 can detect it. Thus
f
d
= 0.125
4.3 (Boolean Difference)
(a) The set of vectors that can detect e/0 is
()
()
()
()
()
()
()
()
()
bc
bc
a
bc
b
c
a
e
b
c
a
e
b
g
e
b
g
g
b
g
g
e
b
g
g
e
e
z
e
z
e
d
d
e
e
z
=
+
=
⋅
+
⋅
=
⋅
⋅
=
⋅
⋅
=
+
⋅
+
⋅
⋅
=
+
⊕
⋅
=
=
⊕
=
⋅
⋅
=
⋅
=
m
1
0
1
Thus, the set of vectors is {011, 111}.
4.8 (D Algorithm)
Initially, we place a
D
on b. The Dfrontier at this time includes {d,
e}. Next, we pick a Dfrontier to propagate the fault effect across.
Suppose we pick d. Then, the decision a=0 is made. At this time, the
Dfrontier becomes {x, e}. We pick the Dfrontier that is closest to a
PO. Thus, we pick x. The next decision is e1=0. This decision implies y=1
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 Fall '08
 Parhami,B
 Vector Motors, 0$, PODEM

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