Homework #2 Solution - UAH - Engineeringgaede/cpe628/08f_cpe628_hw2_solution.pdfHomework #2 Solution...
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The University of Alabama in Huntsville ECE Department CPE 628 01 Fall 2008 Homework #2 Solution 3.2(30 points), 3.4(30 points), 3.7(20 points), 3.8(20 points), Concurrent Fault Simulation Problem(60 points), Deductive Fault Simulation Problem(40 points) 3.2 For the circuit shown, complete the following timing diagram with respect to each timing model given below: (a) Nominal delay – Two-input gate, 1 ns; three-input gate, 1.2 ns; inverter, 0.6 ns. Inertial delay – All gates, 0.3 ns; (b) Rise delay – Two-input gate, 0.8 ns; three-input gate, 1 ns; inverter, 0.6 ns. Fall delay – Two-input gate, 1 ns; three-input gate, 1.2 ns; inverter, 0.8 ns. (c) Minimum delay – Two-input gate, 0.8 ns; three-input gate, 1 ns; inverter, 0.6 ns. Maximum delay – Two-input gate, 1 ns; three-input gate, 1.2 ns; inverter, 0.8 ns. (a) Nominal + Inertial Delays
Transcript of Homework #2 Solution - UAH - Engineeringgaede/cpe628/08f_cpe628_hw2_solution.pdfHomework #2 Solution...
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The University of Alabama in Huntsville ECE Department
CPE 628 01 Fall 2008
Homework #2 Solution 3.2(30 points), 3.4(30 points), 3.7(20 points), 3.8(20 points), Concurrent Fault Simulation Problem(60 points), Deductive Fault Simulation Problem(40 points) 3.2 For the circuit shown, complete the following timing diagram with respect to each timing model
given below: (a) Nominal delay – Two-input gate, 1 ns; three-input gate, 1.2 ns; inverter, 0.6 ns. Inertial delay – All gates, 0.3 ns; (b) Rise delay – Two-input gate, 0.8 ns; three-input gate, 1 ns; inverter, 0.6 ns. Fall delay – Two-input gate, 1 ns; three-input gate, 1.2 ns; inverter, 0.8 ns.
(c) Minimum delay – Two-input gate, 0.8 ns; three-input gate, 1 ns; inverter, 0.6 ns. Maximum delay – Two-input gate, 1 ns; three-input gate, 1.2 ns; inverter, 0.8 ns. (a) Nominal + Inertial Delays
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(b) Rise/Fall Delays
(c) Min-Max Delays
3.4 Redo Problem 3.2a using the nominal-delay event-driven simulation technique. Show all events
and activity lists of each time stamp. Time LE LA Scheduled Events 0 {(B,0)} {D,E,F} {(D,0,1),(E,1,0.6),(F,1,1)} 0.1 {(A,0)} {D} {(D,0,1.1)} 0.4 {(A,1)} {D} {(D,0,1.4)} 0.5 {(C,0)} {F} {(F,0,1.5)} 0.6 {(E,1)} {H} {(H,0,1.8)} 1 {(D,0),(F,1)} {H} {(H,1,2.2)} 1.1 {(D,0)} 1.4 {(D,0)} 1.5 {(F,0)} {H,J} {(H,1,2.7),(J,1,2.1)} 1.8 {(H,0)} {L} {(L,1,2.8)}
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2.1 {(J,1)} {L} {(L,1,3.1)} 2.2 {(H,1)} {L} {(L,0,3.2)} 2.7 {(H,1)} 2.8 {(L,1)} 3.1 {(L,1)} 3.2 {(L,0)} 3.7 For the circuit and two given stuck-at faults shown, use the parallel-pattern single-fault
propagation technique to identify which faults can be detected by the given test patterns.
PPSFP input internal output fault P a b D E F c
P1 1 0 0 0 1 1 P2 0 1 1 0 0 0 fault-free P3 1 1 1 1 0 1 P1 1 0 0 0 1 1 P2 0 1 0 0 0 0 α P3 1 1 0 0 0 0 P1 1 0 0 0 1 1 P2 0 1 1 0 1 1 β P3 1 1 1 1 1 1
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3.8 Repeat Problem 3.7 using parallel fault simulation.
PFS input internal output Pattern fault a b D E F c
fault-free 1 0 0 0 1 1 α 1 0 0 0 1 1 P1
β 1 0 0 0 1 1 fault-free 0 1 1 0 0 0
α 0 1 0 0 0 0 P2
β 0 1 1 0 1 1 fault-free 1 1 1 1 0 1
α 1 1 0 0 0 0 P3 β 1 1 1 1 1 1
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For the circuit shown, use concurrent fault simulation to determine the faults detected by an exhaustive test set.
a
b
c
1
23
g4
d
e
f
h
j
ik
Pattern 1 (a, b, c) = (0, 0, 0)
a
b
c
23
g4
de
fh
j
ik
0
0
0
a/110
0
00
1
d/110
0
b/101
0
e/101
0
00
0
a/10
10
f/11
00
h/10
10
b/110
0
00
0
i/10
11
c/11
01
j/000
1
1
01
k/111
0
b/100
1
d/100
1e/100
1
a/100
1
i/111
0
c/111
0
g/010
0
Faults detected are {c/1, g/0, i/1, k/1}
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Pattern 2 (a, b, c) = (0, 0, 1)
a
b
c
23
g4
de
fh
j
ik
0
0
1
a/110
0
00
1
d/110
0
b/101
0
e/101
0
00
0
a/10
10
f/110
0
h/10
10
b/11
00
01
1
i/11
11
c/00
00
j/001
1
1
10
k/010
1
b/101
1
d/101
1e/101
1
a/101
1
c/010
1
c/111
0
g/111
1
i/111
0
Faults detected are {a/1, b/1, c/0, d/1, e/1, g/1, j/0, k/0} Pattern 3 (a, b, c) = (0, 1, 0)
a
b
c
23
g4
de
fh
j
ik
0
1
0
a/110
0
01
0
d/110
0
b/000
1
e/000
1
10
0
a/11
11
f/000
0
h/11
11
b/00
00
00
0
i/10
11
c/11
01
j/110
1
0
01
k/101
1
b/100
1
d/100
1e/100
1
a/100
1
c/000
1
g/000
0
h/10
11
a/10
11
e/010
1
b/010
1
i/101
1
a/101
1
c/101
1
h/101
1
Faults detected are {g/0}
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Pattern 4 (a, b, c) = (0, 1, 1)
a
b
c
23
g4
de
fh
j
ik
0
1
1
a/011
0
01
0
d/011
0
b/000
1
e/000
1
10
0
a/11
11
f/000
0
h/11
11
b/00
00
01
1
i/10
11
c/01
01
j/111
0
0
11
k/000
1
g/001
0
a/101
1
a/10
11
h/10
11
e/011
0
b/011
0
c/101
1
i/101
1 b/101
1
Faults detected are {b/0, e/0, g/0, j/1} Pattern 5 (a, b, c) = (1, 0, 0)
a
b
c
23
g4
de
fh
j
ik
1
0
0
a/000
1
10
0
d/000
1
b/111
0
e/111
0
01
0
a/00
00
f/111
1
h/00
00
b/11
11
00
0
i/10
11
c/11
01
j/110
1
0
01
k/101
1
g/000
0
h/10
00
a/10
00
e/000
1
b/000
1
b/10
11
f/10
11
d/010
1
a/010
1
i/101
1 b/101
1
c/101
1 f/101
1
Faults detected are {g/0}
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Pattern 6 (a, b, c) = (1, 0, 1)
a
b
c
23
g4
de
fh
j
ik
1
0
1
a/00
01
10
0
d/000
1
b/11
10
e/111
0
01
0
a/00
00
f/111
1
h/00
00
b/11
11
01
1
i/11
11
c/00
00
j/111
0
01
1
k/000
1
g/001
0
i/101
1
f/11
11
b/11
11
a/011
0
d/011
0
b/101
1
c/101
1 f/101
1
c/000
1
Faults detected are {a/0, d/0, g/0, j/1}
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Pattern 7 (a, b, c) = (1, 1, 0)
a
b
c
23
g4
de
fh
j
ik
1
1
0
a/00
10
11
0
d/001
0
b/01
00
e/010
0
11
1
a/01
00
f/001
0
h/01
00
b/001
0
10
1
i/00
00
c/11
11
j/111
0
01
1
k/000
1
g/001
0
b/10
11
f/10
11
d/001
1
a/001
1
c/001
1
i/000
1
Faults detected are {g/0, j/1}
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Pattern 8 abc = (1, 1, 1)
a
b
c
23
g4
de
fh
j
ik
1
1
1
a/001
0
11
0
d/001
0
b/010
0
e/010
0
11
1
a/01
00
f/00
10
h/01
00
b/001
0
11
1
i/01
01
c/00
11
j/111
0
0
11
k/000
1
g/001
0
i/001
1
a/00
11
b/00
11
f/00
11
h/00
11
Faults detected are {g/0, j/1}
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For the circuit shown, use deductive fault simulation to determine the faults detected by the following set of input patterns: (a, b, c, d, f, i , h) = (1, 1, 0, 1, 1, 1), (1, 1, 0, 1, 1, 0), (0, 1, 0, 1, 0, 1), (0, 0, 0, 1, 1, 1), (1, 1, 1, 0, 1, 1), (1, 1, 1, 1, 0, 1). Pattern 1 (a, b, c, d, f, i , h) = (1, 1, 0, 1, 1, 1)
a
d
ih
ce
k
gf
j
n
lm
p
b11
0
11
1
00
0
00
10
00
La = {a/0} Lb = {b/0} Lc = {c/1} Ld = {d/1} Le = {d/1, c/1, e/1} Lf = {f/0} Lg = {d/1, c/1, e/1, g/1} Lh = {h/0} Li = {i/0} Lj = {d/1, c/1, e/1, j/1} Lk = {h/0, i/0, k/1} Ll = {d/1, c/1, e/1, j/1, l/0} Lm = {h/0, i/0, k/1, m/1} Ln = {d/1, c/1, e/1, g/1, n/1} Lp = {c/1, d/1, e/1, g/1, h/0, i/0, k/1, m/1, n/1, p/1} Pattern 2 (a, b, c, d, f, i , h) = (1, 1, 0, 1, 1, 0)
a
d
ih
ce
k
gf
j
n
lm
p
b11
0
10
1
00
0
01
11
01
La = {a/0} Lb = {b/0} Lc = {c/1} Ld = {d/1} Le = {d/1, c/1, e/1} Lf = {f/0} Lg = {d/1, c/1, e/1, g/1} Lh = {h/1} Li = {i/0} Lj = {d/1, c/1, e/1, j/1} Lk = {h/1, k/0} Ll = {d/1, c/1, e/1, j/1, l/0} Lm = {h/1, k/0, d/1, c/1, e/1, j/1, l/0, m/0} Ln = {d/1, c/1, e/1, g/1, n/1} Lp = {h/1, j/1, k/0, l/0, m/0, p/0}
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Pattern 3 (a, b, c, d, f, i , h) = (0, 1, 0, 1, 0, 1)
a
d
ih
ce
k
gf
j
n
lm
p
b01
0
01
1
00
0
01
11
01
La = {a/1} Lb = {b/0} Lc = {c/1} Ld = {d/1} Le = {d/1, c/1, e/1} Lf = {f/0} Lg = {d/1, c/1, e/1, g/1} Lh = {h/0} Li = {i/1} Lj = {d/1, c/1, e/1, j/1} Lk = {i/1, k/0} Ll = {d/1, c/1, e/1, j/1, l/0} Lm = {c/1, d/1, e/1, i/1, j/1, k/0, l/0, m/0} Ln = {c/1, d/1, e/1, g/1, n/1} Lp = {i/1, j/1, k/0, l/0, m/0, p/0} Pattern 4 (a, b, c, d, f, i , h) = (0, 0, 0, 1, 1, 1)
a
d
ih
ce
k
gf
j
n
lm
p
b00
0
11
1
11
1
10
00
1
1
La = {a/1} Lb = {b/1} Lc = {a/1, b/1, c/0} Ld = {d/1} Le = {a/1, b/1, c/0, e/0} Lf = {f/0} Lg = {a/1, b/1, c/0, e/0, g/0} Lh = {h/0} Li = {i/0} Lj = {a/1, b/1, c/0, e/0, j/0} Lk = {h/0, i/0, k/1} Ll = {a/1, b/1, c/0, e/0, j/0, l/1} Lm = {m/1} Ln = {a/1, b/1, c/0, e/0, f/0, g/0, n/0} Lp = {a/1, b/1, c/0, e/0, f/0, g/0, n/0, p/0}
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Pattern 5 (a, b, c, d, f, i , h) = (1, 1, 1, 0, 1, 1)
a
d
ih
ce
k
gf
j
n
lm
p
b11
1
11
0
01
1
10
00
0
0
La = {a/0} Lb = {b/0} Lc = {c/1} Ld = {d/0} Le = {d/0, e/0} Lf = {f/1} Lg = {d/0, e/0, g/0} Lh = {h/0} Li = {i/0} Lj = {d/0, e/0, j/0} Lk = {h/0, i/0, k/1} Ll = {d/0, e/0, j/0, l/1} Lm = {m/1} Ln = {f/1, n/1} Lp = {f/1, m/1, n/1, p/1} Pattern 6 (a, b, c, d, f, i , h) = (1, 1, 1, 1, 0, 1)
a
d
ih
ce
k
gf
j
n
lm
p
b11
1
01
1
01
1
11
00
1
1
La = {a/0} Lb = {b/0} Lc = {c/1} Ld = {d/0} Le = {d/0, e/0} Lf = {f/0} Lg = {d/0, e/0, g/0} Lh = {h/0} Li = {i/1} Lj = {d/0, e/0, j/0} Lk = {i/1, k/0} Ll = {d/0, e/0, j/0, l/1} Lm = {d/0, e/0, j/0, l/1, m/1} Ln = {d/0, e/0, f/0, g/0, n/0} Lp = {f/0, g/0, n/0, p/0}
