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Combinational logic unit failures in two ways:
Failure Mode 1 - Fixed 1 fault ( s − a −1)
. While
the logic unit has AND gate and Boolean algebraic
expression is
Z = x x
(2)
1 2x3
Then any one of element (such as x
1
) take 1, Z will
not change, that is, expression (2) can not find the failure
mode 1 of x
1
, so that x
1
is Fixed 1 fault ( s − a −1)
of Boolean algebraic expression Z . Similarly, x
2
, x
3
is also Fixed 1 fault ( s − a −1)
.
Failure Mode 2 - Fixed 0 fault ( s − a − 0)
. While
the logic unit has OR gate and Boolean algebraic
expression is
Z x + x +
= (3)
1 2
x3
x ) take 0, Z will
Then any one of element (such as
1
not change, that is, expression (3) can not find the failure
mode 2 of x
1
, so that x
1
is Fixed 0 fault ( s − a − 0)
of Boolean algebraic expression Z . Similarly, x
2
, x
3
is also Fixed 0 fault ( s − a − 0)
of Boolean algebraic
expression Z .
2) Calculation of single failure test code
If a , a ,..., a )
(
1 2 n
T is the test code of Fixed 0
fault in measured channel X ( x1,
x2,...,
xn)
, then the
formula established:
df ( X )
X ( x1,
x2,...,
xn)
×
T ( a , ,..., )
= 1
1 a2
a
(4)
n
dx
Where a = 0 or 1.
i
( a1,
a2,...,
an
i
If T ) is the test code of Fixed 1
X ( x , ,..., )
fault in measured channel
1
x2
xn
, then the
formula established:
df ( X )
X ( x1,
x2,...,
xn)
×
T ( a , ,..., )
= 1
1 a2
a
(5)
n
dx
Where a = 0 or 1.
i
i
Application the Property 1 and Property 2 of
Boolean difference, it would be appraised whether the
Fixed 1 fault ( s − a −1)
and Fixed 0 fault ( s − a − 0)
may be measured on the channel X x , x ,..., x ) .
(
1 2 n
Applications formula (4) and (5) , the single failure
test code of the Fixed 1 fault ( s − a −1)
and Fixed 0
fault ( s − a − 0)
on the unit X ( x1,
x2,...,
xn)
would be
calculated. For example, added the single test code
T ( a i
) to the input of the measured unit, if the unit
output Z = 1, then there is no failure of ( s − a −1)
in X x , x ,..., x ) , otherwise there is.
(
1 2 n
Similarly, the multi-fault test code
T ( a1,
a2
,..., an
) can be calculated, not repeat them
due to space limitations.
III. EXAMPLE
There is a petrochemical company using ESD as
DDS in its FCCU-II sets. [6][7] The main logic diagram of
the combination logic unit shown in Fig.1 .[8] Let's
examine the identification of the DDS’ failure.
A.Failure identifiability of combination logic unit
There are three kinds of basic logic unit in fig.1 (Fig.
2). [9][10]
For fig. 2 (a), because of
f ( X ) = x1x2
+ x1x3
+ x2x3
, then
df ( X)
= ( x1x
2
+ x1x
3
+ x2x3
) ⊕(
x1x
2
+ x1x
3
+ x2x3
)
dx
1
= x +
2
+ x3
x2
x3
So that the fault test expressions:
df ( X )
x
1
= x1
( x2
dx1
+ x3
+ x2
x3
) (6-1)
df ( X )
x
1
= x1
( x2
dx1
+ x3
+ x2
x3
) (6-2)
By symmetry,
df ( X )
x
2
= x2
( x1
+ x3
+ x1x3
)
dx
(7-1)
1
df ( X )
x
2
dx1
x2
( x1
+ x3
+ x2
x3
)
df ( X )
x
3
dx1
x3
( x1
+ x2
+ x1x2
)
df ( X )
x
3
x3
( x1
+ x2
+ x1x2
)
dx
1
= (7-1)
= (8-1)
= (8-1)
Similarly, for fig. 2 (b) and fig. 2 (c), may also find
thier fault test expressions. Choosed different value of
x
i
( x
i
= 1 or 0) and fixed others in these pairs of
expressions (for example in expression (6-1)and (6-2)), if
the value of these pairs expression do chuange, then x
i
is the single fault test code, and this set of variables is
called combinational fault test code.
Tab.1 lised out all of the combinational fault test
code of the logic unit of fig.2 (a), fig.2 (b) and fig.2 (c).
B.Faults identification of combination of logic unit
Based on the analysis above, the fault of the DDS
can be identification as follow.
In Figure 1,
x + x
A
= ( x1 x2
+ x1x3
+ x2x3)
x4
+ xB
= x26
x4
x + x + x
B
= ( x5 + x6
+ x7x8
) x9
+ xC
+ xD
= x27x9
x C
= ( x
x + x = x x + x
10
x11
+ x10x12
+ x11x12)
13
24
28
13
C
B
24
D
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