test generation

1. Chapter 3: Test Generation

2. Fault Diagnosis of Digital Systems
• Digital systems, even when designed with highly reliable
components, do not operate for ever without developing
some faults
• When a system ultimately does develop a fault it has to be
detected and located so that its effect can be removed
• Fault detection means the discovery of something wrong in
a digital system or circuit
• Fault location means the identification of the faults with
components, functional modules or subsystems, depending
on the requirements
• Fault diagnosis includes both fault detection and fault
location

3. Fault detection
• Fault detection in a logic circuit is carried out
by applying a sequence of test inputs and
observing the resulting outputs
• the cost of testing includes the generation of
test sequences and their application
• One of the main objectives in testing is to
minimize the length of the test sequence

4. Cont.

5. Test Generation for Combinational
Logic Circuits
• There are several methods available for deriving
tests for combinational circuits
• All these methods are based on the assumption
that the circuit under test is non-redundant (A
circuit is said to be non-redundant if the function
realized by the circuit is not the same as the
function realized by the circuit in the presence of
a fault) and only a single stuck-at fault is present
at any time

6. Cont.
Some of the methods available for
combinational circuits testing include:
• One dimensional path sensitization
• Boolean difference

7. One-dimensional Path Sensitization
• The basic principle involved in “path
sensitizing” is to choose some path from the
origin of the failure to the circuit output
• The path is said to be “sensitized” if the inputs
to the gates along the path are assigned
values so as to propagate the fault along the
chosen path to the output.

8. Example:
• Let us consider the circuit shown below and
suppose that the fault is line X3 s-a-1

9. Solution (the forward trace)
• This process of propagating the effect of the fault from its
original location to the circuit output is known as the
“forward trace”.
• To test for X3 s-a-1, X5 and G2 must be set at 1 and X3 set
at 0 so that if the fault is absent.
• We now have a choice of propagating the fault from G5 to
the circuit output Z via a path through G7G9 or through
G8G9
• To propagate through G7G9 requires the output of G4 and
G8 to be 1
• If G4=1, the output of G7 depends on the output of G5 and
similarly if G8=1, the circuit output depends on G7 only

10. Solution (the backward trace)
• The next phase of the method is the “backward
trace”, in which the necessary gate conditions to
propagate the fault along the sensitized path are
established
• For example, to set G4 at 1 both X1 and G1 must
be set at 1; G1=1 implies X2=0
• In order for G2 to be 1, X4 must be set at 0
• For G8 to be 1, G6 must be 0, which requires
either X2=0 or G3=0
• Since X2 has already been specified as 0, the
output of G6 will be 0

11. Cont.
• It is worth noting that G6 cannot be set at 0 by
making G3=0, since this would imply X3=0, which
is inconsistent with the previous assignment of X3
• Therefore the test X1X2X3X4X5=10001 detects
the fault s-a-1, since the output of Z will be 0 for
the fault-free circuit and 1 for the circuit having
the fault.
• The flaw in the one-dimensional path
sensitization technique is that only one path is
sensitized at a time

12. Boolean Difference
• The basic principle of the Boolean difference is
to derive two Boolean expressions – one of
which represents normal fault-free behavior
of the circuit and the other represents the
logical behavior under an assumed single s-a-1
or s-a-0 fault condition
• These two expressions are then exclusive
ORed; if the result is 1 a fault is indicated.

13. Cont.
• Let F(x)=F(X1,…Xn) be a logic function of n
variables
• If one of the inputs to the logic function, e.g.
input Xi, is faulty, then the output would be
F(X1…Xi…Xn).
• The Boolean difference of F(X) with respect to
Xi is defined as:

14. Cont.
• The function dF(X)/dXi is called the Boolean
difference of F(X) with respect to Xi
• It is easy to see that when F(X1…Xi…Xn) is not
equal to F(X1…Xi’…Xn), dF(X)/dXi = 1 and that
when F(X1…Xi…Xn) is equal to F(X1…Xi’…Xn),
dF(X)/dXi = 0
• the aim is to find input combinations for each
fault occurring on Xi such that dF(X)/dXi = 1.

15. Cont.
• Some useful properties of the Boolean
difference are:

16. Example:
• Consider the logic circuit shown in Fig. 3.3(a).
Find the Boolean difference with respect to
X3.

17. solution
•
This means that a fault on X3 will cause the output to be in error only if (X1X2)’X4,
i.e. if X1 or X2 (or both) are equal to 0 and X4 equal to 1

18. Cont.

19. Example:
• Find the partial Boolean difference associated
with the path X2-l-n-p-F in the fig. below

20. Solution:

21. Cont.
• The Boolean difference method generates all
tests for every fault in a circuit. It is a complete
algorithm and does not require any trial and
error. However, the method is costly in terms
of computation time and memory
requirements.

22. Test Generation for Sequential Logic
Circuits
There two distinctly different approaches to the
problem of finding tests for sequential circuits:
1. By converting a given synchronous sequential
circuit into a one dimensional array of identical
combinational circuits. Most techniques for
generating tests for combinational circuits can
then be applied.
2. By verifying whether or not a given circuit is
operating in accordance with its state table

23. State Table Verification
• In this approach a sequential machine is tested by
performing an “experiment” on it, i.e. by applying
an input signal and observing the output
• Hence, the testing problem may be stated as
follows: given the state table of a sequential
machine, find an input/output sequence pair (X,
Z) such that the response of the machine to X will
be Z if and only if the machine is operating
correctly

24. Checking experiment
• Checking experiment: The application of input sequence X
and the observation of the response, to see if it is Z.
• Checking sequence: the sequence pair (X, Z)
• Checking experiments are classified either as “adaptive” or
“preset”.
• In “adaptive” experiments the choice of the input symbols
is based on the output symbols produced by a machine
earlier in the experiment
• In “preset” experiments the entire input sequence is
completely specified in advance
• . A measure of efficiency of an experiment is its “length”,
which is the total number of input symbols applied to the
machine during the execution of an experiment

25. The derivation of “checking sequence”
• The derivation of “checking sequence” is based on the
following assumptions:
1. The network is fully specified and deterministic. In a
deterministic machine the next state is determined
uniquely by the present state and the present input.
2. The network is strongly connected, i.e. for every pair of
states qi and qj of the network, there exists an input
sequence that takes the network from qi to qj.
3. The network in the presence of faults has no more
states than those listed in its specification. In other
words, no fault will increase the number of states.

26. Example:
• Draw the successor tree for machine(m/c) M1
whose table is shown below
Present state
A
B
C
D
x=0
C,0
C,0
A,1
B,0
x=1
D,1
A,1
B,0
C,1

27. Successor tree

28. Homing experiment
• Homing experiment: defines the final states
• Homing sequence: ones and zeros that define the
final states
• Eg. For machine M1, 0,1 defines the final state
sequence for final state:
Initial state Response to 0,1 Final state
A
0,1
B
B
0,0
B
C
1,1
D
D
0,1
A

29. Homing tree
• Homing tree: its a truncated version of the
successor tree.
• A node becomes a terminal if it has either:
1. The node is associated with an uncertainty
vector whose non-homogenous components
are associated with the same node at
preceding levels
2. The node is associated with a trivial or
homogenous vector

30. Distinguishing experiment
• Distinguishing experiment: defines the initial
state
• Distinguishing sequence: ones and zeros that will
lead to the initial states(from machine M1, 111 is
a distinguishing sequence)
sequence for M1
Initial state Response to 111 Final state
A
110
B
B
111
C
C
011
D
D
101
A

31. Distinguishing tree
• Distinguishing tree: truncated version of successor tree
• A node becomes a terminal if either:
1. The node is associated with an uncertainty vector
whose non-homogenous components are associated
with the same node at preceding levels
2. The node is associated with an uncertainty vector
containing a homogenous non-trivial component
3. The node is associated with a trivial uncertainty vecor

32. Example:
• For machine M2 shown below, find the
shortest homing sequence and determine
whether or not a distinguishing sequence exist
and if any do exist, find the shortest one
state table of machine M2
Present state
A
B
C
D
inputs
x=0 x=1
B,0 D,0
A,0 B,0
D,1 A,0
D,1 C,0

33. Solution (Successor tree)

34. Cont.
• The shortest sequence is 010
Initial state Response to 010 Final state
A
000
A
B
001
D
C
101
D
D
101
D
• The machine has no distinguishing sequence

35. Synchronizing experiment
• It defines the final state regardless of the
output or the initial state
• Synchronizing sequence: ones and zeros that
define the experiment
• Eg. The synchronizing tree for machine M2

36. Synchronizing tree

37. Transfer sequence
• It is the shortest input sequence that takes the
machine from one state to another
• Eg. T.S for machine M2 starting at B

38. Transfer tree
• Shortest transfer sequence to take the
machine from B to C is 011

39. Checking experiment (fault detection
experiment)
• For any strongly connected diagnosable experimental
machine (diagnosable – has at least one distinguishing
sequence), there are three steps in fault detection
• Initialization phase: apply homing sequence followed
by transfer sequence if necessary
• State identification phase: apply distinguishing
sequence to all different states
• Transition verification phase: the machine is made to
go through energy state transition, each state
transition is checked by using distinguishing sequence

40. Fault tolerance design using error
correcting codes
1. Hamming code
2. Horizontal and vertical parity schemes

41. Soft error correction using horizontal
and vertical parity method
• The method can correct all errors detectable
by parity including multiple errors in a single
word
• Eg. Consider the five word memory shown
below using odd parity.

42. Example:
•
•
•
•
XOR all words 00111101 -(1)
Assume word 3 bit 5 is in error
XOR all words except word 3 10111110 -(2)
XOR (1) and (2) to get the original word
1000001

43. Design for testability
There are two approaches:
1. Testability design style
2. Testability measure analysis
The use of testability design style guarantees
that generating test for a circuit will not be
impossible.

44. Testability measure analysis
• This approach computes the difficulty of
controlling and observing each internal node
from primary input or outputs
• The information is used in the design process
to locate potential circuit testing problems
and provide feedback about the effect of
circuit modification on testability.

45. Design techniques
• Adhoc technique: aim at a particular solution
to a particular design problem. May involve
data compression(signature analysis)
• Structure technique: design methodology aim
at generalized solution in testing problems. Eg.
Level sensitivity scan design (LSSD), built-in
logical block observer(BILBO). All allow testing
of sequential circuits using only test patterns
for the combinational portions of the circuit

46. Key concepts
• Predictability: can the circuit be put in a known
state from which all future states can be
predicted?
• Controllability: can the state of given node in a
circuit be controlled from the primary input?
• Observability: can the state of the network be
determined from the primary inputs?
In general, add some control gates and control lines
to increase controllability and add output terminals
for testing purposes (observability)

47. Reed-Muller expansion technique

48. Cont.

49. Example:
• Given the function F(w,x,y)=wx+w’y+x’y’, find
the Reed-Muller expansion and hence draw
the circuit

50. Solution:

51. Cont.
• Find the C’s
Co = Fo = 1
C1=Fo+F4=0
C2=Fo+F2=1
C3=Fo+F1=0
C4=Fo+F2+F4+F6=1
C5=Fo+F1+F4+F5=1
C6=Fo+F1+F2+F3=1
C7=Fo+F1+F2+F3+F4+F5+F6+F7=0

53. Gookyi Agyemanh Nana Dennis
(dennisgookyi@gmail.com)