State-identification Experiments and
Testing of Sequential Circuits
1
Zvi Kohavi and Niraj K. Jha
Experiments
Experiment: application of an input sequence to the input terminals of a
machine
• Simple: performed on a single copy
• Multiple: performed on two or more identical copies
• Adaptive: input symbol at any time instant depends on previous output
symbols
• Preset: entire input sequence is predetermined
• Length: total number of symbols in the experiment
Checking experiment: designed to take the machine through all its
transitions to ascertain if it is working correctly
2
Introductory Example
Example: Consider machine M1 and its responses to 01 and 111
• Output response to 01: uniquely determines the machine’s final state, but
not the initial state
• Output response to 111: uniquely determines the machine’s final and initial
states
3
Uncertainties
Initial uncertainty: minimal subset of set of states S, which is known to
contain the initial state
• For machine M1: (ABCD)
If application of input symbol 1 results in output symbol 1: uncertainty is
(ACD)
•
•
•
•
1-successors of (ACD): uncertainty vector (B)(CD)
Components of uncertainty vector: (B) and (CD)
Trivial uncertainty vector: all components have a single state
Homogeneous uncertainty vector: components contain a single state or
identical repeated states
4
Successor Tree
Successor tree: displays Ii-successor uncertainties for all Ii
• Path: sequence of j branches starting at the highest node and terminating at
the jth level
• Length of the path: j
• Each path describes: an input sequence
Level
(ABCD)
0
0
1
(A)(BCC)
(ACD)(B)
0
0
(AA)(C)(C)
1
(A)(BB)(D)
(A)(BC)(C)
0
1
(A)(A)(C)(C) (A)(B)(B)(D)
1
1
(A)(B)(CD)
0
2
1
(A)(B)(C)(C) (A)(B)(C)(D) 3
5
Homing Experiments
An input sequence Y0 is a homing sequence if the final state of the
machine can be determined uniquely from the machine’s
response to Y0, regardless of the initial state
Homing tree: successor tree in which a jth-level node becomes terminal
when:
1. It is associated with an uncertainty vector whose non-homogeneous
components are associated with some node in a preceding level
2. It is associated with a trivial or homogeneous vector
Example: Consider machine M2
Level
(ABCD)
0
1
(AB)(DD)
0
(AB)(DD)
0
(ABCD)
1
1
(BD)(CC)
0
(A)(D)(DD)
2
1
(AA)(BC)
3
Theorem: A preset homing sequence of length at most (n-1)2 exists for
every reduced n-state machine
6
Synchronizing Experiments
A synchronizing sequence of a machine M is an input sequence that takes
M to a specified final state, regardless of the outputs symbols or
the initial state
In a synchronizing tree: a jth-level node becomes terminal whenever:
1. The node is associated with an uncertainty that is also associated with
some node in a preceding level
2. Some node in the jth level is associated with an uncertainty containing
just a single element
Level
Example: Machine M2 and its synchronizing tree
(ABCD)
0
1
(ABD)
0
(ABCD)
1
1
(ABD)
(BCD)
0
0
2
1
(AD)
Theorem: If a synchronizing sequence for an
n-state machine M exists, then its length is at
most (n-1)2n/2
0
(ABC)
3
1
(BD)
(CD)
0
4
1
7
(D)
(AC)
5
Distinguishing Experiments
An input sequence X0 of machine M is said to be a distinguishing
sequence if the resulting output sequence is different for each
initial state
•
Since the knowledge of the initial state and input sequence is always
sufficient to uniquely determine the final state: every distinguishing
sequence is also a homing sequence
– The converse is not true
Distinguishing tree: successor tree in which a node at the jth level becomes
terminal when:
1. The node is associated with an uncertainty vector whose nonhomogeneous components are associated with some node in a
preceding level
2. The node is associated with an uncertainty vector containing a
homogeneous nontrivial component
3. Some node at the jth level is associated with a trivial uncertainty vector
8
Distinguishing Experiments (Contd.)
Example: Machine M1 and its distinguishing tree
Level
(ABCD)
0
0
1
(A)(BCC)
(ACD)(B)
0
0
(AA)(C)(C)
1
(A)(BB)(D)
1
(A)(BC)(C)
0
1
(A)(B)(CD)
1
0
(A)(A)(C)(C) (A)(B)(B)(D)
2
1
(A)(B)(C)(C) (A)(B)(C)(D) 3
Example: While every machine has a homing sequence, not every machine
has a distinguishing sequence – Consider M2 and its distinguishing tree
Level
(ABCD)
0
1
(AB)(DD)
0
(AB)(DD)
0
(ABCD)
1
1
(BD)(CC)
0
2
1
9
(A)(D)(DD)
(AA)(BC)
3
Shortest Distinguishing Prefix
Example: Consider machine M1 and its response to 111
• Shortest distinguishing prefix for state C: 1
• For state D: 11
• For states A and B: 111
10
Machine Identification
Machine identification: experimentally determining the state table of an
unknown machine
• Input alphabet known
• Upper bound on the number of states known
• Machine must be reduced and strongly connected
Example: Suppose a machine is supposed to have two states and its
response to input sequence X is output sequence Z:
• Corresponding machine M3:
11
Checking Experiments
Given a machine and its state table: determine from terminal experiments
whether the actual machine is isomorphic to the one described by
the state table
•
•
•
•
Machine assumed to be: strongly connected, completely specified and
reduced
Faults assumed to be: permanent
Machine either has a synchronizing sequence: or a reset input exists
that will transfer it to the initial state
Initial assumption: a distinguishing sequence exists
Designing checking experiments: two parts
1. Use the synchronizing sequence or reset input to transfer the machine
into a prespecified state: initial state for the second part of the
experiment
2. Preset experiment: take machine through all possible transitions
1. Machine is caused to display the response of each of its states to the
distinguishing sequence
2. Then, actual state transitions are verified
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Example
Example: Machine M4 and its responses to distinguishing sequences 00
and 01
• Suppose the synchronizing sequence or reset input places M4 into A
• First ascertain: starting state is indeed A and the machine being tested
actually contains four distinct states
• Next: verify every state transition
– Apply the input symbol corresponding to the transition
– Identify it by applying the distinguishing sequence
13
Example (Contd.)
Example (contd.): First stage for checking all state transitions except those
from D to A and A to B and C
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Example (Contd.)
Example (contd.): Complete experiment:
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Design of Diagnosable Machines
Diagnosable machine: one which possesses one or more distinguishing
sequences
Testing table and graph: for machine M2 that does not possess a
distinguishing sequence
0/0
AB
1/0
BC
AC
1/0
BD
1/0
1/0
AD
1/0
1/0
CD
Uncertainty pair
Implied pair
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Definitely Diagnosable Machines
A machine is definitely diagnosable machine of order  if  is the least
integer s.t. every sequence of length  is a distinguishing
sequence for M
• Every node in level  of the distinguishing tree is associated with a trivial
uncertainty vector
Theorem: A machine is definitely diagnosable if and only if its testing graph
G is loop-free and no repeated states exist in the testing table
Corollary: Let the testing table of machine M be free of repeated entries,
and let G be a loop-free testing graph for M
• If the length of the longest path in G is l: then  = l + 1
17
Designing Definitely Diagnosable
Machines
Example: Obtaining definitely diagnosable M2’ from M2
• Assign different output symbols to each transition that may cause a
repeated entry in the testing table
• Open loops in the testing graph by removing one of the arcs
0/0
AB
1/0
BC
AC
1/0
BD
1/0
1/0
AD
1/0
1/0
CD
18
Definitely Diagnosable Machines (Contd.)
For any 2k-state machine: addition of k output terminals is sufficient to
convert it into a definitely diagnosable machine
x
Logic
z
Logic
z
Logic
z1
S
M
(a) Machine M.
x
S
M
(b) Machine M . z1 is only used for
diagnosis purposes.
19
State Table based Test Generation
Functional fault model: faults assumed to be associated with a state
transition
•
•
•
•
•
Single-state-transition (SST) fault model: fault results in the destination
state of the state transition becoming corrupted while retaining its correct
input/output symbols
Test generation based on SST faults: known to detect a very high
percentage of single stuck-at faults in the sequential circuit
Assumption: SST fault does not increase the number of states in the
state table
State transition designated as a four-tuple: <input symbol, source state,
destination state, output symbol>
A state transition can become corrupted: if its destination state, output
symbol or both are faulty
– However, if a test sequence detects a corrupted destination state:
then it also detects the corrupted output symbol or both the corrupted
destination state and output symbol
– Three parts to a test sequence:
» Initialization sequence
» Input symbol of the transition to activate the fault
» State-pair differentiating sequence (SPDS)
20
Test Generation (Contd.)
Fault collapsing: An n-state m-transition machine has m(n-1) SST faults
•
•
•
For each state transition: there are n-1 faulty destination states possible
Suppose the four-tuple <Ik,Sj,Si,Ol> is corrupted to: <Ik,Sj,Si’,Ol> by SST
fault f1 and to <Ik,Sj,Si’’,Ol> by SST fault f2
If SPDS(Si,Si’) also differentiates between Si and Si’’: fault f2 dominates
fault f1, and f2 can be removed from the fault list
Example: Consider machine M5
•
•
•
SPDS(A,B) = SPDS(A,C) = 0 and SPDS(B,C) = 1
Consider <1,C,A,0>: can be corrupted to <1,C,B,0>, <1,C,C,0>, <1,C,D,0>
Since SPDS(A,B) = SPDS(A,C): the first two faulty transitions can be
collapsed into just the first one
21
Test Generation Example
Example: Consider the SST fault that corrupts <0,D,A,0> to <0,D,B,0>
•
•
•
•
We first need transfer sequence T(A,D) = 00 (10 is also a valid sequence)
Fault is activated by x = 0
Finally, SPDS(A,B) = 0 is applied
Hence, one possible test sequence: 0000
22
Sequential Circuit based Test Generation
Extended D-algorithm:
1. Target a fault in some time frame, say time frame 0 in the iterative array
model: use D-algorithm to generate a test vector for it
2. If a D or D’ propagates to a circuit output: no further error propagation
required
3. If D or D’ only propagates to next state lines: add new time frames to the
right until the error signal reaches some circuit output
4. If test vector contains assignments of specific logic values to any present
state lines in time frame 0: add new time frames to the left until no
particular values are required on the present state lines
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Example
Example: Test sequence for x2 s-a-1: {(1,1),(1,0),(1, )}
z
x1
x2
s-a-1
D
y
Y
(a) A sequential circuit.
z0
z -1
x1-1 1
x2-1 1
s-a-1
-1
y
x10 1
D
x20 s-a-1
0
Y
Time frame -1
-1
D
1
y0
Time frame 0
D
0
Y
z1
D
x11 1
x21 s-a-1
Y
y1
1
Time frame 1
(b) Iterative array model.
24
Nine-valued Logic
Five-valued logic {0,1,,D,D’} used in D-algorithm: not adequate for
sequential circuits because it overspecifies the value
requirements at some lines in the circuit
• This may prevent the test generator from obtaining a test sequence even
when one exists
• This problem can be tackled by nine-valued logic: 0/0, 0/1, 0/, 1/0, 1/1,
1/, /0,  /1,  / 
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Example
Example: Unsuccessful test generation with five-valued logic
x1
s-a-1
z
x2
y
D
Y
(a) A sequential circuit.
x1-1
s-a-1
0
x2-1 0
y -1
conflict
x10
G1
s-a-1
D
0
z -1
1
Y
Time frame -1
-1
0
D
0
x20 1
1
D
z0
Y
0
0
y
0
Time frame 0
(b) Application of the extended D-algorithm.
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Example (Contd.)
Example (contd.): Successful test generation with nine-valued logic
•
x1-1
s-a-1
x10
0/
x2-1 0/
y -1
Test sequence: {(0,0),(0,1)}
0/1
z -1
1/
Y
Time frame -1
-1
0/
G1
s-a-1
x20 1/
0/
0/1
0/0
0/
y0
1/0
1/
z0
Y
0
Time frame 0
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Design for Testability
Scan design for sequential circuits: two modes of operation
•
•
•
•
Normal mode: circuit exhibits original behavior
Test mode: its flip-flops are chained together into a shift register
Full-scan: all flip-flops are chained
Partial-scan: a subset of flip-flops are chained
Scan flip-flop and its symbol:
Yi
M
U
X
S
Yi
M
U
X
Yi
yi
D
S
Yi
yi
D
T
T
(a) Scan flip-flop.
(b) Symbol.
Scan chain: only combinational test generation needed
x1
z1
x2
xl
z2
Combinational logic
y1 Y2
Y1
ScanIn
YS
1
T
M
U
X
y2
D
YS
2
M
U
X
yk
Yk
D
S
Y3
YS
k
M
U
X
zm
D
ScanOut
28
Scan-based Test Application
Application of the test set derived for the combinational logic to the
sequential circuit:
1. Make T = 1 to set circuit into the test mode
2. Scan in the state part of the vector through the ScanIn input in the next k
clock cycles. Primary inputs can be fed arbitrary values in these cycles
3. Apply the primary input part of the vector to the primary inputs. Now, all
l+k bits of the test vector have been applied to the combinational logic.
After allowing the logic to settle down, observe the output response at
circuit outputs z1, z2, …, zm
4. Make T = 0 to set the circuit into the normal mode
5. Apply a clock pulse. This results in the values on the next state lines, Y1,
Y2, …, Yk, being latched in the k flip-flops
6. Make T = 1 and observe the values captured in the flip-flops by scanning
them out through ScanOut while repeating the procedure for the next
test vector
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Testing of Scan Designs
Suppose there are n test vectors in the test set
•
•
•
•
A total of k cycles required to scan in the state part
One cycle to capture state response
k-1 cycles required to scan out the captured state
Since the state part of the next state vector is scanned in at the same
time the captured state for the previous vector is being scanned out:
total no. of clock cycles required for testing = n(k+1)+k-1
Example: total no. of clock cycles = 4(2+1)+2-1 = 13
y2
y1
z
x1
z
x1
y2
Y1
x1
1
0
1
0
x2
0
1
1
0
y1
0
1
0
1
y2
1
0
0
1
x2
y1
x2
Y2
y1
D
y2
D
(a) Sequential circuit.
Y1
Y2
(b) Combinational logic.
(c) Stuck-at fault test set.
30
Built-in Self-test (BIST)
Integration of circuitry on-chip to enable the circuit under test (CUT) to test
itself
•
•
•
•
At-speed testing possible: at the normal clock rate
– Detects delay faults
Test pattern generator (TPG)
Response analyzer (RA): compresses output response into a signature
Golden signature: when no fault is present
T
P
G
CUT
R
A
31
Test Pattern Generator
TPG: usually a linear feedback shift register (LFSR)
•
LFSR with degree-k feedback polynomial:
p(x) = xk + b1xk-1 + … + bk-1x + bk
Yk
•
b1
bk-1
bk
yk
+
D
Y2
Yk-1
y2
+
D
Y1
y1
D
Outputs of the k flip-flops directly fed to inputs of a k-input CUT
Example: Three-stage LFSR with p(x) = x3 + x2 + 1
Y3
D
y3 Y2
y2
D
+
Y1
y1
D
32
Feedback Polynomial
A feedback polynomial is said to be primitive if the state diagram
corresponding to the k-stage LFSR consists of two loops:
•
•
Trivial loop: with the all-0 state
Non-trivial loop: with the remaining 2k-1 states
Example: State diagram of the three-stage LFSR based on p(x) = x3+x2+1
•
Thus, p(x) is primitive
111
101
110
001
011
010
100
000
33
LFSR Seed
LFSRs based on primitive polynomials find use in BIST
•
•
•
Test pattern can start with any state in the non-trivial loop
Initial state: seed
LFSR re-seeding: Start from different seeds and apply a few test
patterns from each in order to shorten the test application time
Example: For the circuit below, possible stuck-at test set: (x3,x2,x1) =
{(1,0,1), (1,1,1), (1,0,0), 0,1,0)}
•
•
Choose three-stage LFSR: yi connected to xi
Testing accomplished by: two patterns starting with seed (1,0,1) and two
additional patterns starting with (1,0,0)
111
– Alternative: six clock cycles from (1,0,1) to (0,1,0) 110
101
x1
x3
x2
z
Y3
D
y3 Y2
y2
D
+
Y1
y1
D
001
011
010
100
000
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Response Analyzer
For a k-output CUT, to which n patterns have been applied by the TPG:
need to analyze kn output bits to detect error
•
•
Storing these bits and performing bit-by-bit comparison to error-free
values is expensive in space and time:
– Thus, RAs used to compress the output response
Aliasing: Signature in the presence of a fault is the same as the golden
signature
– Typically, negligible aliasing probability = 1/2m
zk-1
zk
+
bk
D
+
bk-1
z1
D
+
D
b1
Multiple-input signature register
35