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Unit 10: Clocked Sequential Circuits
Since, these are not in the same block in P2 then at least one of the states in (CEFG) is Notes
not equivalent to the others.
State F must be different from C, E and G because its 1- successor, D, is in a different
block than E, C, and G.
• Therefore, P = (ABD)(CEG)(F).
3
• At this point, we know that state F is unique since it is in a block by itself.
• P = (ABD)(CEG)(F).
3
• The process repeats yielding the following:
0-successors of (ABD) are (BDB).
A, B and D are still considered equivalent.
1-successors of (ABD) are (CFG), which are not in the same block,
B cannot be equivalent to A and D since F is in a different block than C and G.
The 0- and 1-successors of (CEG) are (FFF) and (ECG)
C, E, and G must still be considered equivalent.
• Thus, P = (AD)(B)(CEG)(F).
4
• If we repeat the process to check the 0- and 1-successors of the blocks (AD) and (CEG), we
find that
P = (AD)(B)(CEG)(F).
5
• Since P = P and no new blocks are generated, it follows that states in each block are
4
5
equivalent.
A and D are equivalent.
C, E, and G are equivalent.
• The state table can be rewritten, removing the rows for D, E and G and replacing all
occurrences of D with A and all occurrences of E or G with C.
• The resulting state table is as follows:
Table 10.10: Resulting State Table
Next state Output
Present state
w = 0, w = 1 Z
A B C 1
B A F 1
C F C 0
F C A 0
Always clock has low rise/fall times, then both pMOS and nMOS may conduct.
10.5 State Assignment
A 2n-state finite state machine has n!-different possible ways to assign state variable values to
states. Thus, a 16-state finite state machine (which requires at least 4 state variables) has 24 (4!)
different assignment of states to state variable values. A 1024-state finite state machine (which
requires at least 10 state variables) has over 3 million (10!) state assignments. If the state is not
densely encoded in the fewest number of bits, even more encodings are possible!
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