CS E6831 Sequential Logic Circuits 04f

Midterm Exam Solutions

 

1. Constraints to be satisfied are:

 

Long-path-1 P>DLM+D1DQM+D2DQM (1)

 

Long-path-2 P>DLM+D2CQM+T2L+T1T+S1-V (2)

 

Short-path DLm<DLmB=V+T2L+T1T+H1-DCQm (3)

 

From (1) we obtain P>12+2+3=17 (4)

 

From (2) we obtain P>12+3+1+2+2-V=20-V (5)

 

From (3) we obtain DLmB=V+2+2+2-2=V+4 (6)

 

Since the maximum acceptable value of DLmB was given as 6, we find from (6) that, since V=DLmB-4, V cannot be allowed to exceed 2. With this value (5) gives us P>20-2=18. This satisfies (4), and so we have our solution. P=18, V=2.

 

2. Since we are told that the first long-path constraint--(1) in above solution--is satisfied by a substantial margin, we know that V has been made as large as permitted by the constraint on DLmB. (If it could have been made larger, then P could have been reduced until (1) was marginally satisfied.) Since DCQm has been reduced by x, it is evident from the short-path constraint--(3) above--that we have to decrease V by x in order to avoid increasing DLmB. Now, in the second long-path constraint--(2) above--the effect of the reduction in D2CQM is cancelled by the reduction of V, and so P cannot be changed at all. So the answer is that P cannot be reduced and V must be reduced by x.

 

3. (a) Our algorithm shows that state-1 is reachable only from states 2 and 4, and so that table is NOT strongly connected. Table-b IS strongly connected.

 

4. (a) The implications from 56 are 12 and 34, so the answer is (12, 34, 56).

 

(b) The implications from 13 are 26, 35, 24, 46, 15. This indicates that (135, 246) is closed. Inspection of the flow table shows that, for every state, the NS entries for both inputs are either in 135 or in 246. So this partition is also input consistent.

 

5. (a) For every flow table, the trivial partition I is closed. Since, for EVERY partition, q, q<I, and since not every partition is closed, it follows that p being closed and q<p does NOT imply that q is closed.

 

(b) If p and q are closed, then pq (glb of p and q) and p+q (lub of pa and q) are also closed. So it follows that (123, 45, 678)(135, 24, 678)=(13, 2, 4, 5, 678) is closed, and

(123, 45, 678)+(135, 24, 678)=(12345, 678) is closed.

 

6. (a) The product of the partitions or covers corresponding to the component machines must be 0. This means that there cannot be any pair of states that are in common blocks of all the blocks or partitions. The only listed candidate which, when multiplied by (123, 345) yields 0 is (14, 25, 3).

 

(b) The block diagram shows M1 fed only by X, M2 fed by X and M1, and a logic block, generating the output, fed by X, M1 and M2.

 

(c) The flow table for M1 is:

X

0 1

-----

(123) 1 2 1

(456) 2 1 1

----

7. p1 is found by finding the implications of 12 and 45, which are 13, 24, 12. These result in the partition (1234, 5).

 

P2 can be found by considering the flow table columns separately. For X=0, the next-state blocks of (15, 234) map into the blocks of (12, 3, 45). For X=1, the next-state blocks of (1234, 5) map into the blocks of (12, 3, 45). So the product, (15, 234)(1234, 5)=(1, 234, 5) is a partition that, for BOTH input values, maps next-state blocks into (12, 3, 45). This is the desired value of p2.

 

8. Suppose the state assignments for rows i and j are made adjacent (i.e., differ in only one y-variable). Then, if there are k y-variables in the assignment, there will be k K-map adjacencies associated with a common NS entry in rows i and j for some input column. So, in this example, where k=3, we add 3 points each to score for the column-00 and column-11 common entries of 3 and 5, respectively. The number of points for the 1-2 pair in column-10 is k-1, or 2. For each i, j pair of NS entries in adjacent columns of any row, we also add 2 points, and there are two of these in row-3. Finally, for each column in which the outputs are the same in rows i and j, we add one point (there are pairs of this type in columns 00 and 01. This gives us a total score for 1-2 of 14.

 

9. Logic expressions for M1 are:

Y1=Ay1+CyÝ2+BVÝ4VÝ5y1

Y2=AyÝ1+ByÝ1yÝ2+BV4y1

V1=yÝ1yÝ2

 

The flow table for the original machine is:

 

A B C

--------

1 2 2 4

2 2 1 1

3 4 3 5

4 3 2 5

5 3 1 4

--------

 

10. 1234=> {1568, 1479, 278}. A 3-member compatible cannot imply a 4-member compatible, so we know that 123 cannot imply 1568. Since 569 is not a subset of any compatible implied by 1234, a subset of 1234 cannot imply 569. We can't sure that any of the other members of the given set are not implied by 123.