Computer Science 150 Homework Assignment 7
Spring 1997 Solutions
-
A manufacturing error results in a batch of flip-flops with cross-coupled
NOR/NAND gates as shown below:
(a) Show a primitive flow graph for the circuit.
[20]
Let F be the state bit, so put a buffer between the output of the NAND
gate and the input of the NOR gate (only one state bit is necessary because
if you cut the wire at this point, there will be no more feedback loops).
So, F* = ((A + F)' B)' = (A' F' B)'.
Here's the flow table you get from this equation:
STATE(F)
|
NS (IN=AB)
|
OUTPUT(F)
|
|
00
|
01
|
11
|
10
|
|
0
|
1
|
0
|
1
|
1
|
0
|
1
|
1
|
1
|
1
|
1
|
1
|
But, for it to be a primitve flow table, we have to expand state 1 into
4 states (B,C,D,E) and let state 0 be A:
STATE
|
NS(IN=AB)
|
Z
|
|
00
|
01
|
11
|
10
|
|
A
|
B
|
A
|
C
|
-
|
0
|
B
|
B
|
D
|
-
|
E
|
1
|
C
|
-
|
D
|
C
|
E
|
1
|
D
|
B
|
D
|
C
|
-
|
1
|
E
|
B
|
-
|
C
|
E
|
1
|
So, the primitve flow graph (which is essentially an asynchronous state
transition graph looks like:
(b) Show that this device cannot be reset via
its input terminals A and B. [20]
Once you get out of state A, you cannot get back to that state through
any sequence of inputs since there is no arrow going into state A in the
primitive flow graph from part (a).
-
A sequential system is comprised of two feedback
logic functions:
Under what input conditions will the system
oscillate? [20]
The flow table for the equation looks like:
STATE(XY)
|
NS (IN=AB)
|
|
00
|
01
|
11
|
10
|
00
|
00
|
01
|
10
|
01
|
01
|
01
|
01
|
11
|
01
|
10
|
10
|
11
|
10
|
11
|
11
|
01
|
01
|
01
|
01
|
So, it oscialltes under the conditions italicized in the table, when AB==11
and Y=1. This state can be reached from when state XY = 01 and AB
=> 11.