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EE 2730 — HW 6 solutions
1
EE 2730 — Homework 6 solutions
Fall 2008
1.
Brown and Vranesic, Problem 9.3:
Derive the minimal flow table that specifies the
same functional behavior as the flow table in Figure P9.3.
Figure P9.3.
Flow table for Problem 9.3.
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EE 2730 — HW 6 solutions
3
2.
Brown and Vranesic, Problem 9.19:
Example 9.6 describes a simple arbiter for two
devices contending for a shared resource.
Design a similar arbiter for three devices that use a
shared resource.
In case of simultaneous requests, namely, if one device has been granted access
to the shared resource and before it releases its request the other two devices make requests of
their own, let the priority of the devices be Device 1 > Device 2 > Device 3.
Carry out the design through the excitation table, but you need not derive the nextstate
and output expressions or draw the circuit.
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This gives the following flow table.
A state name in parentheses indicates a stable state
(rather than circling as in the text).
next state, for
r
3
r
2
r
1
=
present
state
000 001 010 011 100 101 110 111
output
g
3
g
2
g
1
A
(A)
B
C
B
D
B
C
B
000
B
A
(B)
C
(B)
D
(B)
C
(B)
001
C
A
B
(C)
(C)
D
B
(C)
(C)
010
D
A
B
C
B
(D)
(D)
(D)
(D)
100
By making unreachable transitions into unspecified states, we obtain the following
modified flow table.
next state, for
r
3
r
2
r
1
=
present
state
000 001 010 011 100 101 110 111
output
g
3
g
2
g
1
A
(A)
B
C
—
D
—
—
—
000
B
A
(B)
C
(B)
D
(B)
C
(B)
001
C
A
B
(C)
(C)
D
B
(C)
(C)
010
D
A
B
C
B
(D)
(D)
(D)
(D)
100
No state reduction is possible for this circuit, so the next step is to work on state
assignment.
Labeling the stable states in rowmajor order and labeling the others accordingly
results in the following relabeled flow table.
next state, for
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This note was uploaded on 06/20/2011 for the course EE 2730 taught by Professor Desouza during the Spring '08 term at LSU.
 Spring '08
 DeSouza

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