00 00 01 11 10 01 11 10 x x 1 1 1 1 0 0 x 0 0 1 1 1 1

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Unformatted text preview: mber
the
previous
input
 
 After
state
reduction
 State
reduction
 
 PS 01/1 01/1 =
 ≠
 ≠
 =
 ≠
 =
 ≠
 0
 1
 1
 0
 1
 0
 1
 
 
 10/1,11/0 
S 
 NS 01 10 11 00 S1 S S 0 S1 S S 0 S1 S S 1 S≡S0 Y 01 10 10 10 01 01/1 11 1 1 1 00/0 10/0,11/1 S 1
 S 0
 00/1, 10/1,11/0 01/0 
 Q4)(10
pts)

 
 For
the
circuit
given
left,
complete
the
state
transition
 diagram
where
state
order
is
QAQB.
 
 
 0
 Y
 J
 Clk
 
 Q A
 C
 
 X
 K
 Q’A
 
 T
 Q B
 00/0
 0,1
 C
 
 Q’B
 
 JA=QA+QB,
KA=TB=QA+x,
Y=QA+QB
 
 PS Q AQ B 00 01 10 11 
 FF inputs X=0 X=1 J AK A T B J AK A T B 00 0 01 1 10 0 11 1 11 1 11 1 11 1 11 1 NS Y X=0 X=1 00 11 01 00 01 10 01 00 0 1 1 1 
 11/1
 
 

 
 
 
 
 
 
 
 1
 01/1
 1
 0
 10/1
 0,1
 
 
 Q5)(15
pts)
 You
are
given
the
characteristic
 table
for
AB
FF.
 AB
 Q(t+1)
 
 
 00
 0
 01
 Q(t)
 10
 Q'(t)
 11
 1
 a)
Find
out
characteristic
equation
 
 
 
 A
 Q(t+1):
 
 
 0
 0

 1
 1
 Q(t+1)=Q’A+QB
 0 1
 
 1
 0
 Q
 
 B
 


 
 c)
Implement
AB
FF
using
D
FF
 A
 D:
 QAB
 D
 000
 0
 0

 1
 1
 
 0
 001
 0
 0
 1
 
 1
 0
 Q
 010
 1
 011
 1
 B
 
 100
 0
 



















 101
 1
 D=AQ’+BQ
...
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