Insteadf2heref3adcanbe f1 f1 b b used f f d d f3 f3 a

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Unformatted text preview: ircuit
for
minimal
two
level
OR‐AND
implementation
 c
 F’=bd’+ab’d
⇒
F=(b’+d)(a’+b+d’)
 cd
 F’:
 00
 01
 11
 10
 b'
 ab
 F
 d
 


 

 
x 
 00
 
x
 
 
 
 a’
 x
 

 1

 01
 
x
 b

 b
 
 d’
 
x 
 x
 11
 
1
 
 

 
 
 a
 
 
 
 1
 
1
 x

 10
 

 d
 
 c) Draw
the
circuit
for
minimal
implementation
using
one
EX‐OR
gate
(no
restriction
on
the
 number
of
inputs)
and
other
gates
necessary
 c
 c
 c
 F1:
 cd
 F2:
 F1+F2:
 cd
 cd
 00
 01
 11
 10
 00
 01
 11
 10
 00
 01
 11
 10
 ab
 ab
 ab
 
 

 
1
 00
 
1
 1
 1
 
1
 00
 
1
 1

 1
 1
 00
 1
 
 
 
 
 
 
 
1
 1
 

 0...
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This test prep was uploaded on 04/06/2014 for the course EEE 348 taught by Professor Gozdeakar during the Summer '11 term at Middle East Technical University.

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