Tugas Rangkaian Digital Bab III
Soal 3.
Sederhanakan : A + A . B’+ A’. B =
Penyelesaian : A + A . B’+ A’. B = (A + A . B’) + A’. B
=
A + A’. B (Hukum absorbtif
type a/5a)
= A
+ B (Hukum absorbtif type c/5c)
Soal
4
Bila A=1, B=0, C=1, dan D=0 maka berapakah nilai Q =.........?
Bila A=1, B=0, C=1, dan D=0 maka berapakah nilai Q =.........?
Q = ( A + B” + C” ) ( A” . D” + B” )
Q = A + ( B” . C” ) ( ( A” + D” ) + B” ) ( Hukum Van De Morgan )
Q = ( A’ + B . C ) ( ( A + D ) + D )
Q = ( 0 + 0 ) ( ( 1 + 0 ) + 0 )
= ( 0 ) ( 0 + 1 )
= ( 0 ) ( 1 )
= 0
Q = A + ( B” . C” ) ( ( A” + D” ) + B” ) ( Hukum Van De Morgan )
Q = ( A’ + B . C ) ( ( A + D ) + D )
Q = ( 0 + 0 ) ( ( 1 + 0 ) + 0 )
= ( 0 ) ( 0 + 1 )
= ( 0 ) ( 1 )
= 0
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