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 ( H ukum 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 =.........? 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