Soal-soal dan Jawaban Rangkaian Logika
1.a). Hitunglah :
(i). (8AD)16 = (…….)10 = (…….)2 = (…….)8
(ii).(3AE)16 = (…….)8 = (…….)10= (…….)2
b). Jumlahkan dalam bentuk BCD :
(i). 25 + 32 = ……. (ii). 123 + 382 = …….
c). Bagilah :
(i). 110111 : 10011 = ……. (ii). 10011 : 100 = …….
d). Kurangkanlah dalam bentuk komplemen –1 dan –2 :
(i). 00,111 – 0,1001 (ii). 01011 – 101
e). Pada soal (a). hitunglah (ii)-(I) menggunakan komplemen –1 dan
–2.
Jawab :
a). (I). (8AD)16 = (1000 1010 1101)2
= (4 2 5 5)8
= (1 x 210) + (1 x 27) + (1 x 25) + (1 x
23) + (1 x 22) + (1 x 20)
= (1 + 4 + 8 + 32 + 128 + 1024)
= (1197)10
(ii). (3AE)16 = (0011 1010 1110)2
= (1 6 5 6)8
= (1 x 29) + (1 x 28) + (1 x 27) + (1 x 25) + (1 x
23) + (1 x 22) + (1 x 21)
= (512 + 256 + 128 + 32 + 8 + 4 + 1)10
= (768 + 160 + 13)10
= (941)10
b). (I). 25 + 32 = 57 (ii). 123 + 382 = 505
0010 0101 0001 0010
0011
0011 0010 + 0011 1000
0010 +
0101 0111 0100 1010
0101
5 7 1
+0101 + 5
0101
1111
5
1 +
0000
0
c). (I). 11,100…
100,11…
10011 110111 100 10011
10011
10000
10001
11
10011
100
10
10
0
10
d). (I). Komplemen-1 : Komplemen-
2 :
00,111 0,111
0,111 0,111
0,1001 _ 0,0110 +
0,1001 _ 0,0111 +
1,0100
1,0101
1. +
diabaikan
0,0101
hasilnya 0,0101
(ii). Komplemen-1 :
Komplemen-2 :
01011 01011
01011
101 _ 11010 +
11011 +
100101
100110
1. + diabaikan
110. hasilnya : 110
e).
001110101110 Kompl-1
100010101101 _ 001110101110
01110 101 0010
+
101100000000 (btk
kompl-1)
- 010011111111 (min)
Kompl-2
001110101110 001110101110
100010101101 _ 011101010011 +
(kompl-2)
101100000001 (dlm btk kompl-2)
Kompl-2 = Kompl-1
1. +
101100000001 = Kompl-1 101100000000
1. +
010011111111
2. Suatu persamaan Aljabar Boole, yang mempunyai 4 karakter masukan masing-
masing w,x,y dan z, yang dicirikan oleh fungsi :
P = (m (1,3,5,7,9) + (d (4,11,12,13,14,15)
Carilah format rangkaian logika yang sesederhana mungkin, sehingga
diperoleh rangkaian elektronika yang semurah mungkin dalam
implementasinya.
Jawab :
P = (m (1,2,5,7,9) + (d (4,11,12,13,14,15)
" "W "X "Y "Z "P "
"0 "0 "0 "0 "0 "0 "
"1 "1 "0 "0 "0 "1 "
"2 "0 "1 "0 "0 "0 "
"3 "1 "1 "0 "0 "1 "
"4 "0 "0 "1 "0 "X "
"5 "1 "0 "1 "0 "1 "
"6 "0 "1 "1 "0 "0 "
"7 "1 "1 "1 "0 "1 "
"8 "0 "0 "0 "1 "0 "
"9 "1 "0 "0 "1 "1 "
"10 "0 "1 "0 "1 "0 "
"11 "1 "1 "0 "1 "X "
"12 "0 "0 "1 "1 "X "
"13 "1 "0 "1 "1 "X "
"14 "0 "1 "1 "1 "X "
"15 "1 "1 "1 "1 "X "
" " " " " "
" "0 "0 "1 "1 "
" "0 "0 "X "1 "
" "X "X "X "X "
" "X "0 "1 "1 "
Pilihan-1 P = W
" " " " " "
" "0 "0 "1 "1 "
" "0 "0 "X "1 "
" "X "X "X "X "
" "X "0 "1 "1 "
Pilihan-2 P =
" " " " " "
" "0 "0 "1 "1 "
" "0 "0 "X "1 "
" "X "X "X "X "
" "X "0 "1 "1 "
Pilihan-3 P =
" " " " " "
" "0 "0 "1 "1 "
" "0 "0 "X "1 "
" "X "X "X "X "
" "X "0 "1 "1 "
Pilihan-4 P = W + (YZ)
Pilihan-4 ternyata lebih sederhana daripada pilihan-2 dan pilihan-3 dan
mudah untuk di-implementasikan.
P = W + YZ
3. Suatu sistem Elektronika Digital mempunyai karakter sebagai berikut :
X
A
Input
Output
Y
B
Z
Digunakan sebagai bagian perangkat Automation. Output-A akan bernilai
logika-1 jika keadaan masukan adalah x y z + x y z + x y z + x y z
+ x y z , sedangkan Output-B akan bernilai logika-1, jika keadaan
inputnya masing-masing adalah : x y z + x y z + x y z + x y z.
a. Tentukan masing-masing keadaan output yang paling menguntungkan, dengan
cara:
i. Sum of pruduct (SOP) dan Product of Sum (POS)
ii. Karnaugh Map.
b. Implementasikan Sistem Elektronika Digital yang anda desain
.
Jawab :
X
A
Input
Output
Y
B
Z
ii. Karnaugh Map
Output-A
" " " " " "
" "1 "1 "0 "1 "
" "1 "1 "0 "0 "
A =
Output-B
" " " " " "
" "1 "1 "0 "1 "
" "1 "1 "0 "0 "
B = +
b).
X
A = X + (YZ)
Input
Output
Y
B = YZ + YZ
Z
b). Implementasi Rangkaian Logika
Tabel Kebenaran
"Z "Y "X "A "B "
"0 "0 "0 "1 "1 "
"0 "0 "1 "1 "0 "
"0 "1 "0 "1 "0 "
"0 "1 "1 "1 "1 "
"1 "0 "0 "0 "1 "
"1 "0 "1 "0 "0 "
"1 "1 "0 "0 "0 "
"1 "1 "1 "0 "1 "
4. Desain suatu counter sinkron yang mencacah dari 0000 – 1011.
" "D "C "B "A "
"0 "0 "0 "0 "0 "
"1 "0 "0 "0 "1 "
"2 "0 "0 "1 "0 "
"3 "0 "0 "1 "1 "
"4 "0 "1 "0 "0 "
"5 "0 "1 "0 "1 "
"6 "0 "1 "1 "0 "
"7 "0 "1 "1 "1 "
"8 "1 "0 "0 "0 "
"9 "1 "0 "0 "1 "
"10 "1 "0 "1 "0 "
"11 "1 "0 "1 "1 "
"12 "0 "0 "0 "0 "
TRUTH TABLE J – K FLIP-FLOP
"Qn "Qn+1 "J "K "
"0 "0 "0 "X "
"0 "1 "1 "X "
"1 "0 "X "1 "
"1 "1 "X "0 "
" " " " " "
" "0 "2 "3 "1 "
" "8 "10 "11 "9 "
" "X "X "X "X "
" "4 "6 "7 "5 "
a). Fungsi input J dan K untuk flip-flop D
" " " " " "
" "0 "0 "0 "0 "
" "X "X "X "X "
" "X "X "X "X "
" "0 "0 "1 "0 "
JD = ABC
" " " " " "
" "X "X "X "X "
" "0 "0 "1 "0 "
" "X "X "X "X "
" "X "X "X "X "
KD = AB
b). Fungsi input J dan K untuk flip-flop C
" " " " " "
" "0 "0 "1 "0 "
" "0 "0 "0 "0 "
" "X "X "X "X "
" "X "X "X "X "
JC =
" " " " " "
" "X "X "X "X "
" "X "X "X "X "
" "X "X "X "X "
" "0 "0 "1 "0 "
KC = AB
c). Fungsi input J dan K untuk flip-flop B
" " " " " "
" "0 "X "X "1 "
" "0 "X "X "1 "
" "X "X "X "X "
" "0 "X "X "1 "
JB = A
" " " " " "
" "X "0 "1 "X "
" "X "0 "1 "X "
" "X "X "X "X "
" "X "0 "1 "X "
KB = A
d). Fungsi input J dan K untuk Flip-flop A
" " " " " "
" "1 "1 "X "X "
" "1 "1 "X "X "
" "X "X "X "X "
" "1 "1 "X "X "
JA =
" " " " " "
" "X "X "1 "1 "
" "X "X "1 "1 "
" "X "X "X "X "
" "X "X "1 "1 "
KA = A
Sehingga Implementasi rangkaian lengkapnya adalah :
5. Desain suatu pencaah sinkron yang akan mencacah dari 0011(desimal 3)
hingga 1011(desimal 11). Dan setelah itu kembali lagi ke 0011.
" "D "C "B "A "
"0 "0 "0 "0 "0 "
"1 "0 "0 "0 "1 "
"2 "0 "0 "1 "0 "
"3 "0 "0 "1 "1 "
"4 "0 "1 "0 "0 "
"5 "0 "1 "0 "1 "
"6 "0 "1 "1 "0 "
"7 "0 "1 "1 "1 "
"8 "1 "0 "0 "0 "
"9 "1 "0 "0 "1 "
"10 "1 "0 "1 "0 "
"11 "1 "0 "1 "1 "
"12 "0 "0 "0 "0 "
" " " " " "
" "0 "2 "3 "1 "
" "8 "10 "11 "9 "
" "X "X "X "X "
" "4 "6 "7 "5 "
"Qn "Qn+1 "J "K "
"0 "0 "0 "X "
"0 "1 "1 "X "
"1 "0 "X "1 "
"1 "1 "X "0 "
Flip-flop A :
" " " " " "
" "X "X "X "X "
" "1 "1 "X "X "
" "X "X "X "X "
" "1 "1 "X "X "
JA = 1
" " " " " "
" "X "X "X "X "
" "X "X "1 "1 "
" "X "X "X "X "
" "X "X "1 "1 "
KA = 1
Flip-flop B :
" " " " " "
" "X "X "X "X "
" "0 "X "X "X "
" "X "X "X "X "
" "0 "X "X "1 "
JB = A
" " " " " "
" "X "X "1 "X "
" "X "0 "1 "1 "
" "X "X "X "X "
" "X "0 "1 "X "
KB = A
Flip-flop C :
" " " " " "
" "X "X "1 "X "
" "0 "0 "0 "0 "
" "X "X "X "X "
" "1 "1 "X "X "
JC =
" " " " " "
" "X "X "X "X "
" "X "X "X "X "
" "X "X "X "X "
" "0 "0 "1 "0 "
KC = AB
Flip-flop D:
" " " " " "
" "X "X "0 "X "
" "X "X "X "X "
" "X "X "X "X "
" "0 "0 "1 "0 "
JD = ABC
" " " " " "
" "X "X "X "X "
" "0 "0 "1 "0 "
" "X "X "X "X "
" "X "X "X "X "
KD = AB
-----------------------
Sistem
Elektronika
Digital
Sistem
Elektronika
Digital
Sistem
Elektronika
Digital
