Contoh Soal Sistem Digital Beserta Penyelesaiannya.docx 5e671x
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CONTOH SOAL SISTEM DIGITAL BESERTA PENYELESAIANNYA
TUGAS 3 SOAL. 1
Buatlah table dan gambarkan gerbang AND
2.
3A
3.
A341
4.
25
(10)
=…
(2)
5.
73
(10)
=…
(8)
6.
BC
(16)
7.
1011
8.
10100110 AND 01110101
=…
9.
11010110 NAND AB
=…
(2)
=…
(2)
(16)
10. 7456
(16)
(10)
=…
(10)
= … (2)
(2)
(16)
11. 101,11
=…
+ 0101
+ 7878
=…
(2)
(16)
(16)
(2)
=…
(2)
(2)
(16)
JAWABAN. A
B
OUTPUT
0
1
0
1
0
0
1
1
1
0
0
0
2.
3A
(16)
= 58
(10)
1.
TABEL AND
Gambar AND
= (3 X 161 ) + ( 10 X 160) = 48 + 10 = 58 3.
A341
= 41793
(16)
(10)
= (10 X 163) + ( 3 X 162) + (4 X 161) + (1 X 160) = 40960 + 768 + 64 + 1 = 41793 4.
25
= 11001
(10)
(2)
2 25 = 1 2 12 = 0 2 6 =0 2 5.
73
3=1 = 111
(10)
(8)
8 73 = 1 89 =1 1 6.
7.
BC
= 10111100
(16)
(2)
8
4
2
1
B= 11
1
0
1
1
C= 12
1
1
0
0
BC
101111 00
1011
(2)
+ 0101
(2)
= 10000
(2)
1011 0101 + 10000 8.
10100110 AND 01110101 = 00100100
10100110
(2)
01110101 AND 00100100 9.
11010110 NAND AB
(16)
= 01111101
(2)
11010110 10101011 AND 10000010 NOT 01111101 10. 7456
(16)
+ 7878 (16) = ECCE
(16)
7456 7878 + ECCE 11.
101,11
(2)
= 5,75
(10)
= (1X 22) + (1 X 20) + (1 X 2-1) + (1 X 2-2) = 4 + 1 + 0,5 + 0,25 = 5,75
TUGAS 3 SOAL. 1
Buatlah table dan gambarkan gerbang AND
2.
3A
3.
A341
4.
25
(10)
=…
(2)
5.
73
(10)
=…
(8)
6.
BC
(16)
7.
1011
8.
10100110 AND 01110101
=…
9.
11010110 NAND AB
=…
(2)
=…
(2)
(16)
10. 7456
(16)
(10)
=…
(10)
= … (2)
(2)
(16)
11. 101,11
=…
+ 0101
+ 7878
=…
(2)
(16)
(16)
(2)
=…
(2)
(2)
(16)
JAWABAN. A
B
OUTPUT
0
1
0
1
0
0
1
1
1
0
0
0
2.
3A
(16)
= 58
(10)
1.
TABEL AND
Gambar AND
= (3 X 161 ) + ( 10 X 160) = 48 + 10 = 58 3.
A341
= 41793
(16)
(10)
= (10 X 163) + ( 3 X 162) + (4 X 161) + (1 X 160) = 40960 + 768 + 64 + 1 = 41793 4.
25
= 11001
(10)
(2)
2 25 = 1 2 12 = 0 2 6 =0 2 5.
73
3=1 = 111
(10)
(8)
8 73 = 1 89 =1 1 6.
7.
BC
= 10111100
(16)
(2)
8
4
2
1
B= 11
1
0
1
1
C= 12
1
1
0
0
BC
101111 00
1011
(2)
+ 0101
(2)
= 10000
(2)
1011 0101 + 10000 8.
10100110 AND 01110101 = 00100100
10100110
(2)
01110101 AND 00100100 9.
11010110 NAND AB
(16)
= 01111101
(2)
11010110 10101011 AND 10000010 NOT 01111101 10. 7456
(16)
+ 7878 (16) = ECCE
(16)
7456 7878 + ECCE 11.
101,11
(2)
= 5,75
(10)
= (1X 22) + (1 X 20) + (1 X 2-1) + (1 X 2-2) = 4 + 1 + 0,5 + 0,25 = 5,75
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