Download Eng. Huda M. Dawoud

THE ISLAMIC UNIVERSITY OF GAZA
ENGINEERING FACULTY
DEPARTMENT OF COMPUTER ENGINEERING
DIGITAL LOGIC DESIGN DISCUSSION – ECOM 2012
Eng. Huda M. Dawoud
September, 2015
DIGITAL LOGIC DESIGN
1.1
ECOM 2012
ENG. HUDA M. DAWOUD
List the octal and hexadecimal numbers from 16 to 32. Using A and B for
the last two digits, list the numbers from 8 to 28 in base 12.
Answer:
Decimal
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
Octal
20 21 22 23 24 25 26 27 30 31 32 33 34 35 36 37 40
Hexadec… 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F 20
Decimal 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
25 26 27 28
Base 12
8 9 A B 10 11 12 13 14 15 16 17 18 19 1A 1B 20
21 22 23 24
1.2
What is the exact number of bytes in a system that contains (a) 32K
bytes, (b) 64M bytes, and (c) 6.4G bytes?
Answer:
(a)
(b)
(c)
1.3
32 * 210 = 32768 bytes
64 * 220 = 67108864 bytes
6.4 * 230 = 6871947674 bytes
Convert the following numbers with the indicated bases to decimal:
(a) (4310) 5
(b) (198) 12
(c) (435) 8
(d) (345) 6
Answer:
(a)
(b)
(c)
(d)
4 * 53 + 3 * 52 + 1 * 51 + 0 * 50 = (580)10
1* 122 + 9 * 121 + 8 * 120 = (260)10
4 * 82 + 3 * 81 + 5 * 80 = (285)10
3 * 62 + 4 * 61 + 5 * 60 = (137)10
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DIGITAL LOGIC DESIGN
1.4
ECOM 2012
ENG. HUDA M. DAWOUD
What is the largest binary number that can be expressed with 16 bits?
What are the equivalent decimal and hexadecimal numbers?
Answer:
The largest number of any k-digits binary number is that with all digits
being 1, such that the largest number of 16-digit binary number is
(1111 1111 1111 1111)2 = (65535)10 = (FFFF)16
To calculate the largest number of k bits we can simply use the formula
2k -1
In this question 216 -1 = 65536
1.5
Determine the base of the numbers in each case for the following
operations to be correct:
(a) 14/2 = 5
(b) 54/4 = 13
(c) 24 + 17 = 40.
Answer:
(a)
(b)
(c)
1.6
(1 * b1 + 4 * b0 1) / 2 * b0 1 = 5 * b0 1
(b + 4) / 2 = 5
b/2 + 4/2 = 5  b = 6
(5 * b1 + 4 * b0 1) / 4 * b0 1 = 1 * b1 + 3 * b0 1
(5b + 4) / 4 = 1b + 3
5b/4 + 4/4 = 1b + 3  b = 8
(2 *b + 4) + (b + 7) = 4b  b = 11
The solutions to the quadratic equation x2 - 11x + 22 = 0 are x = 3 and x
= 6. What is the base of the numbers?
Answer:
Notice that the solutions to the equation x2 - 11x + 22 in decimal system
are 8.37 and 2.62, which means that this equation is in another system,
we are asked to get the base of it.
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DIGITAL LOGIC DESIGN
ECOM 2012
ENG. HUDA M. DAWOUD
(x – 3) (x - 6) = x2 – 9x + 18
(x2 - 11x + 22)b = (x2 – 9x + 18)10
(11)b = (9)10
(1 * b1 + 1 * b0 1) = 9
b=8
1.7
Convert the hexadecimal number 64CD to binary, and then convert it
from binary to octal.
Answer:
(64CD)16 = (0110 0100 1100 1101)2
(110 010011 001 101)2 = (62315)8
1.8
Convert the decimal number 431 to binary in two ways: (a) convert
directly to binary; (b) convert first to hexadecimal and then from
hexadecimal to binary. Which method is faster?
Answer:
(a)
Integer Remainder
431/2
215
107
53
26
13
6
3
1
0
1
1
1
1
0
1
0
1
1
431 = (110101111)2
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DIGITAL LOGIC DESIGN
(b)
Integer
ECOM 2012
ENG. HUDA M. DAWOUD
Remainder
431/16
26
F
1
A
0
1
431 = (1AF)16 = (110101111)2
The second method is faster than the first one.
1.9
Convert Express the following numbers in decimal:
(a) (10110.0101) 2
(b) (16.5) 16
(c) (26.24) 8
Answer:
(a)
(b)
(c)
(10110.0101) 2 = 1 * 24 + 0 * 23 + 1 * 22 + 1 * 21 + 0 * 20 + 0 * 2-1
+ 1 * 2-2 + 0 * 2-3 + 1 * 2-4 = 22.3125
(16.5) 16 = 1 * 161 + 6 * 160 + 5 * 16-1 = 22.3125
(26.24) 8 = 2 * 81 + 6 * 80 + 2 * 8-1 + 4 * 8-2 = 22.3125
1.10 Convert the following binary numbers to hexadecimal and to decimal:
(a) 1.10010, (b) 110.010. Explain why the decimal answer in (b) is 4
times that in (a).
Answer:
(a)
(b)
1.100102 = 0001.10012 = 1.916 = 1 + 9/16 = 1.56310
110.0102 = 0110.01002 = 6.416 = 6 + 4/16 = 6.2510
Reason: 110.0102 is the same as 1.100102 shifted to the left by two places.
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DIGITAL LOGIC DESIGN
ECOM 2012
ENG. HUDA M. DAWOUD
1.11 Perform the following division in binary: 111011 ÷ 101.
Answer:
1011.11
101
111011.00
101
01001
101
1001
101
1000
101
0110
101
001
1.12 Add and multiply the following numbers without converting them to
decimal.
(a) Binary numbers 1011 and 101.
(b) Hexadecimal numbers 2E and 34.
Answer:
a) Addition
1
1001
0101
+
1110
Multiplication
1001
0101
×
1001
00000
100100
101101
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DIGITAL LOGIC DESIGN
ECOM 2012
b) Addition
ENG. HUDA M. DAWOUD
Multiplication
1
2E
34
2E
+
34
×
38 +
80
62
B8
+
2A0
600
958
1.13 Do the following conversion problems:
(a) Convert decimal 27.315 to binary.
Answer:
a)
Integer Remainder
Fraction
Integer
27/2
13
6
3
1
0
0.315*2
0.36
0.26
0.52
0.04
0.08
0.16
0.32
0
1
0
1
0
0
0
1
1
0
1
1
27.31510 = 11011.01012
1.14 Obtain the 1’s and 2’s complements of the following binary numbers:
(a) 00010000
(b) 00000000
(c) 11011010
(d) 10101010
Answer:
1st complement
2nd complement
00010000
11101111
11110000
00000000
11111111
00000000
11011010
00100101
00100110
10101010
01010101
01010110
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DIGITAL LOGIC DESIGN
ECOM 2012
ENG. HUDA M. DAWOUD
1.15 Find the 9’s and the 10’s complement of the following decimal numbers:
(a) 25,478,036
(b) 63, 325, 600
(c) 25,000,000
(d) 00,000,000.
Answer:
9’s complement
10’s complement
25478036
74521963
74521964
63325600
36674399
36674400
25000000
74999999
75000000
00000000
99999999
00000000
1.16 (a) Find the 16’s complement of C3DF.
(b) Convert C3DF to binary.
(c) Find the 2’s complement of the result in (b).
(d) Convert the answer in (c) to hexadecimal and compare with the
answer in (a).
Answer:
a)
b)
c)
d)
FFFF – C3DF + 1 = 3C21
1100 0011 1101 1111
0011 1100 0010 0001
3C21
1.17 Perform subtraction on the given unsigned numbers using the 10’s
complement of the subtrahend. Where the result should be negative,
find its 10’s complement and affix a minus sign. Verify your answers.
(a) 4,637 - 2,579
(b) 125 - 1,800
Answer:
a) 10’s complement of 2579 = 9999 - 2579 + 1 = 7421
4637 + 7421 = 12058  here, the result should be positive; we discard
the 1.
result = + 2058
b) 10’s complement of 1800 = 8200
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DIGITAL LOGIC DESIGN
ECOM 2012
ENG. HUDA M. DAWOUD
125 + 8200 = 8325  here, the result should be negative; we find its
10’s complement and affix a minus sign.
result= - 1675
1.18 Perform subtraction on the given unsigned binary numbers using the
2’s complement of the subtrahend. Where the result should be negative,
find its 2’s complement and affix a minus sign.
(a) 10011 – 10010
(b) 100010 – 100110
Answer:
a) 10011 + 01110 = 100001
Result = + 00001, as we discard 1
b) 100010 + 011010 = 111100
Result= - 000100, as the result should be negative.
1.22 Convert decimal 6,514 to both BCD and ASCII codes. For ASCII, an even
parity bit is to be appended at the left.
Answer:
6514 = ( 0110 0101 0001 0100 )BCD
= ( 0 0110110 0110101 0110001 0110100)ASCII
1.23 Represent the unsigned decimal numbers 791 and 658 in BCD, and then
show the steps necessary to form their sum.
Answer:
1 1
1
0111 1001 0001
+
0110 0101 1000
1 1 1 11 1 1
= (1449)BCD
1101 1110 1001
0110 0110
0001 0100 0100 1001
+
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