Logic Design
LOGIC DESIGN ASSIGNMENT 1. A.
B. To convert 85.85 from its decimal form to binary.
2|85 0.85 × 2 = (1). 70
2|42 r 1 0.70 × 2 = (1).40
2|21 r 0 0.40 × 2 = (0).80
2|10 r 1 0.80 × 2 = (1).60
2|5 r 0 0.85₁₀ = .1101₂
2|2 r 1
2|1 r 0 =1010101₂ THEREFORE 85.85₁₀ = 1010101.1101₂
2|0 r 1
To convert 85.85₁₀ to Octal
8|85 0.85 x 8 = (6).80
8|10 r 5 0.80 x 8 = (6).40
8|01 r 2 0.40 x 8 = (3).20
8|00 r 1 85₁₀ = 125₈ 0.20 x 8 = (1).60 .85₁₀ = .6631₈
THEREFORE 85.85₁₀ = 125.6631₈
To convert 105.15₁₀ to Binary
2|105 .15 x 2 = (0).30
2|52 r 1 .30 x 2 = (0).60
2|26 r 0 .60 x 2 = (1).20
2|13 r 0 .20 x 2 = (0).40 .15₁₀= .0010₂
2|06 r 1
2|03 r 0
2|01 r 1
2|00 r 1 105₁₀ = 1101001₂
THEREFORE 105.15₂ = 1101001.0010₂
To convert 105.15₁₀ to Octal
8|105 .15 x 8 = (1).20
8|13 r 1 .20 x 8 = (1).60
8|01 r 5 .60 x 8 = (4). 80 .15₁₀ = 1146₈
8|00 r 1 .80 x 8 = (6).40
105₁₀= 151₈ THEREFORE 105.15₁₀ = 151.1146₈
C. Arithmetic operations using the 2’s complement notation.
67 + (-67)
67₁₀ = 01000011₂
-67₁₀ =10111100₂ +00000001₂ = 10111101₂
01000011₂ + 10111101₂ = 00000000₂ (67) + (-67) = 0
87- 5
87 = 01010111₂
5 = 0000101₂
-5 = 11111010₂ + 00000001₂ = 11111011₂
87- 5 =01010111₂ + 11111011₂ = 01010010₂
-37-(40)
37 = 00100101₂
-37 =11011010₂ + 00000001₂ = 11011011₂
40 = 00101000₂
-37-(40) = 11011011₂ – 00101000₂ = 1011001₂
25-(-25)
25 =00011001₂
-25 = 11100110₂ + 00000001₂ =11100111₂
25-(-25) = 00011001₂ – 11100111₂ = 10110010₂
2. Why is parity bit required? How is it used in correcting single bit errors? Explain with the help of an example.
A parity bit is a bit, with a value of 0 or 1, which is added to a block of data for error detection purposes. It gives the data either an odd or even parity, which is
B. To convert 85.85 from its decimal form to binary.
2|85 0.85 × 2 = (1). 70
2|42 r 1 0.70 × 2 = (1).40
2|21 r 0 0.40 × 2 = (0).80
2|10 r 1 0.80 × 2 = (1).60
2|5 r 0 0.85₁₀ = .1101₂
2|2 r 1
2|1 r 0 =1010101₂ THEREFORE 85.85₁₀ = 1010101.1101₂
2|0 r 1
To convert 85.85₁₀ to Octal
8|85 0.85 x 8 = (6).80
8|10 r 5 0.80 x 8 = (6).40
8|01 r 2 0.40 x 8 = (3).20
8|00 r 1 85₁₀ = 125₈ 0.20 x 8 = (1).60 .85₁₀ = .6631₈
THEREFORE 85.85₁₀ = 125.6631₈
To convert 105.15₁₀ to Binary
2|105 .15 x 2 = (0).30
2|52 r 1 .30 x 2 = (0).60
2|26 r 0 .60 x 2 = (1).20
2|13 r 0 .20 x 2 = (0).40 .15₁₀= .0010₂
2|06 r 1
2|03 r 0
2|01 r 1
2|00 r 1 105₁₀ = 1101001₂
THEREFORE 105.15₂ = 1101001.0010₂
To convert 105.15₁₀ to Octal
8|105 .15 x 8 = (1).20
8|13 r 1 .20 x 8 = (1).60
8|01 r 5 .60 x 8 = (4). 80 .15₁₀ = 1146₈
8|00 r 1 .80 x 8 = (6).40
105₁₀= 151₈ THEREFORE 105.15₁₀ = 151.1146₈
C. Arithmetic operations using the 2’s complement notation.
67 + (-67)
67₁₀ = 01000011₂
-67₁₀ =10111100₂ +00000001₂ = 10111101₂
01000011₂ + 10111101₂ = 00000000₂ (67) + (-67) = 0
87- 5
87 = 01010111₂
5 = 0000101₂
-5 = 11111010₂ + 00000001₂ = 11111011₂
87- 5 =01010111₂ + 11111011₂ = 01010010₂
-37-(40)
37 = 00100101₂
-37 =11011010₂ + 00000001₂ = 11011011₂
40 = 00101000₂
-37-(40) = 11011011₂ – 00101000₂ = 1011001₂
25-(-25)
25 =00011001₂
-25 = 11100110₂ + 00000001₂ =11100111₂
25-(-25) = 00011001₂ – 11100111₂ = 10110010₂
2. Why is parity bit required? How is it used in correcting single bit errors? Explain with the help of an example.
A parity bit is a bit, with a value of 0 or 1, which is added to a block of data for error detection purposes. It gives the data either an odd or even parity, which is
References: 1. http://www.facstaff.bucknell.edu/mastascu/elessonshtml/logic/logic3.html 2. http://courseware.ee.calpoly.edu/~rsandige/KarnaughExplorer.html 3. http://www.ee.surrey.ac.uk/Projects/Labview/minimisation/karnaugh.html 4. http://k-map.sourceforge.net/ 5. http://cse4k12.org/binary/convert_bin2oct.html 6. http://www.robotroom.com/NumberSystems.html 7. http://academic.evergreen.edu/projects/biophysics/technotes/program/2s_comp.htm 8. http://www.cs.cornell.edu/~tomf/notes/cps104/twoscomp.html