# Logic Design

Topics: Parity bit, Interrupt, RAID Pages: 8 (2184 words) Published: March 17, 2013
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 10.70 × 2 = (1).40
2|21 r 0 0.40 × 2 = (0).80 2|10 r 10.80 × 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|850.85 x 8 = (6).80
8|10 r 50.80 x 8 = (6).40
8|01 r 20.40 x 8 = (3).20
8|00 r 1 85₁₀ = 125₈0.20 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 used to validate the integrity of the data. Parity bits are often used in data transmission to ensure that data is not corrupted during the transfer process. For example, every 7 bits of data may include a parity bit (for a total of 8 bits, or one byte). If the data transmission protocol is set to an odd parity, each data packet must have an odd parity. If it is set to even, each packet must have an even parity. If a packet is received with the wrong parity, an error will be produced and the data will need to be retransmitted.

The parity bit for each data packet is computed before the data is transmitted. Below are examples of how a parity bit would be computed using both odd and even parity settings. Odd parity: * Initial value: 1010101 (four 1s) * Parity bit added: 1 * Transmitted value: 10101011 * Result: Odd parity (five 1s)| Even parity: * Initial value: 1010101 (four 1s) * Parity bit added: 0 * Transmitted value: 10101010 * Result: Even parity (four 1s)| The value of the parity bit depends on the initial parity of the data. For example, the binary value 10000000 has an odd parity. Therefore, a 0 would be added to keep the parity odd and a 1 would be added to give the value an even parity. While parity checking is a useful way validating data, it is not a foolproof method. For instance, the values 1010 and 1001 have the same parity. Therefore, if the value 1010 is transmitted and 1001 is received, no error will be detected. This means parity checks are not 100% reliable when validating data. Still, it is unlikely that more than one bit will be incorrect in a small packet of data. As long as only one bit is changed, an error will result. Therefore, parity checks are most reliable when using small packet sizes.

3. Differentiate among the different levels of Redundant Array of Inexpensive Disks (RAID) Mission-critical information, whether it’s your customer database or your Web server, needs unprecedented levels of reliability. One...