Questions 1 through 7 measures Outcome A and questions 8 through 12 measure outc

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Questions 1 through 7 measures Outcome A and questions 8 through 12 measure outcome C NOTE: Please write neatly and number all the answers in the right order of question. utcome C 1. How many bits are needed to count up to a maximum of 511? (2 points) The computer in mars has a fivenary memory-that is each location can hold a frit value, with a frit holding 0, 1, 2, 3, or 4. How many frits are needed to hold a 6-octal_bit number. Give an expression for the number of frits to hold n octal_bits. (6 points) 2· Suppose that the decimal integer values from 0 to 15 are to be transmitted in binary (2 points) a. How many lines will be needed if parallel representation is used? b. How many lines will be needed if serial representation is used? 3. . Represent the decimal value 37 in each of the following ways. (4 points) a, Straight binary b.hex c, octal d' BCD e.ASCIIf.Base 7 In an audio CD, the audio voltage is typically sampled about 44,000 times per second, and the value of each sample is recorded on the CD surface as a binary number. In other words, each recorded binary number represents a single voltage point on the audio signal waveform. (6 points) a. If the binary numbers are six bits in length, how many different voltage values can be represented by a single binary number? Repeat for eight bits and ten bits. b. If ten-bit numbers are used, how many bits will be recorded on the CD in one second. c. If a CD can typically store 5 billion bits, how many seconds of audio can be recorded when ten-bit numbers are used? Represent the statement "X-25/Y" in ASClI code (excluding quotes). Attach an odd parity bit. points)

Explanation / Answer

(1)

We are required to count from 0 to 511, so total 512 discrete values are to be counted.

If number of bits required are n, then it can represent 2n values.

So, here, 512 = 2n

So, n = 9

Thus, it requires 9 bits to count upto 511.

(2)

Octal bit has 8 states from 0 to 7.

Fivenary bit has 5 states from 0 to 4.

Octal number with n bits can represent upto 8n numbers.

So, 6 bit octal number can hold max 86 = 262144.

For, fivenary number, n bit can represent 5n numbers.

So, 262144 <= 5n

5^7 = 78125 and 5^8 = 390625.

So, minimum 8 fivenary bits required.

(OR 262144 = 5n => n = log5(262144) = ln(262144)/ln(5) = 7.75 So, n = 8.)

Formula: fivenary bit required = log5(8n)

(3)

010 = 02 and 1510 = 11112 . So,minimum 4 bits required to represent from 0 to 15.

For Parallel representation, 4 lines required. (Each line transmits 1 bit simultaneously.)

For Series representation, 1 line required. (One line transmits all bits one after one.)

(4)

3710

(a)in binary: 1001012

(b)hex: 2516

(c)octal: 4510

(d)BCD(Binary Coded Decimal): 0011 0111

(e)ASCII: in ASCII 3 = 5110 = 001100112 and 7 = 5510 = 001101112 So, 37 = 00110011 00110111

(f)Base 7: 527

(5)

samples = 44000 per second

(a) n bit numbers can represent 2n voltages.

So, 6 bit represents 26 = 64

8 bit represents 28 = 256

10 bit represents 210 = 1024

(b) If 10 bit numbers used,

data recorded in one second = 10*44000 = 440000 bits per second.

(c) seconds = total bits / bits per sencond

So, seconds = 5000000000/ 440000 = 11363.6363 seconds.

(6)

X=25/Y

in ASCII codes(in base 10) : 88 61 50 53 47 89

in ASCII codes(in binary) : 01011000 00111101 00110010 00110101 00101111 01011001

in ASCII codes(in binary, with first bit as parity bit) : 11011000 10111101 10110010 00110101 10101111 01011001

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