Labs 1.1-1.4

Topics: Binary numeral system, Character encoding, Unicode Pages: 9 (586 words) Published: April 24, 2014
Homework Labs 1.1 - 1.4 Thursday 8:30-12:306/25/2013 Exercise 1.1 Base 10
Mapping for decimal number 2931
10^3 10^210^110^0
2931
2x1000=2000+9x100=900+3x10=30+1x1=1 = 2931

Exercise 1.1.2
Mapping for binary number 110 base 2
421
***
110
===
4 +2 +0= 6

Exercise 1.1.3
Mapping for binary number 11 base 2
21
**
11
==
2 +1= 3

Exercise 1.1.4
Mapping for binary number 10010 base 2
168421
*****
10010
=====
16 +0 +0 +2 +0= 18

Exercise 1.1.5
Mapping for binary number 11100010 base 2
1286432168421
********
11100010
========
128643200020= 226

Exercise 1.1.6
Binary conversion for decimal 156 base 10

156-128=28-16=12-8=4-4=0
128 64 32 168421
1 0011100

Exercise 1.1.7
Binary conversion for decimal 255

255-128=127-64=63-32=31-16=15-8=7-4=3-2=1-1=0
1286432168421
1111111`1

Exercise 1.1.8
Binary conversion for decimal 200

200-128=72-64=8-8=0
1286432168421
11001000

Lab 1.1 Reviews
1)Convert the decimal value 127 to binary.

127-64=63-32=31-16=15-8=7-4=3-2=1-1=0
6432168421
1111111

2)Explain why values 10 base 2 and 0010 base 2 are equivalent- Two digits start over, and next digit to convert from a base 10 integer numeral to its base 2 equivalent. 3)When you multiply 2 by the power of you get 1,2,4,8,16,32 etc. 4)It wouldn’t be that difficult you just have to ally the correct formulas.

Exercise 1.2.1
Binary addition for 110 base 2 + 1001 base 2
1001
+++
110
===
1111

Exercise 1.2.2
Adding binary 110 base 2 + 101 base 2

110
+++
101
===
1111 One plus one means you carry the one its one zero not 10

Exercise 1.2.3
Adding binary 111 base 2 and 111 base 2

Carry11
111
+++
111
===
1110 one plus one plus one is one one to be carried

Exercise 1.2.4
Determine the result of 100 base 2 OR 011 base 2
100
+++
011
===
111

Exercise 1.2.5
Determine the result of 111 base 2 AND 100 base 2
111
+++
100
===
100

Exercise 1.2.6
Determine the result of NOT 1001 base 2

0110

Exercise 1.2.71010 base 2 + 10 base 2
1 Carried
1010
++
10
====
1110

Exercise 1.2.8 1100 base 2 AND 1111 base 2
1100
1111
====
1100

I conclude that when using AND with a string of 1’s the answer is 1 Exercise 1.2.9
1100 base 2 OR 1111 base 2
1100
++++
1111
====
1111

If there is a 1 present the answer is 1.

Lab 1.2 Reviews

10010000 base 2 + 1101110 base 2
1)10010000
+++++++
1101110
========
11111110

11001100 base 2 AND 11111100 base 2
2)11001100
++++++++
11111100
========
11001100

11001100 base 2 OR 11111100 base 2
3)11001100
++++++++
11111100
========
11111100

NOT 11001100 base 2 AND 11111100 base 2
4)11001100
++++++++
11111100
========
11001100
NOT =
00110011

Exercise 1.3.1
Decimal value for Byte 1 = 65,208
Decimal value for Byte 2= 255

Exercise 1.3.2
The decimal value with them combined is 65,463
It is lager when combined.

Exercise 1.3.3

120MB x 1024 bytes = 122,880 bytes

Exercise 1.3.4

16 x 1024 = 16,384MB x 1024 = 16,777,216kb x 1024 = 1.717986918 x 10^10 bytes x 8 = 1.374389535 x 10^011 bits / 32 = 4,294,967,296

Exercise 1.3.5
Binary value of 110110 base 2 in hexadecimal

32168421
110110
_____________________________
6c base 16

Exercise 1.3.6
Hexadecimal f6 base 16 = binary 11110110
Decimal 246

Lab 1.3 Reviews
1)This way you will be familiar with sizing a hard drive for what it will be used for whether...
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