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Number Systems Part 2Numerical Overflow
Right and Left Shifts
Storage Methods
Subtraction
Ranges
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Numerical Overflow
An overflow is when something doesntfit in a certain space
Numeric overflow is when the storagefor a calculation is too small to hold theresult
For example we have an 8 bits register, ifwe add two binary numbers and theresult turns out to be 9 bits it would not
fit in the register
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Example
Lets say we have an 8 bit register
Add the following;
Do we have an overflow?
11111111+
10101010
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Numerical Overflow
When we have a numeric overflow we
will have an error in our calculation
When we have an overflow we would
need to remove the extra bit at the start
of the number
Lets say we had a 7 bit register and the
result of a calculation is 11001100 the
actual answer would be 1001100
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Example
Lets say we have a 7 bit register
Add the following;
Do we have an overflow?
Actual answer =
1101111+
1101101
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Working
Suppose we have a 4 bit registers.
1. Perform the additions
2. Is there and overflow?
3. What is the actual answer?
1110+ 0101+ 1101+
1111 0110 0100
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Answers
1110+ 0101+ 1101+
1111 0110 0100
11101 1011 10001
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What is Bit Shifting?
Bit shifting is the process of moving all thebits in a binary number
We have two shifts1. A right shift
2. A left shift
The right shift would divide the numberwhile the left would multiply it
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Right Shift
The right shift is used to perform divisions
Hence if we where to perform three rightshifts we would be diving the number by 2
three times
For example if we shift 101100112, = 17910right by three places, we get 000101102 =
2210 179 /2 = 89
90/2 = 44
45/2 = 22
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How Right Shifts work
Lets say we have 11001012 and we wish
to perform 2 right shifts
1 1 0 0 1 0 11st Shift right
0 1 1 0 0 1 0 1
LOST
0 ADDED
2nd Shift right
0 0 1 1 0 0 1 0ANS =
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Working
Shift the numbers to the right by thenumber of shifts indicated in the brackets;
1. 1000110 (2)
2. 1110001 (4)
3. 1111000 (1)
4. 01010101 (4)
5. 0000111 (3)
Change to
decimal to check
your answers
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1000110 70 1110001 113 1111000 120
0100011 35 0111000 56 011110060
0010001 17 0011100 28
0001110 14
00001117
01010101 85 0000111 7
00101010 42 0000011 3
00010101 21
00001010 10
000001015
0000001 1
00000000
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Left Shift
The left shift is used to performmultiplications
Hence if we where to perform four left
shifts we would be multiplying the numberby 2 four times
If any bits are lost when shifting left the
result is no longer accurate
For example if we shift 011001012, = 10110left by one place, we get 110010102 = 20210
101 x 2 = 202
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How Left Shifts work
Lets say we have 011001012 and we wishto perform 2 left shifts
0 1 1 0 0 1 0 1
1 1 0 0 1 0 1 00
LOST
1 0 0 1 0 1 0 01Answer no
longer
correct as a
1 was lost
1st Left Shift
2nd Left Shift
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Working
Shift the numbers to the left by thenumber of shifts indicated in the brackets,also state which answers are not correct.
1. 0000110 (4)
2. 1110001 (3)
3. 0011110 (1)4. 01010101 (2)
5. 0000111 (3)
Change correct
answers to
decimal to checkyour answers
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0000110 6 1110001 0011110 30
0001100 12 1100010 011110060
0011000 24
0110000 48
110000096
1000100
0001000
Not correct
01010101 0000111 7
10101010 0001110 14
01010100
Not Correct
0011100 28
011100056
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Storage Methods
When we have a fixed register we mightwant to store out binary number in acertain way
The three ways are using;
1. sign and magnitude (first digit is 0 =positive, first digit is1 = negative)
2. ones complement
3. twos complement.
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Example 1
Lets say we have a register of 8-bits andwe wish to store the number 1410
Since the number is positive, all we needto do is convert it to binary and store it
It would be the same in the three
different storage methods, 1410 =
000011102
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Example 2
Lets say I have an 8 bit register and wantto store -1410
The number now changes since it is
negative, we first change it to binary =000011102
Sign and Magnitude 10001110 Note that the fist digit is 1 because in this
case 14 is a negative number.
Ones Complement 11110001 In this case we simply applied NOT on the
binary value of14.
Twos Complement 11110010 We change the binary number 14 into
twos complement
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Working
Convert the following to;a) Sign and magnitudeb) Ones Complementc) Twos Complement
Size of register is indicated in the brackets.1. 3310 (8)2. -6010 (9)
3. -4410 (7)4. 1010 (5)5. -7410 (10)
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Subtraction
We are able to subtract using binary numbersthanks to Twos Complement
When we need to subtract in binary we first
need to change the negative number to TwosComplement
if we have 2210 we have to change 10 to Twos
Complement then perform an addition
Binary numbers must be of the same length inorder to subtract in binary
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Example
We want to perform the followingsubtraction in binary;
22 - 10110 - 10110+
10 01010 10110
12 101100
Changeto BinaryTwos
Complement
Overflow
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Working
Perform the following subtractions inbinary;
1. 30-20
2. 10050
3. 5025
4. 52
5. 66- 60
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Ranges
A ranges determines the lowest and thehighest values which can be represented
by a certain register
We will be going through two types of
ranges using binary;
1. normal binary2. twos complement.
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Normal Binary
What is the range of numbers which can berepresented using 8-bits?
Since we have 8 Bits, the maximum valuerepresented in binary is 11111111 and
the smallest is 00000000.
If we convert these numbers to decimal
we end up with: 0 to 255
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Example
What is the range of numbers which can berepresented using 5-bits?
Since we have 5 Bits, the maximum valuerepresented in binary is 11111 and the
smallest is 00000.
If we convert these numbers to decimal
we end up with: 0 to 31
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Twos Complement
What is the range of numbers which can berepresented using 8-bits in twos complement?
The first number in twos complement
identifies the number as a negative number,the highest possible positive number is011111112 = 12710
To find the smallest number we convert thepositive number to negative so in this case-12710.
The range is: -127 to 126 (126 due to the 0)
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Example
What is the range of numbers which can berepresented using 10-bits in twos
complement?
01111111112 = 51110
51110 becomes negative to show thesmallest number helps = -51110.
The range is: 511 to 510
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