Transcript EE261 Lecture Notes (electronic)

EE 261 – Introduction to Logic Circuits
Module #2 – Number Systems
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Topics
A.
B.
C.
D.
E.
F.
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Number System Formation
Base Conversions
Binary Arithmetic
Signed Numbers
Signed Arithmetic
Binary Codes
Textbook Reading Assignments
 2.1–2.16
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Practice Problems
 2.1, 2.5, 2.6, 2.7, 2.8, 2.11, 2.12
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Graded Components of this Module
 2 homeworks, 2 discussions, 1 quiz
(all online)
EE 261 – Introduction to Logic Circuits
Module #2
Page 1
EE 261 – Introduction to Logic Circuits
Module #2 – Number Systems
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What you should be able to do after this module
 Convert numbers between bases (decimal, binary, octal, hexadecimal)
 Perform arithmetic (both signed & unsigned) using other bases
 Create and Decode various binary codes (BCD, Gray, ASCII, Parity)
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Module #2
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Number Systems
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Number System
- a system that contains a set of numbers/symbols/characters
and
at least one operation (+, -, …)
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BASE or RADIX
- the number of symbols in the number system
- humans use Base10 (why?)
- computers use Base2 (why?)
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Positional Number System
- we can represent 10 unique quantities w/ our Base10 system
- what if we want more?
- we can add a "leading" digit to our number that has increased "weight”
ex)
0
1
:
9
10
11
:
99
100
101
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Number Systems
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Radix Point
- place in the string of digits at which numbers represent either
whole or fractional. ex) 125.178
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p
- Number of digits to the LEFT of the radix point.
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n
- Number of digits to the RIGHT of the radix point.
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d
- Digits in the system, described with a positional subscript
(NOTE: if the radix point is missing, we assume it is to the right of the #)
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r
- Radix or Base
(Base10, r=10)
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i
- Position - starting at 0
- increasing to the left of the radix point
- negative to the right of the radix point
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Weight
- each digit has a weight based on its position = ri
- we multiply the digit by its weight to find how much value
that digit represents
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Number Systems
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Value
- the value of a number is the sum of each digit multiplied by
its corresponding weight.
p 1
D   di  r i
i  n
Example: Given the BASE 10 (r=10) number 327.2
What is p?
What is n?
Write the digit notation?
p=3
n=1
d2 d1 d0 . d-1
What is the weight for each position?
102 101 100 .10-1
Show the expanded decimal equivalent?
= 3(102) + 2(101 ) +7(100 ) + 2(10-1 )
= 327.2
Why all this framework? Because this generic format works for all Bases.
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Number Systems
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Binary
- A number system with 2 symbols (BASE=2 or r=2).
- The symbols are 0 and 1
- each symbol is called a "bit"
- we’ll give a subscript to the number to indicate its base
ex) 101010
10102
(decimal)
(binary)
- 4 bits are called a "Nibble"
- 8 bits are called a "Byte"
- the leftmost bit in a string is called the Most Significant (MSB) or High Order
- the rightmost bit in a string is called the Least Significant (LSB) or Low Order
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Number Systems
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Binary Systems
B
p 1
i
b

r
i
i  n
Example: Given the Binary number 10011
What is p?
What is n?
What is r?
Write the bit notation?
p=5
n=0
r=2
b4 b3 b2 b1 b0
What is the weight for each position?
24 23 22 21 20
Show the expanded decimal equivalent?
B = 1(24) + 0(23) +0(22 ) + 1(21 ) + 1(20 )
= 1(16) + 0(8) + 0(4) + 1(2) +1(1)
= 1910
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Base Conversions
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Base
- the number of symbols in a number system
- we instinctively know decimal
- we've talked about Binary (two symbol)
- there are other bases of interest in digital systems
- the bases we typically care about are powers of 2 (or associated with Binary)
- these bases are typically used to represent a lot of "bits"
ex) it's hard to describe the value of a 64-bit bus in 1's and 0's
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Octal
- A number system with 8 symbols
- 0,1,2,3,4,5,6,7
- each digit in this system is equivalent to 3-bits
- it is a positional number system
ex)
0
1
:
7
10
11
:
17
20
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Base Conversions
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Hexadecimal
- A number system with 16 symbols
- we use alphabetic characters as symbols in the set above 9
- 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F
- each digit in this system is equivalent to 4-bits
- it is a positional number system
ex)
0
1
:
9
A
B
C
D
E
F
10
11
:
1F
20
:
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Base Conversions
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How the Bases Relate to Each Other
Decimal
Binary
Octal
Hexadecimal
Base10
Base2
Base8
Base16
0
1
2
3
4
5
6
7
10
11
12
13
14
15
16
17
0
1
2
3
4
5
6
7
8
9
A
B
C
D
E
F
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0
1
10
11
100
101
110
111
1000
1001
1010
1011
1100
1101
1110
1111
- these are the commonly used bases in digital systems, we'd like to be able to convert between them
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Base Conversions
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Converting to Decimal
- we sum the products of each digit value with its positional weight
p 1
D   di  r i
i  n
- this works for whole and fractional digits
- this works for Binary to Decimal
- this works for Octal to Decimal
- this works for Hex to Decimal
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Base Conversions
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Base Conversion – Binary to Decimal
- each digit has a “weight” of ri that depends on the position of the digit
- multiply each digit by its “weight”
- sum the resultant products
ex) Convert 1011 to decimal
23
1
22
0
21
1
= 1•(23) + 0• (22)
= 1•(8) + 0• (4)
= 8
+ 0
= 11 Decimal
+ 1• (21)
+ 1• (2)
+ 2
20
1
(weight)
+ 1• (20)
+ 1• (1)
+ 1
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Base Conversions
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Base Conversion – Binary to Decimal with Fractions
- the weight of the binary digits have negative positions
ex) Convert 1011.101 to decimal
23
1
=
=
=
=
22
0
21
1
1•(23) + 0• (22)
1•(8)
+ 0• (4)
8
+ 0
11.625 Decimal
+ 1• (21)
+ 1• (2)
+ 2
20
1
.
+ 1• (20)
+ 1• (1)
+ 1
2-1
1
2-2
0
2-3
1
+ 1• (2-1) + 0• (2-2) + 1• (2-3)
+ 1• (0.5) + 0• (0.25) + 1• (0.125)
+ 0.5
+ 0
+ 0.125
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Base Conversions
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Base Conversion – Hex to Decimal
- the same process as “binary to decimal” except the weights are now BASE 16
- NOTE (A=10, B=11, C=12, D=13, E=14, F=15)
ex) Convert 2BC from Hex to decimal
162
2
161
B
160
C
(weight)
= 2• (162) + B• (161) + C• (160)
= 2•(256) + 11• (16) + 12• (1)
= 512
+ 176
+ 12
= 700 Decimal
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Base Conversions
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Base Conversion – Hex to Decimal with Fractions
- the fractional digits have negative weights (BASE 16)
- NOTE (A=10, B=11, C=12, D=13, E=14, F=15)
ex) Convert 2BC.F to decimal
162
2
161
B
160
C .
16-1
F
(weight)
= 2• (162) + B• (161) + C• (160) + F• (16-1)
= 2•(256) + 11• (16) + 12• (1) + 15• (0.0625)
= 512
+ 176
+ 12
+ 0.938
= 700.938 Decimal
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Base Conversions
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Base Conversion – Decimal to Binary
- the decimal number is divided by 2, the remainder is recorded
- the quotient is then divided by 2, the remainder is recorded
- the process is repeated until the quotient is zero
ex) Convert 11 decimal to binary
2 11
Quotient
5
Remainder
1
LSB
2 5
2
1
2 2
1
0
2 1
0
1
MSB
= 1011 binary
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Base Conversions
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Base Conversion – Decimal to Binary with Fractions
- the fraction is converted to binary separately
- the fraction is multiplied by 2, the 0th position digit is recorded
- the remaining fraction is multiplied by 2, the 0th digit is recorded
- the process is repeated until the fractional part is zero
ex) Convert 0.375 decimal to binary
0.375•2
0.75•2
0.5•2
Product
0.75
1.50
1.00
0th Digit
0
1
1
MSB
LSB
0.375 decimal
= 0.011 binary
finished
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Base Conversions
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Base Conversion – Decimal to Hex
- the same procedure is used as before but with BASE 16 as the divisor/multiplier
ex) Convert 420.625 decimal to hex
1st, convert the integer part…
Quotient
Remainder
16 420
26
4
16 26
1
10
16 1
0
1
LSB
MSB
= 1A4
2nd, convert the fractional part…
Product
0.625•16
10.00
0th Digit
10
MSB
= 0.A
420.625 decimal = 1A4.A hexadecimal
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Base Conversions
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Base Conversion – Octal to Decimal / Decimal to Octal
- the same procedure is used as before but with BASE 8 as the divisor/multiplier
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Base Conversions
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Base Conversion of "Powers of 2"
- converting bases that are powers of 2 are simple due to straight forward mapping
- An Octal digital represents 3 bits
- A Hex digit represents 4 bits
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Base Conversions
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Converting Binary to Hexadecimal
- every 4 binary bits represents one HEX digit
- begin the groups of four at the LSB
- if necessary, fill with leading 0’s to form the groups of four
ex) Convert 111010100 from Binary to Hex
0001 1101 0100
1
D
4
Hex, notice that we had to fill with
MSB 0's to get groups of four…
- This works for fractions too. The only difference is that the grouping starts at the Radix Point
ex) Convert 0.11 from Binary to Octal
0.1100
0.C
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Base Conversions
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Converting Binary to Octal
- every 3 binary bits represents one OCTAL digit
- begin the groups of three at the LSB
- if necessary, fill with leading 0’s to form the groups of three
ex) Convert 111010100 from Binary to Octal
111 010 100
7
2
4
Octal
ex) Convert 10100 from Binary to Octal
010 100
1
4
Octal, note we had to fill with 0's on the MSB side
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Base Conversions
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Converting Hex to Binary
- each HEX digit is made up of four binary bits
ex) Convert ABC Hex to Binary
A
= 1010
B
C
1011
1100
= 101010111100
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Base Conversions
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Converting Octal to Binary
- each Octal digit is made up of three binary bits
ex) Convert 567 Octal to Binary
5
= 101
6
7
110
111
= 101110111
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Base Conversions
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Terminology
NIBBLE
= 4 bits
BYTE
= 8 bits
- you should be familiar with converting Binary Nibbles to Hex & Dec
- there is a table on page 28 of your textbook which lists the basic conversions
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Binary Arithmetic
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Addition and Subtraction
- same as BASE 10 math, remember borrows and carries
Addition Table
0
+ 0
0
0
+ 1
1
1
+ 0
1
1
+ 1
10
1
0
1
1
1
Carry
Subtraction Table
0
- 0
0
10
0
- 1
1
-
0
Need to "Borrow" from a more significant bit
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Binary Arithmetic
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Binary Addition
- same as BASE 10 addition
- need to keep track of the carry bit for a given system size (n), i.e., 4-bit, 8-bit, 16-bit,…
ex) Add the binary numbers 1011 and 1001
11
1011
1001
+_____
1 0100
Carry Bit
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Binary Arithmetic
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Binary Subtraction
- same as BASE 10 subtraction
- need to keep track of the borrow bit for a given system size (n), i.e., 4-bit, 8-bit, 16-bit,…
ex) Subtract the binary numbers 1010 and 0011
1 10
0 10 0 10
1 0 1 0
0 0 1 1
-________
0 1 1 1
Borrow bits : if necessary, we could assume a borrow from an even higher significant bit
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Signed Numbers
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Negative Numbers
- So far, we've dealt with Positive numbers
- The real world has Negative numbers
- We need a method to represent negatives in binary
- However, our number system doesn't have a "-", just 0's and 1's
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Sign Bit
- We will use the MSB to represent a "+" or "-"
0 = Positive
1 = Negative
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Signed Numbers
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Signed Magnitude Representation
- a negative number system (there is more than one)
- uses the MSB as the sign bit
- the remaining LSB's represent the number
ex)
•
85 dec
+85 dec
-85 dec
= 1010101
= 01010101
= 11010101
(additional sign bit adds one bit to the number)
Is this Efficient?
- The number of unique combinations that a number can represents is given by:
N = rx
N = # of unique combinations
x = # of digits in the number (p + n)
r = base
ex) 2-bits: N=22 = 4
7-bits: N=27 = 128
8-bits: N=28 = 256
(00, 01, 10, 11)
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Signed Numbers
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Signed Magnitude Range
- The Range of Signed Magnitude is given by:
-(2x-1 - 1) < NSM < (2x-1 -1)
ex) if we use 8-bits, the range is:
-(28-1 - 1) < NSM < (28-1 -1)
-127 < NSM < +127
- This is only 255 unique numbers
- But, if we are using 8-bits, there should be 28 unique numbers, or 256?
- This is because in Signed Magnitude, there are two representations for Zero
0 0000000 = +0
1 0000000 = -0
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Signed Numbers
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Advantages of Signed Magnitude
- Very easy to understand and use
- the binary number is simply represented in binary, then a sign bit is added
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Disadvantages of Signed Magnitude Representation
- We loose a possible number by having +0, and -0.
- This also creates a gap in our number list, which makes simple math harder
- Addition/Subtraction in general are more difficult
Algorithm :
if ( signs are the same)
- add and give same sign
else if (signs are different)
- compare magnitudes
- subtract smaller from larger
- give the result the sign of larger number
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Signed Numbers
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Complement Numbers
- Complementing a Binary Number means:
0 changes to 1
1 changes to 0
- using this technique, number can be a "complemented" to find the negative representation,
then the arithmetic becomes much simpler
- the ones we care about are
1) "Radix Complement" (2's complement for binary)
2) "1's Complement
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Signed Numbers
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Radix Complement
- i.e., 10's complement, 2's complement
- Technique to get the Radix Complement:
- subtract current number from rx
Dr-comp = rn - Dorig
- Rules:
- The MSB is still the sign bit (0="+", 1="-")
- Now that we're subtracting, we can't have arbitrary number of bits…
x is predetermined and fixed
- A simpler method is :
1) Complement all digits in Dorig
2) Add 1 to the result, ignore any carry-out
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Signed Numbers
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2's Complement
- this is Radix Complement on a BASE2 number system
- straight forward to use the "complement and add 1" technique
ex) Give the 8-bit, 2's complement representation of -1710
1710 =
000100012
(Notice that the MSB is the sign bit, 0=+)
Step 1:
11101110
(we first complement all bits)
Step 2:
+ 1
11101111
(we then add 1)
(Notice the Sign Bit is negative)
- these 8-bits represent -1710 in 2's complement
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Signed Numbers
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2's Complement Checking
- We can convert back to find Original magnitude and do checking
ex) Give the 8-bit, 2's complement representation of -1710
-1710 =
11101111 2's Comp
(Notice that the MSB is the sign bit, 1="-")
Step 1:
00010000
(we first complement all bits)
Step 2:
+ 1
00010001
(we then add 1)
(Notice the Sign Bit is positive)
- these 8-bits represent +1710,Which is what we originally started with,
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Signed Numbers
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2's Complement Range
- We need to know how many numbers we can represent using this system
- Notice that we are still using a bit for the sign…
- BUT, we don't duplicate Zero in this system
ex) 00000000 = 010
10000000 = -12810
- The Range of 2's Comp is given by:
-(2x-1) < N2Comp < (2x-1 -1)
ex) if we use 8-bits, the range is:
-(28-1) < N2Comp < (28-1 -1)
-128 < N2Comp < +127
- There are now 256 unique numbers, this is a more efficient use of bits
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Signed Numbers
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1's Complement
- In 2's complement, the codes are asymmetric
Lowest 2's Comp # = 1000 00002Comp = -12810
Highest 2's Comp # = 0111 11112Comp = +12710
- This is good because we have all 256 possible numbers that 8-bits can give us
- 1's Comp is similar, but it gives us symmetry around Zero
- To find the 1's Comp, we subtract current number from (rx -1)
D1Comp = (rn -1) - Dorig
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Signed Numbers
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1's Complement
- Or we can use the simple way
1) Complement all the #'s (Don't add 1)
ex) Give the 8-bit, 1's complement representation of -1710
1710 =
Step 1:
000100012
(Notice that the MSB is the sign bit, 0=+)
111011101Comp
(complement all bits)
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Signed Numbers
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1's Complement Range
- The Range of 1's Complement is given by:
-(2x-1 - 1) < NSM < (2x-1 -1)
- Once again, we have two values for Zero
00000000 = +0
11111111 = -0
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Signed Numbers
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Negative Representation
- We have covered 3 different "signed" codes
- You need to KNOW THE CODE you are using
ex) Represent -2210
1) Signed Magnitude
: 10010110SM
2) 2's Complement
: 111010102Comp
3) 1's Complement
: 111010011Comp
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Signed Arithmetic
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Two’s Compliment Arithmetic
- Two's complement has advantages when going into Hardware
- Two's complement addition is straight forward because the numbers are in
sequential order (+1) from their least significant (1000 0000 )
to their most significant (0111 1111)
- There is only one value for Zero, so "avoiding the gap" isn't necessary
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Signed Arithmetic
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Two’s Compliment Addition
- Addition of two’s compliment numbers is performed just like
standard binary addition.
- However, the carry bit is ignored
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Signed Arithmetic
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Two’s Compliment Subtraction
- We can build a subtraction circuit out of the same hardware as an adder
- 2's Comp inherently adds negative numbers so
- to subtract, we can just complement one number, and add it.
ex) Subtract 8dec from 15dec
15dec = 0000 1111
8dec = 0000 1000 -> two’s compliment -> invert
add1
Now Add: 15 + (-8) =
1111 0111
+
1
----------------1111 1000 = -8dec
%0000 1111
%1111 1000
+_________
1 0000 0111 = 7dec
Disregard Carry
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Signed Arithmetic
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Two’s Compliment Overflow
- If a two’s compliment arithmetic operation results in a number that is outside
the range of representation (i.e., 8bits : -128 < N < +127), an “overflow”
has occurred.
ex) -100dec – 100dec- = -200dec (can’t represent)
- There are three cases when overflow occurs
1) Sum of like signs results in answer with opposite sign
2) Negative – Positive = Positive
3) Positive – Negative = Negative
- Boolean logic can be used to detect these situations.
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Signed Arithmetic
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Remember 2's Comp Range
- The Range of 2's Comp is given by:
-(2x-1) < N2Comp < (2x-1 -1)
ex) if we use 8-bits, the range is:
-(28-1) < N2Comp < (28-1 -1)
-128 < N2Comp < +127
•
What goes on in real Hardware?
- A generic, Binary adder circuit is created
- the user must be aware of when 2's complement is being used
- additional circuitry checks for "Overflow"
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Binary Codes
•
Codes
- a string of x-bits that represent information
- we've seen codes already using the binary # system
1) Signed Magnitude
2) 2's Complement
3) 1's Complement
- the same information can be encoded differently
- it's up to the engineer to KNOW THE CODE that is being used
Code Word - the term to represent the discrete string of bits that make up the information
ex) 4-bit code words in
a stream of information
1111
0100
0100
word
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Binary Codes
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Binary Coded Decimal (BCD)
- Sometimes we wish to represent an individual decimal digit as
a binary representation (i.e., 7-segment display to eliminate a decoder)
- We do this by using 4 binary digits.
Decimal
0
1
2
3
4
5
6
7
8
9
BCD
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
ex) Represent 17dec
Binary = 10111
BCD = 0001 0111
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Binary Codes
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BCD Addition
- If we are using BCD, traditional addition doesn’t work.
12
+ 08
1A
= correct answer in BCD is 20. Traditional addition only considers
8-bit addition carry. In BCD we need to consider 4-bit addition carry.
- The solution is to add 6 to every sum greater than 9
12
08
1A
+ 6
20
= When a nibble addition results in “A” or greater, we add 6
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Binary Codes
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Gray Code
- There are applications where we want to count, but we only want one bit to
transition each time we increment counts
- Reduce Power & Noise in Digital Electronics
- Electromechanical devices
- Such a is called a "Gray Code"
ex) 3-bit Gray Code
Decimal
0
1
2
3
4
5
6
7
Binary
000
001
010
011
100
101
110
111
Gray
000
001
011
010
110
111
101
100
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Binary Codes
•
ASCII
- American Standard Code for Information Interchange
- English Alphanumeric characters are represented with a 7-bit code
- See Table 2-11 in text
ex) ‘A’ = $40
‘a’ = $61
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Binary Codes
•
Parity Codes
- there are times when information can be corrupted during transmission
- we can include an "Error Checking" bit along with the original data
- this "Error Checking" bit contains additional information about the original data
- this can be used by the receiver to monitor whether an error in the data occurred
PARITY
- the number of 1's in the information are counted,
- the parity bit represents whether there are an EVEN or ODD number of 1's
EVEN PARITY
= 0, if there are an EVEN number of 1's in the information
= 1, otherwise
ODD PARITY
= 0, if there are an ODD number of 1's in the information
= 1, otherwise
EE 261 – Introduction to Logic Circuits
Module #2
Page 52
Binary Codes
•
Parity Example
Information
000
001
010
011
100
101
110
111
EVEN Parity
0
1
1
0
1
0
0
1
EE 261 – Introduction to Logic Circuits
ODD Parity
1
0
0
1
0
1
1
0
Module #2
Page 53
Module Overview
•
Topics
- # systems & bases
- # system conversions
- to & from decimal
- to and from binary/oct/hex
- binary arithmetic
- addition (carries)
- subtraction (borrows)
- negative #'s in binary
- Signed Magnitude
- 2's Complement
- 1's Complement
- 2's complement arithmetic
- complement & add
- range & overflow
- Codes
- BCD, Gray, ASCII, Parity
EE 261 – Introduction to Logic Circuits
Module #2
Page 54