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Lecture 2Binary Arithmetic
Lecture 2Binary Arithmetic
TopicsTopics
Terminology Fractions Base-r to decimal Unsigned Integers Signed magnitude Two’s complement
August 26, 2003
CSCE 211 Digital Design
– 2 – CSCE 211H Fall 2003
OverviewOverview
Last Time: Readings 1.1, 1.2 2.1-2.3 Last Time: Readings 1.1, 1.2 2.1-2.3 Analog vs Digital Conversion Base-r to decimal Conversion decimal to Base-r
New: Readings 2.4-2.10, 3.1New: Readings 2.4-2.10, 3.1 Conversion of Fractions base-r decimal Unsigned Arithmetic Signed Magnitude Two’s Complement Excess-1023
VHDLVHDL Introductory Examples
– 3 – CSCE 211H Fall 2003
Review Last Time: Base-r to decimalReview Last Time: Base-r to decimal
Converting base-r to decimal by definitionConverting base-r to decimal by definition
ddnndd
n-1n-1dd
n-2n-2…d…d
2 2dd
1 1dd
0 0 = d = d
nnrrnn + d + d
n-1n-1rrn-1n-1… d… d
22rr22 +d +d
1 1rr1 + 1 + dd
0 0rr00
ExampleExample
4F0C 4F0C = = 4*164*1633 + F*16 + F*1622 + 0*16 + 0*1611 + C*16 + C*1600
== 4*4096 + 15*256 + 0 + 12*1 4*4096 + 15*256 + 0 + 12*1
= = 16384 + 3840 + 1216384 + 3840 + 12
== 2023620236
– 4 – CSCE 211H Fall 2003
Review Last Time: decimal to Base-rReview Last Time: decimal to Base-r
Repeated division algorithmRepeated division algorithm
Justification:Justification:
ddnndd
n-1n-1dd
n-2n-2…d…d
2 2dd
1 1dd
0 0 = d = d
nnrrnn + d + d
n-1n-1rrn-1n-1… d… d
22rr22 +d +d
1 1rr1 + 1 + dd
0 0rr00
Dividing each side by r yieldsDividing each side by r yields
(d(dnndd
n-1n-1dd
n-2n-2…d…d
2 2dd
1 1dd
0 0) / r = d) / r = d
nnrrn-1n-1 + d + d
n-1n-1rrn-2n-2… d… d
22rr11+d+d
1 1rr0 + 0 + dd
0 0rr-1-1
So dSo d 0 0 is the remainder of the first division is the remainder of the first division
((q((q11) / r = d) / r = d
nnrrn-2n-2 + d + d
n-1n-1rrn-3n-3… d… d
33rr11+d+d
2 2rr0 + 0 + dd
1 1rr-1-1
So dSo d 1 1 is the remainder of the next division is the remainder of the next division
and dand d 2 2 is the remainder of the next division is the remainder of the next division
……
– 5 – CSCE 211H Fall 2003
Review Last Time: decimal to Base-rReview Last Time: decimal to Base-r
Repeated division algorithm ExampleRepeated division algorithm Example
Convert 4343 to hexConvert 4343 to hex
4343/16 = 271 remainder = 74343/16 = 271 remainder = 7
271/16 = 16 remainder = 15271/16 = 16 remainder = 15
16/16 = 1 remainder = 0 16/16 = 1 remainder = 0
1/16 = 0 remainder = 11/16 = 0 remainder = 1
So 4343So 43431010
= 10F7 = 10F71616
To check the answer convert back to decimalTo check the answer convert back to decimal
10F7 = 1*1610F7 = 1*1633 + 15*16 + 7*1 = 4096 + 240 + 7 = 4343 + 15*16 + 7*1 = 4096 + 240 + 7 = 4343
– 6 – CSCE 211H Fall 2003
Review Last Time: Hex to BinaryReview Last Time: Hex to Binary
One hex digit = four binary digitsOne hex digit = four binary digits
ExampleExample
3FAC = 0011 1111 1010 1100 3FAC = 0011 1111 1010 1100 (spaces just for readability)(spaces just for readability)
Binary to hex four digits = one (from right!!!)Binary to hex four digits = one (from right!!!)
ExampleExample
111101001111010 = 0011 1101 0011 1010111101001111010 = 0011 1101 0011 1010
= 3 D A= 3 D A
– 7 – CSCE 211H Fall 2003
Hex to Decimal FractionsHex to Decimal Fractions
.d.d-1-1dd
-2-2dd
-3-3…d…d
–(n-2) –(n-2)dd
–(n-1) –(n-1)dd
-n -n= d= d
-1-1rr-1-1 + d + d
-2-2rr-2-2…d…d
-(n-1)-(n-1)rr-(n-1)-(n-1) +d +d
1 1rr-n-n
ExampleExample
.1EF.1EF1616
= 1*16-1 + E*16-2 + F*16-3 = 1*16-1 + E*16-2 + F*16-3
= 1*.0625 + 14*.003906025 + 15*2.4414e-= 1*.0625 + 14*.003906025 + 15*2.4414e-44
= .117201… (probably close but not = .117201… (probably close but not right)right)
– 8 – CSCE 211H Fall 2003
Decimal Fractions to hexDecimal Fractions to hex
.d.d-1-1dd
-2-2dd
-3-3…d…d
–(n-2) –(n-2)dd
–(n-1) –(n-1)dd
-n -n= d= d
-1-1rr-1-1 + d + d
-2-2rr-2-2…d…d
-(n-1)-(n-1)rr-(n-1)-(n-1) +d +d
1 1rr-n-n
Multiplication by r yields Multiplication by r yields
r *(.dr *(.d-1-1rr-1-1 + d + d
-2-2rr-2-2…d…d
-(n-1)-(n-1)rr-(n-1)-(n-1) +d +d
1 1rr-n-n
= d= d-1-1rr00 + d + d
-2-2rr-1-1…d…d
-(n-1)-(n-1)rr-(n-2)-(n-2) +d +d
1 1rr-(n-1)-(n-1)
Whole number part = dWhole number part = d-1-1rr00
Multiplying again by r yields dMultiplying again by r yields d-2-2rr0 0 as the whole number as the whole number
partpart
… … till fraction = 0till fraction = 0
– 9 – CSCE 211H Fall 2003
Example: Hex Fractions to decimalExample: Hex Fractions to decimal
Convert .3FA to decimalConvert .3FA to decimal
.3FA16 = 3*16-1 + F*16-2 + A*16-3.3FA16 = 3*16-1 + F*16-2 + A*16-3
= 3*.0625 + 15*.00390625 +10* (1/4096)= 3*.0625 + 15*.00390625 +10* (1/4096)
= .191162109= .191162109
– 10 – CSCE 211H Fall 2003
Unsigned integersUnsigned integers
What is the biggest integer representable using n-bits(n What is the biggest integer representable using n-bits(n digits)?digits)?
What is its value in decimal?What is its value in decimal?
Special casesSpecial cases
8 bits8 bits
16 bits16 bits
32 bits32 bits
– 11 – CSCE 211H Fall 2003
Arithmetic with Binary NumbersArithmetic with Binary Numbers
10010110 10010110 1001011010010110 11011101
+00110111+00110111 - 00110111- 00110111 x 101x 101
Problems with 8 bit operationsProblems with 8 bit operations
1001011010010110
++ 1001011010010110
– 12 – CSCE 211H Fall 2003
Signed integersSigned integers
How do we represent?How do we represent?
Signed-magnitudeSigned-magnitude
Excess representationsExcess representations
w bits w bits 0 <= unsigned_value < 2 0 <= unsigned_value < 2ww
In excess-B we subtract the bias (B) to get the value.In excess-B we subtract the bias (B) to get the value.
exampleexample
– 13 – CSCE 211H Fall 2003
Complement Representations of Signed integersComplement Representations of Signed integers
One’s complementOne’s complement
Two’s complementTwo’s complement
– 14 – CSCE 211H Fall 2003
Two’s Complement OperationTwo’s Complement Operation
One’s complement + 1 orOne’s complement + 1 or
Find rightmost 1, complement all bits to the left of it.Find rightmost 1, complement all bits to the left of it.
ExamplesExamples
0100111001001110 0000000100000001 0000000000000000 0000001000000010
– 15 – CSCE 211H Fall 2003
Two’s Complement RepresentationTwo’s Complement Representation
Consider a two’s complement binary numberConsider a two’s complement binary number
ddnndd
n-1n-1dd
n-2n-2…d…d
2 2dd
1 1dd
0 0
If dIf dn n , the sign bit = 0 the number is positive and its , the sign bit = 0 the number is positive and its
magnitude is given by the other bits.magnitude is given by the other bits.
If dIf dn n , the sign bit = 1 the number is negative and take its , the sign bit = 1 the number is negative and take its
two’s complement to get the magnitude.two’s complement to get the magnitude.
Weighted Sum InterpretationWeighted Sum Interpretation
0 10 1
1 21 2
……
n-1 2 n-1 2 n-2n-2
n -2 n -2 n-1n-1
– 16 – CSCE 211H Fall 2003
Two’s Complement RepresentationTwo’s Complement Representation
Consider a two’s complement binary numberConsider a two’s complement binary number
ddnndd
n-1n-1dd
n-2n-2…d…d
2 2dd
1 1dd
0 0
If dIf dn n , the sign bit = 0 the number is positive and its , the sign bit = 0 the number is positive and its
magnitude is given by the other bits.magnitude is given by the other bits.
If dIf dn n , the sign bit = 1 the number is negative and take its , the sign bit = 1 the number is negative and take its
two’s complement to get the magnitude.two’s complement to get the magnitude.
Weighted SumWeighted Sum
0 10 1 Example 10010011 = Example 10010011 = 1 21 2
……
n-1 2 n-1 2 n-2n-2
n -2 n -2 n-1n-1
– 17 – CSCE 211H Fall 2003
Two’s Complement RepresentationTwo’s Complement Representation
What is the 2’s complement representation in 16 bits of What is the 2’s complement representation in 16 bits of –5?–5?
+7?+7?
-1?-1?
00
-2-2
– 18 – CSCE 211H Fall 2003
Arithmetic with Signed IntegersArithmetic with Signed Integers
Signed Magnitude AdditionSigned Magnitude Addition
if the signs are the same add the magnitude if the signs are the same add the magnitude
if the signs are different subtract the smaller from if the signs are different subtract the smaller from the larger and use the sign of the largerthe larger and use the sign of the larger
Subtraction?Subtraction?
Two’s complementTwo’s complement
Just add signs take care of themselvesJust add signs take care of themselves
– 19 – CSCE 211H Fall 2003
Overflow in Two’s ComplementOverflow in Two’s Complement
– 20 – CSCE 211H Fall 2003
Binary Code DecimalBinary Code Decimal
– 21 – CSCE 211H Fall 2003
Representations of CharactersRepresentations of Characters
– 22 – CSCE 211H Fall 2003
Basic GatesBasic Gates
ANDAND
OROR
NOTNOT
– 23 – CSCE 211H Fall 2003
Basic GatesBasic Gates
NANDNAND
NORNOR
XORXOR
– 24 – CSCE 211H Fall 2003
Half Adder CircuitHalf Adder Circuit
– 25 – CSCE 211H Fall 2003
Full AdderFull Adder