digital logic lecture 3 binary arithmetic by zyad dwekat the hashemite university computer...

Digital LogicLecture 3

Binary Arithmetic

ByZyad DwekatThe Hashemite UniversityComputer Engineering Department

The Hashemite University 2

Outline Introduction. Numbers range. Sign extension. Overflow detection. Unsigned numbers addition and

subtraction. Signed numbers addition and

subtraction. Examples.


Introduction Computers perform the arithmetic operations in

Binary. Computers perform arithmetic on fixed-size

numbers which is called finite-precision arithmetic that is the operands and the result are stored in a fixed size memory storage and cannot be exceeded.

The rules for finite-precision arithmetic are different from the rules of ordinary arithmetic (which are performed by humans in decimal).

The general design rule is to simplify operations as possible in order to get simpler and lower cost logic circuits implementation for these operations.


Numbers in Computers Two types of numbers are found in computers:

Unsigned: the concept of +ve and –ve numbers does not exist. Simply can be considered to contain +ve numbers only where all bits within the number can be used to represent the magnitude of the number (no bits are reserved).

Signed: the number contains a sign representation. So, you have +ve and –ve numbers. One bit (which is the MSB) is reserved for the sign. So, the magnitude of the number occupies (n – 1) bits (for an n bit number).

Unsigned numbers have exactly one representation in Binary systems.

Signed numbers have many different representation systems where we will deal with only 3 of them.

Signed and unsigned concepts are applicable for all numbering systems (decimal, octal, hexa, binary, …).


Signed Binary Numbers In decimal system the sign is

represented using – and + symbols. However, such symbols cannot be used

by hardware where every thing must be binary (either 0 or 1).

So, the MSB of a binary number is reserved for the sign (i.e. sign bit): 0 +ve number. 1 -ve number.


Signed Binary Numbers Representation Signed-magnitude system:

the MSB represents the sign, the remaining bits represent the number value (the value is the same as in unsigned numbers).

Negates the number by reversing its sign. Hard to be used for binary operations in computers.

Signed-complement system: the MSB gives an indication about the sign but not

reserved for it. Negates a number by taking its complement (either

1’s or 2’s complement). Taking the complement of the positive number

includes the sign bit which becomes 1 (-ve) in all cases.

Convenient to be used by computers (actually the 2’s complement is the most common).


Complements The complement of a number B is the number that

when added to B you get either the quantity rn or (rn – 1) based on the complement type.

Two types of complement systems: Diminished radix complement: or simply (r – 1)’s

complement (r is the radix). Radix complements: or simply r’s complement.

In binary they are called: 1’s and 2’s complement. In decimal: 9’s and 10’s complement. In octal: 6’s and 7’s complement. In hexa: 15’s and 16’s complement. We are mainly interested in binary and decimal

complements (to see how arithmetic operations are performed by computers).


Diminished Radix Complement In this system we have: B + (-B) = (rn -

1) So, in general for number N which

contains n digits:(r – 1)’s complement = (rn - 1) – N

Very Important Note: rn is computed in decimal then before subtraction you convert it into the proper numbering system that you are working with. OR just put a 1 followed by n zeros.


Binary 1’s Complement – Simple Rule

Simply invert each bit in the binary number: 1 0 and 0 1

Examples:

Number 1’s complement

01101 10010

11010001 00101110

01 10


Floating Point Diminished Radix Complement Remove the radix point

temporarily and find the (r – 1)’s complement then put the floating point in its original relative location in the result (the same location found in the original number).


Floating Point Diminished Radix Complement -- Examples

Number (r – 1)’s Complement

101.12 10112

1’s Complement = 01002

(final result) 01.002

11101.112 11101112




Radix Complement In this system we have: B + (-B) = rn

So, for number N which contains n digits:r’s complement = rn – N

So:r’s complement = 1 + (r - 1)’s complement

The rule of rn calculations is the same as in the (r-1)’s complement.

Special rule for 2’s complement in Binary: Start from the LSB and leave all 0’s without change in

addition to the first 1 you encounter. After that invert all remaining bits (0 1 and 1 0).

For floating point apply the same rule used by (r – 1)’s complement: Remove the radix point temporarily and find the r’s complement then put the floating point in its original relative location in the result.


Radix Complement -- Examples

Number r’s Complement

10010002 01110002

11101.112 11101112




Remember ...

Negative of the negative is positive.

So, applying complement twice for a number you return to the original number (none complemented one).


Number Ranges I Unsigned System:

Has one zero Number range: 0 to 2n -1

Signed-magnitude and the signed 1’s complement: Have two zeros (-ve and +ve zero). Symmetric values for both +ve and –ve

directions (i.e. mirror). Number range: -(2n – 1 - 1) to +(2n – 1 - 1).

Signed 2’s complement: Has one +ve zero (more natural). Asymmetric (there is an additional value in the

negative direction). Number range: - 2n – 1 to +(2n – 1 - 1).


Number Ranges II

Note that:

The number of values that can be represented by an n bit number = 2n

(same for signed and unsigned numbers)


Number Format and Arithmetic Operations In life we are working with:

Integer numbers: no fractions. Real numbers: a floating point exists.

We need special representation for floating point numbers since computers deals with fixed size numbers represented in binary.

Many standards exist to represent real numbers in binary, e.g. NIST, IEEE 754 standard.

Arithmetic operations: Addition. Subtraction. Multiplication. Division.

In this lecture we will deal with addition and subtraction (only) for signed and unsigned Integer numbers and floating point numbers in ordinary representation (contain radix point as we write in daily life).


Arithmetic Operations -- Notes

Big Picture:1. Make sure that the two number you are adding os

subtracting are of the same size (have the same number of bits). This is done by sign extension as shown in the

next slide.2. Perform the required operation (addition or

subtraction).3. Check if there is overflow or not (see next slides).

If there is overflow, then the operation is incorrcet, no answer.

If there is not overflow, then the operation is correct, take the answer.


Sign Extension Concept I If you are given the following two

unsigned numbers to be added which they are not of equal number of bits what will you do?11101 + 00111010

Sol: Remember in the unsigned system there is

no –ve numbers. So, to reserve the value of the number you must add zeros to the left of the number (since they have no value)

So it becomes 00011101


Sign Extension Concept II If you are given the following two signed

numbers (using the 2’s complement) to be added which they are not of equal number of bits what will you do?11101 + 00111010

Sol: 11101 is a –ve number since MSB = 1. you

must extend the sign bit to reserve the sign of the number.

So it becomes 11111101


OverflowArithmetic operation may not obtain the correct answer in some cases. This happens when the result is out of range that the reserved bits can represent.For example, when 8-bit binary code is used to represent a number in the 2’s complement system, the range of numbers that can be represented is from -128 to +127. If the result of an operation lies outside this range, overflow occurs.Overflow occurs because of the finite precision arithmetic of the computer hardware.

Computers checks for overflow and set a flag in case of over flow.

The user can check this flag to know the current overflow status.


Overflow Detection Overflow cases:

Unsigned numbers: Addition: an overflow occurs if Cout from the MSB is not

equal to 0. Subtraction: as in signed numbers as shown below.

For signed numbers three cases exist this depends on the used system to represent –ve numbers:

Sign-magnitude system will not be studied 1’s and 2’s complement system Cin != Cout means

there is an overflow (Cin is the carry into the MSB). This is checked in the 1’s complement before

performing the end-around carry step.


Binary Addition and subtraction

We will study them as follows: Unsigned Addition and subtraction:

deal with unsigned number systems. Signed Addition and subtraction: deal

with signed number systems both 1’s and 2’s complement systems.


Binary Addition – Basic

0+ 0 0

0+ 1 1

1+ 0 1

1+ 11 0

Carry Bit

(a) (b)

(c) (d)

1+ 1 1 11

(e)

Carry Bit

Remember Counting Principles


Unsigned Binary Addition --Examples

10011001+ 00101100 11000101

0101 + 1001 1110

1011 + 0101 10000

1010 + 0100 1110

1011 + 1100 10111

(a) (b) (c)

(d) (e)

All calculations are right, no overflow, (Cout = 0)!!


Unsigned Addition – Other Systems

Pay attention to the used numbering system.

Simply convert the number into decimal or binary (if you are working in octal or hexa) and perform addition then convert the result back to the original numbering system that you started from.


Unsigned Binary Subtraction I All subtraction operations can be converted

into addition by taking complements (to reserve the simplicity rule):A – B = A + (-B)

Remember we are dealing with unsigned numbers (no –ve numbers).

So, (-B) in the above example is the complement of B not the –ve quantity of B.

Such general rule is applied for all numbering systems.


Unsigned Binary Subtraction II So, perform unsigned subtraction using

complements: r’s complement unsigned subtraction (2’s

complement in binary). (r – 1)’s complement unsigned subtraction

(1’s complement in binary). Different rules are applied for each

method. The basic method is the r’s complement

unsigned subtraction.


r’s Complement Unsigned Subtraction

To subtract M – N: Find the r’s complement of N and add it to

M. If a carry out produced then it must be

discarded which means that M >= N. If no carry out produced this means that M

< N, in this case the result of the subtraction is – (N – M), i.e. –ve of the difference. So, find the r’s complement of the result and place a – sign in front of it.


Examples I

Ex. 1: Compute 2510 – 2510 using 2’s complement, n = 8.

000110012 = (2510)

+ 111001112 = (-2510)

---------------------------

1000000002 = (010)discard

The calculation is right, no overflow, (Cin = Cout)!!


Examples II

Ex: Compute 11510 – 3910 using 2’s complement, n = 8.

11510 - 3910 = 11510 + (-3910)

-3910 = 110110012 (2’s)

011100112 = (11510)+ 110110012 = (-3910)--------------------------- 1010011002 = (7610)discard



Examples IIIEx. 3: Compute 01012 – 11002 using 2’s complement, n = 4.

– 11002 = 01002 (2’s)

01012

+ 01002

-------------

10012 The calculation is wrong !! Cin != Cout, Overflow


Examples IVEx. 4: Compute 00012 – 01102 using 2’s complement, n = 4.

– 01102 = 10102 (2’s)

00012

+ 10102

-------------

10112 => no carry outThe answer is the 2’s complement of 1011= - (0101)



(r – 1)’s Complement Unsigned Subtraction To subtract M – N:

Find the (r – 1)’s complement of N and add it to M.

If a carry out produced then it must be discarded (which means that M >= N) and then add 1 to the result. This method is called end-around carry.

If no carry out produced this means that M < N, in this case the result of the subtraction is – (N – M), i.e. –ve of the difference. So, find the (r - 1)’s complement of the result and place a – sign in front of it.


Examples I

Ex. 1: Compute 2510 – 2510 using 1’s complement, n = 8.

000110012 = (2510)

+ 111001102 = (-2510)

---------------------------

111111112

Final answer = - (1’s of 111111112) = - 000000002



Examples IIEx. 2: Compute 11510 – 3910 using 1’s complement, n = 8.

11510 - 3910 = 11510 + (-3910)

-3910 = 110110002 (1’s)

011100112 = (11510)+ 110110002 = (-3910)--------------------------- 1010010112

+ 12

--------------------------- 010011002 = (7610)

discardThe calculation is right, no overflow, (Cin = Cout)!!


Examples IIIEx. 3: Compute 01012 – 11002 using 1’s complement, n = 4.

– 11002 = 00112 (1’s)

01012

+ 00112

-------------

10002 No carry out

The calculation is wrong !! Cin != Cout, Overflow


Example IV

Ex. 4: Compute 112 – 10.112 using 2’s complement where n = 8.

– 00000010.112 = 11111101.012 (2’s)

00000011.002

+ 11111101.012

-------------

100000000.012 discardThe calculation is right, no overflow, (Cin = Cout)!!


Examples VEx. 5: Compute 00012 – 01102 using 1’s complement, n = 4.

– 01102 = 10012 (1’s)

00012

+ 10012

-------------

10102 => no carry outThe answer is the 1’s complement of 1011= - (0100)



Signed Binary Arithmetic Which system to use?

We will use the complement signed system. Using signed complements we need one circuit to

perform both addition and subtraction because subtraction can be done by addition.

No sign checking of operands is required simpler in hardware.

Also, overflow detection is very easy using this system.

2’s complement is preferred over 1’s complement because no duplication of zero exist.


2’s Complement Signed Binary Addition

Represent –ve numbers in 2’s complement (if not already in binary).

Add numbers and if a carry out obtained discard it.

If the result is –ve (i.e. the MSB is 1) then it is in the 2’s complement form.


1’s Complement Signed Binary Addition

Represent –ve numbers in 1’s complement (if not already in binary).

Add numbers and if a carry out obtained discard it and add 1 to the result (end-around carry method).

If the result is –ve (i.e. the MSB is 1) then it is in the 1’s complement form.


Signed Binary Subtraction Convert the subtraction into addition by

taking the complement of the second operand.

Then, apply the same rules of addition for both types of complements.

Note that arithmetic operations for both signed and unsigned numbers are the same but the difference lies in the interpretation of the results.


Example I

Find the following using 2’s complement (signed):

115 - 39 = 115 + (-39)

First: convert into binary

115(10) = 01110011(2) (2‘s)

-39(10) = 11011001(2) (2’s)

Second: perform subtraction using 2’s complement:

01110011 = (115)

+ 11011001 = (-39)

---------------------------

101001100 = (76)discard



Example II

Find the following using 2’s complement (signed):

68 + 39

First: convert into binary:

68(10) = 01000100(2) (2‘s)

99(10) = 01100011(2) (2’s)

Second: add using 2’s complement:

01000100 = (68)

+ 01100011 = (99)

---------------------------

10100111 = (-89)

The calculation is wrong !! Cin != Cout

The actual answer (i.e. 167) exceeds the largest limit (i.e. 127) =>OVERFLOW


Example III

Find the following using 2’s

complement (signed):

-78 - 52 = (-78) + (-52)

First: convert into binary:

-78(10) = 10110010(2) (2‘s)

-52(10) = 11001100(2) (2’s)

Second: subtract using 2’s complement :

10110010 = (-78)

+ 11001100 = (-52)

---------------------------

101111110 = (126)

The calculation is wrong !! Cin != Cout overflow

discard


Examples IVEx. : Compute 00012 – 01102 using 1’s complement, n = 4, signed.

– 01102 = 10012 (1’s)

00012

+ 10012

-------------

10102 => no carry outThe answer is 1011



Examples VEx. 5: Compute 00012 – 01102 using 2’s complement, n = 4, signed.

– 01102 = 10102 (2’s)

00012

+ 10102

-------------

10112 => no carry outThe answer is 1011



Additional Notes

This lecture covers the following material from the textbook: Chapter 1: Sections 1.5 – 1.6

digital logic lecture 3 binary arithmetic by zyad dwekat the hashemite university computer...

Documents