CSI-2111 Computer Architecture Ipage 2-1 2. Revision Objective : To examine basic concepts of:...

CSI-2111 Computer Architecture I page 2-1

2. Revision

Objective: To examine basic concepts of:

– 2.1 Numbering Systems– 2.2 Binary Numbers– 2.3 Boolean Algebra – 2.4 Logic Gates – 2.5 Adders– 2.6 Timing Diagram


2.1 Number System (base B)

Number is represented in terms of Positional Weighting:

(N)B = dn-1 Bn-1 + dn-2 Bn-2+...+ d1 B1 + d0 B0 · d-1 B-1+...+ d-m B-m

Integral Part. Fractional Part• B = base• dk = digit in position k, -m ≤ k ≤ n-1• Bk = weight of position k, -m ≤ k ≤ n-1 • n = number of integral digits in N• m = number of fractional digits in N


The most known systems

B = 10 (Decimal)• digits : (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)

B = 2 (Binary)• digits: (0, 1)

B = 8 (Octal)• digits: (0, 1, 2, 3, 4, 5, 6, 7)

B = 16 (Hexadecimal)• digits: (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D,

E, F)


Conversions (I)

How to convert from one number system to the other (change of base B)?

Algorithm 1: – favorable for conversions to the

decimal system

(N1)A (N2)10


Conversions (II)

Algorithm 2: Use of successive divisions and multiplications– favorable for conversions from the

decimal system

(N1)10 (N2)B

(34.625)10 (?)2


Conversions (II) - Example

(34.625)10 (100010.101)2

Integral Part: Fractional Part:

34 ÷ 2 = 17 r 0 0.625 x 2 = 1 .2517 ÷ 2 = 8 r 1 0.25 x 2 = 0 .5 8 ÷ 2 = 4 r 0 0.5 x 2 = 1 4 ÷ 2 = 2 r 0 2 ÷ 2 = 1 r 0 1 ÷ 2 = 0 r 1


Binary Hexadecimal

Binary Hexadecimal– Form groups of 4 bits starting at

binary point.– Each group of 4 bits represents a

hexadecimal digit.

Hexadecimal Binary– Convert each hexadecimal digit to its

binary equivalent (4 bits).


Binary Octal

Binary Octal– Form groups of 3 bits starting at

binary point.– Each group of 3 bits represents an

octal digit.

Octal Binary– Convert each octal digit to its binary

equivalent (3 bits).


2.2 Binary Numbers

Fixed Point Representation.

– (N)2 has an implicit binary point in a fixed position.

– Notion of complement:• Complement(d) = (Base - 1) - d


Complements of Binary Numbers

1’s Complement : (1CF)

2’s Complement : (2CF)

2’s Complement of 1001011 = ?


Unsigned Binary Integers (UBI)

N is represented in terms of positional weighting: 0 ≤ N ≤ 2n – 1

dn-1dn-2dn-3 ... d2d1d0

No sign

N = (00011010110)2(7.4) = (?)10– Word of 11 digits– Fixed point representation, with 4 digits for

the fractional part


Signed Binary Integers

3 ways of representing ± N with n bits:

– Sign-Magnitude Form (SMF)

– 1’s Complement Form (1CF)

– 2’s Complement Form (2CF)


Comparison of SMF, 1CF, and 2CF

Sign Magnitude / 2’s Complement / 1’s Complement(4 bits word length)

Binary SMF 2CF 1CF0111 +7 +7 +70110 +6 +6 +60101 +5 +5 +50100 +4 +4 +40011 +3 +3 +30010 +2 +2 +20001 +1 +1 +10000 +0 0 +01111 -7 -1 -01110 -6 -2 -11101 -5 -3 -21100 -4 -4 -31011 -3 -5 -41010 -2 -6 -51001 -1 -7 -61000 -0 -8 -7


Binary Addition (by complement)

The bit sign is treated like any other bit (they are added!)

The subtraction is performed by addition; the negative numbers are treated like numbers to add.

Addition by 1CF Addition by 2CF


Overflow

An overflow occurs when the operands have the same sign and the result has a sign different from that of the operands.

Ex. 7 + 3; 4 bits word; SMF

+7 0 111+3 0 011(-2) 1 010


2.3 Boolean Algebra

Two elements: 0 and 1 Elementary operators:

{AND, OR, NOT} Representation and algebraic

simplification of Boolean functions and their realization using logic gates will be studied.


Boolean Functions & Truth Tables

Any Boolean function defined over n variables, each taking a Boolean constant value (0 or 1).

The Truth Table represents the function f, with all 2n combinations of 1’s and 0’s of its variables.

Each Boolean function is defined by its truth table and is represented by inter-connected logic gates.


Functions and Logic GatesNom Symbole Fonction algébrique Table de vérité

ET(AND) F = A • B

A B F0 0 00 1 01 0 01 1 1

OU(OR) F = A + B

A B F0 0 00 1 11 0 11 1 1

NON(NOT) F = A’

A F0 11 0

Délai(Buffer) F = A

A F0 01 1

NON-ET(NAND) F = (A • B)’

A B F0 0 10 1 11 0 11 1 0

NON-OU(NOR) F = (A + B)’

A B F0 0 10 1 01 0 01 1 0

OU-Exclusif(XOR)

F = A B, ou encoreF = (A•B’) + (A’•B)

A B F0 0 00 1 11 0 11 1 0

Équivalence(XNOR)

F = A B, ou encoreF = (A’•B’) + (A•B)

A B F0 0 10 1 01 0 01 1 1


Laws of Boolean Algebra

Together with – Postulates (or axioms)– Theorems

Manipulations (proof)– Algebraic– Tabular

Simplifications– Algebraic


Proofs and Simplifications

Algebraic Proofs

Tabular Proofs

Algebraic Simplifications


Examples *

Algebraic Proof:(A • B’) + B = A + B

= B + (A • B’) (Commutative, OR)= (B + A) • (B + B’) (Distributive, OR)= (B + A) • (1) (Complementation, OR)

= (B + A) (Identity Element, AND)

= A + B (Commutative, OR)


Examples *

Proof by Truth Table:(A • B’) + B = A + B

A B (A • B’) + B A + B0 0 0 00 1 1 11 0 1 11 1 1 1


Examples *

Algebraic Simplification

Example: f(ABC) = AC’ + A’B + A’B’C’ = ?

= AC’ + A’B + A’BC’ + A’B’C’ (Absorption, OR)= AC’ + A’B + A’C’ (B + B’) (Distributive, AND)= AC’ + A’B + A’C’ (1) (Complementation, OR)= AC’ + A’B + A’C’ (Identity Element, AND)= A’B + AC’ + A’C’ (Commutative, OR)= A’B + C’A + C’A’ (Commutative, AND (2 times))= A’B + C’ (A+A’) (Distributive, AND)= A’B + C’ (1) (Complimentation, OR)= A’B + C’ (Identity Element, AND)


Representation: Canonical Forms

Canonical Sum of Products Form (CSOP or m)– Sum of mintermes – Ex.: f (A, B) = (A’•B) + (A•B) = m(1, 3 )

Canonical Product of Sums Form (CPOS or M)

– Product of maxtermes – Ex.: f (A, B) = (A+B’) • (A’+B) = M(1, 2)


Examples of Canonical Forms

According to De Morgan’s law:Mi' = mi and mi' = Mi

Say f(A, B, C) defined by: A B C f 0 0 0 0 0 0 1 1 0 1 0 0 0 1 1 0 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 1


Examples of Canonical Forms *

f(A, B, C) defined by:A B C f0 0 0 00 0 1 10 1 0 00 1 1 01 0 0 11 0 1 11 1 0 11 1 1 1

= A’B’C + AB’C’ + AB’C + ABC’ + ABC= m (1, 4, 5, 6, 7) is the CSOP form

of f


Examples of Canonical Forms *

f(A, B, C) defined by: A B C f 0 0 0 0 0 0 1 1 0 1 0 0 0 1 1 0 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 1

= (A+B+C) • (A+B’+C)• (A+B’+C’)= M (0, 2, 3) is the CPOS form of f.


Equivalence of Canonical Forms*

f(A, B, C) = m (0, 4, 5, 7) = M (?)

M (1, 2, 3, 6)

f(A, B, C, D) = M (2, 3, 5, 6, 7) = m (?)

m (0, 1, 4) ???m (0, 1, 4, 8, 9, 10, 11, 12, 13, 14,

15) !!!


Canonical Forms (encore!)

A function f is not necessarily represented in a canonical form.

f(A, B, C) = A’B’C + AB’+ BC’

How to obtain the canonical forms of such functions?– Algebraic method


Algebraic Method*

f(A, B, C) = A’B + C’ + ABC=A’B(C+C’) + C’(A+A’)(B+B’) +ABC=A’BC + A’BC’ + ABC’ + AB’C’ +

A’BC’ +A’B’C’ + ABC

=A’BC + A’BC’ + ABC’ + AB’C’ +A’B’C’ + ABC

= m(3, 2, 6, 4, 0, 7)


Remarks on Boolean Functions

Single representation (POS or SOP).

Logical equivalence.

How many possible functions for N Boolean variable?

Functions with 2 Boolean variables


Functionally Complete Sets

Together of operators being able to represent all the functions

{AND, NOT, OR}, {NOR}, {NAND}

POS form with {NOR} SOP form with {NAND}


2.4 Logic Gates

The logic gates implement the switching functions

A gate with N inputs represents a function with N Boolean variables

One comes across OR, AND, NOR, and NAND gates with N inputs


Synthesis with Logic Gates

To implement a switching function with logic gates

f(A, B, C) = (A + (BC)')'

BC

A f


Analysis with Logic Gates

To find the functionality of the circuit made up of logic gates

f(A, B, C) = (A + (BC)')'

BC

A f


Synthesis with Single Gates

More economical than {AND, OR, NOT}

SOP form with {NAND}

POS form with {NOR}– Similar to SOP.

AB A+B

A1

B1

A+B


Return to NOR-NAND *

f(A, B, C, D) = AB’ +A’C + D To implement with NAND ( ) only:

= ((AB’ +A’C + D)’)’= ((AB’)’ . (A’C)’ . (D)’)’= (AB’)’ (A’C)’ (D)’= (AB’) (A’ C) (D)’



f(A, B, C, D) = AB’ +A’C + D To implement with NOR ( only:

= ((AB’)’)’ + ((A’C)’)’ + D= (A’+B)’ + (A+C’)’ + D= (A’B) +(AC’) + D= (((A’B) +(AC’) + D)’)’= ((A’B) (AC’) D)’= ((A’B) (AC’) D) 0



g(A, B, C, D) = (A+B’).(A’+C) . D Implement with NAND ( ) only:

= ((A+B’)’)’. ((A’+C)’)’ . D= (A’B)’ . (AC’)’ . D= (A’B) . (AC’) . D= (((A’B) . (AC’) . D)’)’= ((A’B) (AC’) D)’= ((A’B) (AC’) D)



g(A, B, C, D) = (A+B’).(A’+C) . D Implement with NOR ( only:

= (((A+B’).(A’+C).D)’)’= ((A+B’)’ + (A’+C)’ + D’)’= (A+B’)’ (A’+C)’ D’= (AB’) (A’ C) D’


2.5 Adders

Classic combinational circuits Various common circuits

– Half-Adders– Elementary adder– Parallel full-adder– Elementary subtracter– Adder-substracter


Adding words of several bits?

Parallel full-adder of 4 bits EA = Elementary Adder

EA

B4 A4

C4

S4

EA

B3 A3

C3

S3

EA

B2 A2

C2

S2

EA

B1 A1

C1

S1

R5


2.6 Logic Timing Diagram


Complementary Reading

In Mano and Kime– Sections 1.2 and 1.3

• Numbers and binary arithmetic

– Sections 2.1, 2.2, 2.3, 2.6 and 2.7• Boolean algebra, logic gates, canonical forms

– Section 3.8, except « Carry Lookahead Adder »

• Adders

– Sections 3.9 and 3.10• Subtracters

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