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### ece15_2_2012_6

Course: ECE 15a, Spring 2012
School: UCSB
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Word Count: 1570

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negative Representing numbers A negative number is usually indicated by its complement. 2s complement is the most common. ECE 15A Fundamentals of Logic Design Lecture 2 Example: +11 -11 Malgorzata Marek-Sadowska : : 00001011 10001011 (signed magnitude) 11110100 (signed ones complement) 11110101 (signed twos complement) Signed 2s complement has only one representation for 0 (+) Electrical and Computer...

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negative Representing numbers A negative number is usually indicated by its complement. 2s complement is the most common. ECE 15A Fundamentals of Logic Design Lecture 2 Example: +11 -11 Malgorzata Marek-Sadowska : : 00001011 10001011 (signed magnitude) 11110100 (signed ones complement) 11110101 (signed twos complement) Signed 2s complement has only one representation for 0 (+) Electrical and Computer Engineering Department UCSB 2 One's complement format - 8 bit arithmetic One's Complement to Decimal Change the number N to binary, ignoring the sign. Add 0s to the left of the binary number to make a total of 8 bits If the sign is positive, do nothing. If the sign is negative, complement every bit (i.e. change from 0 to 1 or from 1 to 0) In this way we compute (28 -1) - N Convert the following 1's complement representation to decimal: a) 11110001 Since the sign bit is 1, complement the number: 00001110 Convert to decimal: 000011102 = 1410 Put a negative sign in front: -14 b) 00011010 -> 26 3 Two's complement format - 8 bit arithmetic Example Most computers today use 2's complement representation for negative numbers. The 2's complement of a negative number N is obtained by adding 1 to the 1's complement. Write 25 in one's complement 00011001 25 Write -25 in one's complement 8 Since the number is negative, complement each bit 11100110 4 -25 5 This is the same as computing 2 - N For -13 00001101 base integer 11110010 1's complement +1 11110011 2's complement 6 1 Example Two's Complement to Decimal If the sign bit is 0, convert the binary number to decimal. If the sign bit of N is 1 Write -25 in two's complement format. 00011001 11100110 11100111 25 one's complement two's complement Compute 2s complement of N convert the binary number to decimal put a minus sign in front 7 8 Complements Example Convert the following 2's complement representation to decimal: 11100011 Compute 2s complement: 11100011 -> (1s complement: 00011100) -> (Add 1: 00011101) (change to decimal) -> 29 -> (put in front) -> -29 Used to simplify the subtraction Arithmetic subtraction (+-A) - (+B) = (+-A) + (-B) (+-A) - (-B) =(+-A) + (+B) 9 Example (1-s complement) Application: 11000000 -00100111 The carry out of the MSB is added to LSB (end around carry) 10 Example (1-s complement) <=> 1 1 0 0 0 0 0 0 + 1 1 0 1 1 0 0 0(ones complement) 1 1 0 0 11 0 0 0 + 1 (end around carry) 100 1 1 0 0 1 The difference Application: 10110011 -01101101 The carry out of the MSB is added to LSB (end around carry) <=> 101 10011 +10010010 1 0 1 0 0 0 1 0 1(ones complement) 1 (end around carry) ??? 01 0 0 0 1 1 0 The difference 11 12 2 Suppose Example M = 10110011 (negative) N = 00001011 (positive) Example Suppose Compute M-N M = 10110011 (negative) N = 00001011 (positive) Compute M-N 1s complement If M is in 1s complement, then the actual value of M is -1001100 2s complement Subtraction in 1s complement: 10110011 11110100 (original M in 1s complement) (1s complement of N) If M is in 2s complement, then the actual value of M is -01001101 10110011 11110101 (original M in 1s complement) (2s complement of N) So, the result of M-N is: So, the result of M-N is: 110100111 - 1001100 - 0001011 - 1010111 Subtraction in 2s complement: 110101000 - 1001101 - 0001011 - 1011000 +1 10101000 Changing to 2s complement Changing to 1s complement 10101000 10101000 13 14 Overflow Today: The Algebra of Sets 0 The Algebra of Sets is Carry into sign bit Carry out of the sign bit 11 0 1010011 + 0 1011100 1 0101111 carry 1 Sign bit an example of Boolean Algebra useful in manipulating digital circuits 01 1 1010011 + 1 0011100 0 1101111 We will investigate the nature of sets and the way in which they may be combined Sign bit Adding two positive numbers results in a negative number! Adding two negative numbers results in a positive number! 15 16 Definitions Definitions Elements are basic objects Collections of objects constitute sets Example: Basic objects S3 Universal set, denoted by 1: Consists of all elements under consideration S1 S4 Sets The algebra will be developed as an algebra for sets, not for elements of sets Every set is a subset of 1; 0 is a subset of every other set. 0 and 1 are not numbers! m Example, continues: Basic objects Universal set 1 S5 S2 S4 m is a member of S4 m p S3 ={p} S3 S4 S5 Complement of a set S6 Null denoted set, by 0, Contains no elements S6 17 S6 (also denoted as S6) S2 S2 r k u S1 S2 = {r,k,u} S1 S2 S1 S4 = S5 18 3 Example: red, black and yellow books The red (R) books and some of the black books (B) are in English (E); The reminder of the black books (B) are in German (G); Yellow books (Y) are in French (F). B R Rules of forming new sets The union of sets X and Y is a set X+Y consisting of all elements which are either in X or in Y. Example: Y E R B G B R E Y R+Y = B E G F R+E=E G Y=F F 19 20 Rules of forming new sets Theorem The intersection of sets X and Y is a set X Y consisting of all elements which are both in X and in Y. Example: If m 1 and X,Y are arbitrary sets, then m is a member of only one of the sets: XY, XY, XY, XY. E m X or m X, but not to both. m Y or m Y, but not to both. If m X then mXY or m XY, but not to both. If mX then mXY or m XY, but not to both. G F BY R E B is the set of black bound books which are in English R Y is a null set R E=R For any set X: X + X = 1; X X = 0 E G F 21 Symbols in common use Meaning 22 Venn diagrams Symbolic notation X Y , X Y , X Y XY , X Y , X Y Union of sets X and Y Intersections of X and Y Complement of X X ', X Y The set of points interior to the rectangle is taken as the universal set. X, ~ X XY Example B R G Y Say in words, what are: Y+G; RB, G(B+R); B+BR F Show that (G+R) = GR; (GB)=G+B XY X+Y E (X+Y) (all blue) (X+Y)=XY X Y X X (all blue) Y Y (all blue) XY XY (all blue) 23 24 4 Venn diagrams for X+YZ=(X+Y)(X+Z) Y X Y X Z Z YZ Y X X+YZ Z Each of the laws given here may be justified intuitively using Venn diagrams Y X Z X+Y Commutative laws (1a) XY = YX X+Z (1b) X+Y = Y+X Associative laws Y X Fundamental laws of the algebra of sets (Also valid in any Boolean algebra) (2a) X(YZ) = (XY)Z Z (2b) X+(Y+Z) = (X+Y)+Z Distributive laws (3a) X(Y+Z) = XY+XZ (X+Y)(X+Z) (3b) X+YZ= (X+Y)(X+Z) 25 Fundamental laws of the algebra of sets (Also valid in any Boolean algebra) 26 Fundamental laws of the algebra of sets (Also valid in any Boolean algebra) Laws of Tautology (4a) XX = X Laws of De Morgan (8a) (XY) = X+Y (4b) X+X = X Laws of Absorbtion (5a) X(X+Y) = X (9a) 0X=0 (10a) 1X=X (11a) 0=1 (5b) X+XY = X Laws of Complementation (6a) XX=0 (8b) (X+Y) = XY Operations with 0 and 1 (6b) X+X=1 (9b) 1+X=1 (10b)0+X=X (11b) 1=0 Expressions such as 2X or X2 will never appear in the algebra of sets. Double Laws of Complementation Principle of duality: If in any identity, each union is replaced by intersection, each intersection by union, 0 by 1, and 1 by 0, then the resulting equation is also an identity. (7) (X)=X 27 28 Monomials and polynomials Factors A monomial is either a single letter representing a set (with or without a prime), or a product of two or more of such symbols representing the intersection of sets. In any expression representing an intersection of sets, each such set is called a factor of the set of intersection. X, Y, XYZ are examples of monomials The factors of X(Y+Z) are X and Y+Z. A factor is linear if it is a letter (possibly with a prime) or a sum of such letters. The polynomial represents the union of the sets corresponding to separate terms X+Z is linear (X+Y), X+YZ are not linear X+Y+XYZ is an example of a polynomial 29 30 5 Expanding expressions into polynomials Theorem Expand (X+Y)(Z+W) into a polynomial (X+Y)(Z+W)=(X+Y)Z+(X+Y)W by (3a) [x(y+z)=xy+xz] =Z(X+Y)+W(X+Y) by(1a) [xy=yx] =ZX+ZY+WX+WY (by 3a) Factor AC+AD+BC+BD into linear factors AC+AD+BC+BD=A(C+D)+B(C+D) (by 3a) =(A+B)(C+D) (by 1a and 3a) It can be shown, that any expression can be factored into linear factors by repeated application of the (3b). [x+yz=(x+y)(x+z)] For any sets X and Y, X+XY=X+Y Proof: (X+XY) = (X+X)(X+Y) (by 3b) [x+yz=(x+y)(x+z)] = 1(X+Y) (6b) [x+x=1] = X+Y 31 X Y X Y XY 32 At home Get ready for discussion #1: Do the following problems from CHR: 2.1- 2.3; 2.5; 2.6; 2.11. Read before lecture #3: CHR, pp.62-68. Homework #1 is posted, due Monday, January 23 at noon. You can read more on the algebra of sets in R. Serrel, Elements of Boolean Algebra for the Study of Information-Handling Systems, in Proc. of the IRE, vol. 41, issue 10, Oct. 1953, pp.1366-1380. This publication can be found on IEEE Xplore: http://www.library.ucsb.edu/eresources/databases/ Scroll down to I and find IEEE Xplore 33 6
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