1 논리공학 고려대학교 전기전자전파공학부 김종국

2  1 or 0, yes or no, true or false …  디지털 시스템을 위한 기초  거의 모든 전기전자제품은 디지털 시스템 e.g., 핸드폰, 전자시계, 컴퓨터 …  이진수  필요성  십진수에서 이진수 변환  불 대수  논리회로 설계 (combinational circuits)  논리회로 간소화 / 최적화 논리 공학

3  자동판매기 (vending machine)? 논리 공학 input ( 입력 ) output ( 출력 ) ?

4 이진수  a 5 a 4 a 3 a 2 a 1 a 0.a -1 a -2 a -3 = a n r n +a n-1 r n a 2 r 2 +a 1 r+a 0 +a -1 r -1 +a -2 r a -m r -m (power series) 7392 = 7 × × × × 10 0 ( ) 2 = = 1Kilo = 1Mega = 1Giga

× 이진수

6 수체계 변환  베이스 r 수에서 십진수로 변환  Power series 를 이용  Term 들을 더함 ( ) 2 = 1×2 4 +1×2 3 +0×2 2 +1×2 1 +0×2 0 +1× ×2 -2 = (26.75) 10  십진수에서 베이스 r 수로 변환  정수 부분과 소수 부분을 따로함  정수 부분 : 숫자를 베이스 r 로 나누고 이후 몫을 다시 r 로 나누고 나머지를 읽으면 됨  소수 부분은 ?  ( ) 10 = (?) 2  ( ) 10 = (?) 8

7 수체계 변환  Ex 1-1) Convert decimal 41 to binary. answer : (41) 10 = (a 5 a 4 a 3 a 2 a 1 a 0 ) 2 = (101001) 2 Integer Quotient RemainderCoefficient 41/2 =20+½a 0 = 1 20/2 =10+0a 1 = 0 10/2 =5+0a 2 = 0 5/2 =2+½a 3 = 1 2/2 =1+0a 4 = 0 1/2 =0+½a 5 = 1 IntegerRemainder answer =

8  Ex 1-2) Convert decimal 153 to octal.  Ex 1-3) Convert (0.6875) 10 to binary. 수체계 변환 = (231) 8 IntegerFractionCoefficient ×2 = a -1 = ×2 = a -2 = ×2 = a -3 = ×2 = a -4 = 1 answer:(0.6875) 10 = (0.a -1 a -2 a -3 a -4 ) 2 = (0.1011) 2

9 수체계 변환  ( ) 10 = ( ) 2  ( ) 10 = ( ) 8

10 Binary Logic  definition of binary logic  logic gates

11 Theorems and Properties of Boolean Algebra  operator precedence 1. parentheses 2. NOT 3. AND 4. OR

12 Boolean Algebra  commutative law: the order doesn’t affect the result  X + Y = Y + X  XY = YX  associative law: parentheses can be removed  (X + Y) + Z = X + (Y + Z )  (XY)Z = X(YZ)  distributive law:  X(Y+Z) = XY + XZ  X + YZ = (X + Y)(X + Z)  DeMorgan’s theorem: complement of an expression  (X + Y) = XY  (XY) = X + Y  can be extended to three or more variables

13 Boolean Functions

14 Simplification (Minimization)?

15  ex 2-1) Simplify the following Boolean functions to a minimum number of literals. 1. x(x'+y) = xx' + xy = 0 + xy = xy. 2. x +x'y = (x+x')(x+y) = 1(x+y) = x + y. 3. (x+y)(x+y') = x + xy + xy' + yy' = x(1+y+y') = x. 4. xy + x'z + yz = xy + x'z + yz(x+x') = xy + x'z + xyz + x'yz = xy(1+z) + x'z(1+y) = xy + x'z Boolean Functions – Algebraic Manipulation

16  (A + B + C)'= (A+x)' let B+C=x = A'x' by theorem 5(a)(DeMorgan) = A'(B+C)' substitute B+C=x = A'(B'C') by theorem 5(a)(DeMorgan) = A'B'C' by theorem 4(b)(associative) => (A+B+C+D+…+F)' = A'B'C'D'…F' (ABCD…F)' = A' +B'+ C' + D' + … + F' Complement of a Function

17  ex 2-2) Find the complement of the functions F 1 =x'yz'+x'y'z, F 2 =x(y'z'+yz). F 1 ' = (x'yz'+x'y'z)' = (x'yz')'(x'y'z)' = (x+y'+z)(x+y+z') F 2 ' = [x(y'z'+yz)]' = x'+(y'z'+yz)' = x'+(y'z')'(yz)' = x'+(y+z)(y'+z')  ex 2-3) Find the complement of the functions F 1 And F 2 in Ex 2-2 by taking their duals and complementing each literal. 1. F 1 = x'yz' + x'y'z. The dual of F 1 is (x'+y+z')(x'+y'+z) Complement each literal : (x+y'+z)(x+y+z')=F 1 ' 2. F 2 = x(y'z'+yz). The dual of F 2 is x+(y'+z')(y+z) 이다. Complement each literal : x'+(y+z)(y'+z')=F 2 ' Complement of a Function

18 Simplification Example  F = XYZ + XYZ + XY = XZ(Y + Y) + XY = XZ + XY  circuit simplification  less gates  (a) 2 NOT, 3 AND, and 1 OR  (b) 1 NOT, 2 AND, and 1 OR  less fan-in

19 Minterm and Maxterm (1)  minterm ( a product term) and Maxterm (a sum term)  all variables appear exactly once, either complemented or uncomplemented  show exactly one combination of the binary variables  2 n distinct terms for n variables

20 Minterm and Maxterm (2)  minterm  term: x, y, z are such values, when do they give us 1?  maxterm  term: x, y, z are such values, when do they give us 0?  j is the designation number for terms: M j = m j c  m 3 = (XYZ) = X + Y + Z = M 3

21 f 1 = x'y'z+xy'z'+xyz = m 1 +m 4 +m 7 f 2 = x'yz+xy'z+xyz'+xyz = m 3 +m 5 +m 6 +m 7 f 1 = (x+y+z)(x+y'+z)(x'+y+z')(x'+y'+z) = M 0 M 2 M 3 M 5 M 6 f 2 = (x+y+z)(x+y+z‘)(x+y'+z)(x'+y+z) = M 0 M 1 M 2 M 4 Canonical and Standard Forms

22  conversion between canonical forms F(A, B, C) = ∑(1, 4, 5, 6, 7) F' (A, B, C) = ∑(0, 2, 3) = m 0 + m 2 + m 3 F = (m 0 +m 2 +m 3 )' = m 0 'm 2 'm 3 ' = M 0 M 2 M 3 = ∏(0, 2, 3), m j ' = M j ex) F = xy + x'z F(x, y, z) = ∑(1, 3, 6, 7) F(x, y, z) = ∏(0, 2, 4, 5) Canonical and Standard Forms

23 Gray Code  gray code of three variables  1 bit difference between neighbors  convert from the lower bit order  if the converted binary number already exists, then try to convert the bit that is one higher

24 Karnaugh Map (K-Map)  simplify Boolean functions of up to four variables (5 or 6 variables can be drawn, but cumbersome to use)  simplified expressions are in sum-of-products or product-of- sums (two-level implementations)  diagram of squares, one indicating one minterm

25 Karnaugh Map (K-Map) Algorithm  pick the largest 2 n element group (n = 0, 1, …)  the already-grouped element can be regrouped for another group  all elements in a group cannot be a member of another group

26 Two-Variable Map m1+m2+m3 = x'y +xy' +xy = x +y

27 Three-Variable Map

28 Three-Variable Map Examples

29 Four-Variable Map

30 Four-Variable Map Examples Ex 3-5) Simplify the Boolean function, F(w, x, y, z) = Σ(0, 1, 2, 4, 5, 6, 8, 9, 12, 13, 14)

31 Product-of-Sum  from sum-of-products to product-of-sums  complement the function (taking the dual)  combine squares marked with 0’s  change the function, which was expressed in sum-of- products to product-of-sums

32 Product of Sums Simplification  ex 3-8) Simplify the Boolean function, F(A, B, C, D) = Σ(0, 1, 2, 5, 7, 9, 10) a) sum of products F = B'D' + B'D' +A'C'D' b) product of sum F' = AB + CD + BD' F = (A' +B')(C' +D')(B' +D)

33  F( x, y, z) = Σ(1, 3, 4, 6) = Π(0, 2, 5, 7) F = x'z +xz' F' = xz +x'z' F = (x'+z)(x + z') Product of Sums Simplification

34 So Far…  simplification  Boolean algebra  K-map  group 1’s: sum-of-products (minterms)  group 0’s: product-of-sums (maxterms)

35 Don’t care condition  unspecified minterms of a function  marked with a cross or a “d”  provides further simplification of the function  “d” can become 0 or 1

36 Don’t-Care Conditions Example  ex 3-9) Simplify the Boolean function, F(w, x, y, z) = Σ(1,3,7,11,15) Don’t-care conditions, d(w, x, y, z) = Σ(0, 2, 5) F(w, x, y, z) = yz + w'x' = Σ(0, 1,,2, 3, 7, 11, 15) F(w, x, y, z) = yz + w'z = Σ(1, 3, 5, 7, 11, 15)

37  Extension to Multiple Inputs - The NAND and NOR operators are not associative. (x↓y)↓z≠x↓(y↓z) (x↓y)↓z= [(x+y)'+z]' = (x+y)z'= xz' + yz' x↓(y↓z)= [x+(y+z)'] ' = x'(y+z)= x'y + x'z x↓y↓z= (x+y+z)' x↑y↑z= (xyz)' Digital Logic Gate F = [(ABC)'(DE)']' = ABC + DE

38 NAND Implementation  NAND Circuit

39  F = ((AB)'(CD)')' = AB + CD  ex 3-10) Implement the following Boolean function with NAND gates: F(x, y, z) = Σ(1, 2, 3, 4, 5, 7) = xy' + x'y + z NAND Implementation

40 NAND Implementation

41 NOR Implementation  F = (AB' + A'B)(C + D')

42 Exclusive-OR Function  x  y = xy' + x'y (x  y)' = (xy' + x'y)' = xy + x'y' x  0=x x  1=x' x  x=0 x  x'=1 x  y'=x'  y=(x  y)‘  A  B = B  A (A  B)  C = A  (B  C) = A  B  C

43  parity generation and checking P = x  y  z C = x  y  z  P Exclusive-OR Function Application