Lecture 4 Nand, Nor Gates, CS147 Circuit Minimization and Karnaugh Maps Prof. Sin-Min Lee Department of Computer Science

Boolean Algebra to Logic Gates Logic circuits are built from components called logic gates. The logic gates correspond to Boolean operations +, *, ’. Binary operations have two inputs, unary has one OR + AND * NOT ’

Logic Circuits ≡ Boolean Expressions All logic circuits are equivalent to Boolean expressions and any boolean expression can be rendered as a logic circuit. AND-OR logic circuits are equivalent to sum-of-products form. Consider the following circuits: A B C A Y=AB*+B*C B ABC C A B C Y AB*C Y A*B y=ABC+AB*C+A*B

NAND and NOR Gates NAND and NOR gates can greatly simplify circuit diagrams. As we will see, can you use these gates wherever you could use AND, OR, and NOT. A B AB 1 NAND A B AB 1 NOR

XOR and XNOR Gates XOR is used to choose between two mutually exclusive inputs. Unlike OR, XOR is true only when one input or the other is true, not both. A B AB 1 A B A B 1 XOR XNOR

NAND and NOR as Universal Logic Gates Any logic circuit can be built using only NAND gates, or only NOR gates. They are the only logic gate needed. Here are the NAND equivalents:

NAND and NOR as Universal Logic Gates (cont) Here are the NOR equivalents: NAND and NOR can be used to reduce the number of required gates in a circuit.

Example Problem A hall light is controlled by two light switches, one at each end. Find (a) a truth function, (b) a Boolean expression, and (c) a logic network that allows the light to be switched on or off by either switch. Let x and y be the switches: x y f(x,y) 1 (What kind of gate has this truth table?

x y f(x,y) 1 Example (cont) One possible equation is the complete sum-of-products form: f(X,Y) = XY* + X*Y Use The Most Complex Machine xLogicCircuit Module to implement the equation.

Hint 2 : Use 2 NAND gates to build 2 NOT gates How to use NAND gates to build an OR gate? Truth Table A B C D Q A B C D Q 1 Hint 1 : Use 3 NAND gates Hint 2 : Use 2 NAND gates to build 2 NOT gates Hint 3 : Put the 3rd NAND gate after the 2 “NOT” gates

Hint 2 : Use 3 NAND gates to build an OR gate How to use NAND gates to build a NOR gate? Truth Table A B C D Q E A B C D E Q 1 Hint 1 : Use 4 NAND gates Hint 2 : Use 3 NAND gates to build an OR gate Hint 3 : Use a NOR gate to build a NOT gate Hint 4 : Put the “NOT” gate after “OR” gate

Since functions can be represented by a sum of product form of minterms, any function can be shown in the map by placing each a 1 in each square which represents a minterm in the function. EXAMPLE: Draw the map for f = X + Y 1. Determine which minterms are needed by using truth table. X Y f = X + Y minterm 0 0 0 m0 0 1 1 m1 1 0 1 m2 1 1 1 m3

2. Draw the map and place a 1 in each square for required minterms. Y X 1 1 X 1 f = X + Y = m1 + m2 + m3 = X´ Y + X Y´ + X Y

Use the 2-Variable Karnaugh Map for Minimization Example: Given f (X,Y) =  (0, 2) Find: Simplified sum of products Y Y X m0 m1 X m2 m3

2-Variable Karnaugh Map 1. Place the 1’s corresponding to the minterms on the map. f (X,Y) =  (0, 2) Y Y X 1 X 1

2-Variable Karnaugh Map 2. Now group the 1’s into columns or rows; in this case we can group them in the first column. f (X,Y) =  (0, 2) Y Y X 1 X 1 3. This column is Y´ so the simplified function f (X,Y) = Y´

2-Variable Karnaugh Map Example 2: Given f (X,Y) =  (0, 2, 3) Find: Simplified sum of products Y Y X m0 m1 X m2 m3

2-Variable Karnaugh Map Example 3: Given f (X,Y) =  (0, 1) Find: Simplified sum of products Y Y X m0 m1 X m2 m3

Three Variable Map m0 m1 m3 m2 m4 m5 m7 m6 Y YZ X 0 0 0 1 1 1 10 0 0 0 1 1 1 10 X´ Y´ Z´ X´ Y´ Z X´ Y Z X´ YZ´ 1 X Y´ Z´ XY´Z X Y Z X Y Z´ X Z

Three Variable Map There are 2N = 23 = 8 squares. The minterms are arranged so that only one variable changes from 0 to 1 or from 1 to 0 as you move from square to square in the vertical or horizontal direction. For any two adjacent squares, only one literal changes from complemented to non-complemented (normal). From this property the left and right ends of the map are “adjacent.”

Y YZ X 0 0 0 1 1 1 10 X´Y´Z´ X´ Y´ Z X´ Y Z X´ Y Z´ X 1 X Y´Z´ X Y´ Z X Y Z X YZ´ Z Using the distributive and complement properties, any two adjacent minterms can be combined and simplified to a single term with one less literal.

Combining terms on the Karnaugh Map simplifies Boolean functions. Example: combine adjacent squares for X´ Y´ Z and XY´Z X´Y´ Z + XY´Z = Y´ Z ( X´ + X) = Y´ Z · 1 = Y´ Z Example: Combine adjacent squares for X Y´Z´ and X YZ´ XY´Z´ + X YZ´ = X Z´ ( Y´ + Y) = X Z´ · 1 = X Z´ Y YZ X 0 0 0 1 1 1 10 X´Y´Z´ X´ Y´ Z X´ Y Z X´ Y Z´ X 1 X Y´Z´ X Y´ Z X Y Z X YZ´ Z

Example: Simplify f = X´Y´ Z´ + X Y Z + X´Y´ Z + X´ Y Z Y YZ X 0 0 0 1 1 1 10 1 1 1 1 X 1 Z f = X´Y´ + Y Z

Even if the function is not in its simplest form, we can still use the map to simplify it further. Example: Simplify f = X´ Y´ + Y´ Z + X´ Z + X Y Z Y YZ X 0 0 0 1 1 1 10 1 1 1 1 1 X 1 Z f = Z + X´ Y´ (by further grouping of minterms)

3-Variable Karnaugh Map Example: Given f (X,Y,Z) =  (0, 2, 3, 4, 7) Find: Simplified sum of products YZ Y X m0 m1 m3 m2 m5 m7 m6 m4 X Z

3-Variable Karnaugh Map Example: Given f (X,Y,Z) =  (0, 2, 3, 4, 7) Simplified sum of products: f = YZ + YZ + XY YZ Y X 1 1 1 1 1 X Z

Truth Table f (X,Y,Z) =  (0, 2, 3, 4, 7) Simplified sum of products: f = Y´Z´ + YZ + X´Y The truth table for f: X Y Z f 0 0 0 1 0 0 1 0 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 0 1 1 0 0 1 1 1 1

f (X,Y,Z) =  (0, 2, 3, 4, 7) = Y´Z´ + YZ + X´Y Y Z Y f Z X Y

3-Variable Karnaugh Map Example 2: Given f (X,Y,Z) =  (2, 3, 4, 5) Find: Simplified sum of products YZ Y X m0 m1 m3 m2 m5 m7 m6 m4 X Z

3-Variable Karnaugh Map Example 2: Given f (X,Y,Z) =  (1, 2, 5, 6, 7) Find: Simplified sum of products YZ Y X m0 m1 m3 m2 m5 m7 m6 m4 X Z

3-Variable Karnaugh Map Example 3: Given f (X,Y,Z) = X´Z + X´Y + XY´Z + YZ Find: “Sum of minterms” expression YZ Y X m0 m1 m3 m2 m5 m7 m6 m4 X Z

Exclusive OR XOR f = X´Y + XY´ = X  Y X Y f 0 0 0 0 1 1 1 0 1 1 1 0

Exclusive OR f (X,Y) =  (1, 2) = X  Y Y Y X 1 1 X

XOR Truth Table f (X,Y,Z) = X  Y  Z =  (1, 2, 4, 7) The truth table for f: X Y Z f 0 0 0 0 0 0 1 1 0 1 0 1 0 1 1 0 1 0 0 1 1 0 1 0 1 1 0 0 1 1 1 1

Exclusive OR X Y f (X,Y,Z) = X  Y  Z Z

Four Variable Map Y YZ WX 0 0 0 1 1 1 10 m0 m1 m3 m2 m4 m5 m7 m6 00 01 0 0 0 1 1 1 10 m0 m1 m3 m2 m4 m5 m7 m6 W x´y´z 00 01 m12 m13 m15 m14 m8 m9 m11 m10 X 11 w x y´z W 10 Z

Four Variable Map N = 4 variables 2N = 24 = 16 square (minterms) Row and column are numbered using a reflected-code sequence. The minterm number can be obtained by concatenation of the row and column number . Example: Row 4 = 10, Column 2 = 01 giving 1001 = 9 decimal for W X´ Y´ Z.

1 Y YZ 0 0 0 1 1 1 10 wx´y´z wx´y´z´ wx´y´z´ w´x´y z´ 00 01 X 11 W 10 0 0 0 1 1 1 10 wx´y´z 1 wx´y´z´ wx´y´z´ w´x´y z´ 00 01 X 11 W 10 Z Notice that top and bottom edges and right and left edges are “adjacent.”

1 square = a term with 4 literals 16 square = a function equal to 1 Y YZ 0 0 0 1 1 1 10 WX 00 01 X 11 W 10 Z

Simplify: f (W, X, Y, Z) = W X´ Z + W X Z + W´ Y Z f = W Z + Y Z f = Z ( W + Y) Y YZ 0 0 0 1 1 1 10 WX 00 01 X 11 W 10 Z

Simplify: f (W, X, Y, Z) =  ( 0, 1, 4, 5, 6, 8, 9, 12, 13, 14) f = Y´ + WZ´ + XZ´ Y YZ 0 0 0 1 1 1 10 WX 00 01 X 11 W 10 Z 8 square 4 square 4 square

Simplify: f = W´ X´ Y´ + X´Y Z´ + W´ X Y Z´ + W X´ Y´ f = X´ Z´ + X´ Y´ + W´ Y Z´ Y YZ 0 0 0 1 1 1 10 WX 00 01 X 11 10 Z

Simplification of kmap Generate all PIs Find EPIs If EPIs can cover all minterms, then it is answer. Otherwise choose some non-essential PIs (which has less cost) such that all minterms are cover.

Step 1 Generate PIs Blue circle are PIs They are the largest circle you can drawn on kmap

Step 2 Find EPIs Red circle are EPIs Minterm 5, 14, 11 can only be cover by these 3 red circle

Step 3 EPIs cannot cover minterm 7 Choose between green/blue circle to cover minterm 7 Green is chosen as it is larger Less cost

Final Final result is obtained x3x4’ + x2’x3 + x1’x3 + x2x3’x4

Complement and Product of Sums On the K Map a 1 was placed in the squares which represented minterms for a function. The squares not used represent the complement of the function. Mark these squares with 0, combine them and read the complement as a sum of products function, f´. Using De Morgan’s Law on f´ will give f, the original Function. It will be in a product of sums form. Example: Determine the Product of sums form for f ( W, X, Y, Z) =  (0, 1, 2, 5, 8, 9, 10) f´ = W X + YZ + X Z´ from the map f = (W´ + X´) ( Y´ + Z´) ( X´ + Z) by De Morgan’s

Complement and Product of Sums Also, can determine the product of sums form directly from The map by recognizing that the 0’s on the map represent maxterms. Remember that a 1 on the edge of the map Represents a complemented literal for maxterms. Thus, for f ( W, X, Y, Z) =  (0, 1, 2, 5, 8, 9, 10)……. (combine 0’s on k-map)… f = ( Y´ + Z´) (W´ + X´) ( X´ + Z)

Y YZ 0 0 0 1 1 1 10 WX 1 1 0 1 00 0 1 0 0 01 11 0 0 0 0 W 10 1 1 0 1 Z f = ( Y´ + Z´ ) ( W´ + X´ ) ( X´ + Z)