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ex Gray Codes = reflected binary codes Karnaugh Maps: find (minimal) - PowerPoint PPT Presentation

Dec 08, 2023 •602 likes •687 views

But first Recall: sum of products logical sum (OR) of products (AND) 1. Basic combinational building blocks of inputs or their complements (NOT). 2. Logic for arithmetic Construct with: A B C M 1 code detector per 1-valued output

  1. But first… Recall: sum of products logical sum (OR) of products (AND) 1. Basic combinational building blocks of inputs or their complements (NOT). 2. Logic for arithmetic Construct with: A B C M • 1 code detector per 1-valued output row 0 0 0 0 • 1 large OR of all code detector outputs 0 0 1 0 Common combinational circuits: encoders, decoders, multiplexers, 0 1 0 0 adders, Arithmetic Logic Unit Is it minimal? 0 1 1 1 1 0 0 0 1 0 1 1 (printed together, separate sets of slides online) 1 1 0 1 1 1 1 1 ex Gray Codes = reflected binary codes Karnaugh Maps: find (minimal) sums of products CD gray code Alternate binary encoding order 00 01 11 10 A B C D F(A, B, C, D) designed for electromechanical switches and counting. 0 0 0 0 0 00 0 0 0 0 0 0 0 1 0 0 0 1 0 0 01 0 0 0 1 0 0 1 1 0 AB 00 01 11 10 0 1 0 0 0 11 1 1 0 1 0 1 0 1 0 0 1 2 3 0 1 1 0 1 10 1 1 1 1 0 1 1 1 0 000 001 011 010 110 111 101 100 1 0 0 0 1 1. Cover exactly the 1s by drawing a (minimum) number of 1 0 0 1 1 0 1 2 3 4 5 6 7 maximally sized rectangles whose dimensions (in cells) 1 0 1 0 1 are powers of 2. (They may overlap or wrap around!) 1 0 1 1 1 2. For each rectangle, make a product of the inputs (or 1 1 0 0 1 How many bits change when incrementing? complements) that are 1 for all cells in the rectangle. 1 1 0 1 1 ( minterms ) 1 1 1 0 1 3. Take the sum of these products. 1 1 1 1 0

  2. ex Voting again with Karnaugh Maps Toolbox: Building Blocks Processor datapath Microarchitecture Instruction Decoder Memory Arithmetic Logic Unit A B C M Adders Registers 0 0 0 0 Multiplexers 0 0 1 0 Demultiplexers Digital Logic Flip-Flops 0 1 0 0 Encoders Latches Decoders 0 1 1 1 1 0 0 0 Gates 1 0 1 1 1 1 0 1 Devices (transistors, 1 1 1 1 etc.) ex Multiplexers Decoders Select one of several inputs as output. Decodes input number, asserts corresponding output. n -bit input (an unsigned number) D 0 2 n outputs D 1 Built with code detectors. D 2 8-to-1 2 n data inputs D 3 F 1 data output MUX D 4 D 0 D 5 D 0 B 0 D 1 D 6 B 0 D 1 D 7 D 2 B 1 D 2 B 1 D 3 D 3 A B C n selector lines

  3. ex Build a 2-to-1 MUX from gates 8-to-1 MUX If S=0, then F=D 0 . D 0 2-to-1 If S=1, then F=D 1 . F MUX D 1 1. Construct the truth table. S 2. Build the circuit. Costume idea: MUX OX MUX + voltage source = truth table Buses and Logic Arrays A bus is a collection of data lines A B C M treated as a single logical signal. 0 0 0 0 0 = fixed-width value 0 0 1 0 1 0 1 0 0 2 Array of logic elements applies 8-to-1 0 1 1 1 3 same operation to each bit in a bus. M MUX 1 0 0 0 4 = bitwise operator 1 0 1 1 5 1 1 0 1 6 1 1 1 1 7 A B C

  4. Addition: Carry in ex ex 1-bit full adder A start small with a 1-bit (half) adder + Sum B Carry out n-bit addition: Sum i = A i + B i + CarryOut i-1 Need a bigger adder! A B Carry in Carry out Sum Carry in 0 0 0 A B Carry out Sum 0 0 1 A A 0 0 0 1 0 Sum Sum 0 1 0 1 1 B B 1 0 1 0 0 1 1 1 0 1 Carry out Carry out 1 1 0 1 1 1 n-bit ripple-carry adder Carry in Processor Components A 0 + Sum 0 B 0 A 1 + Sum 1 4 1 3 2 B 1 A 2 + Instruction Sum 2 ALU Registers Memory Fetch and B 2 Decode … A n-1 + Sum n-1 B n-1 Carry out There are faster, more complicated ways too…

  5. 1 ex Arithmetic Logic Unit (ALU) 1-bit ALU for bitwise operations Build an n-bit ALU from n 1-bit ALUs. Each bit i in the result is computed from the corresponding bit i in the two inputs. a few bits Condition Codes Op A B Result (sign, overflow, carry-out, zero) 0 0 0 Operation word Operand A 0 0 1 ALU words A Result 0 1 0 0 Operand B MUX 0 1 1 Result 1 0 0 B 1 Operation 1 0 1 a few bits 1 1 0 Hardware unit for arithmetic and bitwise operations. 1 1 1 Carry in 2 Operation 1-bit ALU A 0 n-bit ripple carry adder 0 Result 0 MUX 1 Carry in Sum + B 0 2 A 0 + Operation Sum 0 A 1 B 0 Carry in 2 0 Result 1 MUX A A 1 1 + 0 Sum 1 Sum + B 1 2 B 1 MUX Result A 2 + .... n-bit ALU 1 … Sum 2 B 2 Sum … + A n-1 2 0 B Result n-1 MUX A n-1 + 1 Sum n-1 Sum + B n-1 B n-1 2 Carry out Carry out Carry out

  6. Operation Carry in 2 ALU conditions Operation 2 A 0 0 Add subtraction A 0 Result 0 MUX 0 1 Result 0 Extra ALU outputs MUX + Sum B 0 2 1 describing properties of result. How can we control ALU inputs B 0 + 0 2 or add minimal new logic A 1 Zero Flag: 1 0 to compute A-B ? Result 1 MUX 1 if result is 00...0 else 0 1 A 1 Sum + 0 Sign Flag: B 1 2 Result 1 MUX 1 if result is negative else 0 1 .... … B 1 + Carry Flag: 0 2 1 1 if carry out else 0 .... .... … (Signed) Overflow Flag: A n-1 0 Result n-1 1 if signed overflow else 0 MUX 1 A n-1 Sum + 0 B n-1 2 Result n-1 MUX Implement these. Carry out 1 + B n-1 0 2 1 Carry out Negate B Operation Invert A A NAND B 2 A 0 0 Controlling the ALU 0 1 Result 0 MUX 1 A NOR B B 0 + 0 2 1 Condition Codes ALU control lines Function A 1 0 0 1 0000 AND A<B Result 1 Operand A MUX 1 0001 OR ALU + Result B 1 0 2 0010 add 1 Operand B A==B 0110 subtract .... .... .... … 1100 NOR Control Lines A n-1 0 0 1 Result n-1 MUX How can we control ALU inputs 1 or add minimal new logic B n-1 + 0 ex to compute each? 2 1 Carry out

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