Sana'a University Faculty of Engineering Department of Mechatronics LOGIC SYSTEM DESIGN Lab Manual 2015 Prepared by: Eng. Zakarya EXPERIMENT 1 VERIFICATION AND INTERPRETATION OF TRUTH TABLES Objective: Verification and interpretation of truth tables for AND, OR, NOT, NAND, NOR Exclusive OR (EX-OR), Exclusive NOR (EX-NOR) Gates. Theory: Logic gates are electronic circuits which perform logical functions on one or more inputs to produce one output. There are seven logic gates. When all the input combinations of a logic gate are written in a series and their corresponding outputs written along them, then this input/ output combination is called Truth Table. Procedure: 1- Connect the inputs to the input switches. 2- Connect the output to the switches of O/P LEDs. Apply various combinations of inputs according the truth tables. Inverter Gate (NOT Gate – 7404LS) 2-Input AND Gate (7404LS) 2-Input OR Gate (7432LS) 2-Input NAND Gate (7400LS) 2-Input NOR Gate (7402LS) 2-Input EX-OR Gate (7486LS) 3-Input NAND Gate (7410LS) 2-Input NAND Gate (CD4011) 2-Input NOR Gate (CD4001) 4-Input NAND Gate (7420LS) 2-Input EX-NOR (IC 74266) EXPERIMENT 2 REALIZATION OF LOGIC FUNCTIONS Objective: Realization of logic functions with the help of universal gates-NAND Gate. Theory: NAND gate is actually a combination of two logic gates: AND gate followed by NOT gate. So its output is complement of the output of an AND gate. This gate can have minimum two inputs, output is always one. By using only NAND gates, we can realize all logic functions: AND, OR, NOT, X-OR, X-NOR, NOR. So this gate is also called universal gate. NAND gates as NOT gate: Y = (A.A)’ => Y = (A)’ NAND gates as AND gate: Y = ((A.B)’)’ => Y = (A.B) NAND gates as OR gate From DeMorgan’s theorems: (A.B)’ = A’ + B’ NAND gates as NOR gate: Y = (A + B)’ NOR gates as NOT gate: Y = (A+A)’ => => (A’.B’)’ = A’’ + B’’ = A + B Y = (A)’ NOR gates as OR gate: Y = ((A+B)’)’ NOR gates as AND gate From DeMorgan’s theorems: (A+B)’ = A’B’ NOR gates as X-OR gate: NOR gates as NAND gate: => => Y = A’B+ AB’ Y = (AB)’ Y = (A+B) (A’+B’)’ = A’’B’’ = AB