Lecture 12 - This Week Logic System Design Truth Tables,...

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1 ECTE290 Spring 2007 Lecture 12 2 This Week ± Logic System Design ± Truth Tables, minterms ± Karnaugh Maps ± Positive Logic ± Voltage Transformers 3 Example: Problem Description ± Consider a buzzer which sounds when ± The lights are on and ± The door is open and ± No key is in the ignition Variable Value Situation A1 Lights are on 0 Lights are off B1 Door is open 0 Door is closed C1 Key is in ignition 0 Key is out of ignition P1 Buzzer is on 0 Buzzer is off Alarm system Active B A C P How to design a logic circuit to realize this logic? Desired logic Truth table K-Map Logic function Logic circuit 4 Example cont. ± Truth Table ± A Truth Table can be used to show the relationships between : ± the 3 inputs and ± the single output ± Implementation as a circuit using logic gates 000 0 001 0 010 0 011 0 100 0 101 0 110 1 111 0 ABC P A B P C lights door keys buzzer
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2 5 Example of a Truth Table ± Three switches “A”, “B” and “C” are used to generate an output signal high (logic 1) for certain combinations of switching positions else it is low (logic 0). ± The output signal is always high except when all switches are OFF (logic 0) or Switch A and Switch C are OFF and Switch B is ON (logic 1). ± List all possible combinations of switch states A, B and C in the blank truth table shown here. ± Complete the truth table as per instructions mentioned above (determine the output signal for various A, B and C switch states). A B C Output A B C Output 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 0 1 0 1 1 1 1 1 6 Revision on Basic Gates ± AND P = A.B ± OR P=A+B ± Not P = A ABP 000 010 100 111 A B P 011 101 A B P AP 01 10 A P 7 ± NAND P = A.B ± NOR P=A+B ± XOR P = A B Revision on Basic Gates 001 110 A B P A B P A B P 8 Truth Tables and Minterms ± Any boolean function can be written in terms of minterms ± Consider a simple truth table: 1 1 1 0 0 1 0 1 0 1 0 0 F B A ± We have 4 entries (rows) in this table ± We form the minterm for each entry in the table by AND ing A and B or their possible complement permutations Example: For the first minterm, m 0 , it evaluates true when A=0 and B=0 so we say: m 0 B A m = 0
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3 9 Sum of Product Expression of Truth Table We can form all the available minterms as shown from 0 to n-1 where here n=4! 1 1 1 0 0 1 0 1 0 1 0 0 F B A B A m = 0 B A m = 1 B A m = 2 B A m = 3 We form the Sum of Product (SOP) expression from those rows where a ‘1’ is present. Here: B A B A m m F + = + = 3 0 10 Buzzer Example ± The buzzer sounds only under this condition A.B.C 0 1 1 1 1 0 1 1 0 1 0 1 0 0 0 1 0 1 1 0 0 0 1 0 0 1 0 0 0 0 0 0 P C B A Alarm system Active B A C P Variable Value Situation A 1 Lights are on 0 Lights are off B 1 Door is open 0 Door is closed C 1 Key is in ignition 0 Key is out of ignition P 1 Buzzer is on 0B u z z e r i s o f f ± Consider a buzzer which sounds when : ± The lights are on and ± The door is open and ± No key is in the ignition •The truth table 11 Example: OR Gate We form the Sum of Product (SOP) expression from those rows where a ‘1’ is present.
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This note was uploaded on 10/27/2009 for the course ECTE 290 taught by Professor Zhengli during the Three '09 term at University of Wollongong, Australia.

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Lecture 12 - This Week Logic System Design Truth Tables,...

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