Tutorial 6 sol

# Tutorial 6 sol - EE 2000 Logic Circuit Design, Semester A,...

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EE 2000 Logic Circuit Design, Semester A, 2007/08 Tutorial 6 Question 1: Figure below shows a simple locker. The output signal Z will be 1 when either 0, 1, 2, 3, 4, 5, 8, 9, A is entered form the hexadecimal keypad. Otherwise, Z is equal to 0. A 3 A 2 A 1 A 0 is a 4 digit binary number output from the keypad. Block A decodes these singals and output the signal Z . (a) Write down the truth table of Block A. (b) Find the simplest SOP and POS expression of Block A. (c) Design the circuit of Block A using minimum number of AND, OR and NOT gates. Ans: (a) Hex A 3 A 2 A 1 A 0 Z 0 0 0 0 0 1 1 0 0 0 1 1 2 0 0 1 0 1 3 0 0 1 1 1 4 0 1 0 0 1 5 0 1 0 1 1 6 0 1 1 0 0 7 0 1 1 1 0 8 1 0 0 0 1 9 1 0 0 1 1 A 1 0 1 0 1 B 1 0 1 1 0 C 1 1 0 0 0 D 1 1 0 1 0 E 1 1 1 0 0 F 1 1 1 1 0 (b) SOP = 31 21 32 20 A AA A +++ POS = 3103221 () ( ) ( ) A AAAAAA ++ + +

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(c) Cost of SOP = 4 AND + 1 OR + 4 NOT Cost of POS = 3 OR + 1 AND + 4 NOT Therefore, POS has less cost then SOP. So it’s better to implement the POS expression. Question 2: You have learnt how to construct a BCD-to-Excess-3-Code converter in last lecture. Now, instead of using two-level AND-OR circuit, construct
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## This note was uploaded on 01/11/2011 for the course EE 2000 taught by Professor Vancwting during the Fall '07 term at City University of Hong Kong.

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Tutorial 6 sol - EE 2000 Logic Circuit Design, Semester A,...

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