Computer Architecture: Logic Design Problem Set Solution

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Description

Problem 1:

Simplify the following equations using Boolean theorems. Check the correctness using a truth table and a K-Map. (3 points each for the Boolean simplifications, 1 point for the truth table, 3 points for the K-map, 3 point for the resultant circuit)

  1. = +

  1. = ++( + )

lu2178s67y3k tmp 23af2184a9a68516 lu2178s67y3k tmp d3e0981daf1a51ca lu2178s67y3k tmp d3e0981daf1a51ca lu2178s67y3k tmp 8222ffd0e432a0c7

Problem 2:

Write the sum of products canonical representation (minterms) for each of the following truth tables. Y is the output of the circuits.

lu2178s67y3k tmp 3c1a60dfa98762b lu2178s67y3k tmp 3c1a60dfa98762b lu2178s67y3k tmp 621f4a163dbc835a lu2178s67y3k tmp 3c1a60dfa98762b lu2178s67y3k tmp f76f2803d854fe03 lu2178s67y3k tmp f76f2803d854fe03 lu2178s67y3k tmp f76f2803d854fe03 lu2178s67y3k tmp f76f2803d854fe03 lu2178s67y3k tmp f76f2803d854fe03

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Problem 3:

Minimize each of the circuits in Problem 2 and implement them using a simple combinational circuit using two-input AND and OR gates, and/or a NOT gate.

Problem 4:

Design the simplest sum of products circuit that implements: lu2178s67y3k tmp c9fce71521cee5a1 !, !, !lu2178s67y3k tmp 23fd3ca36fc5e01b = lu2178s67y3k tmp 63efd78e8fe001fc (0,1,2,3,4,6,7). Use Boolean algebra to simplify the function and confirm your solution using K-Maps.

Problem 5:

Consider the following truth table:

lu2178s67y3k tmp 1eb6fb444e28109c lu2178s67y3k tmp 79d493e4026441cf

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( !, !, !)

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Show its sum of products (SoP) and product of sums (PoS). Also, use Boolean algebra to find the minimum-cost SoP form.

Consider the following truth table:

lu2178s67y3k tmp 3c1a60dfa98762b lu2178s67y3k tmp b36bb4546f61fb29

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!

( !, !, !)

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Use K-Maps to design the simplest sum of products circuit.

Problem 6:

Find a minimal Boolean Equation for the truth table given below. Remember to take advantage of the don’t care entries. Draw the resultant circuit.

lu2178s67y3k tmp defc1342369ee65a lu2178s67y3k tmp defc1342369ee65a lu2178s67y3k tmp defc1342369ee65a lu2178s67y3k tmp defc1342369ee65a lu2178s67y3k tmp defc1342369ee65a lu2178s67y3k tmp defc1342369ee65a

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Problem 7:

Consider the following circuit. What is its truth table?

lu2178s67y3k tmp 48945a93b7020567

What is the minimal-cost expression for this true table? Find the simplest SoP and confirm your solution using K-Maps.

Problem 8:

The following circuit (with NOT and 2-input AND gates) implements the 2-4 decoder.

lu2178s67y3k tmp c4fdb85fe4b33f5d

Design a 3-8 decoder circuit with NOT and 2-input AND gates. How many gates (NOT and 2-input AND gates) are needed to build the n-2n decoder?

Problem 9:

Give the Boolean expression for the function performed by the following circuit:

lu2178s67y3k tmp 31c3876353f44a78

Problem 10:

Implement the following truth table as described below:

lu2178s67y3k tmp 5b1a9306d6c9d2d

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  1. Using a 3:8 decoder and one other logic gate

  1. Using a 8:1 multiplexer

  2. Using a 4:1 multiplexer and one inverter

  3. A 2:1 multiplexer and two other logic gates

Problem 11:

Identify the Boolean equation performed by the circuit below and minimize the circuit and implement it using simple gates.

lu2178s67y3k tmp 56d641037d70c9db

Problem 12:

Design a 4-to-16 decoder using five 2-to-4 decoder

Problem 13:

Design a 32-to1 multiplexer (MUX) using

  1. 8-to-1 MUX and 2-to-4 decoders.

  2. 4-to-1 MUX and 2-to-4 decoders.

Problem 14:

Given a 3-input Boolean expression lu2178s67y3k tmp 21e486d1efda12be !, !, !lu2178s67y3k tmp 9fa7732bc06c61ed = lu2178s67y3k tmp 900f1070cb6d3944 (0,2,4,6,7).

  1. Implement this expression using only 2-4 decoders and OR gates.

  1. Implement this expression using only 4-1 MUX.

Problem 15:

Implement the function lu2178s67y3k tmp 997e52a55f8ff645 !, !, !, !, !lu2178s67y3k tmp 9104ce07acecbd3a = ! ! ! ! + ! ! + ! ! + ! ! + ! ! ! by using a 4-to-1 multiplexer and as few other gates as possible.

lu2178s67y3k tmp d3e0981daf1a51ca lu2178s67y3k tmp d3e0981daf1a51ca lu2178s67y3k tmp d3e0981daf1a51ca lu2178s67y3k tmp d3e0981daf1a51ca

Problem 16:

Consider the Boolean function lu2178s67y3k tmp 48f2a4b4242c6c7 !, !, !lu2178s67y3k tmp 6e2515f5f35d28e1 = ! ! + ! ! + ! ! !.

lu2178s67y3k tmp d3e0981daf1a51ca lu2178s67y3k tmp d3e0981daf1a51ca lu2178s67y3k tmp d3e0981daf1a51ca lu2178s67y3k tmp d3e0981daf1a51ca

  1. Use a 3-to- 8 decoder plus logic gates to implement this function.

  2. Use an 8-input multiplexer to implement this function.

Problem 17:

Gray codes have a useful property in that consecutive numbers differ in only a single bit position. The 3-bit gray code is given below.

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lu2178s67y3k tmp 965b0d9f26408da

Design a 3-bit modulo 8 Gray code counter FSM that has no inputs but produces three outputs. When reset the output should be 000. On each clock edge the output should advance to the next Gray code. After reaching 100, it should repeat with 000. Implement the circuit using combinational logic and D-flip flops.

Problem 18:

Design an FSM that detects a stream of two consecutive 1’s in a stream of 0’s and 1’s. Implement it using a combination of sequential and combinational logic.

INPUT:0110111010….

OUTPUT:0010011000….