Fa10_HW3 Solution

# Fa10_HW3 Solution - CS231 Spring 2010 Homework 3 Solution...

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CS231 Spring 2010 Homework 3 Solution 1. (10 points) Consider the Boolean function: F(x,y,z) = x’yz’ + xz’ + yz’ (a) Implement the function using a 3-to-8 Active-Low Decoder from LogicWorks and as few additional gates as possible. Solution: S2=X; S1=y; S0=Z F=AND(Q0’,Q1’,Q3’,Q5’,Q7’) (b) Implement the function using a MUX and as few additional gates as possible. Solution: S1=X; S0=Y; D3=D2=D1=Z’; D0=0 2. (10 points) Consider the function f(x,y,z) = Σ m(2,3,4,7) (a) Implement f(x,y,z) using only a demultiplexer and a single OR gate. Draw the circuit, being sure to label all inputs and outputs. Choose the demultiplexer size you wish. Note: The demultiplexer is the inverse of the multiplexer, in that it takes a single data input and n address inputs. It has 2n outputs. The address inputs determine which data output is going to have the same value as the data input. The other data outputs will have the value 0.

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(b) Implement f(x,y,z) using a multiplexer 3. (10 points) Implement a 8-to-1 multiplexer with data inputs D0, D1, D2, . .., D7 and with select inputs A, B, C (complemented inputs are not available). The selection should be done as usual; e.g., ABC = 011 will select the input D3, ABC = 110 will select the input D6, etc. The only components available to you are four 2-to-1 multiplexers with enable, a 2-to-4 decoder and one additional gate (AND, OR, or XOR). Be sure to label all inputs and outputs.
4. (10 points) A combinational circuit is defined by the following three Boolean functions: F1 = (X+Y)’ + XYZ F2 = (XZ)’Y F3 = (X’+Y)’+X’Y’Z’ Design the circuit with a decoder and external OR gates. F1 =(X+Y)’ + XYZ = X’Y’ + XYZ = X’Y’Z’ + X’Y’Z + XYZ =

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## This note was uploaded on 11/08/2010 for the course CS 231 taught by Professor - during the Spring '08 term at University of Illinois at Urbana–Champaign.

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Fa10_HW3 Solution - CS231 Spring 2010 Homework 3 Solution...

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