hw3_solution_cs231_sp08_uiuc.html

hw3_solution_cs231_sp08_uiuc.html -

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Sheet1 Page 1 <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title>CS231 Spring 2008 Homework 3 Solution</title> <link rel="stylesheet" type="text/css" href="/~cs231/style.css" /> <style type="text/css"> ol li { margin-top: 1em </style> </head> <body> <?xml version="1.0" encoding="us-ascii"?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <h2 align="center"> CS231 Spring 2008<br> Homework 3 Solution </h2> <hr> <ol> <!-- ========== Problem 1 ========== --> <li> (10pts) Show how to build a 4-to-16 decoder, using <b> only</b> 2-to-4 decoders (no gates). Your decoder should accept four inputs S3-S0, and have sixteen outputs D15-D0.<br> <br> <b>Solution:</b><br> <br> <img border="0" src="1a.png" width="342" height="330"><br> <br> <!-- ========== Problem 2 ========== --> <hr><li> (10pts) Circuit Design: Consider the Boolean function F(x,y,z) = xy + z'(y + x)(z + x')<br> <br> <ol type=a> <li>Simplify this expression to a minimum number of literals using the Boolean identities from lecture. Specify the name or formula of the identity used at each step.<br> <br> <b>Solution:<br> </b><br> = xy + z' ( yz + x'y + xz + xx' )&nbsp
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Sheet1 Page 2 by x' * x = 0<br> by Distributive<br> by x' * x = 0<br> by Distributive<br> by (x + x'y) = (x + y)<br> by Distributive <br> <li>Implement the function using a minimum number of AND, OR, and NOT gates.<br> <br> <b>Solution:<br> </b><br> <img border="0" src="2b.png"><br> <li>Implement the function using a minimum number of NAND gates.<br> <br> <b>Solution:<br> </b><br> The minimum number of NAND gates you need is 6. <br> <img border="0" src="2c.png"><br> <li>Implement the function using a 3-to-8 Active-Low decoder and only one multi-input AND gate.<br> <br> <b>Solution:</b><br> <br> <img border="0" src="2d.png"><br> <br></ol> <!-- ========== Problem 3 ========== --> <hr><li> (10 pts) Circuits using Decoders:<br> <ol type =a > <li>Draw a circuit for the following function F using a 3-to-8 decoder (note this is not Active-Low) and one OR gate.<br> <br> <b>Solution:</b><br> <br> <img border="0" src="3a.png"><br> <br> <li>Draw a circuit for the following function G using only one (active-low) 2-to-4 decoder from LogicWorks.<br> Err:510 (<b>Hint</b>: Since there is only one zero in the K-maps, consider under what condition will a certain output pin of 2-4 decoder output zero.)<br>
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Sheet1 Page 3 <br> <b>Solution:</b><br> <br> <img border="0" src="3b.png"><br>
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This note was uploaded on 04/10/2008 for the course CS 231 taught by Professor - during the Spring '08 term at University of Illinois at Urbana–Champaign.

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hw3_solution_cs231_sp08_uiuc.html -

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