HW4_solution - ECE 2504 Introduction to Computer...

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1 ECE 2504 Introduction to Computer Engineering, Fall 2008 Homework 4 Due date: 25 Sep 08, 11:59pm Show all work. Use Logicworks to draw any circuit diagrams. 1. (6 pts) Using the Boolean identities, simplify the following Boolean expressions as much as possible. Show all steps. a. a’bc’ + b’cd’ + b’c’d’ + abd + a’b’c’ a’c’(b+b’) + b’d’(c+c’) + abd a’c’ + b’d’ + abd b. ((uw’) + ((t+(u’t)’)’v)’)’•((tv’)’+(u’+w)’) ((uw’)’ • ((t+(u’t)’)’v)•((tv’)’+(u’+w)’) ((u’+w) • ((t+u+t’)’v)) • ( (t’+v) + (uw’) ) ((u’+w) • (u’v)) • (t’ + v + uw’) (u’u’v+u’vw) • (t’ + v + uw’) (u’v(1+w)) • (t’ + v + uw’) u’v • (t’ + v + uw’) t’u’v +u’vv + uu’vw’ t’u’v + u’v + 0 t’u’v + u’v c. y’z+xy’+xy’z’+x’y’ y’z+xy’(1+z’)+x’y’ y’z+xy’+x’y’ y’z+y’(x+x’) y’z+y’ y’(z+1) y’ 2. (6 pts) Given the Boolean expression F = (x+y’+z’)•(x+z)•(x+y’+z) a. Derive a Boolean expression for the complement F’. F’ = ( (x+y’+z’)•(x+z)•( x+y’+z) )’ = (x+y’+z’)’ + (x+z)’ + (x+y’+z)’ F’ = x’yz + x’z’ + x’yz’ b. Show that F + F’ = 1. F + F’ = (x+y’+z’)•(x+z)•( x+y’+z) + x’yz + x’z’ + x’yz’ = (x+y’+z’)•(x•( x+y’+z) +z•( x+y’+z)) + x’yz + x’z’ + x’yz’ = (x+y’+z’)•(xx+xy’+xz) + ( xz+y’z+zz)) + x’yz + x’z’ + x’yz’ = (x+y’+z’)•(x+xy’+xz+xz+y’z+z)) + x’yz + x’z’ + x’yz’ = (x•(x+xy’+xz+y’z+z)+y’•(x+xy’+xz+y’z+z)+z’•(x+xy’+xz+y’z+z)) + x’yz + x’z’ + x’yz’ = (xx+xxy’+xxz+xy’z+xz+xy’+xy’y’+xy’z+y’y’z+y’z+xz’+xy’z’+xzz’+y’zz’+zz’) + x’yz + x’z’ + x’yz’ = (x+xy’+xz+xy’z+xz+xy’+xy’+xy’z+y’z+y’z+xz’+xy’z’+xzz’+y’zz’+zz’) + x’yz + x’z’ + x’yz’ = x+xy’+xz+xy’z+xz+xy’+xy’+xy’z+y’z+y’z+xz’+xy’z’+0+0+0+ x’yz + x’z’ + x’yz’ = x(1+y’+z+y’z+z+y’+y’+y’z+z’+y’z’)+x’yz + x’z’ + x’yz’ +y’z = x(1)+x’y(z+z’) + x’z’ + y’z = x+x’y(1) + x’z’ + y’z
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2 = x + x’y + x’z’ + y’z
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This note was uploaded on 06/15/2009 for the course ECE 2504 taught by Professor Klcooper during the Fall '08 term at Virginia Tech.

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HW4_solution - ECE 2504 Introduction to Computer...

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