hw_3_solution_2009 - Homework#3 Solutions ECE 15a Winter...

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Homework #3 Solutions ECE 15a Winter 2009 Solution 1: (a). f = (u+v) (uv’ + u’w)’ = (u+v) [ ( u’+v ) ( u+w’) ] = (u+v) (u’+v) (u+w’) = (uv + u’v) (u+w’) = v (u+w’) = uv(w+w’) + (u+u’)vw’ = uvw + uvw’ + uvw’ + u’vw’ = uvw + uvw’ + u’vw’ (b). g = xy’z + (x’+y) (x+z’) = xy’z + x’z’ + xy = xy’z + x’ (y+y’) z’ + xy (z+z’) = xy’z + x’yz’ + x’y’z’ + xyz + xyz’ (c). h =(x+z) (x+y’) (x’+z’) = (xz’+x’z) (x+y’) = xz’x + xz’y’ + x’zx + x’zy’ = xz’ + xz’y’ + 0 + x’y’z = x (y+y’)z’ + xz’y’ + x’y’z = xyz’ + xy’z’ + x’y’z
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Solution 2: (a). f = x’ + y’ = x’ (y+y’) (z+z’) + (x+x’) y’ (z+z’) expanding the above expression and neglecting redundant terms we get, = x’yz + x’y’z + x’yz’ + x’y’z’ + xy’z + xy’z’ (b). g = x’z’ + x’z = x’ (z+z’) = x’ = x’ (y+y’) (z+z’) = x’yz + x’yz’ + x’y’z + x’y’z’ Solution 3: x y z f 1 1 0 1 0 1 0 1 For all other x, y and z, f is 0 0 0 f = xyz’ + x’yz’ Solution 4: (a). f = ( u+v ) ( uv + u’w )’ = (u+v) [ (u’+v’) (u+w’) ] = (u+v + ww’) [ (u’+v’+ww’) (u+w’+vv’) ] = (u+v+w) (u+v+w’) . . . . . .
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