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hw_3_solution_2009

Course: ECE 15A, Winter 2009
School: UCSB
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Word Count: 633

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#3 Homework Solutions ECE 15a Winter 2009 Solution 1: (a). f = (u+v) (uv + uw) = (u+v) [ ( u+v ) ( u+w) ] = (u+v) (u+v) (u+w) = (uv + uv) (u+w) = v (u+w) = uv(w+w) + (u+u)vw = uvw + uvw + uvw + uvw = uvw + uvw + uvw (b). g = xyz + (x+y) (x+z) = xyz + xz + xy = xyz + x (y+y) z + xy (z+z) = xyz + xyz + xyz + xyz + xyz (c). h =(x+z) (x+y) (x+z) = (xz+xz) (x+y) = xzx + xzy + xzx + xzy = xz + xzy + 0 + xyz = x...

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#3 Homework Solutions ECE 15a Winter 2009 Solution 1: (a). f = (u+v) (uv + uw) = (u+v) [ ( u+v ) ( u+w) ] = (u+v) (u+v) (u+w) = (uv + uv) (u+w) = v (u+w) = uv(w+w) + (u+u)vw = uvw + uvw + uvw + uvw = uvw + uvw + uvw (b). g = xyz + (x+y) (x+z) = xyz + xz + xy = xyz + x (y+y) z + xy (z+z) = xyz + xyz + xyz + xyz + xyz (c). h =(x+z) (x+y) (x+z) = (xz+xz) (x+y) = xzx + xzy + xzx + xzy = xz + xzy + 0 + xyz = x (y+y)z + xzy + xyz = xyz + xyz + xyz 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, = xyz + xyz + xyz + xyz + xyz + xyz (b). g = xz + xz = x (z+z) = x = x (y+y) (z+z) = xyz + xyz + xyz + xyz Solution 3: x y z f 1 1 0 0 1 1 0 0 1 0 For all other x, y and z, f is 0 f = xyz + xyz Solution 4: (a). f = ( u+v ) ( uv + uw ) = (u+v) [ (u+v) (u+w) ] = (u+v + ww) [ (u+v+ww) (u+w+vv) ] = (u+v+w) (u+v+w) . . . . . . Expanding and cancelling out redundant terms, = (u+v+w) (u+v+w) (u+v+w) (u+v+w) (u+v+w) (b). g = xyz + xyz + xy = xy (z+z) + xy = xy + xy = x (y+y) =x = x + yy + zz = (x+y+z) (x+y+z) (x+y+z) (x+y+z) (c). h = (xy + xyz + xzt + t) = [xy(z+z) (t+t) + xyz (t+t) + x(y+y)zt + (x+x)(y+y)(z+z)t] Expanding and neglecting redundant terms, = [xyzt + xyzt xyzt + + xyzt + xyzt] = (x+y+z+t) (x+y+z+t) (x+y+z+t) (x+y+z+t) (x+y+z+t) Solution 5: (a). x (x+y) = (x+yy+zz) (x+y+zz) = (x+y+z) (x+y+z) (x+y+z) (x+y+z) (x+y+z) (x+y+z) (b). x + xz = x + z ----------- Using, a + ab = a + b = x + z + yy = (x+y+z) (x+y+z) Solution 6: (a). f = uv + uv + uv = v (u+u) + uv = v + uv = (v+u) ---------- Using, a + ab = a + b Solution 7: (u+v+w) (u+v+w) (u+v+w) (u+v+w) (u+v+w) = (001) (110) (011) (100) (111) = M (1, 6, 3, 4, 7) = m (0, 2, 5) = (000) (010) (101) = (xyz) + (xyz) + (xyz) Solution 8: (cd) (a+b) + ab = (cd + cd) (a+b) + ab = acd + acd + bcd + bcd + ab = bcd + bcd + ab ------------- Eliminating terms using Consensus theorem. Solution 9: xz (x+y+z+w) (x+z+w+v) (x+z+w+v) = xz (x+y+z+w) [ (x+z+w + v) (x+z+w + v) ] Using, (X+Y) (X+Y) = Y = xz (x+y+z+w) [ (x+z+w) ] = xz [ (x+z+w) + y ] [ (x+z+w ) ] = xz [ (x+z+w) + y.(x+z+w) ] = xz (x+y+w) + xyz (x+y+w) = xwz + xwyz = xwz (1+y) = xwz Solution 10: ab + ac + bc = ab + ac + bc LHS: ab + ac + bc = ab(c+c) + a(b+b)c + (a+a)bc = abc + abc + abc + abc + abc + abc ------------------ 1 RHS: ab + ac + bc = ab (c+c) + a (b+b)c + (a+a)bc = abc + abc + abc + abc + abc + abc ------------------- 2 From 1 and 2, LHS = RHS. Therefore, ab + ac + bc = ab + ac + bc
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