For the following, find all minimum sum of products expressions (number of solut

Question

For the following, find all minimum sum of products expressions (number of solutions in parenthesis).

f(w,x,y,z)=S(igma)m=(1,3,6,8,11,14)+S(igma)d(2,4,5,13,15) (3 solutions)

Please be sure to show the K Map and all 3 solutions. I am only able to find 2 from my k map, so I think that I may have done my map wrong. Please help.

Explanation / Answer

I would expand it to a SOP, then reduce it with a K-map. Given: f = ((w'+x)' + (x'+z)' + (y'+z)')' Apply DeMorgan's: f = (w'+x) & (x'+z) & (y'+z) Multiply out the terms: f = (w'x' + xx' + w'z + xz)(y' + z) f = (w'x' + 0 + w'z + xz)(y' + z) f = (w'x' + w'z + xz)(y' + z) f = w'x'y' + w'x'z + w'zy' + w'zz + xzy' + xzz Simplify: f = w'x'y' + w'x'z + w'y'z + w'z + xy'z + xz At this point, I'd plot it on a K-map to get the minimal SOP. To show what would happen, I'll do the boolean algebra. Reduce: f = w'x'y' + w'x'z + (w'y'z + w'z) + xy'z + xz f = w'x'y' + w'x'z + w'z(y' + 1) + xy'z + xz f = w'x'y' + w'x'z + w'z(1) + xy'z + xz f = w'x'y' + w'x'z + w'z + xy'z + xz f = w'x'y' + (w'x'z + w'z) + xy'z + xz f = w'x'y' + w'z(x' + 1) + xy'z + xz f = w'x'y' + w'z(1) + xy'z + xz f = w'x'y' + w'z + xy'z + xz f = w'x'y' + w'z + (xy'z + xz) f = w'x'y' + w'z + xz(y' + 1) f = w'x'y' + w'z + xz(1) f = w'x'y' + w'z + xz To verify, note that the truth table is identical for both the original and reduced equation: w x y z => f 0 0 0 0 => 1 0 0 0 1 => 1 0 0 1 0 => 0 0 0 1 1 => 1 0 1 0 0 => 0 0 1 0 1 => 1 0 1 1 0 => 0 0 1 1 1 => 1 1 0 0 0 => 0 1 0 0 1 => 0 1 0 1 0 => 0 1 0 1 1 => 0 1 1 0 0 => 0 1 1 0 1 => 1 1 1 1 0 => 0 1 1 1 1 => 1

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---