Question

Can anyone please help me to simplify the two Z1 and Z2 expressions using boolean identities? then help me to draw the logisgm circuit.

(POS) Boolean algebra Z1 = (X1∙X2∙~X3∙X4) ∙ (X1∙~X2∙X3∙~X4) ∙ (~X1∙X2∙X3∙X4) ∙ (~X1∙X2∙3∙X4) ∙ (~X1∙~X2∙~X3∙X4) ∙ (~X1∙~X2∙~X3∙~X4)

Kmap simplify algebra: Z1 = (X2.~X3.X4)+(~X1.X2.X4)+(~X1.~X2.~X3)+(X1.~X2.X3.~X4)

(POS) Boolean algebra Z2 = (X1∙X2∙X3∙X4) ∙ (X1∙X2∙X3∙~X4) ∙ (X1∙~X2∙X3∙~X4) ∙ (X1∙~X2∙~X3∙~X4) ∙ (~X1∙X2∙X3∙~X4) ∙ (~X1∙X2∙~X3∙~X4) ∙ (~X1∙~X2∙X3∙~X4)

Kmap simplify algebra: Z1 = (X3.~X4)+(X1.X2.X3)+(X1.~X2.~X4)+(~X1.X2.~X4)

please help me to simplify these two;

Z1 = (X2.~X3.X4)+(~X1.X2.X4)+(~X1.~X2.~X3)+(X1.~X2.X3.~X4)

Z1 = (X3.~X4)+(X1.X2.X3)+(X1.~X2.~X4)+(~X1.X2.~X4)

Answer 1

DE Take any as "1" symbol example NX3 as to z = ( x + x2 + x3 +X4) (X* xo'+ x2 + x4) (x + X2 + x 2 + x4) (x +X , fXg & Xy cxit x2 + x₂ + xy) ( x +X2' + X₃ + X 3 Add to Set as (Double negation Law (Al) =A7 Send

X 2 + x3 + X4) (x1 + x2 + x3 +X4) cx + X2 + X₂ + X4) (x + X 2 + X3 & Xy) cxit x2 + x₂ + x4) ( x) +22² + X 3 + X, )))

bemorgan's law : (A+B+C+D)' Alpic!! (ABCD]!= A'+8!+c'+D! +x, X2 X3 x yt x, X2 X₃ xy + x, X2 X 3 Xy) Idempotent Law A TAZA) 2 = (x, X2 X3 xy + x, x 2 x 3 xy + x, X 2 X 3 X y + X, X, X3 x + X, X, X, 2,-(( xix, xs xiy)+(x'x, xx,]+(2,X5X5X50* 486ytky pregation law At Ale 1] 2 . , X3 x 4 x 7, Gy, 1 , Y, X, X, X, Y, neukte Demorgan's lawo! (A+B )!= A' B! ( x + 7, 7 + x + x + x + x + Y) - ( x + x + 1 Replace "," with a symbol. - (X, FX2 + 2x2 + xy) (X + NX 2 + x3 + NX y) (NX, + X 7 X tax) (NX U & NX 24 ~43) is the simplified expression.

Z2 = (x1+X at Xz+xy ) CX , tx g + X z w x y ) ( x tw X 2 + x zt w X4) (X, NX NXzt wxy) (NX + x + x + xy ) (NX,X tNX & X4) (NX ANX 2 + X3 & Nxy) Relplace we with "," symbol. = (x1 + x2 + x3 + xy) C %, + Ag+ xs+ xyl) (x, + xz' + x3 + xy'). exy + xy +xz' + xy (x;' + x2 + x2 + xy CX, ' +x2+xz'txu) (x, '+ X2' & Xz +X41) Double negation lawo! (All-A.

JUU (2])=(((x + x2 + x3 + xy) (X1 + Xg+ *g+ xy!) (x, + xy' + x3 + x4) ex, +x) +xz'+xu (x,+ x2 + x3 + xyl) CX,'+x2+xg'+xy!) (X,'t X2' + x3 + xy'))')"

+ x, xo'x g' xy + xi x;'*xy + xy X2 xs xy)' =(x) 'X, ' xxxy + xixx8xXy + x, X;\x3Xy ****y (w/x3+xx, 1 X2 =(x'xx' xx' x' + x,' x, xx xy + x, xx' xg xy + x3}x^{x}; cx'=x) + X, CX2 A Negation law : AtAler. -(,'X2' x' xy' + x,' X2 X3 Xy +X, X' xs xy + xs xy(x,'4 %,) =(xixe' ki xy + X1 X₂ X3 X 4 X X, X₂ X3 X4 + X X4 )' = (x3'(x,'X2' xy + xy) +x/x283Xy + x, Xy'x3 xy)! Absorption law : At ABC A+ BC OL A TABEC A' +BC - (xixz' x 3! + x3 xy + x,' x, X3 X4 + x, x' X3 Xy) ! (x'x,' ' + X4 (x3 + x,' x2x3) + X1 X2' xz Xy) =(x,! x 2 x3 + x, x 2 x 4 + X3 X y + X, X2 X3 X4 ) =(x,' X2 X3 + X1 X₂ XQ + x 4 (X3'+ X1 X2 X3)) ! (x, x 2 x3 + X, X ₂ X y + x, X2' xy + X3 x ) The above expression is simplified expression.

Demorgan's law : (A+B) = A'B' (AB)' = A'ABI = ( x1 + x2 + x3) CX,+xy' + x4) (x3 + x2 + x4) (x3 + x4) Replace an with N. =X, + x2 + x3)(x + NX 2 + w X4) (ax, + X z twxy) CX 3 tuX 4) is expression. the simplified The above expression