# hw3soln - a F:Qxcﬁﬁlr’é For the following circuits...

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Unformatted text preview: a) F. :Qxcﬁﬁlr’é For the following circuits, put them into minimized Product of Sums and Sum of Products forms by using a K- map. Note: to get to Product of Sums, use the K—map to get the Sum of Products of the function’s inverse, and then use DeMorgan’s Law. a.) F=llA®Bj*C+Dj*iA*E+C*D' Remember, (-9 = XOR . Also, to evaluate complex equations you can use a truth table, and evaluate subfunctions ﬁrst. b.) Z*§*C*D+_§*D(Z+E)+Z*B(E+5)+A*§*C*D P2=F2+D P3 =A§+E§ Fx’iTﬁrl-Zg _ ,— F +ﬂrHBD+PrBB+Eé6 F 2 C .,_ ﬂ —+D)(,q+8+ D) P : C (A +8+6X3+3 POM i c:— ccﬁ+§><em<é+z+6> A 4—input majority gate is, a function that is TRUE when most of its inputs are TRUE, and FALSE when most of its inputs are FALSE. Ambiguous situations should be treated as “Don’t Cares”. If there is more than one possible solution to the K—map, pick one. a.) Produce the Product of Sums and Sum of Products for the 4-input majority gate. b.) Produce the Product of Maxterms and Sum of Minterms form. Note that you can use Don’t Cares to help make each implementation smaller, which means they might be used differently in the POM and SoM forms respectively. W I 442/48 +CD -or— AC +80 l mox— AD+BC w—- __,.__.'-—M- M : (A+B)CC+D> ﬂax-4 (An 6) U340) ——ox» For each of the Boolean expressions given below, use K—maps to provide a minimized sum of products and product of sums expressions (2 K—maps per circuit). First, identify all of the prime implicants (write out the product term), then identify the essential prime implicants, and lastly provide a minimized sum of product expression. a_)F=E([email protected])+AB+AC§ 1K85+§S¢+A8 «mesa. b.) F=D(Z+3)+B(C+D)+ZB(C+D)+§CD =2¥D 48!) +ec+80+2§ 355+é‘co c.) F = ABE+§AE+§CE+ZBD (iv/7mg: 751(8) 8 Z) ARIEL; ESVI‘IiQei'. 85} f? L D {OWN BE +Ec+ﬂr8 —. of .. B "c". + EC + A C- H” inn" W ”‘ e“ ' ,«rm’éb— AF“ Fb A PrimeszAB QUAD 6‘9ch- ‘P 5' __) ‘Cﬁ wait; D 2 J ESSMf—jqi BDJALD)ABQ Esyﬂxﬁﬁhgﬁ) CDJEDJBDJ BC jig—En" ‘ D Coucv: BO‘tAEB +AEC (my: ﬁgwbdggseoﬂd . Pb = (B+D)CE+C+ E3303? +13% 0. EBD+§CE+ —ow-— _ a” EBD+§CE+AB¢+ABD M “””""’:’” * “WM? - W— V— — A ‘XCA-tB—rQC/q +é’ +"D)(21,"+e'+a) COL,” F¢:BD+E§E+ABD+ABC 71.48va __ % ‘7 E —0r~ E: ED+EZD +1185 moc— FCICB+D)(A+L~+D)CI4~+§+D)(A +B+LD ’0’" 5;: ED “56““3 “'39 PC: (BJDXMHD) (mew) “4&5 a.) F=2(B®C)+AB+ACE b.) F =D(Z+B)+B(C+D)+ZB(C+D)+§CD c.) F : ABE + 5A5 + ECE + ZED Using the expressions from problem 3, provide two efficient implementations, one consisting of NOR gates only and the second consisting of NAND gates only. You may use inverters as necessary. Thus, the correct answer for this problem will be 6 ciICuit diagrams. I 0\) Fa: BE+E¢+A< F“ 2C3+C>CA+§*E3 B B —-——- ; 5.427" L C F FF; F W ~“ at A33?) cm {3’ B \i AM VJ cur C H a ‘ —e— ”‘d—W—A—FV’ P1: C {’C inn-5 5‘45 I‘ rfg>+D)( ﬁat-r1?)ng "tg’fcy L33 F5 = ﬁm <9 +’AD+E>I>+B< P: B ...
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hw3soln - a F:Qxcﬁﬁlr’é For the following circuits...

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