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# Verfahren von Quine-McCluskey

## Tags: Boolsche Algebra, Schaltalgebra, Sonstig

18:23 Uhr, 15.05.2019

Hallo,

ich will mit Hilfe des Verfahrens von Quine-McCluskey zu der folgenden Schaltfunktion die jeweiligen Primimplikanten bestimmen und daraus eine disjunktive Minimalform ermitteln. Falls vorhanden, soll man auch alle weiteren disjunktiven Minimalformen angeben.

Die Funktion ist $f\phantom{\rule{0.12em}{0ex}}\left({x\phantom{\rule{0.12em}{0ex}}}_{1},{x\phantom{\rule{0.12em}{0ex}}}_{2},{x\phantom{\rule{0.12em}{0ex}}}_{3},{x\phantom{\rule{0.12em}{0ex}}}_{4},{x\phantom{\rule{0.12em}{0ex}}}_{5}\right)={\stackrel{‾}{x\phantom{\rule{0.12em}{0ex}}}}_{1}{\stackrel{‾}{x\phantom{\rule{0.12em}{0ex}}}}_{2}{\stackrel{‾}{x\phantom{\rule{0.12em}{0ex}}}}_{3}{\stackrel{‾}{x\phantom{\rule{0.12em}{0ex}}}}_{4}{\stackrel{‾}{x\phantom{\rule{0.12em}{0ex}}}}_{5}\vee {\stackrel{‾}{x\phantom{\rule{0.12em}{0ex}}}}_{1}{x\phantom{\rule{0.12em}{0ex}}}_{2}{\stackrel{‾}{x\phantom{\rule{0.12em}{0ex}}}}_{3}{\stackrel{‾}{x\phantom{\rule{0.12em}{0ex}}}}_{4}{\stackrel{‾}{x\phantom{\rule{0.12em}{0ex}}}}_{5}\vee {\stackrel{‾}{x\phantom{\rule{0.12em}{0ex}}}}_{1}{\stackrel{‾}{x\phantom{\rule{0.12em}{0ex}}}}_{2}{\stackrel{‾}{x\phantom{\rule{0.12em}{0ex}}}}_{3}{\stackrel{‾}{x\phantom{\rule{0.12em}{0ex}}}}_{4}{x\phantom{\rule{0.12em}{0ex}}}_{5}\vee {\stackrel{‾}{x\phantom{\rule{0.12em}{0ex}}}}_{1}{x\phantom{\rule{0.12em}{0ex}}}_{2}{\stackrel{‾}{x\phantom{\rule{0.12em}{0ex}}}}_{3}{x\phantom{\rule{0.12em}{0ex}}}_{4}{\stackrel{‾}{x\phantom{\rule{0.12em}{0ex}}}}_{5}$ $\vee {\stackrel{‾}{x\phantom{\rule{0.12em}{0ex}}}}_{1}{x\phantom{\rule{0.12em}{0ex}}}_{2}{\stackrel{‾}{x\phantom{\rule{0.12em}{0ex}}}}_{3}{\stackrel{‾}{x\phantom{\rule{0.12em}{0ex}}}}_{4}{x\phantom{\rule{0.12em}{0ex}}}_{5}\vee {x\phantom{\rule{0.12em}{0ex}}}_{1}{x\phantom{\rule{0.12em}{0ex}}}_{2}{\stackrel{‾}{x\phantom{\rule{0.12em}{0ex}}}}_{3}{x\phantom{\rule{0.12em}{0ex}}}_{4}{\stackrel{‾}{x\phantom{\rule{0.12em}{0ex}}}}_{5}\vee {\stackrel{‾}{x\phantom{\rule{0.12em}{0ex}}}}_{1}{x\phantom{\rule{0.12em}{0ex}}}_{2}{x\phantom{\rule{0.12em}{0ex}}}_{3}{x\phantom{\rule{0.12em}{0ex}}}_{4}{\stackrel{‾}{x\phantom{\rule{0.12em}{0ex}}}}_{5}\vee {\stackrel{‾}{x\phantom{\rule{0.12em}{0ex}}}}_{1}{x\phantom{\rule{0.12em}{0ex}}}_{2}{x\phantom{\rule{0.12em}{0ex}}}_{3}{x\phantom{\rule{0.12em}{0ex}}}_{4}{x\phantom{\rule{0.12em}{0ex}}}_{5}\vee {x\phantom{\rule{0.12em}{0ex}}}_{1}{x\phantom{\rule{0.12em}{0ex}}}_{2}{\stackrel{‾}{x\phantom{\rule{0.12em}{0ex}}}}_{3}{x\phantom{\rule{0.12em}{0ex}}}_{4}{x\phantom{\rule{0.12em}{0ex}}}_{5}$

Es gilt dass $\stackrel{‾}{x\phantom{\rule{0.12em}{0ex}}}={x\phantom{\rule{0.12em}{0ex}}}^{0},x\phantom{\rule{0.12em}{0ex}}={x\phantom{\rule{0.12em}{0ex}}}_{1}$ und somit bekommen wir:
$f\phantom{\rule{0.12em}{0ex}}\left({x\phantom{\rule{0.12em}{0ex}}}_{1},{x\phantom{\rule{0.12em}{0ex}}}_{2},{x\phantom{\rule{0.12em}{0ex}}}_{3},{x\phantom{\rule{0.12em}{0ex}}}_{4},{x\phantom{\rule{0.12em}{0ex}}}_{5}\right)={x\phantom{\rule{0.12em}{0ex}}}_{1}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{2}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{3}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{4}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{5}^{0}\vee {x\phantom{\rule{0.12em}{0ex}}}_{1}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{2}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{3}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{4}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{5}^{0}\vee {x\phantom{\rule{0.12em}{0ex}}}_{1}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{2}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{3}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{4}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{5}^{1}\vee {x\phantom{\rule{0.12em}{0ex}}}_{1}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{2}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{3}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{4}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{5}^{0}$ $\vee {x\phantom{\rule{0.12em}{0ex}}}_{1}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{2}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{3}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{4}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{5}^{1}\vee {x\phantom{\rule{0.12em}{0ex}}}_{1}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{2}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{3}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{4}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{5}^{0}\vee {x\phantom{\rule{0.12em}{0ex}}}_{1}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{2}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{3}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{4}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{5}^{0}\vee {x\phantom{\rule{0.12em}{0ex}}}_{1}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{2}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{3}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{4}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{5}^{1}\vee {x\phantom{\rule{0.12em}{0ex}}}_{1}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{2}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{3}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{4}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{5}^{1}$

Wir berechnen die Gewichte dieser Terme:
$g\phantom{\rule{0.12em}{0ex}}\left({x\phantom{\rule{0.12em}{0ex}}}_{1}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{2}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{3}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{4}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{5}^{0}\right)=0$
$g\phantom{\rule{0.12em}{0ex}}\left({x\phantom{\rule{0.12em}{0ex}}}_{1}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{2}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{3}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{4}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{5}^{0}\right)=1$
$g\phantom{\rule{0.12em}{0ex}}\left({x\phantom{\rule{0.12em}{0ex}}}_{1}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{2}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{3}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{4}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{5}^{1}\right)=1$
$g\phantom{\rule{0.12em}{0ex}}\left({x\phantom{\rule{0.12em}{0ex}}}_{1}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{2}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{3}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{4}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{5}^{0}\right)=2$
$g\phantom{\rule{0.12em}{0ex}}\left({x\phantom{\rule{0.12em}{0ex}}}_{1}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{2}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{3}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{4}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{5}^{1}\right)=2$
$g\phantom{\rule{0.12em}{0ex}}\left({x\phantom{\rule{0.12em}{0ex}}}_{1}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{2}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{3}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{4}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{5}^{0}\right)=3$
$g\phantom{\rule{0.12em}{0ex}}\left({x\phantom{\rule{0.12em}{0ex}}}_{1}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{2}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{3}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{4}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{5}^{0}\right)=3$
$g\phantom{\rule{0.12em}{0ex}}\left({x\phantom{\rule{0.12em}{0ex}}}_{1}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{2}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{3}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{4}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{5}^{1}\right)=4$
$g\phantom{\rule{0.12em}{0ex}}\left({x\phantom{\rule{0.12em}{0ex}}}_{1}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{2}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{3}^{0}{x\phantom{\rule{0.12em}{0ex}}}_{4}^{1}{x\phantom{\rule{0.12em}{0ex}}}_{5}^{1}\right)=4$

Als nächstes ist folgendes beim Verfahren:
Die Minterme werden nach aufsteigendem Gewicht in die dritte Spalte einer Tabelle eingetragen. In der ersten Spalte der Tabelle steht das Gewicht des jeweiligen Minterms, in der zweiten sein Dezimalwert als seine Bezeichnung, wenn der Minterm als n-stellige Dualzahl interpretiert wird.

Diesen teil habe ich nicht so richtig verstanden. Könnt ihr mir erklären wie man jetzt weiter machen kann?

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