Given the following function generate the lowest cost two level circuit using nand gates only
\(f(x_{3},x_{2},x_{1},x_{0})=\sum(0,2,8,9,10,13)+D(1,3,6,7)\)
My attempt to solution:
I used Karnaugh's map to simplify to
\(f=\bar x_{0}\bar x_{1}+x_{0}\bar x_{1}\bar x_{2}+x_{0}\bar x_{2}x_{3}+\bar x_{1}x_{2}\bar x_{3}\)
\(\bar f=\bar{\bar x_{0}\bar x_{1}+x_{0}\bar x_{1}\bar x_{2}+x_{0}\bar x_{2}x_{3}+\bar x_{1}x_{2}\bar x_{3}}\)
applied demorgan's theorem
\(\bar f= x_{0}x_{1}\cdot\bar x_{0} x_{1} x_{2}\cdot\bar x_{0} x_{2}\bar x_{3}\cdot x_{1}\bar x_{2}x_{3}\)
Guys l get stuck here.......i don't know if this is the lowest cost two level circuit ?
\(f(x_{3},x_{2},x_{1},x_{0})=\sum(0,2,8,9,10,13)+D(1,3,6,7)\)
My attempt to solution:
I used Karnaugh's map to simplify to
\(f=\bar x_{0}\bar x_{1}+x_{0}\bar x_{1}\bar x_{2}+x_{0}\bar x_{2}x_{3}+\bar x_{1}x_{2}\bar x_{3}\)
\(\bar f=\bar{\bar x_{0}\bar x_{1}+x_{0}\bar x_{1}\bar x_{2}+x_{0}\bar x_{2}x_{3}+\bar x_{1}x_{2}\bar x_{3}}\)
applied demorgan's theorem
\(\bar f= x_{0}x_{1}\cdot\bar x_{0} x_{1} x_{2}\cdot\bar x_{0} x_{2}\bar x_{3}\cdot x_{1}\bar x_{2}x_{3}\)
Guys l get stuck here.......i don't know if this is the lowest cost two level circuit ?
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