I have to do synthesis of combinational logic circuit from this expression first I have to do K map (karnaugh) and I want to know how can I correctly do it. The expression is (A+'B)('C+D)'(A+C) Tip: A+C is in complement.
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de Morgan’s Theorem '(A+C)='A.'B – Naman Jain Dec 05 '19 at 08:57
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How can I write in Kmap? – Dec 05 '19 at 11:22
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The same way you write for other expressions. Suppose in all (A+'B) you will write $0$ in Kmap. Similarly In all 'A and all 'C you write $0$. Think of 'A and 'C and a separate POS. – Naman Jain Dec 05 '19 at 12:02
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I dont understand – Dec 05 '19 at 12:06
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I have solved it here http://www.32x8.com/pos4_____A-B-C-D_____m_0-1___________option-2_____899788975274834292792 – Naman Jain Dec 05 '19 at 12:07
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(A+'B)('C+D)'(A+C)=(A+'B)('C+D)('A'C)=$(A+'B)('C+D)('A)('C)$ So there are 4 products in your expression – Naman Jain Dec 05 '19 at 12:09
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Hmm okay I think I got it – Dec 05 '19 at 12:13
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I suppose you need to go through basics of Kmap – Naman Jain Dec 05 '19 at 12:20
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@NamanJain Is $'A$ a common notation for not $A$, I never see this before, is there any reference for it $\dots$ – Ethan Dec 09 '19 at 21:09
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Yes it's a symbol for not – Naman Jain Dec 10 '19 at 02:58