p2.1 - X 2 X 3 0 X X X 1 X 2 X 3 1 0 X 0 1 X X 1 X 2 X 3 f...

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Daniel Mihai Problem 2.1 The Quine McCluskey algorithm for f 1 : Since we have only one prime implicant, it is essential. f 1 = g G ± ± The Quine McCluskey algorithm for f 2 : TTF FTF TFT FTT g G ± ± g ² F T F T g G g ² ±± F F T F Both prime implicants are essential. f 2 = g G ± ± g ² + g G g ² ±±± The Quine McCluskey algorithm for f 3 : X 1 X 2 X 3 0 X 0 0 0 X 0 1 X 0 X 1 X 1 X 2 X 3 f 1 0 0 0 1 0 1 0 1 0 0 1 1 0 1 1 1 X 1 X 2 X 3 0 X 0 0 0 X 0 1 X 0 X 1 X 1
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Unformatted text preview: X 2 X 3 0 X X X 1 X 2 X 3 1 0 X 0 1 X X 1 X 2 X 3 f 1 1 0 0 1 0 1 0 1 1 0 1 1 0 1 1 1 X 1 X 2 X 3 1 1 0 X 0 1 0 X 1 X 1 X 2 X 3 f 1 1 1 0 1 0 0 1 1 1 0 1 1 0 1 1 1 TTF FFT TFT FTT g G g ± g ² ³³³ T F F F g ± ³³³ g ² F T T F g G ³ ³ g ² F T F T All the prime implicants are essential. f 3 = g G g ± g ² ³³³ + g ± ³³³ g ² + g G ³ ³ g ² X 1 X 2 X 3 f 1 f 2 f 3...
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This note was uploaded on 04/14/2010 for the course CS 320102 taught by Professor M.k during the Spring '10 term at Jacobs University Bremen.

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p2.1 - X 2 X 3 0 X X X 1 X 2 X 3 1 0 X 0 1 X X 1 X 2 X 3 f...

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