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TAMU ECEN 248 - 2.10 - 5 points - Texas A&M University

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2.10 (5 points) Let f = m(1, 2, 3, 4, 5, 6, 7). The canonical sum-of-products for f is given by f = x1' x2' x3 + x1' x2 x3' + x1' x2 x3 + x1 x2' x3' + x1 x2' x3 + x1 x2 x3' + x1 x2 x3It can be manipulated as follows: f = m(4, 5, 6, 7) + m(2, 3, 6, 7) + m(1, 3, 5, 7) = (x1 x2' x3' + x1 x2' x3 + x1 x2 x3' + x1 x2 x3) + (x1' x2 x3' + x1' x2 x3 + x1 x2 x3' + x1 x2 x3) + (x1' x2' x3 + x1' x2 x3 + x1 x2' x3 + x1 x2 x3) = x1 ( x2' x3' + x2' x3 + x2 x3' + x2 x3) + x2 (x1' x3' + x1' x3 + x1 x3' + x1 x3) + x3 (x1' x2' + x1' x2 + x1 x2' + x1 x2 ) = x1 ( x2' (x3' + x3) + x2 (x3' + x3)) + x2 (x1' (x3' + x3) + x1 (x3' + x3)) + x3 (x1' (x2' + x2) + x1 (x2' + x2)) = x1 + x2 + x3 2.11 (5 points) Let f = M(0, 1, 2, 3, 4, 5, 6). The canonical sum-of-products for f is given by f = (x1 + x2 + x3) (x1 + x2 + x3') (x1 + x2' + x3) (x1 + x2' + x3') (x1' + x2 + x3)• (x1' + x2 + x3') (x1' + x2' + x3) It can be manipulated as follows: f = M(0, 1, 2, 3) • M(0, 1, 4, 5) • M(0, 2, 4, 6) = ((x1 + x2 + x3) (x1 + x2 + x3') (x1 + x2' + x3) (x1 + x2' + x3'))• ((x1 + x2 + x3) (x1 + x2 + x3') (x1' + x2 + x3) (x1' + x2 + x3'))• ((x1 + x2 + x3) (x1 + x2' + x3) (x1' + x2 + x3)(x1' + x2' + x3)) = ((x1 + x2 + x3) (x1 + x2 + x3') (x1 + x2' + x3) (x1 + x2' + x3'))' ' • ((x1 + x2 + x3) (x1 + x2 + x3') (x1' + x2 + x3) (x1' + x2 + x3'))' ' • ((x1 + x2 + x3) (x1 + x2' + x3) (x1' + x2 + x3)(x1' + x2' + x3))' ' = (x1' x2' x3' + x1' x2' x3 + x1' x2 x3' + x1' x2 x3)'• (x1' x2' x3' + x1' x2' x3 + x1 x2' x3' + x1 x2' x3)'• (x1' x2' x3' + x1' x2 x3' + x1 x2' x3' + x1 x2 x3')' obtained by using DeMorgan's Theorem = (x1' (x2' x3' + x2' x3 + x2 x3' + x2 x3))'• (x2' (x1' x3' + x1' x3 + x1 x3' + x1 x3))'• (x3' (x1' x2' + x1' x2 + x1 x2' + x1 x2))' = x1' ' • x2' ' • x3' ' = x1 x2 x32.12 (5 points) Derivation of the minimum sum-of-products expression: 2.13 (5 points) Derivation of the minimum sum-of-products expression: 2.14 (5 points) The simplest POS expression is derived as 2.15 (5 points) The simplest POS expression is derived as 2.16 (a) (5 points) Location of all minterms in a 3-varialbe Venn diagram:(b) (5 points) We have Therefore, we get the Venn Diagram for f as follows: , x3which implies that the minimal sum-of-products for f is given by f = x1 x2 + x3. 2.17 (5 points) x3x1which implies that the minimal sum-of-products is given by f = x1 x3' + x2' x3. 2.18 (5 points) In Figure P2.1a, it is possible to represent only 14 minterms. It is impossible to represent the minterms x1' x2' x3 x4 and x1 x2 x3' x4'. In Figure P2.1b it is impossible to represent the minterms x1 x2 x3' x4' and x1 x2 x3 x4'. 2.19 (5 points) Locations of minterms: x4x3x2x1x1x2x3m0m1m2m3m4m5m6m7m8m9m10m11m12m13m14m15(5 points) Representation of f: x4x3x2x1x1x2x3 2.20 (5 points) The simplest SOP implementation of the function is 2.21 (5 points) The simplest SOP implementation of the function is Another possibility is 2.22 (5 points) The simplest SOP implementation of the function is 2.23 (5 points) The simplest SOP implementation of the function is2.24 (5 points) The simplest sum-of-products expression for f is derived as follows: f = x1 x3' x4' + x2 x3' x4 + x1 x2' x3' = x3' (x1 x4' + x2 x4 + x1 x2') = x3' (x2 x4 + (x1 x4' + x1 x2')) = x3' (x2 x4 + x1 (x4' + x2')' ') = x3' (x2 x4 + x1 (x2 x4)' ) obtained by using DeMorgan's Theorem Letting g = x2 x4, we get f = x3' (g + x1 g') = x3' (g + x1) obtained by using 16a = x3' (x2 x4 + x1) = x2 x3' x4 + x1 x3' 2.25 (5 points) f = x1' x3' x5' + x1' x3' x4' + x1' x4 x5 + x1 x2' x3' x5 = x1' (x3' x5' + x3' x4' + x4 x5) + x1 x2' x3' x5 = x1' ((x3' x5' + x4 x5) + x3' x4') + x1 x2' x3' x5 = x1' ((x3' x5' + x4 x5) + x3' x4') + x1 x2' x3' x5Letting x = x5', y = x3', and z = x4, we get f = x1' ((x y + x' z) + x3' x4') + x1 x2' x3' x5 = x1' ((x y + y z + x' z) + x3' x4') + x1 x2' x3' x5 obtained by using the consensus property = x1' ((x5' x3' + x3' x4 + x5 x4) + x3' x4') + x1 x2' x3' x5 = x1' ((x3' x5' + x3' x4 + x4 x5) + x3' x4') + x1 x2' x3' x5 = x1' (x3' x5' + x3' x4 + x4 x5 + x3' x4') + x1 x2' x3' x5 = x1' (x3' (x5' + x4 + x4') + x4 x5) + x1 x2' x3' x5 = x1' x3' + x1' x4 x5 + x1 x2' x3' x5 = x1' x4 x5 + x3' (x1' + x1 x2' x5) = x1' x4 x5 + x3' (x1' + x2' x5) = x1' x3' + x1' x4 x5 + x2' x3' x5 An alternative manipulation by using DeMorgan's Theorem f = x1' x3' (x4' + x5')' ' + x1' x4 x5 + x1 x2' x3' x5 = x1' x3' (x4 x5)' + x1' x4 x5 + x1 x2' x3' x5Letting g = x4 x5, we get f = x1' x3' g' + x1' g + x1 x2' x3' x5 = x1' (x3' g' + g) + x1 x2' x3' x5 = x1' (x3' +g) + x1 x2' x3' x5 = x1' x3' + x1' x4 x5 + x1 x2' x3' x5 = x1' x4 x5 + x3' (x1' + x1 x2' x5) = x1' x4 x5 + x3' (x1' + x2' x5) = x1' x3' + x1' x4 x5 + x2' x3' x52.26 (5 points) The simplest product-of-sums expression for f is derived as follows: f = (x1' + x3' + x4') (x2' + x3' + x4) (x1 + x2' …


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TAMU ECEN 248 - 2.10 - 5 points - Texas A&M University

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