Combinational Digital Design. Discussion. Chapter 3. Boolean Algebra (Continued)
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1 The Islamic University of Gaza Engineering Faculty Department of Computer Engineering Fall 2017 ECOM 2013 Basma M. Al-Sousi Khaleel I. Shaheen Combinational Digital Design Discussion Chapter 3 Boolean Algebra (Continued)
2 Multiplying Out and Factoring Expressions Ex: List three laws or theorems which are useful when multiplying out or factoring expressions. 1. X (Y + Z) = XY + XZ 2. (X + Y) (X + Z) = X + YZ 3. (X + Y) (X + Z) = XZ + X Y هذه القوانين الثالثة تستخدم إليجاد SOP و.POS أول قانونين هما قانون التوزيع. في القانون الثالث الحظ أن المتغير المقترن مع X في الشق األول يكون مقترن مع X في الشق الثاني والعكس صحيح. الحظ كذلك أن القانون الثالث يكون باالتجاهين لذلك يستخدم لحساب.multiplying out و حساب factoring Ex: Factor each of the following expressions: ab + b c = (a + b ) (b + c) ab c + bd = (a + b) (c + b) (b + d) abc + (ab) d, suppose x = ab. xc + x d = (x + d) (x + c) (ab + d) ((ab) + c) = (a + d) (b + d) (a + b + c) Ex: Multiply out the following expression: F = (x + y + z) (w + x + y) (w + x + y ) (w + y + z ) Firstly, we apply equation 2 as follows: (x + y' + zw) (w' + y + x'z') Then by applying equation 3 we get the following: F = y(x+zw) + y'(w'+x'z') xy + wyz + x'y'z' + w'y' 2
3 Ex (3.6 a): Multiply out to obtain a sum of products: F = (W + X + Z ) (W + Y ) (W + X + Z ) (W + X ) (W + Y + Z) الخطوة األولى في حل أسئلة multiplying out هي البحث عن terms يمكن تبسيطها سريعا ومباشرة باستخدام قوانين شبتر 2. هنا نالحظ أنه يمكن التخلص من ( Z W) + X + باستخدام قانون Absorption حيث أن (W + X + Z ) (W + X ) = (W + X ) الخطوة الثانية هي البحث عن عامل مشترك في المعادلة لتطبيق القانون الثانية. F = (W + Y ) (W + X + Z ) (W + X ) (W + Y + Z) F = (W + Y (X + Z )) (W + X (Y + Z)) الخطوة الثالثة هي تطبيق القانون الثالث. F = W X Y + W X Z + WY X + WY Z Ex (3.7 b): Factor to obtain a product of sums F = A C D + ABD + A CD + B D الخطوة األولى في حل أسئلة factoring هي البحث عن terms يمكن تبسيطها سريعا ومباشرة باستخدام قوانين شبتر 2. في هذا المثال ليس هناك ما يمكن تبسيطه مباشرة. الخطوة الثانية هي البحث عن عامل مشترك في المعادلة لتطبيق القانون األول. F = D'(A'C' + AB) + D (A'C + B') الخطوة الثالثة نطبق القانون الثالث. 3
4 F = D'(A' + B) (A + C') + D(A' + B') (C + B') نالحظ هنا لم نحصل على POS من أول مرة لذلك نعيد تطبيق القانون الثالث ونوزع حتى نحصل على المطلوب. F = (A' + B' + D') (C + B' + D') (A' + B + D) (A + C' + D) Exclusive-OR and Equivalence Operations Ex (3.8): Write an expression for F and simplify. F = AB [(A D)' + D)] = AB [AD + A'D' + D] = AB (A'D' + D) = AB (A' + D) = (AB)'(A' + D) + AB(A' + D)' = (A' + B')(A' + D) + AB(AD') = A' + B'D + ABD' {Using (X + Y) (X + Z) = X + YZ} = A' + B'D + BD' {Using X + X'Y = X + Y} Ex (3.9): Is the following distributive law valid? Prove your answer. A BC = (A B) (A C) قانون التوزيع ال ينطبق على بوابة XOR وإلثبات ذلك يكفي أن تجد مثال واحد يثبت ذلك. Let A = 1, B = 1, C = 0. Left side of the equation: A BC = 1 1 * 0 = 1 0 = 1 4
5 Right side of the equation: (A B) (A C) = (1 1) (1 0) = 0 * 1 = 0 Ex (3.10 a): Reduce to a minimum sum of products (three terms) نالحظ الشق األيسر ال يساوي الشق األيمن. F = (X + W) (Y Z) + XW = (X + W) (Y'Z + YZ') + XW' = XY'Z + WY'Z + XYZ' + WYZ' + XW' = WY'Z + WYZ' + XW' {using consensus theorem} Ex (3.10 b): Reduce to a minimum sum of products (four terms): F = (A BC) + BD + ACD = A'BC + A(BC)' + BD + ACD = A'BC + A(B' + C') + BD + ACD = A'BC + AB' + AC' + BD + ACD = A'BC + AB' + AC' + AD + BD + ACD {add consensus term AD deletes ACD} = A'BC + AB' + AC' + BD {remove consensus term AD} Ex (3.11): Simplify algebraically to a minimum sum of products (five terms): F = (A + B + C + E ) (A + B + D + E) (B + C + D + E ) هنا ال يوجد تبسيط مباشر. إذن الخطوة األولى نبحث عن عامل مشترك. هنا يوجد 'B A + = [A + B' + (C + E') (D' + E)] (B' + C' + D' + E') = (A + B' + D'E' + CE) (B' + C' + D' + E') مرة أخرى نبحث عن عامل مشترك هنا 'B = B' + (A + D'E' + CE) (C' + D' + E') 5
6 = B' + AC' + AD' + AE' + C'D'E' + D'E' + D'E' + CD'E {distributive law} = B' + AC' + AD' + AE' + CD'E + D'E' + CD' {add consensus term CD'} = B' + AC' + AD' + AE' + D'E' + CD' = B' + AC' + AE' + D'E' + CD' Proving Validity of an Equation Ex (3.12): Prove algebraically that the following equation is valid: F = A CD E + A B D + ABCE + ABD = A B D + ABD + BCD E عندما يطلب إثبات معادلة ما هناك أكثر من طريقة للحل: 1. بناء جدول الصحة Truth Table لكل من شق ي المعادلة وإثبات إذا ما كان ناتج كل شق من شق ي المعادلة يساوي ناتج الشق اآلخر لكل االحتماالت الممكنة. طبعا هذا الحل سهل ولكنه متعب ويأخذ وقت خصوصا إذا كان عدد المدخالت كبيرا. 2. التعديل على أحد شق ي المعادلة من خالل تطبيق قوانين التبسيط حتى نصل إلى الشق اآلخر. تبسيط كل شق من شق ي العبارة أو المعادلة حتى الوصول إلى نتيجة واحدة فهذا.3 يعني أن طرفي المعادلة متكافئان. في هذا السؤال سنبدأ الحل من الشق األيسر للوصول إلى الشق األيمن. F = A'CD'E + A'B'D' + ABCE + ABD = A'CD'E + BCD'E + A'B'D' + ABCE + ABD {add consensus term BCD'E} = A'B'D' + ABD + BCD'E = Right Side the equation is valid. 6
7 Ex (3.13): Simplify each of the following expressions: a) F = KLMN + K L MN + MN = KLMN' + MN {using absorption X + XY = X} = M (KLN' + N) = M (KL + N) {using elimination X + X'Y = X + Y} = KLM + MN b) KL M + MN + LM N = KL'M' + N' (M + LM') {using elimination} = KL'M' + N' (M + L) = KL'M' + MN' + LN' c) (K + L ) (K + L + N) (L + M + N ) = [L' + K (K' + N)] (L' + M + N') {using distributive X + YZ = (X + Y) (X + Z)} = [L' + KN] (L' + M + N') = (L' + K) (L' + N) (L' + M + N') = (L' + K) [L' + N (M + N')] = (L' + K) [L' + NM] {using elimination} = (L' + K) (L' + N) (L' + M) d) (K + L + M + N) (K + M + N + R) (K + M + N + R ) KM {using uniting (X + Y) (X + Y ) = X} = (K' + L' + M' + N) (K' + M' + N) KM = [(K' + M') + N (N + L')] KM = [(K' + M') + N] KM {using absorption X (X + Y) = X} = (K' + M' + N) KM {using elimination} = (M' + N) KM {using elimination again} = N K M Ex (3.14 a): Factor to obtain a product of sums: F = K L M + KM N + KLM + LM N (four terms) 7
8 = K'L'M + KM'N + KLM + LM'N' = M' (KN + LN') + M (K'L' + KL) = M' [(K + N') (L + N)] + M [(K' + L) (K + L')] {using (X + Y) (X + Z) = XZ + X Y} = [M + (K + N') (L + N)] [M' + (K' + L) (K + L')] {using previous rule again} = (M + K + N') (M + L + N) (M' + K' + L) (M' + K + L') {using distributive law} Ex (3.16 a): Eliminate the exclusive OR, and then factor to obtain a minimum product of sums F = (KL M) + M N = (KL)'M + KLM' + M'N' = (K' + L') M + KLM' + M'N' = M (K' + L') + M' (KL + N') = (M' + K' + L') (M + N' + KL) = (M' + K' + L') (M + N' + K) (M + N' + L) Ex (3.29): The following circuit is implemented using two half-adder circuits. The expressions for the half-adder outputs are S = A B, and C = AB. Derive simplified sum-of-products expressions for the circuit outputs SUM and Co. SUM = (X Y) Ci = (XY' + X'Y) C i = (XY' + X'Y) C i ' + (XY' + X'Y)' Ci = XY'C i + X'Y C i + [(X' + Y) (X + Y') C i ] = XY' C i + X'Y C i + XY C i + X'Y' C i 8
9 = Ci C o = XY + Ci (X Y) = XY + C i (XY' + X'Y) = XY + XY'C i + X'YC i = X (Y + Y'C i ) + X'YC i = X (Y + C i ) + X'YC i = XY + XC i + X'YC i = Y (X + X'C i ) + XC i = Y (X + C i ) + XC i = XY + XC i + YC i Good Luck 9
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