Spring 2010 CPE231 Digital Logic Section 1 Quiz 1A. Convert the following numbers from the given base to the other three bases listed in the table:


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1 Section 1 Quiz 1A Convert the following numbers from the given base to the other three bases listed in the table: Decimal Binary Hexadecimal C6.5 Simplify the following Boolean expressions to expressions containing a minimum number of literals Name: Number:
2 Section 2 Quiz 1A Convert the following numbers from one notation to another = = 3B = AC9 16 = For each expression below, use DeMorgan s theorem to obtain an equivalent expression which contains ANDs and ORs of the inputs (e.g. A) and their complements (e.g. A ). There should be no complements (bars) in the final expression except those over the inputs. Name: Number:
3 Section 1 Quiz 1B Convert the following numbers from the given base to the other three bases listed in the table: Decimal Binary Hexadecimal F3C7.A Simplify the following Boolean expressions to expressions containing a minumium number of literals (a) A B C + A C (b) A B D + A C D + B D Name: Student No.: 
4 Section 1 Quiz 1B Convert the following numbers from the given base to the other three bases listed in the table: Decimal Binary Hexadecimal F3C7.A Simplify the following Boolean expressions to expressions containing a minimum number of literals (a) (b) Name: Student No.: 
5 Section 2 Quiz 1B Convert the following numbers from one notation to another = = DEAF = ED5 16 = For each expression below, use DeMorgan s theorem to obtain an equivalent expression which contains ANDs and ORs of the inputs (e.g. A) and their complements (e.g. A ). There should be no complements (bars) in the final expression except those over the inputs. Name: Number:
6 Digital Logic () Quiz 1 Form A Solutions are in RED COLOR رقم التسجيل : الاسم : الشعبة: رقم ====================================================================== Instructions: Time 15 minutes. Closed books and notes. No calculators. No questions are allowed. ======================================================================= Q1. Convert (325) 10 to binary. (8 points) (325) 10 = OR: = = = 1 (325) 10 = = = Q2. Convert (A63) 16 to binary. (4 points) (A63) 16 = Q3. Analyze the following circuit to find Boolean Functions F(X,Y,Z) are being implemented. Then simplify your answer if possible. (16 points) F(X,Y,Z) = X YZ +X YZ + XZ Before simplification = X Y(Z +Z ) + XZ By identity 14 = X Y.1 + XZ By identity 7 = X Y + XZ By identity 2 F(X,Y,Z) = X Y + XZ After simplification 1
7 Q4. Evaluate the following binary addition. (4 points) carry Q5. From the truth table, write the function F(X,Y,Z) in the SOP terms then draw circuit implementing the followin function using AND, Or, and NOT gates without simplification. (18 points). (8 points for the function and 10 points for the circuit) X Y Z F minterm X Y Z X Y Z XY Z XYZ XYZ F(X,Y,Z) = X Y Z + X Y Z + XY Z + XYZ + XYZ 2
8 Digital Logic () Quiz 1. Form B Solutions are in RED COLOR رقم التسجيل : الاسم : الشعبة: رقم ====================================================================== Instructions: Time 15 minutes. Closed books and notes. No calculators. No questions are allowed. ======================================================================= Q1. Convert (295) 10 to binary. (8 points) (295) 10 = OR: = = = = 1 (295) 10 = = = Q2. Evaluate the following binary addition. (4 points) Carry Q3. Analyze the following circuit to find Boolean Functions F(X,Y,Z) are being implemented. Then simplify your answer if possible. (16 points) (6 points for the function and 10 points for simplification) F(X,Y,Z) = X(Y + Z)(X+Y+Z ) Before simplification = X [Y (X+Y+Z ) + Z(X+Y+Z )] = X [XY +0+Y Z + XZ+YZ+0] = X [XY +Y Z + XZ+YZ] = X.XY +XY Z +X.XZ+XYZ = XY +XY Z + XZ+XYZ = XY (1+Z ) + XZ(1+Y) = XY + XZ F(X,Y,Z) = XY + XZ After simplification
9 Q4. Convert (B32) 16 to binary. (4 points) (B32) 16 = Q5. From the truth table, write the function F(X,Y,Z) in the SOP terms then draw circuit implementing the followin function using AND, OR, and NOT gates without simplification. (18 points) (6 points for the function and 12 points for the circuit) X Y Z F minterm X YZ X YZ XY Z XYZ F(X,Y,Z) = X YZ + X YZ + XY Z + XYZ 4
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