1 Advanced Mathematics • Digital Logic Digital circuit can be made by the repeated use of which gate? Digital circuit can be made by the repeated use of which gate? सेव करें Save सेव्ड Saved A. OR OR B. None of the given options None of the given options C. NOT NOT D. NAND NAND उत्तर और व्याख्या देखें Show answer and explanation सही उत्तर: D Correct Answer: D NAND is a universal gate. Any Boolean function and therefore any digital circuit can be implemented using NAND gates alone. NAND is a universal gate. Any Boolean function and therefore any digital circuit can be implemented using NAND gates alone.
2 Advanced Mathematics • Digital Logic Which logic gate has the following truth table? Inputs (A,B) and output C: (0,0)->0, (0,1)->1, (1,0)->1, (1,1)->1. Which logic gate has the following truth table? Inputs (A,B) and output C: (0,0)->0, (0,1)->1, (1,0)->1, (1,1)->1. सेव करें Save सेव्ड Saved A. OR with 2 inputs OR with 2 inputs B. NAND NAND C. AND with 2 inputs AND with 2 inputs D. NOR NOR उत्तर और व्याख्या देखें Show answer and explanation सही उत्तर: A Correct Answer: A The output is 1 whenever at least one input is 1, which is the truth table of a two-input OR gate. The output is 1 whenever at least one input is 1, which is the truth table of a two-input OR gate.
3 Advanced Mathematics • Digital Logic Which is called the universal gate? Which is called the universal gate? सेव करें Save सेव्ड Saved A. AND AND B. OR OR C. NAND NAND D. XNOR XNOR उत्तर और व्याख्या देखें Show answer and explanation सही उत्तर: C Correct Answer: C NAND is a universal gate because every Boolean function can be implemented using NAND gates alone. NAND is a universal gate because every Boolean function can be implemented using NAND gates alone.
4 Advanced Mathematics • Digital Logic The number of Boolean functions that can be generated by n variables is: The number of Boolean functions that can be generated by n variables is: सेव करें Save सेव्ड Saved A. 2^(n-1) 2^(n-1) B. 2^n 2^n C. 2^(2^n) 2^(2^n) D. 2^(2n) 2^(2n) उत्तर और व्याख्या देखें Show answer and explanation सही उत्तर: C Correct Answer: C There are 2^n possible input combinations, and each can independently map to 0 or 1. Hence the number of Boolean functions is 2^(2^n). There are 2^n possible input combinations, and each can independently map to 0 or 1. Hence the number of Boolean functions is 2^(2^n).
5 Advanced Mathematics • Digital Logic The logical expression y = A + A^cB is equivalent to: The logical expression y = A + A^cB is equivalent to: सेव करें Save सेव्ड Saved A. y = A^c + B y = A^c + B B. y = A + B y = A + B C. y = AB y = AB D. y = A^cB y = A^cB उत्तर और व्याख्या देखें Show answer and explanation सही उत्तर: B Correct Answer: B Using the absorption identity, A + A^cB = (A + A^c)(A + B) = 1(A + B) = A + B. Using the absorption identity, A + A^cB = (A + A^c)(A + B) = 1(A + B) = A + B.
6 Advanced Mathematics • Digital Logic The minimized form of the logical expression A^cB^cC^c + A^cBC^c + A^cBC + ABC^c is: The minimized form of the logical expression A^cB^cC^c + A^cBC^c + A^cBC + ABC^c is: सेव करें Save सेव्ड Saved A. A^cC^c + BC^c + A^cB A^cC^c + BC^c + A^cB B. (AC)^c + (BC)^c + (AB)^c (AC)^c + (BC)^c + (AB)^c C. A^cC + B^cC + A^cB^c A^cC + B^cC + A^cB^c D. AC^c + B^cC + AB^c AC^c + B^cC + AB^c उत्तर और व्याख्या देखें Show answer and explanation सही उत्तर: A Correct Answer: A Grouping minterms 0 and 2 gives A^cC^c, 2 and 6 gives BC^c, and 2 and 3 gives A^cB. Hence the minimized sum is A^cC^c+BC^c+A^cB. Grouping minterms 0 and 2 gives A^cC^c, 2 and 6 gives BC^c, and 2 and 3 gives A^cB. Hence the minimized sum is A^cC^c+BC^c+A^cB.
7 Advanced Mathematics • Digital Logic The inputs of a NAND gate are connected together. The resulting circuit is: The inputs of a NAND gate are connected together. The resulting circuit is: सेव करें Save सेव्ड Saved A. OR OR B. XOR XOR C. NOT NOT D. AND AND उत्तर और व्याख्या देखें Show answer and explanation सही उत्तर: C Correct Answer: C If both NAND inputs equal A, the output is (A.A)^c=A^c. Therefore the tied-input NAND acts as a NOT gate. If both NAND inputs equal A, the output is (A.A)^c=A^c. Therefore the tied-input NAND acts as a NOT gate.