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Home ‣ Digital Electronics ‣ Boolean Algebra and Logic Simplification Comments
  • Question

  • The NAND or NOR gates are referred to as "universal" gates because either:


  • Options
  • A. can be found in almost all digital circuits
  • B. can be used to build all the other types of gates
  • C. are used in all countries of the world
  • D. were the first gates to be integrated

  • Correct Answer
  • can be used to build all the other types of gates 

    • Boolean Algebra and Logic Simplification problems

    Search Results

    • 1. Output will be a LOW for any case when one or more inputs are zero for a(n):

    • Options
    • A. OR gate
    • B. NOT gate
    • C. AND gate
    • D. NOR gate
      • Discuss
    • 2. The output of a NAND gate is LOW if ________.

    • Options
    • A. all inputs are LOW
    • B. all inputs are HIGH
    • C. any input is LOW
    • D. any input is HIGH
      • Discuss
    • 3. If a signal passing through a gate is inhibited by sending a LOW into one of the inputs, and the output is HIGH, the gate is a(n):

    • Options
    • A. AND
    • B. NAND
    • C. NOR
    • D. OR
      • Discuss
    • 4. How many entries would a truth table for a four-input NAND gate have?

    • Options
    • A. 2
    • B. 8
    • C. 16
    • D. 32
      • Discuss
    • 5. The logic gate that will have HIGH or "1" at its output when any one (or more) of its inputs is HIGH is a(n):

    • Options
    • A. OR gate
    • B. AND gate
    • C. NOR gate
    • D. NOT operation
      • Discuss
    • 6. Convert the following SOP expression to an equivalent POS expression.

    • Options
    • A.
    • B.
    • C.
    • D.
      • Discuss
    • 7. For the SOP expression , how many 0s are in the truth table's output column?

    • Options
    • A. zero
    • B. 1
    • C. 4
    • D. 5
      • Discuss
    • 8. An OR gate with schematic "bubbles" on its inputs performs the same functions as a(n)________ gate.

    • Options
    • A. NOR
    • B. OR
    • C. NOT
    • D. NAND
      • Discuss
    • 9. Converting the Boolean expression LM + M(NO + PQ) to SOP form, we get ________.

    • Options
    • A. LM + MNOPQ
    • B. L + MNO + MPQ
    • C. LM + M + NO + MPQ
    • D. LM + MNO + MPQ
      • Discuss
    • 10. How many gates would be required to implement the following Boolean expression before simplification? XY + X(X + Z) + Y(X + Z)

    • Options
    • A. 1
    • B. 2
    • C. 4
    • D. 5
      • Discuss

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