Answer Happy

1. Solve the Boolean expression(𝑌∙0)𝑎𝑛𝑑((𝑋∙𝑌)+0)
2. If only two input AND gates are available, draw the circuitto
implement theBoolean expression𝑌=𝐴∙𝐵∙𝐶∙𝐷
3. Use De Morgan’s theorem to express Y=A∙B in a different
form.
4. Draw the circuit and write out the truth table for the Boolean
expression𝑌=(𝐴∙𝐵)∙(𝐴+𝐵̅̅̅̅̅̅̅̅).5. How can Y= (A ∙B)+C be
implemented only from NORgates?
6. Draw the time diagram for the AND gate. The order of the input
signals states is arbitrary.
7. Minimize the expression: 𝑌=𝐴̅𝐵̅𝐶̅+𝐴𝐵̅𝐶̅+𝐴𝐵𝐶̅using Boolean
theorems.
8. Draw the truth table and the Karnaugh map for the
expression𝑌=𝐴̅𝐵+𝐴𝐵̅.
9. Write out the SOP and POS Boolean expressions forthree-input OR
gate.
10. Write the minimizes SOP expression for ∏(0,2,3,8,14,15)using
the Karnaugh map. Create the logic circuit (scheme).