Witryna17 maj 2024 · Using an 8:1 Multiplexer to Implement a 4-input Logical Function. Log in to Reply. Elizabeth Simon says: August 26, 2024 at 7:07 pm. The standard design flow from Karnaugh map to AND-OR logic to NAND logic that you used here works well (and is relatively easy if you know the trick but does not work well for converting to NOR logic. Witrynastep 1: Take the inputs of the circuit to implement as the select lines for the multiplexer. That is for your convenience just write the select line variables above the input …
multiplexer - How do you implement the following …
WitrynaLet it be generalized for any system we need to implement using a multiplexer. step 1: Take the inputs of the circuit to implement as the select lines for the multiplexer. That is for your convenience just write the select line variables above the input variables. step 2: Have a look at the output sop for the given circuit. Mux only has one ... WitrynaQuestion: Question 2 Select all true statements regarding multiplexers Can be impemented using POS 1 All inputs are active low Can be implemented using only NAND gates Are abbreviated as MUX Cannot be use to select data Can be implemented using smaller multiplexers All multiplexer have control pin E tenable! Can be … east anglian beaches
IMPLEMENTATION OF BOOLEAN EXPRESSION AND LOGIC FUNCTION USING …
WitrynaAnswer (1 of 2): First let's simplify given boolean expression. Y=(A\oplus B)C+\overline{A}BC = (\overline{A}B+A \overline{B})C+ \overline{A}BC = \overline{A}BC+ A\overline{B}C+ \overline{A}BC = \overline{A}BC+A\overline{B}C This boolean expression is of three variable so at least one 4:1 MU... Witryna14 gru 2024 · Step 2: To find number of select lines and input lines of the Multiplexer. For n variable Boolean function, the number of select lines of multiplexer (MUX) would be (n-1). As we know that for a 2:1 MUX number of select lines would be 1. In this case there are two variables A & B. Therefore, Number of select lines would be n-1 = 1. Witryna13 gru 2024 · Step 4: To draw the circuit for implementing 2-input NAND Gate using 2:1 MUX. As seen from the implementation table, to design a 2-input NAND Gate, connect the input I0 of the 2:1 multiplexer to 1 and the input I1 to ‘A/’ . In this way a 2 input NAND Gate can be implemented using a 2:1 multiplexer. Hope this post on " 2-Input … c\u0027est ma terre christophe mae youtube