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Boolean Algebra Examples

Reminder from the previous lesson:

https://create.kahoot.it/l/#quiz/5cc426bb-e172-40ff-8759-451ca1a7e9c9

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Boolean Algebra Example No1

Construct a Truth Table for the logical functions at points C, D and Q in the following circuit and identify a single logic gate that can be used to replace the whole circuit.

 Please press  Button text: EXAMPLE_1

SOLUTION:

First observations tell us that the circuit consists of a 2-input NAND gate, a 2-input EX-OR gate and finally a 2-input EX-NOR gate at the output. As there are only 2 inputs to the circuit labelled A and B, there can only be 4 possible combinations of the input ( 22 ) and these are: 0-0, 0-1, 1-0 and finally 1-1. Plotting the logical functions from each gate in tabular form will give us the following truth table for the whole of the logic circuit below.

 

 Please press  Button text:SOLUTION_1

From the truth table above, column C represents the output function generated by the NAND gate, while column D represents the output function from the Ex-OR gate. Both of these two output expressions then become the input condition for the Ex-NOR gate at the output.

CONCLUSIONS:

It can be seen from the truth table that an output at Q is present when any of the two inputs A or B are at logic 1. The only truth table that satisfies this condition is that of an OR Gate. Therefore, the whole of the above circuit can be replaced by just one single 2-input OR Gate.

 

Boolean Algebra Example No2

Find the Boolean algebra expression for the following system.

  Please press  Button text: EXAMPLE_1

The output of the system is given as Q = (A.B) + (A+B), but the notation A+B is the same as the De Morgan´s notation A.B, Then substituting A.B into the output expression gives us a final output notation of Q = (A.B)+(A.B), which is the Boolean notation for an Exclusive-NOR Gate as seen in the previous section.

 Please press  Button text:SOLUTION_2

 

The output of the system is given as Q = (A.B) + (A+B), but the notation A+B is the same as the De Morgan´s notation A.B, Then substituting A.B into the output expression gives us a final output notation of Q = (A.B)+(A.B), which is the Boolean notation for an Exclusive-NOR Gate as seen in the previous section.

 Please press  Button text:SOLUTION_3

CONCLUSIONS:

Then, the whole circuit above can be replaced by just one single Exclusive-NOR Gate and indeed an Exclusive-NOR Gate is made up of these individual gate functions.

 

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Clilstore ReminderEXAMPLE_1SOLUTION_1EXAMPLE_2SOLUTION_2SOLUTION_3EVALUATION

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