This is a Clilstore unit. You can .

**Reminder from the previous lesson:**

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

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 ( 2^{2} ) 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****.**

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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Short url: https://clilstore.eu/cs/6120