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Electronics Tutorial about Boolean Algebra
 

Boolean Algebra Examples

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Boolean Algebra Examples

Boolean Algebra Example No 1

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.

Example Circuit No1

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.

InputsOutput at
ABCDQ
00100
01111
10111
11001

From the truth table above, column C represents the output function from the NAND gate and 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. 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 No 2.

Find the Boolean algebra expression for the following system.

Example Circuit No2

The system consists of an AND Gate, a NOR Gate and finally an OR Gate. The expression for the AND gate is A.B, and the expression for the NOR gate is A+B. Both these expressions are also separate inputs to the OR gate which is defined as A+B. Thus the final output expression is given as:

Example No2

The output of the system is given as Q = (A.B) + (A+B), but the notation A+B is the same as the De Morganīs 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.

InputsIntermediatesOutput
BA A.B A + B Q
00011
01000
10000
11101

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 gates.

Boolean Algebra Example No 3.

Find the Boolean algebra expression for the following system.

Example Circuit No3

This system may look more complicated than the others to analyse but it also consists of simple AND, OR and NOT gates. Again as with the previous example we can write the Boolean notation for each logic function to give us a final expression for the output at Q.

Example No3 Answer

The output from the 3-input AND gate is only a "1" when ALL the inputs are at logic level "1" (A.B.C). The output from the lower OR gate is only a "1" when one or both inputs B or C are at logic level "0". The output from the 2-input AND gate is a "1" when input A is a "1" and inputs B or C are at "0". Then the output at Q is only a "1" when inputs A.B.C equal "1" or A is equal to "1" and both inputs B or C equal "0", A.(B+C). Then by using "de Morgan's theorem" inputs B and input C cancel out as to produce an output at Q they can be either at logic "1" or at logic "0". Then this just leaves input A as the only input needed to give an output at Q as shown in the table below.

InputsIntermediatesOutput
CBA A.B.C B C B+C A.(B+C) Q
000011100
001011111
010001100
011001111
100010100
101010111
110000000
111100001

Then, the whole circuit above can be replaced by just one single input labelled A thereby reducing a circuit of 6 individual logic gates to just one single piece of wire, (or Buffer). This type of circuit analysis using Boolean Algebra can quickly identify any un-necessary logic gates thereby reducing the number of gates required, the power consumption of the circuit and of course the cost.


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