Boolean Algebra Examples
We have seen throughout this section that digital logic
functions can be defined and displayed as either a Boolean Algebra
expression or as a logic gate truth table. So here are a few examples of
how we can use Boolean Algebra to simplify larger digital logic circuits.
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.
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.
| Inputs |
Output at |
| A |
B |
C |
D |
Q |
| 0 |
0 |
1 |
0 |
0 |
| 0 |
1 |
1 |
1 |
1 |
| 1 |
0 |
1 |
1 |
1 |
| 1 |
1 |
0 |
0 |
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.
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.
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:
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.
| Inputs |
Intermediates |
Output |
| B |
A |
A.B |
A + B |
Q |
| 0 |
0 |
0 |
1 |
1 |
| 0 |
1 |
0 |
0 |
0 |
| 1 |
0 |
0 |
0 |
0 |
| 1 |
1 |
1 |
0 |
1 |
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.
Boolean Algebra Example No3
Find the Boolean algebra expression for the following system.
This system may look more complicated than the other two to analyse but again, the logic circuit just consists of simple
AND,
OR and
NOT gates connected together.
As with the previous Boolean examples, we can simplify the
circuit by writing down the Boolean notation for each logic gate
function in turn in order to give us a final expression for the output
at
Q.
The output from the 3-input
AND gate is only at logic “1” when
ALL the gates inputs are HIGH 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).
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.
| Inputs |
Intermediates |
Output |
| C |
B |
A |
A.B.C |
B |
C |
B+C |
A.(B+C) |
Q |
| 0 |
0 |
0 |
0 |
1 |
1 |
1 |
0 |
0 |
| 0 |
0 |
1 |
0 |
1 |
1 |
1 |
1 |
1 |
| 0 |
1 |
0 |
0 |
0 |
1 |
1 |
0 |
0 |
| 0 |
1 |
1 |
0 |
0 |
1 |
1 |
1 |
1 |
| 1 |
0 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
| 1 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
1 |
| 1 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
| 1 |
1 |
1 |
1 |
0 |
0 |
0 |
0 |
1 |
Then we can see that the entire logic circuit above can be replaced by just one single input labelled
A thereby reducing a circuit of six individual logic gates to just one single piece of wire, (or
Buffer). This type of circuit analysis using
Boolean Algebra
can be very powerful and quickly identify any unnecessary logic gates
within a digital logic design thereby reducing the number of gates
required, the power consumption of the circuit and of course the cost.
