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.

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.

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.

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.

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Hello everybody this is a very nice site if you want to learn about these things!

tommorow am writing logic and i dont’s where to start revising it

Let A be an 8×8 Boolean matrix (i.e. every entry is 0 or 1). If the sum of the entries in A is 51, prove that there is a row i and a column j in A such that the entries in row i and in column j add up to more than 13. Further, show that there are at least 4 such pairs of rows and columns.

thanks alot

you are far too helpful

English only

Thank you very much that is helpful.

these basic logic gates are very useful to understand

excellant

Where can i get the soft copy?

Hello Odion, This website is free for anyone to use, there is no hard or soft copy available.

i love circuits !