Activity 2.1.3 Truth Tables & Logic Expressions
[pic]
Jireh Alberto
|Activity 2.1.3 Truth Tables & Logic Expressions |
Introduction
The first step in designing a new product is creating a design specifications document. These design specifications detail all of the features and limitations of the new product.
In digital electronics, the process of translating these design specifications into a functioning circuit starts with the creation of a truth table. A truth table is simply a list of all possible binary input combinations that could be applied to a circuit and the corresponding binary outputs that the circuit produces. Once the truth table is complete, a Boolean expression can easily be written directly from the truth table.
In this activity you will learn how to translate design specifications into truth tables and, in turn, write un-simplified logic expressions from these truth tables.
In future activities we will learn how to use Boolean algebra as well as a graphical technique called Karnaugh mapping to simplify these logic expressions.
Equipment
0. Paper & Pencil
0. Calculator
Procedure
Before we tackle the more challenging aspects of translating a written specification into a truth-table, let’s take a look at the process of writing an un-simplified logic expression from a completed truth table.
Example:
|A |B |Z |Minterms | | |
| | | | | |[pic] |
|0 |0 |0 | | | |
|0 |1 |1 |[pic] | | |
|1 |0 |1 |[pic] | | |
|1 |1 |0 | | | |
Using the above example as a guide, write the un-simplified logic expression for each of the following truth tables. Though it is not required, you may find it helpful to first write the Minterm expression for every place containing a one in the output function.
|F1 = ? F1=MN+MN |F2 =? F2=XY+XY+XY |
|M |X |
|N |Y |
|F1 |F2 |
| | |
|0 |0 |
|0 |0 |
|1 |1 |
| | |
|0 |0 |
|1 |1 |
|0 |0 |
| | |
|1 |1 |
|0 |0 |
|0 |1 |
| | |
|1 |1 |
|1 |1 |
|1 |1 |
| | |
| | |
| | |
| | |
| | |
|F3 =? F3=PQR+PQR+PQ |F4 =? F4=JKL+JKL+JKL+JKL |
|P |J |
|Q |K |
|R |L |
|F3 |F4 |
| | |
|0 |0 |
|0 |0 |
|0 |0 |
|0 |0 |
| | |
|0 |0 |
|0 |0 |
|1 |1 |
|1 |1 |
| | |
|0 |0 |
|1 |1 |
|0 |0 |
|0 |0 |
| | |
|0 |0 |
|1 |1 |
|1 |1 |
|0 |1 |
| | |
|1 |1 |
|0 |0 |
|0 |0 |
|1 |1 |
| | |
|1 |1 |
|0 |0 |
|1 |1 |
|0 |0 |
| | |
|1 |1 |
|1 |1 |
|0 |0 |
|1 |0 |
| | |
|1 |1 |
|1 |1 |
|1 |1 |
|0 |1 |
| | |
| | |
|F5 =? F5=WXYZ+WXYZ+WXYZ+WXYZ+WXYZ+WXYZ | |
|W | |
|X | |
|Y | |
|Z | |
|F5 | |
| | |
|0 | |
|0 | |
|0 | |
|0 | |
|0 | |
| | |
|0 | |
|0 | |
|0 | |
|1 | |
|0 | |
| | |
|0 | |
|0 | |
|1 | |
|0 | |
|1 | |
| | |
|0 | |
|0 | |
|1 | |
|1 | |
|0 | |
| | |
|0 | |
|1 | |
|0 | |
|0 | |
|1 | |
| | |
|0 | |
|1 | |
|0 | |
|1 | |
|1 | |
| | |
|0 | |
|1 | |
|1 | |
|0 | |
|0 | |
| | |
|0 | |
|1 | |
|1 | |
|1 | |
|0 | |
| | |
|1 | |
|0 | |
|0 | |
|0 | |
|0 | |
| | |
|1 | |
|0 | |
|0 | |
|1 | |
|0 | |
| | |
|1 | |
|0 | |
|1 | |
|0 | |
|1 | |
| | |
|1 | |
|0 | |
|1 | |
|1 | |
|0 | |
| | |
|1 | |
|1 | |
|0 | |
|0 | |
|1 | |
| | |
|1 | |
|1 | |
|0 | |
|1 | |
|0 | |
| | |
|1 | |
|1 | |
|1 | |
|0 | |
|1 | |
| | |
|1 | |
|1 | |
|1 | |
|1 | |
|0 | |
| | |
Now that you have mastered the process of writing an un-simplified logic expression from a completed truth table, let’s reverse the process. For the following logic expressions, create a corresponding truth table. Note that some terms in the logic expression may map to more than one place in the truth table.
1. [pic]
|C |D |F1 |
|0 |0 |1 |
|0 |1 |1 |
|1 |0 |0 |
|1 |1 |0 |
2. [pic]
|R |S |T |F2 |
|0 |0 |0 |1 |
|0 |0 |1 |0 |
|0 |1 |0 |1 |
|0 |1 |1 |1 |
|1 |0 |0 |1 |
|1 |0 |1 |0 |
|1 |1 |0 |0 |
|1 |1 |1 |0 |
3. [pic]
|A |B |C |F3 |
|0 |0 |0 |1 |
|0 |0 |1 |0 |
|0 |1 |0 |1 |
|0 |1 |1 |0 |
|1 |0 |0 |1 |
|1 |0 |1 |1 |
|1 |1 |0 |0 |
|1 |1 |1 |0 |
Now that you understand the mechanics of converting from a truth table to a logic expression (and vice-versa), let’s look at a problem that begins with some written specifications. Your new car has an audio alarm that buzzes whenever the door is open and the key is in the ignition or when the key is in the ignition and the seatbelt is not buckled.
1. Using the following variable names and assignment conditions, complete the truth table shown below that captures the functionality of this audio alarm.
|D: Door → 0=Door Open / 1=Door Close | |
|K: Key → 0=Key Not in Ignition / 1=Key in Ignition | |
|S: Seat → 0=Not Buckled / 1=Buckled | |
|B: Buzzer → 0=Buzzer Off / 1=Buzzer On | |
|D |K |S |B |
|(Door) |(Key) |(Seat Belt) |(Buzzer) |
|0 |0 |0 |0 |
|0 |0 |1 |0 |
|0 |1 |0 |1 |
|0 |1 |1 |1 |
|1 |0 |0 |0 |
|1 |0 |1 |0 |
|1 |1 |0 |1 |
|1 |1 |1 |0 |
2. Using the truth table you just created, write the un-simplified logic expression for the buzzer (i.e., variable B). Be sure that your answer is in the Sum-of-Products form.
B=DKS+DKS=DKS
In a future activity, you will see that this logic expression can be simplified to:
[pic]
Note: Some of you may have been able to extract this simplified logic expression directly from the specifications. If so, you may be asking yourself, why did I have to go through the process of creating the truth table and writing the un-simplified logic expression first? This was a simple problem. As you progress through this class, the design problems will become more difficult The process that you are learning now will become invaluable in solving more challenging problems.
|Your Aunt Mary would like you to design a digital logic circuit that monitors the conditions in a room of her | |
|house where her seven cats live. The room’s temperature is monitored by a temperature sensor. The room’s | |
|humidity is monitored by a humidity sensor. The air quality of the room is monitored by a special sensor that | |
|measures fur-ball density. The three sensors output a one (1) to indicate an out-of-range condition. Located | |
|outside of the room is an ALERT light that is on (1) whenever two or more sensors are out-of-range. | |
1. Create a truth table that captures the functionality of this room monitoring system.
|T |H |F |L |
|(temp. monitor) |(humidity monitor) |(fur-ball monitor) |(alarm light) |
|0 |0 |0 |0 |
|0 |0 |1 |0 |
|0 |1 |0 |0 |
|0 |1 |1 |1 |
|1 |0 |0 |0 |
|1 |0 |1 |1 |
|1 |1 |0 |1 |
|1 |1 |1 |1 |
2. Using the truth table that you just created, write the un-simplified logic expression for the ALERT light. Be sure that your answer is in the Sum-of-Products form.
L=THF+THF=THF+THF
Conclusion
1. A digital logic circuit with (2) inputs has (4) input combinations. One with (3) inputs has (8) combinations. One with (4) inputs has (16) combinations.
How many input combinations would a digital logic circuit have if it has (5) inputs? How about (6) input?
• 32 & 64
Mathematically express the relationship between the number of input (N) and the number of input combinations (C).
• C=2^N
2. Write the un-simplified logic expression for the truth table shown below.
Z=XY+XY
|X |Y |Z |
|0 |0 |1 |
|0 |1 |1 |
|1 |0 |1 |
|1 |1 |1 |
Did you see the short cut? (Hint: Z is always a one.)
Going Further – Optional
The un-simplified logic expression for the truth table shown below is
[pic]; Another way to write this would be[pic].
|A |B |C |
|0 |0 |1 |
|0 |1 |1 |
|1 |0 |0 |
|1 |1 |1 |
Explain why this is true. This is true because the only time C is zero is when A is one and B is zero.
-----------------------
Alarm on when:
Door0 & Key1
Or
Key1&Seat0
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