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<h1 align=center>EG1004 Lab 6: Digital Logic</h1>
<!--= NI ELVIS Tutorial Video =
General Engineering has created a video that discusses breadboard wiring and digital logic circuits: [https://stream.nyu.edu/media/EG+1003+Lab+5A+NI+ELVIS+Tutorial/1_36717xpx NI ELVIS Tutorial Video]. This video provides information on the educational NI-ELVIS II + prototyping board that will aid in designing and wiring the physical circuit. <span style="color: red;"> '''Material covered in the video will be on the Lab 8 quiz. Viewing the video is a mandatory pre-lab assignment. ''' </span>-->
= Objective =


<h2>1 OBJECTIVES</h2>
The experimental objective of this lab is to design a combinational logic circuit for a given problem statement, and to activate it under specific conditions and test it using LabVIEW. After testing, it will be built on an NI-ELVIS II+ prototyping board.


<p>The experimental objective of this lab is to design a
= Overview =
combinational logic circuit that will
activate under specific conditions and test it using LabVIEW. After testing, it
will be built on the digital trainer.</p>


<p>Your goal is to become familiar with the
The first step in understanding the digital circuits that control the function of electronic devices is the mastery of <b>Boolean logic</b>. George Boole, an English mathematician, established modern symbolic logic in 1854 with the publication of his paper, "Laws of Thought." Boolean logic is the foundation of digital circuitry. Boole's method of logical inference allows conclusions to be drawn from any proposition involving any number of terms. Boolean logic demonstrates that the conclusions found are logically contained in the original premises (Encyclopedia Britannica, 2003).
principles of digital logic and digital logic circuits. Understanding how logic gates work is critical
to this process.</p>


<h2>2 OVERVIEW</h2>
== Truth Tables ==


<p>The first step in understanding the digital circuits that
In Boolean logic, there are only two values, true and false, represented by the numbers 1 and 0, respectively. These values are combined in equations to yield results that also have these two values. The equations are represented by <b>truth tables</b> that show the inputs to the equation and the outputs produced for those inputs. The rows of the table contain all the possible combinations of 1s and 0s for the inputs. The number of rows is determined by the number of possible combinations.
control the function of elec­tronic devices is the mastery of <b><i>Boolean </i></b>logic.
George Boole<b><i>, </i></b>an English mathemati­cian, established modern
symbolic logic in 1854 with the publication of his paper, <b><i>&quot;Laws of
Thought.&quot; </i></b>Now called <b><i>Boolean algebra, </i></b>it is the
foundation of digital circuitry. Boole's method of logical inference enables us
to draw conclusions from any proposition
involving any number of terms. Boolean logic demonstrates that our con­clusions
are logically contained in our original premises<sub>.</sub><sup>1</sup></p>


<p>In Boolean logic, there are only two values, <b><i>true</i></b>
To illustrate a truth table and additional concepts that will be introduced, an example problem statement will be given. Think of an ATM that has three options: print a statement, withdraw money, or deposit money. The ATM will charge a fee to (1) withdraw money or (2) print a statement without depositing money. The intent of the problem is to develop a Boolean equation and logic circuit that will determine after which possible combination of actions will someone get charged a fee. First, a truth table should be made for all the possible combinations of inputs. The inputs are the ATM's three functions. Let variable <math>P</math> stand for printing a statement, <math>W</math> for withdrawing money, and <math>D</math> for depositing money. There is one output, which is whether or not the ATM will charge a fee. The output will be denoted by <math>C</math>. The truth table in Table 1 shows all the possible combinations of the inputs and their corresponding outputs.
and <b><i>false</i></b>, represented by the numbers 1 and 0, respectively.
These values are combined in equations to yield results that also have these
two values. The equations are represented by <b><i>truth tables</i></b> that
show the inputs to the equation and the outputs produced for those inputs. The
rows of the table con­tain all the possible combinations of 1s and 0s for the
inputs. The number of rows is determined by
the number of possible combinations.</p>


<p>Boolean logic is applied to digital circuitry through the use
<center>
of simple logic gates. There are symbols for each of these types of gates, and
Table 1: Truth Table for the ATM Example
the connections between them are represented by lines running from the output
{| class="truthtable"
of one gate to the input of the other. A line can connect only one output to
|-
each input. There are seven of these gates: the <b><i>NOT, AND, OR, NAND,
!colspan="3"|Inputs!!Output
NOR, XOR, </i></b>and <b><i>XNOR </i></b>gates. We will limit our discussion to
|-
the first three.</p>
!<math>P</math>!!<math>W</math>!!<math>D</math>!!<math>C</math>
|-
|0||0||0||0
|-
|0||0||1||0
|-
|0||1||0||1
|-
|0||1||1||1
|-
|1||0||0||1
|-
|1||0||1||0
|-
|1||1||0||1
|-
|1||1||1||1
|-
|}
</center>


<p>The <b><i>NOT </i></b>gate is the simplest of these three. It
== Logic Gates ==
is an inverter. It takes one only input and produces its opposite as output. Its
symbol looks like this:</p>


<p>[[Image:lab_logic_1.jpg]]</p>
Boolean logic is applied to digital circuitry through the use of simple <b>logic gates</b>. There are symbols for each of these gates, and the connections between them are represented by lines running from the output of one gate to the input of another. A line can connect only one output to each input. There are seven of these gates: the NOT, AND, OR, NAND, NOR, XOR, and XNOR gates. Only the first three will be used in this lab (Figure 1).


<p>The truth table for a NOT gate has one input, which we’ll call
[[Image:Lab 8 Figure 1.jpg|frame|center| Figure 1: Logic Gate Symbols]]
A, and one output. The symbol for the operation is a horizontal bar over the
variable, so the truth table looks like this:</p>


<table cellspacing=0>
The <b>NOT gate</b> is the simplest of these three. It is an inverter. It has one input and produces its opposite as the output. For example, if a 1 value is put into a NOT gate, a 0 value is outputted, as seen in Table 2. The symbol for the operation is a horizontal bar over the variable. The truth table for a NOT gate is shown in Table 2.
<tr>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt">
<p class=caption><b>A</b></p>
  </td>


<td style="border-bottom:solid windowtext 1.0pt">
<center> Table 2: Truth Table for a NOT Gate
<p align=center>[[Image:lab_logic_3.gif]]</p>
{| class="truthtable"
  </td>
|-
</tr>
!<math>A\,</math>!!<math>\overline{A}\,</math>
<tr>
|-
<td style="border-right:solid windowtext 1.0pt">
|0||1
<p class=entry>0</p>
|-
  </td>
|1||0
<td>
|-
<p class=entry>1</p>
|}
  </td>
</center>


</tr>
The <b>AND gate</b> performs a multiplication operation on its inputs. If all the inputs are true, the output is also true. But if either of the inputs is false, the output is also false. An AND gate can have two or more inputs, but for this lab, it will have two inputs (denoted by A and B in Table 3). The symbol for the AND operation is a dot (·) or the two inputs one after the other with nothing between them. The truth table for an AND gate is shown in Table 3.
<tr>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>
  </td>
<td>
<p class=entry>0</p>
  </td>
</tr>
</table>


<p>The <b><i>AND
<center>Table 3: Truth Table for an AND Gate
</i></b>gate performs an <b><i>and</i></b><b><i> </i></b>operation on its inputs.
{| class="truthtable"
Like English, if all the inputs are true, then the output is also true.
|-
However, if either of the inputs is false, then the output is also false. An
!<math>A\,</math>!!<math>B\,</math>!!<math>AB\,</math>
AND gate can have two or more inputs, but for this lab, we’ll only use two
|-
inputs. The symbol for an AND gate looks like this:</p>
|0||0||0
|-
|0||1||0
|-
|1||0||0
|-
|1||1||1
|-
|}
</center>


<p>[[Image:lab_logic_4.jpg]]</p>
An <b>OR gate</b> performs an addition operation on its inputs. If either of the inputs is true, the output is also true. But if all the inputs are false, the output is also false. An OR gate can have two or more inputs, but for this lab, it will have two inputs (denoted by A and B in Table 4). The symbol for the OR operation is a plus (+). The truth table for an OR gate is shown in Table 4.


<p>This gate has two inputs, which we’ll call A and B, and one
<center>Table 4: Truth Table for an OR Gate
output. The symbol for the AND operation is a dot(·)  or just has the two inputs one after the other
{| class="truthtable"
with nothing between them. The truth table looks like this:</p>
|-
!<math>A\,</math>!!<math>B\,</math>!!<math>A+B\,</math>
|-
|0||0||0
|-
|0||1||1
|-
|1||0||1
|-
|1||1||1
|-
|}
</center>


<table cellspacing=0>
== Boolean Equations ==
<tr>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt">
<p class=caption><b>A</b></p>


  </td>
A truth table is used to write a <b>Boolean equation</b> for a problem. All the combinations that yield an output of 1 are kept, and the equation is written. This is called a Sum of Products solution. Only the combinations that yield an output of 1 are kept because the Boolean equation intends to represent a quantitative function for when the result will have a value of true (when a fee is charged in the example of the ATM machine). The combinations that yield an output of 0 are essentially discarded because there is no interest in when the result has a false value. The Boolean equation for the ATM example can be seen in Figure 2.
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt">
<p class=caption><b>B</b></p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt">
<p class=caption><b>AB</b></p>
  </td>
</tr>
<tr>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>


  </td>
<center><math>C = \overline{P} W \overline{D} + \overline{P} W D + P \overline{W}\,\overline{D} + P W \overline{D} + P W D\,</math><br>
<td style="border-right:solid windowtext 1.0pt">
Figure 2: Boolean Equation</center><br>
<p class=entry>0</p>
  </td>
<td>
<p class=entry>0</p>
  </td>
</tr>
<tr>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>


  </td>
For each of the combinations (rows) resulting in true in Table 1, the terms of the Boolean equation are determined by multiplying the variables together (the AND operation). If an input is true, the variable can be used as is; for an input that is false, the variable is inverted using the horizontal bar (the NOT operation). Since the output is true if any of the combinations are true, the equation is formed by using addition operations (the OR operation) on all the terms formed by the product of each combination.
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>
  </td>
<td>
<p class=entry>0</p>
  </td>
</tr>
<tr>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>


  </td>
== Karnaugh Maps (K-Maps) ==
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>
  </td>
<td>
<p class=entry>0</p>
  </td>
</tr>
<tr>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>


  </td>
A truth table showing every possible combination of inputs and their corresponding outputs yields a Boolean equation that is the solution to a problem in digital logic. However, this equation can be simplified using a <b>Karnaugh map (K-map)</b>. A K-map identifies and eliminates all the conditions that do not contribute to the solution. The resulting simplified Boolean equation is used to build the digital circuit and will be a combination of the logic gates described above.
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>
  </td>
<td>
<p class=entry>1</p>
  </td>
</tr>
</table>


<p>Finally, the <b><i>OR </i></b>gate performs an <b><i>or</i></b> operation on its inputs.
A K-map is a two-dimensional representation of the truth table that shows the common characteristics of the inputs. For an equation with three inputs, usually all the combinations of the first two inputs are shown as four columns and the values for the third input are shown as two rows. For four inputs, all the combinations of the third and fourth inputs are shown as four rows. Only one value can change at a time in adjacent rows or columns. For example, in Table 5, the columns change from 00 to 01 to 11 to 10. Table 5 illustrates the K-map for the ATM example.
Like English, if either of the inputs is true, then the output is also true.
However, if ALL the inputs are false, the output is also false. An OR gate can
have two or more inputs, but for this lab, we’ll only use two inputs. The
symbol for an OR gate looks like this:</p>


<p>[[Image:lab_logic_7.jpg]]</p>
<center> Table 5: K-Map for the ATM Example
{| class="kmap"
|-
!!!!!0 0!!0 1!!1 1!!1 0
|-
!!!!!<math>\overline{P} \, \overline{W}\,</math>!!<math>\overline{P} W</math>!!<math>P W\,</math>!!<math>P \overline{W}</math>
|-
!0!!<math>\overline{D}\,</math>
|0||1||1||1
|-
!1!!<math>D\,</math>
|0||1||1||0
|-
|}
</center>


<p>This gate has two inputs, which we’ll call A and B, and one
== Simplified Boolean Equations ==
output. The symbol for the OR operation is a plus(+).
The truth table looks like this:</p>


<table cellspacing=0>
To develop the <b>simplified Boolean equation</b>, look for the cells that have the 1 value in common. Look for cells that can physically be boxed together, where the number of cells in the box is a power of 2 (1, 2, 4, or 8 and so on). The boxes are allowed to overlap so the same cells can be used more than once. Look for the biggest boxes first, and keep finding smaller boxes until all the cells with a 1 in them are used. This means that sometimes just single cells will remain at the end. The cells in the K-map for the ATM example can be grouped like what is shown in Figure 3.
<tr>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt">
<p class=caption><b>A</b></p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt">
<p class=caption><b>B</b></p>
  </td>


<td style="border-bottom:solid windowtext 1.0pt">
[[Image:Kmap_Boxes.gif|250px|center]]
<p class=caption><b>A+B</b></p>
<center>Figure 3: K-Map Boxes for the ATM Example</center><br>
  </td>
</tr>
<tr>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>
  </td>


<td>
In this example, all the cells containing a 1 were covered with two boxes. This means is that if the conditions for either box are true, the output will be true. This also means that the simplified Boolean equation will only have two terms. The term for each box can be determined by observing which variables are constant throughout each cell of that box. For the red box, only <math>W</math> remains constant through the four cells. <math>D</math> switches from the top row to the bottom row, and <math>P</math> switches from the left column to the right column. For the blue box, <math>P</math> and <math>\overline{D}</math> remains constant through the two cells, while <math>W</math> switches from the left cell to the right cell. The two boxes yield the simplified Boolean equation shown in Figure 4.
<p class=entry>0</p>
  </td>
</tr>
<tr>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>
  </td>


<td>
<center><math>C = W + P\overline{D}\,</math></center>
<p class=entry>1</p>
<center>Figure 4: Simplified Boolean Equation for the ATM Example</center><br>
  </td>
</tr>
<tr>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>
  </td>


<td>
This equation meets the conditions in the truth table in Table 1 and achieves the desired result with the minimum number of logic elements.
<p class=entry>1</p>
  </td>
</tr>
<tr>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>
  </td>


<td>
== Combinational Logic Circuits ==
<p class=entry>1</p>
  </td>
</tr>
</table>


<p>The truth table attached to each of these gates indicates
Lastly, a <b>combinational logic circuit</b> can be created using the simplified equation. First, a NOT operation is performed on <math>D</math> by inputting and outputting <math>D</math> through a NOT gate. Then, the <math>P</math> and inverted <math>D</math> are inputted into an AND gate, as denoted by them being multiplied in Figure 4. Finally, the result of the AND operation between the <math>P</math> and inverted <math>D</math> and the <math>W</math> are inputted into an OR gate, as denoted by the addition function in Figure 4. The final combinational logic circuit can be seen in Figure 5.
the circumstances under which the gate will return a value of <b><i>true. </i></b>We'll
use these tables to write a <b><i>Boolean equation </i></b>for our problem. All
the combina­tions that yield an output of 1 are kept, and the equation is
written. This is called a <b><i>Sum of Products</i></b> solution.</p>


<p>Note: If we use zeroes instead of ones, we'll get a Product of Sums solution, which is
[[Image:lab_logic_16.gif|frame|center|Figure 5: Combinational Logic Circuit for the ATM Example]]
beyond the scope of this description.</p>


<p>We can make a Boolean equation that is a solution to a problem
= Materials and Equipment =
in digital logic by forming a truth table of our own, where we show every
possible combination of inputs and what their corresponding outputs should be. Once
that is accomplished, we will simplify our equation using a Karnaugh
Map (K-Map). This tool is used to devise a simplified Boolean equation which
identi­fies and removes all the conditions
that do not contribute to the solution. This final equation is the one
used to build the digital circuit, and will be a combination of the gates
described above.</p>


<p>A K-Map is a two dimensional representation of the truth table that allows us
* A 7432 IC (4 dual-input OR gates)
to see common characteristics of the inputs, and allows us to simplify the digital
* A 7408 IC (4 dual-input AND gates)
logic which would implement the truth table. For an equation with three inputs, we
* A 7404 IC (6 single-input NOT gates)
usually show the all the combinations of first two inputs as four columns, and the third
* A lab PC with LabVIEW 2019 software
input values as two rows. For four inputs, we would show all the combinations of the
* An NI-ELVIS II+ with prototyping board
third and fourth inputs as four rows. Only one value can change at a time in adjacent
* Electrical leads
rows or columns. For example, consider the following truth table, where A, B, and C
<!--
are the inputs and O is the output:
= Sample Problem =
An ATM has three options: print a statement, withdraw money, or deposit money. The ATM will charge a $1 fee to:
# Withdraw money
# Print a statement without depositing money


<table cellspacing=0>
== Truth Table ==
<tr>
The inputs are the ATM's three functions. Let variable P stand for printing a statement, W for withdrawing money, and D for depositing money. There is one output, which is whether or not the ATM session has a cost. The output is C. The truth table in Table 6 shows the combinations of the inputs and their corresponding outputs.
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt">
{| class="truthtable"
<p class=caption><b>A</b></p>
|+ style="caption-side: bottom;" | Table 6: Truth table.
  </td>
|-
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt">
!colspan="3"|Inputs!!Output
<p class=caption><b>B</b></p>
|-
  </td>
!P!!W!!D!!C
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt">
|-
<p class=caption><b>C</b></p>
|0||0||0||0
  </td>
|-
|0||0||1||0
|-
|0||1||0||1
|-
|0||1||1||1
|-
|1||0||0||1
|-
|1||0||1||0
|-
|1||1||0||1
|-
|1||1||1||1
|-
|}


<td style="border-bottom:solid windowtext 1.0pt">
In Table 6, a 1=true and 0=false.
<p class=entry><b>O</b></p>
  </td>
</tr>
<tr>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>
  </td>


<td>
The inputs are P=print, W=withdraw, D=deposit. The output is 0=false (do not charge) or 1=true (charge $1.00).
<p class=entry>0</p>
  </td>
</tr>
<tr>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>
  </td>


<td>
== Boolean Equation ==
<p class=entry>0</p>
The combinations that yield an output of 1 are kept. For each of these combinations, determine the input values and multiply those variables together. If an input is true, it can be used as is; for an input that is false, invert it using the horizontal bar NOT operation. Since the output is true if any of the combinations are true, the equation is formed by using addition operations on all the terms formed by the product of each combination. The Boolean equation in Figure 5 is created.
  </td>
</tr>
<tr>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>
  </td>


<td>
<math>C = \overline{P} W \overline{D} + \overline{P} W D + P \overline{W}\,\overline{D} + P W \overline{D} + P W D\,</math>
<p class=entry>0</p>
  </td>
</tr>
<tr>
<tr>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>
  </td>


<td>
Figure 5: Boolean Equation
<p class=entry>1</p>
  </td>
</tr>
<tr>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>
  </td>


<td>
== Karnaugh Map ==
<p class=entry>0</p>
A table like the one in Table 7 that maps out all the possible combinations is drawn. Only one variable can change at a time between adjacent columns. For this example, the columns will change from 00 to 01 to 11 to 10, as shown in Table 7. A TA will explain this further. The cells that are true are grouped in rectangles of two, four, or eight cells. This shows the values that these cells have in common. These common values yield the simplified Boolean equation. The K-map in Table 7 corresponds to the Boolean equation created.
  </td>
</tr>
<tr>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>
  </td>


<td>
Compare this K-map in Table 7 with the equation to see why some cells are ''1'' and the others are ''0''.
<p class=entry>1</p>
  </td>
</tr>
<tr>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>
  </td>


<td>
{| class="kmap"
<p class=entry>0</p>
|+ style="caption-side: bottom;" | Table 7: K-map.
  </td>
|-
</tr>
!!!!!0 0!!0 1!!1 1!!1 0
<tr>
|-
<td style="border-right:solid windowtext 1.0pt">
!!!!!<math>\overline{P} \, \overline{W}\,</math>!!<math>\overline{P} W</math>!!<math>P W\,</math>!!<math>P \overline{W}</math>
<p class=entry>1</p>
|-
  </td>
!0!!<math>\overline{D}\,</math>
<td style="border-right:solid windowtext 1.0pt">
|0||1||1||1
<p class=entry>1</p>
|-
  </td>
!1!!<math>D\,</math>
<td style="border-right:solid windowtext 1.0pt">
|0||1||1||0
<p class=entry>1</p>
|-
  </td>
|}


<td>
== Simplified Boolean Equation ==
<p class=entry>1</p>
Form the largest rectangles of two, four, and eight cells. There is a rectangle of four cells (2 rows by 2 columns) plus an overlapping rectangle of two cells (1 row by 2 columns), as shown in Table 8.
  </td>
</tr>
</table>


<p>This table looks complex, but it can be simplified. We'll make the K-Map for this
[[Image:Kmap_Boxes.gif|thumb|250px|frame|left|Table 8: K-map with boxes]]
truth table:</p>
<p>**************************************************************************</p>


<table cellspacing=0 align=center>
Always try to group the greatest number of common values in powers of 2.
<tr>
<td><p>&nbsp;</p></td>
<td style="border-right:solid windowtext 1.0pt"><p>&nbsp;</p></td>
<td><p class=caption>0</p></td><td style="border-right:solid windowtext 1.0pt"><p class=caption>0</p></td>
<td><p class=caption>0</p></td><td style="border-right:solid windowtext 1.0pt"><p class=caption>1</p></td>
<td><p class=caption>1</p></td><td style="border-right:solid windowtext 1.0pt"><p class=caption>1</p></td>
<td><p class=caption>1</p></td><td style="border-right:solid windowtext 1.0pt"><p class=caption>0</p></td>
</tr>
<tr>
<td style="border-bottom:solid windowtext 1.0pt"><p class=caption>&nbsp;</p></td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt"><p class=caption>&nbsp;</p></td>
<td style="border-bottom:solid windowtext 1.0pt"><p align=center><b><u>&nbsp;&nbsp;</u><br>A</b></p></td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt"><p align=center><b><u>&nbsp;&nbsp;</u><br>B</b></p></td>
<td style="border-bottom:solid windowtext 1.0pt"><p align=center><b>&nbsp;&nbsp;</u><br>A</b></p></td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt"><p align=center><b>&nbsp;&nbsp;<br>B</b></p></td>
<td style="border-bottom:solid windowtext 1.0pt"><p align=center><b>&nbsp;&nbsp;<br>A</b></p></td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt"><p align=center><b>&nbsp;&nbsp;<br>B</b></p></td>
<td style="border-bottom:solid windowtext 1.0pt"><p align=center><b>&nbsp;&nbsp;<br>A</b></p></td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt"><p align=center><b><u>&nbsp;&nbsp;</u><br>B</b></p></td>
</tr>
<tr>
<td style="border-bottom:solid windowtext 1.0pt;border-left:solid windowtext 1.0pt">


<p class=caption>0</p>
To create a simplified Boolean equation, group all the combinations together and eliminate those that contain both values of a variable. For example, in the big rectangle in the middle, W is always true, but the rectangle also includes both the true and false values of P and D. This means that if W is true, the values of P and D do not matter since all their combinations are included. Similarly, for the small rectangle, if P is true and D is false, the value of W does not matter since both its true and false values are included in the rectangle. Putting this together, if W is true, or P is true and D is false, the output should be true, and this includes all the input conditions required to produce a true output. A simplified Boolean equation for this problem is shown in Figure 6.
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt">
<p>[[Image:lab_logic_13.gif]]</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt" colspan=2>
<p class=entry>0</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt" colspan=2>
<p class=entry>1</p>
  </td>


<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt" colspan=2>
<math>C = W + P\overline{D}\,</math>
<p class=entry>1</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt" colspan=2>
<p class=entry>1</p>
  </td>
</tr>
<tr>
<td style="border-bottom:solid windowtext 1.0pt;border-left:solid windowtext 1.0pt">
<p class=caption>1</p>
  </td>


<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt">
Figure 6: Simplified Boolean equation
<p class=caption>D</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt" colspan=2>
<p class=entry>0</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt" colspan=2>
<p class=entry>1</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt" colspan=2>
<p class=entry>1</p>


  </td>
This equation in Figure 6 is much simpler than the original equation, but is fully equivalent to it.
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt" colspan=2>
<p align=center>0</p>
  </td>
</tr>
</table>


== Combinational Logic Circuit ==
Construct a combinational logic circuit using the simplified equation. If a NOT operation is performed on D, and an AND operation is performed on the result then P, the second term in the equation, is found. This result combined with performing an OR operation with W produces the result.


<h2>3 YOUR ASSIGNMENT</h2>
This is shown in Figure 7.


<h3>Individual Lab Report</h3>
[[Image:lab_logic_16.gif|frame|center|Figure 7: Combinational logic circuit diagram for the simplified equation.]]
-->
= Problem Statement =


<p>Follow the lab report guidelines laid out in the page called
Farmer Georgi owns a 350-acre dairy farm in upstate New York. In addition to milk and butter, Farmer Georgi sells fresh eggs at the Union Square Greenmarket in Manhattan. It is imperative that Farmer Georgi protects the hen that produces eggs for his business.
[[Specifications for Writing Your Lab Reports]] in the <i>Technical Communication</i>
section of this manual. As you write, the following discussion points should be
addressed in the appropriate section of your lab report:</p>


<ul>
Farmer Georgi has two barns, one hen, and a supply of corn. A fox has been attempting to eat the hen by hiding in a barn. The hen can move freely from one barn to the other. Farmer Georgi sometimes stores corn in one barn and sometimes in the other, but he never stores it in both at the same time. The hen would like to eat the corn and the fox would like to eat the hen. Farmer Georgi needs an alarm system that uses digital logic circuits to protect the hen and the corn. The design should use the fewest logic gates and input variables possible. The alarm will sound if:
<li>Describe the problem you are solving in your introduction.</li>


<li>Describe how <b><i>AND, OR, </i></b>and <b><i>NOT </i></b>gates work.</li>
# The fox and the hen are in the same barn
# The hen and the corn are in the same barn


<li>What is a truth table? How does it contribute to obtaining a Boolean equation?</li>
= Procedure =


<li>What is a K-map? How does it to contribute to obtaining a simplified Boolean equation?
== 1. Test the Integrated Circuits (ICs) ==
To demonstrate the simplification, show what the equation would look like if you used wrote
terms for each of the true values, and the simplified equation by applying the K-Map. Discuss
how much simpler the circuitry would be by applying K-Maps.</li>


<li>Describe how a digital logic circuit is built using these tools.</li>
Before performing the lab, the integrated circuits (IC) used in wiring the alarm system on the NI-ELVIS board must be tested to determine if they are functioning. If the procedure is completed without verifying that the ICs work, problems may occur in the circuit that require additional time and troubleshooting. The system for checking the ICs uses an Arduino board to send and receive signals from the AND/OR/NOT chips and determines if they are functioning properly.


<li>How does the use of a combinational logic circuit contribute to advances in technology?</li>
First, load the chip onto the Arduino board (Figure 6) and start the Serial Monitor on the Arduino program. The code will cycle through the basic inputs (0, 1 for NOT; 00, 01, 10, 11 for AND/OR) and then send the outputs of each gate in the chip back to the Arduino board. The code compares those results to the expected result. It displays each input/output pair and displays the final result indicating if the chip is functional or defective. The goal is to determine if the gates are outputting the correct results. <b>After, remove the chips carefully from the Arduino board.</b>


<li>Describe the design changes that would be necessary if one of the barns used a bell as
[[Image:IC_Tester.JPG|thumb|500px|center|Figure 6: IC Testing Setup]]
an alarm and the other used a horn.</li>
</ul>


<h3>Team PowerPoint Presentation</h3>
== 2. Procedure ==


<p>Follow the presentation guidelines laid out in the page called
# The two barns must be assigned a numeric equivalent before the truth table is prepared. Use 1 for Barn 1 and 0 for Barn 2 (1 does not mean true and 0 does not mean false for the inputs). For the alarm output, use 0 to indicate that the alarm should be off, and 1 to indicate that the alarm should be on.
[[EG1004 Lab Presentation Format]] in the <i>Introduction to Technical Presentations</i>
# On a sheet of Lab Notes paper provided by the TAs, create a truth table that includes the three inputs and one output. Assign the input variables. Include all possible scenarios for the hen, the corn, and the fox.
section of this manual. When you are preparing your presentation, consider the following
# Compute the output column. To do this, analyze the three inputs and logically determine if the alarm will sound in each scenario. Place a 1 in the output column if the alarm will sound and a 0 if it will not.
points:</p>
# Note all the combinations that produce a 1 in the output column. Create a Boolean equation from this table that includes each of the inputs that produced a true output.
# Create a K-map. Draw a map that lists all the possible combinations from the Boolean equation. Use the Boolean equation to fill in the 1s and 0s on the K-map.
# Box the pairs of 1s appropriately. The 1s may only be boxed in powers of 2 starting with the largest possible combination and working down to the smallest. The outcomes that do not contribute to the solution are to be discarded.
# This process yields a simplified Boolean equation. Write this simplified Boolean equation down.
# Draw a combinational logic circuit of the simplified Boolean equation.
# Have a TA approve the steps to obtain the combinational logic circuit for accuracy.
# Build the logic circuit in LabVIEW 2019 using the combinational logic circuit that was drawn. Open LabVIEW and select New VI. Go to Window > Tile Left and Right.
# Place three switches on the front panel that represent the hen, the corn, and the fox. Place a Boolean indicator on the front panel to represent the output.
# Go to the Functions palette > Modern tab > Programming > Boolean and select the AND, NOT, and OR gates necessary for the circuit.
# Once the LabVIEW program is complete, have a TA check and approve the circuit.
# Build the circuit on a NI-ELVIS II+ prototyping board. On the board, identify each of the three IC chips as an AND, OR, or NOT gate. To do this, read the number on the chip and match it with the numbers shown in Figure 7. Look for the notch at one end of each chip to orient the chip to the diagrams in Figure 7 and to match the pins.
#: [[Image:lab_logic_17.jpg|frame|center|Figure 7: IC Numbers and Pins]]
# Place the IC chips over the bridge in the NI-ELVIS board (Figure 8).
#: [[Image:IC Chip in Breadboard.jpg|250px|thumb|center|Figure 8: IC Chip in Breadboard]]
# Power the chips. Insert one end of an electrical lead into Row 54 (marked +5V) of the NI-ELVIS board. Insert the other end of the lead into Pin 14 of the IC chip. Repeat this process for all three chips.
# Ground the chips using the same method. Insert one end of another electrical lead into Row 53 (marked Ground) of the NI-ELVIS board and the other end into Pin 7 of the IC chip. Repeat this process for all three chips.
# Look at the top right corner of the breadboard to find the Digital Input/Output (DIO) holes. Select one of the DIO rows to use as one of the input variables and begin wiring the circuit based on the simplified Boolean equation. Assign the other DIO rows for the remaining input variables.
# Continue this process until the entire circuit is wired. Insert one end of a lead into the final output of the NI-ELVIS circuit. Attach the other end of the lead to any row for an LED on the board.
# Plug in the NI-ELVIS II+ and connect the USB cable to the lab PC.
# Power on the NI-ELVIS II+ by toggling the switches on the back of the device and top of the device. When the pop-up window appears, select NI-ELVISmx Instrument Launcher and click OK (Figure 9).
#: [[Image:lab_logic_21.png|300px|thumb|center|Figure 9: New Data Acquisition Device Dialog Window]]
# Click on the DigOut Icon on the Instrument Launcher window (Figure 10).
#: [[Image:Lab 8 Figure 10.png|frame|center|Figure 10: NI ELVISmx Instrument Launcher Dialog Window]]
# The NI ELVISmx Digital Writer window should appear (Figure 11). Select the appropriate range for the Lines to Write corresponding to the DIO rows used in the wiring. For example, if the fox, hen, and corn are wired to DIO 21, 22, and 23, select 16&ndash;23 for the Lines to Write.
#: [[Image:lab_logic_23.png|300px|thumb|center|Figure 11: Digital Writer Window]]
# Click Run and toggle the switches corresponding to the fox, hen, and corn. Observe the LED for the alarm light up based on the inputs to each barn.
# Have a TA check the circuit.


<ul>
The lab work is now complete. Please clean up the workstation. Return all unused materials to a TA.
<li>How does digital logic impact the world today?</li>
<li>If you can look into the future, how will digital logic change the world?</li>
</ul>


<h2>4 MATERIALS AND EQUIPMENT</h2>
= Assignment =
<ul>
<li>7432 IC (4 dual-input OR gates)</li>


<li>7408 IC (4 dual-input AND gates)</li>
== Team Lab Report ==


<li>7404 IC (6 single input NOT gates)</li>
{{Labs:Lab Report}}


<li>Lab PC with LabVIEW Software</li>
* Describe the problem statement regarding Farmer Georgi
* Describe how AND, OR, and NOT gates work
* Include the truth table, Boolean equation, K-map, simplified Boolean equation, and combinational logic circuit developed during the lab
* What is a truth table? How does it contribute to obtaining a Boolean equation?
* What is a K-map? How does it contribute to obtaining a simplified Boolean equation? To demonstrate the simplification, show what the equation would look like if terms were written for each of the true values and the simplified equation by applying the K-map. Discuss how much simpler the circuitry is as a result of applying the K-map
* Describe how to construct a digital logic circuit using Boolean equations and K-maps
* Explain why the alarm system did or did not work
* Why is there a need for the minimization of a logic design and what are the advantages?
* How does the use of a combinational logic circuit contribute to advances in technology?
* Discuss other possible design/procedural improvements
* Describe how the final combinational logic circuit was obtained
* Include a screenshot of the front and back panels of the LabVIEW VI


<li>Digital Logic Trainer</li>
{{Labs:Lab Notes}}


<li>Appropriate Wiring</li>
== Team PowerPoint Presentation ==
</ul>


<p><b>Remember: </b><i>You are required to
{{Labs:Team Presentation}}
take notes. Experimental details are easily forgotten unless written down. You
should keep a laboratory notebook for this purpose. Use your lab notes to write
the Procedure section of your lab report. You <b>must </b>attach a copy of your
lab notes to the <b>WC</b> copy of your lab report. Keeping careful notes is an
essential component of all scientific practice.</i></p>


<h2>5 SAMPLE PROBLEM</h2>
* How does digital logic impact the world today?
* Looking into the future, how will digital logic change the world?
* Include the truth table, Boolean equation, K-map, simplified Boolean equation, and combinational logic circuit developed during the lab


<h3>Problem Statement</h3>
= References =


<p>An ATM has three options: to print a statement, withdraw money,
<i>Boole, George</i>, Encyclopedia Britannica, 2003. Encyclopedia Britannica Online. Retrieved July 29, 2003 http://www.britannica.com/eb/article?eu=82823
or deposit money. The ATM will charge you $1 if you:</p>


<ol type="a">
{{Laboratory Experiments}}
<li>Want to withdraw money,</li>
<li>Want to print a statement without withdrawing money</li>
</ol>
 
<ol>
<li><b>Truth Table: </b>The inputs
here are what you can do with the ATM. Let a variable P stand for printing a
statement, W for withdrawing money, and D for depositing money. There is one
output, which is whether or not the ATM session has a cost. Call the output C.
We’ll now take all the combinations of the inputs and show the corresponding
outputs:</li>
<table cellspacing=0 align=center>
<tr>
<td colspan=3 style="border-right:solid windowtext 1.0pt;border-bottom:solid windowtext 1.0pt">
<p class=caption>Inputs</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt">
<p class=caption>Output</p>
  </td>
</tr>
 
<tr>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt">
<p class=caption>P</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt">
<p class=caption>W</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt">
<p class=caption>D</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt">
 
<p class=caption>C</p>
  </td>
</tr>
<tr>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>
  </td>
 
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>
  </td>
<td>
<p class=entry>0</p>
  </td>
</tr>
<tr>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>
  </td>
 
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>
  </td>
<td>
<p class=entry>0</p>
  </td>
</tr>
 
<tr>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>
  </td>
<td>
 
<p class=entry>1</p>
  </td>
</tr>
<tr>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>
  </td>
 
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>
  </td>
<td>
<p class=entry>1</p>
  </td>
</tr>
<tr>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>
  </td>
 
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>
  </td>
<td>
<p class=entry>1</p>
  </td>
</tr>
 
<tr>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>
  </td>
<td>
 
<p class=entry>0</p>
  </td>
</tr>
<tr>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>
  </td>
 
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>0</p>
  </td>
<td>
<p class=entry>1</p>
  </td>
</tr>
<tr>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>
  </td>
 
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=entry>1</p>
  </td>
<td>
<p class=entry>1</p>
  </td>
</tr>
 
</table>
 
<p class=caption>Figure 1: Truth table</p>
 
<p>In the above table 1=true (on), 0=false
(off)</p>
 
<p><b>Inputs: </b>P=print, W=withdraw, D=deposit</p>
 
<p><b>Output: </b>0=false (do not charge), 1=true (charge $1.00)</p>
 
<li><b>Boolean Equation</b> The
combinations that yield an output of <b><i>1</i></b> are kept. Our approach
will be, for each output of <b><i>1</i></b>, to determine the input values and
to AND those variables together. If an input is true, it can be used as is. For
an input that is false, we’ll invert it using the horizontal bar NOT operation
we defined above. Since the output is true if any of the input condition
combinations are true, we’ll form the result by using an OR operation on all
the terms formed by the AND operations. The following Boolean equation
is created:</li>
 
<p>[[Image:lab_logic_10.gif]]</p>
 
<p>You should compare this equation with the truth table to make sure you understand how this works.</p>
 
<li><b>Karnaugh Map (K-Map):</b> Begin by drawing a table like the one below that
maps out all the possible combina­tions. When deciding where to place
each of the letters, keep in mind that <b>you can only change one variable at a time between adjacent
columns</b>. For this example, the columns will chage from 00, to 01, to 11, to 10, as shown in
Figure 2. Your TA will explain this further. The cells that are true are
grouped in boxes of 2, 4, or 8 cells. This allows us to see what these cells
have in common, and use that for the
simplified Boolean equation. The K-Map in Figure 2 corresponds to the
Boolean equation we just created.</li>
 
<p>You should compare this K-Map with the equation to see why some cells are <i>1</i> and the others are <i>0</i>.</p>
 
<table cellspacing=0 align=center>
<tr>
<td>
<p class=caption>&nbsp;</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=caption>&nbsp;</p>
  </td>
<td>
<p class=caption>0</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
 
<p class=caption>0</p>
  </td>
<td>
<p class=caption>0</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=caption>1</p>
  </td>
<td>
<p class=caption>1</p>
 
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=caption>1</p>
  </td>
<td>
<p class=caption>1</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=caption>0</p>
  </td>
 
</tr>
<tr>
<td style="border-bottom:solid windowtext 1.0pt">
<p class=caption>&nbsp;</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt">
<p class=caption>&nbsp;</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt">
<p>[[Image:lab_logic_11.gif]]</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt">
<p>[[Image:lab_logic_12.gif]]</p>
 
  </td>
<td style="border-bottom:solid windowtext 1.0pt">
<p>[[Image:lab_logic_11.gif]]</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt">
<p class=caption>W</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt">
<p class=caption>P</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt">
 
<p class=caption>W</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt">
<p class=caption>P</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt">
<p>[[Image:lab_logic_12.gif]]</p>
  </td>
</tr>
<tr>
<td style="border-bottom:solid windowtext 1.0pt;border-left:solid windowtext 1.0pt">
 
<p class=caption>0</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt">
<p>[[Image:lab_logic_13.gif]]</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt" colspan=2>
<p class=entry>0</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt" colspan=2>
<p class=entry>1</p>
  </td>
 
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt" colspan=2>
<p class=entry>1</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt" colspan=2>
<p class=entry>1</p>
  </td>
</tr>
<tr>
<td style="border-bottom:solid windowtext 1.0pt;border-left:solid windowtext 1.0pt">
<p class=caption>1</p>
  </td>
 
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt">
<p class=caption>D</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt" colspan=2>
<p class=entry>0</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt" colspan=2>
<p class=entry>1</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt" colspan=2>
<p class=entry>1</p>
 
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt" colspan=2>
<p align=center>0</p>
  </td>
</tr>
</table>
 
<p class=caption>Figure 2: K-Map</p>
 
<li><b>Simplified Boolean Equation:</b> Next, we try to form the biggest
boxes we can of 2, 4, and 8 cells. In the middle of Figure 3, you can see a
box of four cells (2 rows by 2 columns), plus an overlapping box of two cells
(1 row by 2 columns).</li>
 
<table cellspacing=0 align=center>
<tr>
<td>
<p class=caption>&nbsp;</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=caption>&nbsp;</p>
  </td>
<td>
<p class=caption>0</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
 
<p class=caption>0</p>
  </td>
<td>
<p class=caption>0</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=caption>1</p>
  </td>
<td>
<p class=caption>1</p>
 
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=caption>1</p>
  </td>
<td>
<p class=caption>1</p>
  </td>
<td style="border-right:solid windowtext 1.0pt">
<p class=caption>0</p>
  </td>
 
</tr>
<tr>
<td style="border-bottom:solid windowtext 1.0pt">
<p class=caption>&nbsp;</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt">
<p class=caption>&nbsp;</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt">
<p>[[Image:lab_logic_11.gif]]</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt">
<p>[[Image:lab_logic_12.gif]]</p>
 
  </td>
<td style="border-bottom:solid windowtext 1.0pt">
<p>[[Image:lab_logic_11.gif]]</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt">
<p class=caption>W</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt">
<p class=caption>P</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt">
 
<p class=caption>W</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt">
<p class=caption>P</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt">
<p>[[Image:lab_logic_12.gif]]</p>
  </td>
</tr>
<tr>
<td style="border-bottom:solid windowtext 1.0pt;border-left:solid windowtext 1.0pt">
 
<p class=caption>0</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt">
<p>[[Image:lab_logic_13.gif]]</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt" colspan=2>
<p class=entry>0</p>
  </td>
<td style="border-top:solid windowtext 2.0pt;border-bottom:solid windowtext 1.0pt;border-left:solid windowtext 2.0pt;border-right:solid windowtext 1.0pt" colspan=2>
<p class=entry>1</p>
  </td>
 
<td style="border-top:solid windowtext 2.0pt;border-bottom:solid windowtext 3.0pt;border-left:solid windowtext 2.0pt;border-right:solid windowtext 3.0pt" colspan=2>
<p class=entry>1</p>
  </td>
<td style="border-top:solid windowtext 2.0pt;border-bottom:solid windowtext 3.0pt;border-right:solid windowtext 3.0pt" colspan=2>
<p class=entry>1</p>
  </td>
</tr>
<tr>
<td style="border-bottom:solid windowtext 1.0pt;border-left:solid windowtext 1.0pt">
<p class=caption>1</p>
  </td>
 
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt">
<p class=caption>D</p>
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt" colspan=2>
<p class=entry>0</p>
  </td>
<td style="border-bottom:solid windowtext 3.0pt;border-left:solid windowtext 2.0pt;border-right:solid windowtext 1.0pt" colspan=2>
<p class=entry>1</p>
  </td>
<td style="border-bottom:solid windowtext 3.0pt;border-right:solid windowtext 3.0pt" colspan=2>
<p class=entry>1</p>
 
  </td>
<td style="border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt" colspan=2>
<p align=center>0</p>
  </td>
</tr>
</table>
 
<p class=caption><b>Figure 3: K-Map with boxes</b></p>
 
<p><b>Note:</b> Always try to group the greatest number of neighbors in powers of 2.</p>
 
<p>To create a simplified Boolean equation, you must first group all
the combinations together and eliminate those that contain both values of a
variable. For example, for the big box in the middle, W is
always true, but the box includes both the true and false values of P and D.
This means that if W is true, the values of P and D don’t matter since all
their combinations are included. Similarly, for the small box, if P is true and
D is false, the value of W doesn’t matter since both its true and false values
are included in the box. Putting this together, if W is true, or P is true and
D is false, the output should be true, and this covers all the input conditions
required to make a true output. We can now write this as a simplified Boolean
equation:</p>
 
<p>[[Image:lab_logic_14.gif]]</p>
 
<p>As you can see, this equation is much simpler than the equation we started with, but is fully equivalent to it.</p>
<li><b>Combinational Logic Circuit:</b> We can now
construct a logic circuit from the simplified equation. Looking at the
equation, we can see that if we perform a NOT operation on D, and do an AND
operation on the result and P, we get the second term in the equation. Then, if
we take this result and do an OR operation with W, we get the overall result.</li>
 
<p>This is shown in Figure 4.</p>
 
<p align=center>[[Image:lab_logic_16.gif]]</p>
 
<p class=caption>Figure 4: Combinational Logic Circuit for Simplified Equation</p>
</ol>
 
== <h2>6 PROCEDURE</h2> ==
 
<h3>Problem Statement</h3>
 
<p>Farmer Georgi owns a 350-acre dairy farm in upstate New York.
In addition to milk and butter, Farmer Georgi sells fresh eggs at the Union
Square Greenmarket in Manhattan. It is imperative that Farmer Georgi protects
the hen that lays his golden eggs.</p>
 
<p>He has two barns, one hen, and a supply of corn. A fox has been
attempting to eat the hen. The hen can move freely from one barn to the other.
Farmer Georgi sometimes stores corn in one barn, and sometimes in the other,
but he never stores it in both at the same time. The hen would like the corn;
the fox would like to eat the hen. Farmer Georgi has hired you to design an
alarm system that uses digital logic circuits to protect the hen and the corn.
Your design should use the fewest gates and input variables possible. The alarm
will sound if:</p>
 
<ol type="a">
<li>The fox and the hen are in the same barn, or</li>
<li>The hen and the corn are in the same barn.</li>
</ol>
 
<h3>Finding a Solution</h3>
 
<p><b>Note:</b> The two barns must be assigned a numeric equivalent before you prepare your
truth table for Step 1. For example, use <b><i>0</i> </b>for Barn <b><i>1</i></b> and 1 for Barn 2. For
the alarm output, use <b><i>0</i></b> to indicate the alarm should be
off, and <b><i>1</i> </b>to indicate that the alarm should be on.</p>
 
<ol>
<li>On a sheet of lined lab note paper from the EG website, create a truth table that
includes three inputs and one output. Assign the input variables. Make sure you
include all possible scenarios for the hen, the corn, and the fox.</li>
 
<li>Compute the output column. To do this, analyze the three inputs and determine whether
the alarm would sound in each scenario. Place a <b><i>1 </i></b>in the output
column if the alarm will sound and a <b><i>0 </i></b>if it will not.</li>
 
<li>Note all the combinations that produce a <b><i>1 </i></b>in the output column.</li>
 
<li>Create a Boolean equation from this table that includes each of the inputs that
produced a positive output.</li>
 
<li>Create a <b><i>K-Map. </i></b>Draw a map that lists all the possible combinations. Use
the Boolean equation to fill in the <b><i>1's</i></b> on the K-Map.</li>
 
<li>Circle the pairs of <b><i>1's. </i></b>The <b><i>1's </i></b>may only be circled in
powers of 2 starting with the largest possible combination and working down to
the smallest. The unfavorable outcomes are discarded.</li>
 
<li>This process yields the <b><i>simplified Boolean equation. </i></b>Write this
simplified equation down.</li>
 
<li>Draw a schematic diagram of your simplified Boolean equation.</li>
 
<li>Have your TA <b>sign </b>your work.</li>
 
<li>After you have created your
circuit, it must be tested. Build the circuit in LabVIEW. Make sure you use
your simplified Boolean equation. Open LabVIEW and select New VI. Pull down the Window menu, select Tile Left and
Right.</li>
 
<li>Place three switches on the front
panel that represent the hen, the corn, and the fox. Place one Boolean
indicator to represent the output.</li>
 
<li>Open the Functions palette. Select Boolean and choose the AND, NOT, and
OR gates necessary for your circuit. Your TA will assist you in this process.</li>
 
<li>Once you have completed your
LabVIEW program, have your TA <b>check </b>your work.</li>
 
<li>Now we will build the circuit on the <b><i>Digital Logic Trainer</i></b>.</li>
 
<p><b><font color=#ff0000>Warning:</font></b><i> Don't plug in the
Digital Logic Trainer until instructed to do so.</i></p>
 
<li>On the trainer, identify each of
the three black chips as an <b><i>AND, OR, </i></b>or <b><i>NOT </i></b>gate.
To do this, read the number on the chip itself and match it with the list in
the <b><i>Materials and Equipment </i></b>section of this lab. Use the pinouts
in Figure 5 to wire your gates.</li>
 
<p align=center>[[Image:lab_logic_17.jpg]]</p>
 
<p class=caption>Figure 5: IC Pinouts</p>
 
<li>Before we begin, we must connect
our gates to a power supply and ground them. Insert one end of a wire into the
small breadboard that is above the larger breadboard on the logic trainer.
Insert the wire into the hole on the small breadboard marked <b><i>5V. </i></b>Insert
the other end of the wire into the hole on the breadboard nearest pin <b><i>14 </i></b>on
the gate. Repeat this process for all three gates.</li>
 
<p align=center>[[Image:lab_logic_20.gif]]</p>
 
<p class=caption>Figure 6: Breadboard Layout</p>
 
<p>The Breadboard is set up in the following manner:</p>
 
<ol type="a">
<li>The top two and bottom two rows of slots are connected horizontally; however there
is a break in the middle separating the rows, as shown above.</li>
<li>The rest of the breadboard is connected vertically.</li>
</ol>
 
<p><b><font color=#ff0000>Warning:</font></b><i> There
are different breadboards available in the Lab, some only having one top </i><i>and bottom row. If you
have any problems, ask your Lab TA for help.</i></p>
 
<li>Ground the circuit using the same method. Insert one end of a wire into the hole
marked <b><i>ground </i></b>on the small breadboard, and the other end into pin
<b><i>7 </i></b>of the logic gate. Repeat this process for all three gates.</li>
 
<li>Select one of the three variables and begin to wire this circuit based on your
simplified Boolean equation. Insert a wire into the one of the first three
switches on the small breadboard. Insert the other end of the wire into the
first input on the appropriate gate.</li>
 
<li>Continue this process until you have wired the entire circuit. Insert one end of the
wire into the final output of your simplified Boolean equation. Attach the
other end to a <b><i>Logic Indicator</i></b>.</li>
 
<li>Have your TA <b>check </b>your circuit.</li>
 
<li>With your TA’s permission, plug in the trainer and turn it on. Using your original
truth table, throw each switch in combination with the other switches to
simulate all the scenarios. Make sure the alarm (represented by the LEDs) sounds when it is supposed to.</li>
 
<li>Unplug the Digital Logic Trainer. Take apart your circuit <b>leaving </b>the chips on
the breadboard. Return the wires to the kit.</li>
</ol>
 
<p>Your lab work is now complete. Please
clean up your workstation. Return all unused materials to your TA. Refer to section <b><i>3 Your
Assignment </i></b>for the instructions you need to prepare your lab report.</p>
 
<h2>Footnotes</h2>
 
<p><sup>1</sup> <i>Boole</i><i>, George, </i>Encyclopedia Britannica, 2003. Encyclopedia Britannica Online.
Retrieved July 29<sup>th</sup>, 2003 <i>http://www.britannica.com/eb/article?eu=82823</i></p>
[[Main_Page | Return to Table of Contents]]

Latest revision as of 17:14, 25 April 2020

Objective

The experimental objective of this lab is to design a combinational logic circuit for a given problem statement, and to activate it under specific conditions and test it using LabVIEW. After testing, it will be built on an NI-ELVIS II+ prototyping board.

Overview

The first step in understanding the digital circuits that control the function of electronic devices is the mastery of Boolean logic. George Boole, an English mathematician, established modern symbolic logic in 1854 with the publication of his paper, "Laws of Thought." Boolean logic is the foundation of digital circuitry. Boole's method of logical inference allows conclusions to be drawn from any proposition involving any number of terms. Boolean logic demonstrates that the conclusions found are logically contained in the original premises (Encyclopedia Britannica, 2003).

Truth Tables

In Boolean logic, there are only two values, true and false, represented by the numbers 1 and 0, respectively. These values are combined in equations to yield results that also have these two values. The equations are represented by truth tables that show the inputs to the equation and the outputs produced for those inputs. The rows of the table contain all the possible combinations of 1s and 0s for the inputs. The number of rows is determined by the number of possible combinations.

To illustrate a truth table and additional concepts that will be introduced, an example problem statement will be given. Think of an ATM that has three options: print a statement, withdraw money, or deposit money. The ATM will charge a fee to (1) withdraw money or (2) print a statement without depositing money. The intent of the problem is to develop a Boolean equation and logic circuit that will determine after which possible combination of actions will someone get charged a fee. First, a truth table should be made for all the possible combinations of inputs. The inputs are the ATM's three functions. Let variable stand for printing a statement, for withdrawing money, and for depositing money. There is one output, which is whether or not the ATM will charge a fee. The output will be denoted by . The truth table in Table 1 shows all the possible combinations of the inputs and their corresponding outputs.

Table 1: Truth Table for the ATM Example

Inputs Output
0 0 0 0
0 0 1 0
0 1 0 1
0 1 1 1
1 0 0 1
1 0 1 0
1 1 0 1
1 1 1 1

Logic Gates

Boolean logic is applied to digital circuitry through the use of simple logic gates. There are symbols for each of these gates, and the connections between them are represented by lines running from the output of one gate to the input of another. A line can connect only one output to each input. There are seven of these gates: the NOT, AND, OR, NAND, NOR, XOR, and XNOR gates. Only the first three will be used in this lab (Figure 1).

Figure 1: Logic Gate Symbols

The NOT gate is the simplest of these three. It is an inverter. It has one input and produces its opposite as the output. For example, if a 1 value is put into a NOT gate, a 0 value is outputted, as seen in Table 2. The symbol for the operation is a horizontal bar over the variable. The truth table for a NOT gate is shown in Table 2.

Table 2: Truth Table for a NOT Gate
0 1
1 0

The AND gate performs a multiplication operation on its inputs. If all the inputs are true, the output is also true. But if either of the inputs is false, the output is also false. An AND gate can have two or more inputs, but for this lab, it will have two inputs (denoted by A and B in Table 3). The symbol for the AND operation is a dot (·) or the two inputs one after the other with nothing between them. The truth table for an AND gate is shown in Table 3.

Table 3: Truth Table for an AND Gate
0 0 0
0 1 0
1 0 0
1 1 1

An OR gate performs an addition operation on its inputs. If either of the inputs is true, the output is also true. But if all the inputs are false, the output is also false. An OR gate can have two or more inputs, but for this lab, it will have two inputs (denoted by A and B in Table 4). The symbol for the OR operation is a plus (+). The truth table for an OR gate is shown in Table 4.

Table 4: Truth Table for an OR Gate
0 0 0
0 1 1
1 0 1
1 1 1

Boolean Equations

A truth table is used to write a Boolean equation for a problem. All the combinations that yield an output of 1 are kept, and the equation is written. This is called a Sum of Products solution. Only the combinations that yield an output of 1 are kept because the Boolean equation intends to represent a quantitative function for when the result will have a value of true (when a fee is charged in the example of the ATM machine). The combinations that yield an output of 0 are essentially discarded because there is no interest in when the result has a false value. The Boolean equation for the ATM example can be seen in Figure 2.


Figure 2: Boolean Equation


For each of the combinations (rows) resulting in true in Table 1, the terms of the Boolean equation are determined by multiplying the variables together (the AND operation). If an input is true, the variable can be used as is; for an input that is false, the variable is inverted using the horizontal bar (the NOT operation). Since the output is true if any of the combinations are true, the equation is formed by using addition operations (the OR operation) on all the terms formed by the product of each combination.

Karnaugh Maps (K-Maps)

A truth table showing every possible combination of inputs and their corresponding outputs yields a Boolean equation that is the solution to a problem in digital logic. However, this equation can be simplified using a Karnaugh map (K-map). A K-map identifies and eliminates all the conditions that do not contribute to the solution. The resulting simplified Boolean equation is used to build the digital circuit and will be a combination of the logic gates described above.

A K-map is a two-dimensional representation of the truth table that shows the common characteristics of the inputs. For an equation with three inputs, usually all the combinations of the first two inputs are shown as four columns and the values for the third input are shown as two rows. For four inputs, all the combinations of the third and fourth inputs are shown as four rows. Only one value can change at a time in adjacent rows or columns. For example, in Table 5, the columns change from 00 to 01 to 11 to 10. Table 5 illustrates the K-map for the ATM example.

Table 5: K-Map for the ATM Example
0 0 0 1 1 1 1 0
0 0 1 1 1
1 0 1 1 0

Simplified Boolean Equations

To develop the simplified Boolean equation, look for the cells that have the 1 value in common. Look for cells that can physically be boxed together, where the number of cells in the box is a power of 2 (1, 2, 4, or 8 and so on). The boxes are allowed to overlap so the same cells can be used more than once. Look for the biggest boxes first, and keep finding smaller boxes until all the cells with a 1 in them are used. This means that sometimes just single cells will remain at the end. The cells in the K-map for the ATM example can be grouped like what is shown in Figure 3.

Kmap Boxes.gif
Figure 3: K-Map Boxes for the ATM Example


In this example, all the cells containing a 1 were covered with two boxes. This means is that if the conditions for either box are true, the output will be true. This also means that the simplified Boolean equation will only have two terms. The term for each box can be determined by observing which variables are constant throughout each cell of that box. For the red box, only remains constant through the four cells. switches from the top row to the bottom row, and switches from the left column to the right column. For the blue box, and remains constant through the two cells, while switches from the left cell to the right cell. The two boxes yield the simplified Boolean equation shown in Figure 4.

Figure 4: Simplified Boolean Equation for the ATM Example


This equation meets the conditions in the truth table in Table 1 and achieves the desired result with the minimum number of logic elements.

Combinational Logic Circuits

Lastly, a combinational logic circuit can be created using the simplified equation. First, a NOT operation is performed on by inputting and outputting through a NOT gate. Then, the and inverted are inputted into an AND gate, as denoted by them being multiplied in Figure 4. Finally, the result of the AND operation between the and inverted and the are inputted into an OR gate, as denoted by the addition function in Figure 4. The final combinational logic circuit can be seen in Figure 5.

Figure 5: Combinational Logic Circuit for the ATM Example

Materials and Equipment

  • A 7432 IC (4 dual-input OR gates)
  • A 7408 IC (4 dual-input AND gates)
  • A 7404 IC (6 single-input NOT gates)
  • A lab PC with LabVIEW 2019 software
  • An NI-ELVIS II+ with prototyping board
  • Electrical leads

Problem Statement

Farmer Georgi owns a 350-acre dairy farm in upstate New York. In addition to milk and butter, Farmer Georgi sells fresh eggs at the Union Square Greenmarket in Manhattan. It is imperative that Farmer Georgi protects the hen that produces eggs for his business.

Farmer Georgi has two barns, one hen, and a supply of corn. A fox has been attempting to eat the hen by hiding in a barn. The hen can move freely from one barn to the other. Farmer Georgi sometimes stores corn in one barn and sometimes in the other, but he never stores it in both at the same time. The hen would like to eat the corn and the fox would like to eat the hen. Farmer Georgi needs an alarm system that uses digital logic circuits to protect the hen and the corn. The design should use the fewest logic gates and input variables possible. The alarm will sound if:

  1. The fox and the hen are in the same barn
  2. The hen and the corn are in the same barn

Procedure

1. Test the Integrated Circuits (ICs)

Before performing the lab, the integrated circuits (IC) used in wiring the alarm system on the NI-ELVIS board must be tested to determine if they are functioning. If the procedure is completed without verifying that the ICs work, problems may occur in the circuit that require additional time and troubleshooting. The system for checking the ICs uses an Arduino board to send and receive signals from the AND/OR/NOT chips and determines if they are functioning properly.

First, load the chip onto the Arduino board (Figure 6) and start the Serial Monitor on the Arduino program. The code will cycle through the basic inputs (0, 1 for NOT; 00, 01, 10, 11 for AND/OR) and then send the outputs of each gate in the chip back to the Arduino board. The code compares those results to the expected result. It displays each input/output pair and displays the final result indicating if the chip is functional or defective. The goal is to determine if the gates are outputting the correct results. After, remove the chips carefully from the Arduino board.

Figure 6: IC Testing Setup

2. Procedure

  1. The two barns must be assigned a numeric equivalent before the truth table is prepared. Use 1 for Barn 1 and 0 for Barn 2 (1 does not mean true and 0 does not mean false for the inputs). For the alarm output, use 0 to indicate that the alarm should be off, and 1 to indicate that the alarm should be on.
  2. On a sheet of Lab Notes paper provided by the TAs, create a truth table that includes the three inputs and one output. Assign the input variables. Include all possible scenarios for the hen, the corn, and the fox.
  3. Compute the output column. To do this, analyze the three inputs and logically determine if the alarm will sound in each scenario. Place a 1 in the output column if the alarm will sound and a 0 if it will not.
  4. Note all the combinations that produce a 1 in the output column. Create a Boolean equation from this table that includes each of the inputs that produced a true output.
  5. Create a K-map. Draw a map that lists all the possible combinations from the Boolean equation. Use the Boolean equation to fill in the 1s and 0s on the K-map.
  6. Box the pairs of 1s appropriately. The 1s may only be boxed in powers of 2 starting with the largest possible combination and working down to the smallest. The outcomes that do not contribute to the solution are to be discarded.
  7. This process yields a simplified Boolean equation. Write this simplified Boolean equation down.
  8. Draw a combinational logic circuit of the simplified Boolean equation.
  9. Have a TA approve the steps to obtain the combinational logic circuit for accuracy.
  10. Build the logic circuit in LabVIEW 2019 using the combinational logic circuit that was drawn. Open LabVIEW and select New VI. Go to Window > Tile Left and Right.
  11. Place three switches on the front panel that represent the hen, the corn, and the fox. Place a Boolean indicator on the front panel to represent the output.
  12. Go to the Functions palette > Modern tab > Programming > Boolean and select the AND, NOT, and OR gates necessary for the circuit.
  13. Once the LabVIEW program is complete, have a TA check and approve the circuit.
  14. Build the circuit on a NI-ELVIS II+ prototyping board. On the board, identify each of the three IC chips as an AND, OR, or NOT gate. To do this, read the number on the chip and match it with the numbers shown in Figure 7. Look for the notch at one end of each chip to orient the chip to the diagrams in Figure 7 and to match the pins.
    Figure 7: IC Numbers and Pins
  15. Place the IC chips over the bridge in the NI-ELVIS board (Figure 8).
    Figure 8: IC Chip in Breadboard
  16. Power the chips. Insert one end of an electrical lead into Row 54 (marked +5V) of the NI-ELVIS board. Insert the other end of the lead into Pin 14 of the IC chip. Repeat this process for all three chips.
  17. Ground the chips using the same method. Insert one end of another electrical lead into Row 53 (marked Ground) of the NI-ELVIS board and the other end into Pin 7 of the IC chip. Repeat this process for all three chips.
  18. Look at the top right corner of the breadboard to find the Digital Input/Output (DIO) holes. Select one of the DIO rows to use as one of the input variables and begin wiring the circuit based on the simplified Boolean equation. Assign the other DIO rows for the remaining input variables.
  19. Continue this process until the entire circuit is wired. Insert one end of a lead into the final output of the NI-ELVIS circuit. Attach the other end of the lead to any row for an LED on the board.
  20. Plug in the NI-ELVIS II+ and connect the USB cable to the lab PC.
  21. Power on the NI-ELVIS II+ by toggling the switches on the back of the device and top of the device. When the pop-up window appears, select NI-ELVISmx Instrument Launcher and click OK (Figure 9).
    Figure 9: New Data Acquisition Device Dialog Window
  22. Click on the DigOut Icon on the Instrument Launcher window (Figure 10).
    Figure 10: NI ELVISmx Instrument Launcher Dialog Window
  23. The NI ELVISmx Digital Writer window should appear (Figure 11). Select the appropriate range for the Lines to Write corresponding to the DIO rows used in the wiring. For example, if the fox, hen, and corn are wired to DIO 21, 22, and 23, select 16–23 for the Lines to Write.
    Figure 11: Digital Writer Window
  24. Click Run and toggle the switches corresponding to the fox, hen, and corn. Observe the LED for the alarm light up based on the inputs to each barn.
  25. Have a TA check the circuit.

The lab work is now complete. Please clean up the workstation. Return all unused materials to a TA.

Assignment

Team Lab Report

Follow the lab report guidelines laid out in the EG1004 Writing Style Guide in the Technical Writing section of the manual. The following points should be addressed in the appropriate section of the lab report.

  • Describe the problem statement regarding Farmer Georgi
  • Describe how AND, OR, and NOT gates work
  • Include the truth table, Boolean equation, K-map, simplified Boolean equation, and combinational logic circuit developed during the lab
  • What is a truth table? How does it contribute to obtaining a Boolean equation?
  • What is a K-map? How does it contribute to obtaining a simplified Boolean equation? To demonstrate the simplification, show what the equation would look like if terms were written for each of the true values and the simplified equation by applying the K-map. Discuss how much simpler the circuitry is as a result of applying the K-map
  • Describe how to construct a digital logic circuit using Boolean equations and K-maps
  • Explain why the alarm system did or did not work
  • Why is there a need for the minimization of a logic design and what are the advantages?
  • How does the use of a combinational logic circuit contribute to advances in technology?
  • Discuss other possible design/procedural improvements
  • Describe how the final combinational logic circuit was obtained
  • Include a screenshot of the front and back panels of the LabVIEW VI

Remember: Lab notes must be taken. Experimental details are easily forgotten unless written down. EG1004 Lab Notes paper can be downloaded and printed from the EG1004 Website. Use the lab notes to write the Procedure section of the lab report. At the end of each lab, a TA will scan the lab notes and upload them to the Lab Documents section of the EG1004 Website. One point of extra credit is awarded if the lab notes are attached at the end of the lab report. Keeping careful notes is an essential component of all scientific practice.

Team PowerPoint Presentation

Follow the presentation guidelines laid out in the EG1004 Lab Presentation Format in the Technical Presentations section of the manual. When preparing the presentation, consider the following points.

  • How does digital logic impact the world today?
  • Looking into the future, how will digital logic change the world?
  • Include the truth table, Boolean equation, K-map, simplified Boolean equation, and combinational logic circuit developed during the lab

References

Boole, George, Encyclopedia Britannica, 2003. Encyclopedia Britannica Online. Retrieved July 29, 2003 http://www.britannica.com/eb/article?eu=82823