EXPERIMENT_33
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Feb 20, 2024
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EXPERIMENT 3
Boolean Laws & Rules and
DeMorgan’s Theorem
OBJECTIVES:
Learn and verify Boolean laws and rules.
Learn and prove DeMorgan’s theorem
Use Xilinx Vivado simulation tools to test combinational circuits. DISCUSSION: A Boolean equation derived directly from a truth table or from a problem statement usually is not
in the simplest form. To have an efficient equivalent logic circuit, the Boolean equation
representing the logic design must be in the simplest from. Boolean equations can be simplified
using Boolean algebra, DeMorgan’s theorem, or/and Karnaugh maps. In this experiment, we will
verify Boolean distribute law, absorption rule and DeMorgan’s theorem.
PROCEDURE: Part I
1.)
Prove this following Boolean distributive law using VHDL code.
A
(
B
+
C
+
D
)
≡ AB
+
AC
+
AD
X
1
=
A
(
B
+
C
+
D
)
X
2
=
AB
+
AC
+
AD
2.)
Define the input and output ports as following:
P a g e 1 | 12
3.)
Type the VHDL code for expressions between the “begin” and “end Behavioral”
.
4.)
Use the xdc template from: https://www.xilinx.com/support/documentation/university/Vivado-
Teaching/HDL-Design/2015x/Basys3/Supporting%20Material/Basys3_Master.xdc
Copy and P a g e 2 | 12
paste the whole text into your constraints file. Uncomment by deleting the # signs in front of the lines of switches and LEDs that we’re going to use. Modify the port names to match with port names defined in the design source as follows:
5.)
Run synthesis, run implementation, and generate bitstream
. If running synthesis or implementation failed, double check your design file (.vhd). If bitstream generation failed, double check your constraints file (.xdc)
6.)
Program the target board.
7.)
Toggle the switches, observe and verify your results. Fill out the truth table.
8.)
Create a simulation source and modify the code as following: P a g e 3 | 12
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9.)
Run the simulation. Capture the waveform from 0 ns to 160 ns.
RESULTS: Part I
P a g e 4 | 12
D
C
B
A
X1
X2
Schematic Diagram
0
0
0
0
0
0
0
0
0
1
0
0
0
0
1
0
0
0
0
0
1
1
1
1
0
1
0
0
0
0
0
1
0
1
1
1
0
1
1
0
0
0
0
1
1
1
1
1
1
0
0
0
0
0
1
0
0
1
1
1
1
0
1
0
0
0
1
0
1
1
1
1
1
1
0
0
0
0
Boolean Equation
1
1
0
1
1
1
X1 = A(B+C+D)
X2 = AB + AC + AD
1
1
1
0
0
0
1
1
1
1
1
1
Questions: Part I
1.)
Attach 2 photos of your BASYS 3 board results showing the following input conditions:
D = LOW, C = HIGH, B = HIGH, A = LOW
D = HIGH, C = HIGH, B = HIGH, A = HIGH
P a g e 5 | 12
2.)
Attach a screenshot of the simulation result showing the waveform from 0 ns to 160 ns.
P a g e 6 | 12
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PROCEDURE: Part II (simulation only)
1.) Prove this following Boolean absorption rules and DeMorgan’s theorem using only
simulation.
a.)
Boolean absorption rules:
W
1
=
A
+
A B
W
2
=
A
+
B
X
1
=
A
+
AB
X
2
=
A
+
B
b.)
DeMorgan’s Theorem:
Y
1
=
A
+
B
Y
2
=
A
B
Z
1
=
AB
Z
2
=
A
+
B
2.)
Define the input and output ports as following:
P a g e 7 | 12
3.)
Type the VHDL code for expressions between the “begin” and “end Behavioral”
.
P a g e 8 | 12
4.)
Create a simulation source and modify the code as following: P a g e 9 | 12
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5.)
Run the simulation. Capture the waveform from 0 ns to 40 ns.
6.)
Fill out the truth table
RESULTS: Part II
sB
sA
sW1
sW2
sX1
sX2
sY1
sY2
sZ1
sZ2
0
0
0
0
1
1
1
1
1
1
0
1
1
1
0
0
0
0
1
1
1
0
1
1
1
1
0
0
1
1
1
1
1
1
1
1
0
0
0
0
QUESTIONS: Part II
1) Attach a screenshot of the simulation result showing the waveform from 0 ns to 40 ns.
P a g e 10 | 12
2) According to your results, are the absorption rules and DeMorgan’s Theorem proven?
Yes they are proven because de Morgan’s theorem states that a the complement of the product of
all the terms is equal to the sum of the complement of each term therefore A
(B+C+D)≡AB+AC+AD
3) Apply Boolean laws and rules and DeMorgan’s theorem to simplify the following Boolean equations. Draw the simplified schematic diagrams. P a g e 11 | 12
P a g e 12 | 12
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