Lab 3:
C
hapter 4, Gates and Circuits
[52 marks] In all of the problems, you must show your work to qualify for the mark.
Type your answer in this worksheet after each question. Export the completed worksheet to P
D
F and Submit the PDF version with D2L.
Learning Objectives
A
t the end of this lab, you should be able to:
· draw circuit diagrams for
B
oolean expressions
· write truth tables for Boolean expressions
· write Boolean expressions for logic circuits
· experiment with adders
· simplify Boolean expressions
Lab Readings
Chapter 4 – Gates and Circuits
Lab
6
Logic Circuits, Lab6_Manual and the related Applet are included on D2L
Lab
Q
uestions
1
. [4] Do Exercise 1 from Lab6_Manual
Draw a circuit with two inputs A, B and one output that corresponds to AxorB. Run the circuit with all possible input values as given in the table below to complete the truth table for the XOR gate:
0
01
0
0
0
1
0
A 
B 
A xor B 

0 
1  
Insert 4 screenshots of the circuit corresponding to each of the 4 input combinations listed above
2. [4] Do Exercise 2 from Lab6_Manual
Note: Correct Exercise 2 part 2 as follows:
Add four switches, three AND gates, and an output. Arrange them so two switches feed into one AND gate while the other two switches feed into another AND. Then the two outputs of these AND gates feed into the third AND gate.
a) Attach a screenshot of your circuit below this line.
b) Draw the corresponding Truth Table below this line.
0 00
0
0
0
1
0
1
0
0
0
0
0
0
0
0
0
1
1
0
1
0
0
1
0
0
0
1
1
1
0
0
1
1
0
0
1
0
1
1
0
0
1
1
1
0
1
0
0
1
0
0
A 
B 
C 
D 
OUTPUT 
3. [6] Do Exercise 4 from Lab6_Manual
Insert a screenshot of your final “Full Adder” circuit. See the text book chapter 4 page 109. Note that this circuit is to be built with 2 truth tables and Not with gates. The Truth Tables are also given below.
In the text the two truth tables are combined in one diagram.
Note. Also see the Truth Table Example in the LOGIC GATE SIMULATOR Applet and Lab6_Manual .
6
AB
0
0
0
0
1
1
1
0
1
0
1
1
0
10
0
1
1
0
1
0
1
1
0
0
11
1
1
Carryin 
Sum 

0 

B
Carryin
0
0
0
0
0
0
1
0
0
1
0
0
01
1
1
1
0
0
0
10
1
1
1
0
1
1
1
1
1
Carryout 
4. [4] Given the following Boolean expression:
X = (A’B + C)
1. Draw the circuit diagram for X (use the applet and insert a screen shot below this line).
1. Write the truth table for X below this line.
AB
0
0
1
1
0
1
0
1
1
0
0
0
0
1
1
1
1
0
1
0
1
1
0
1
00
0
1
1
1
1
1
C 
OUTPUT  
5. [4] Given the following Boolean expression:
Y = (AB’ + BC) D
1. Draw the circuit diagram for Y (use the applet and insert a screen shot below this line).
1. Write the truth table for Y below this line.
AB
C
OUTPUT
0
0
1
0
0
1
0
0
1
0
0
0
1
0
0
0
0
00
1
1
0
1
1
0
0
1
1
0
0
0
0
1
1
1
0
1
0
1
1
0
1
1
1
0
0
11
1
1
1
1
0
1
0
0
0
1
0
1
0
D 
6. [4] Given the following logic diagram:
1. Write the Boolean expression for Q below this line.
Q = (AB+A+B)
1. Write the truth table for intermediate values and Q below this line.
AB
0
0
1
1
0
0
0
1
0
1
1
1
Q 
7. [4] Given the following circuit diagram:
1. Write a Boolean expression for Q below this line.
Q= AB+ (B+C)(BC)
1. Write the truth table for intermediate outputs (4 of them) and Q below this line.
AB
C
Q
0
0
0
0
0
0
1
0
0
1
0
0
1
0
0
0
0
1
1
1
1
0
1
0
1
1
0
1
1
1
1
1
8. [8, 2 each] Using properties of Boolean algebra, simplify the following Boolean expressions so they could be built with the minimum number of gates.
1. X = A + BC + AB + ABC + B
1. Y = AB + B(AC + BC + ABC’ + A)
W = ABC’ + AB’C’ + B’CD + A’C + BC
Z = (A + B’)’ + (ABC’)’ + A(B + A’C)’
9. [6] Given the following Boolean expression:
W = A’B + B’C + CB + ABC’
1. Simplify the expression as much as possible using the properties of Boolean algebra so it could be built with the minimum number of gates.
1. Create the truth table for W and create the truth table for the simplified expression (this will validate your simplified expression).
10. [8] The following table describes two truth tables, one with inputs A, B, C, output X, the other with inputs A, B, C, and output Y. Write the Boolean expressions for X and Y (they are two separate outputs) in terms of A, B , and C, simplify them.
00
0
0
1
0
0
1
1
0
01
0
1
1
0
1
1
0
0
10
0
1
1
1
0
1
1
0
11
0
1
1
1
1
0
1
A  B  C 
X 
Y 
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