EMT 1250 Lab report 4 combinational logic circuits PDF

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EMT 1250 Lab report 4 combinational logic circuits...

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Lab report 4 combinational logic circuits

Objective: In this lab experiment we learn how to use the basic laws, rules, DeMorgan's theorems, and theorems of Boolean algebra to manipulate and simplify an expression and fill out its truth table. Required materials: - 5V DC Power Supply - Digital Trainer (Logic Probe) - Breadboard - DIP Switch - 7400 NAND gate - 7402 NOR gate - 7404 Inverter - 7408 AND gate - 7432 OR gate Stimulation and experimental: Part 1 1. Firstly, we build a circuit on multisim following the figure 4-1.

2. Then we found the boolean equation for the diagram: X = A (B+C) + C

3. After that we fill in the truth table and measure voltages of Va, Vb, Vc and Vx using the input commands from table 4-1. Table 4-1 Voltages Measured VA(V) VB(V)

Truth Table

VC(V)

A

B

C

X

Vx(V)

0

0

0

0

0

0

0

0

0

0

5V

0

0

1

1

5V

0

5V

0

0

1

0

0

0

0

5V

5V

0

1

1

1

5V

5V

0

0

1

0

0

0

0

5V

0

5V

1

0

1

1

5V

5V

5V

0

1

1

0

1

5V

5V

5V

5V

1

1

1

1

5V

4. Simplify the expression and circuit for Figure 4-1. Verify that the simplified circuit is equivalent to the original by showing that the truth tables are identical. 4-1) The simplified expression is : X = A(B+C) = AB + AC + C = AB + C(1+A) = AB +C X = AB +C 4-2) the simplified circuit diagram:

4-3) Truth table for the simplified circuit and measured voltages for each input/output: Table 4-2 Voltages Measured VA(V) VB(V) VC(V)

Truth Table A

B

C

X

Vx(V)

0

0

0

0

0

0

0

0

0

0

5V

0

0

1

1

5V

0

5V

0

0

1

0

0

0

0

5V

5V

0

1

1

1

5V

5V

0

0

1

0

0

0

0

5V

0

5V

1

0

1

1

5V

5V

5V

0

1

1

0

1

5V

5V

5V

5V

1

1

1

1

5V

4-4) The truth tables for the original circuit and simplified circuit have the same voltages for input/output.

Part 2 1. Secondly, we build another circuit on multisim following the figure 4-2.

2. Then we find the boolean expression for the diagram: X = NOT(AB) x NOT(B+C) 3. After that follow the instruction from the truth table for A, B and C and fill out our input and output voltages. Table 4-3 Voltages Measured VA(V) VB(V) VC(V)

Truth Table A

B

C

X

Vx(V)

0

0

0

0

0

0

1

5V

0

0

5V

0

0

1

0

0

0

5V

0

0

1

0

0

0

0

5V

5V

0

1

1

0

0

5V

0

0

1

0

0

1

5V

5V

0

5V

1

0

1

0

0

5V

5V

0

1

1

0

0

0

5V

5V

5V

1

1

1

0

0

4. Then simplify the expression and circuit, and verify that the simplified circuit is equivalent to the original by showing that the truth tables are identical. 4-1. The simplified expression is: X = NOT(B) x NOT(C) 4-2. Simplified circuit diagram:

4-3) Truth table for the simplified circuit and measured voltages for each input/output: Table 4-4 Voltages Measured VA(V) VB(V) VC(V)

Truth Table A

B

C

X

Vx(V)

0

0

0

0

0

0

1

5V

0

0

5V

0

0

1

0

0

0

5V

0

0

1

0

0

0

0

5V

5V

0

1

1

0

0

5V

0

0

1

0

0

1

5V

5V

0

5V

1

0

1

0

0

5V

5V

0

1

1

0

0

0

5V

5V

5V

1

1

1

0

0

4-4) The truth tables for the original circuit and simplified circuit have the same voltages for input/output. Results and Discussions: 1. Why is Boolean algebra used for combinational logic circuits? Ans: Boolean Algebra specifies the relationship between Boolean variables which is used to design combinational logic circuits using Logic Gates. 2. What are three laws for Boolean algebra? Answer with simple examples. Ans: Commutative Law, associative Law, distributive Law 3. Why is De Morgan’s theorem important in the simplification of the Boolean equation? Ans: De Morgan's theorems prove very useful for simplifying Boolean logic expressions because of the way they can 'break' an inversion, which could be the complement of a complex Boolean expression. 4. Using De Morgan’s theorem, you can prove that a NAND gate is equivalent to an __________ (OR or AND) gate with inverted inputs. Ans: NOT(AB) = NOT(A) + NOT(B) A

B

NOT(AB)

NOT(A)+NOT(B)

0

0

1

1

0

1

1

1

1

0

1

1

1

1

0

0

5. Using De Morgan’s theorem, you can prove that a NOR gate is equivalent to an __________ (OR or AND) gate with inverted inputs. Ans: NOT(A +B) = NOT(A) x NOT(B)

A

B

NOT(AB)

NOT(A)+NOT(B)

0

0

1

1

0

1

0

0

1

0

0

0

1

1

0

0

Conclusions: In this lab experiment, we have learned about boolean expression, how to simplify logic circuits and expressions using Boolean Algebra and DeMorgan’s theorems. We learned how a boolean expression has the same truth table values as the simplified expression. References: -

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