Activity 9 - Complement of a Function

THE COMPLEMENT OF A FUNCTION

OBJECTIVES

To plot the given Boolean function 𝐹 = 𝐴′𝐷 + 𝐵′𝐷 + 𝐴𝐵′𝐷 in a Karnaugh Map and simplify it into sum-of-products form.

To obtain the complement of the given Boolean function and express it in sum-of-products form.

To design and construct a circuit implementing both the given Boolean function and its complement using the least number of NAND gates possible.

To obtain the truth table for the given Boolean function and its complement, and then verify that the two circuits are indeed complements of each other.

MATERIALS

Alexan Digital Trainer two 7400 quad two-input NAND gate insulated connecting wires cutter or scissors

PROBLEM ANALYSIS

At the onset, we see that the given Boolean function

𝐹 = 𝐴′𝐷 + 𝐵′𝐷 + 𝐴𝐵′𝐷

consists of three variables 𝐴, 𝐵, and 𝐷. Now even though 𝐹 = 𝐹(𝐴, 𝐵, 𝐷) is not expressed in sum-of-minterms form—but rather, in sum-of-products form—it is still possible to use the Karnaugh map (or K-map) to obtain the minterms of 𝐹 and then simplify the function to an expression with a minimum number of terms. To do this, we use a three-variable K-map, as shown in Figure 9-1. The area in the map covered by the given Boolean function 𝐹 includes all the squares marked with 1’s in Figure 9-1. As expressed, 𝐹 consists of two terms with two literals each and one term with three literals. Each term with two literals is represented by two squares in a three-variable map. That is, 𝐴′𝐷 is represented in squares 001 and 011 while 𝐵′𝐷 is represented in squares 001 and 101. The three-literal term 𝐴𝐵′𝐷 on the other hand, belongs in square 101.

A

BD

00 01 11 10

0

𝑚0

0 𝑚1

1 𝑚3

1 𝑚2

0

1 𝑚4

0 𝑚5

1 𝑚7

0 𝑚6

0

Figure 9-1. Karnaugh map for 𝐹 = 𝐴′𝐷 + 𝐵′𝐷 + 𝐴𝐵′𝐷

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As indicated by the three 1’s in the K-map of Figure 9-1, the given Boolean function 𝐹 has a total of three minterms: 1, 3, and 5. Therefore, 𝐹 can be expressed in sum-of-minterms form as

𝐹 𝐴, 𝐵, 𝐷 = ∑ 1, 3, 5 .

Furthermore, we see from the K-map of Figure 9-2 that 𝐹 can actually be minimized into a function with two terms instead of three:

𝐹 𝐴, 𝐵, 𝐷 = 𝐴′𝐷 + 𝐵′𝐷 .

This is the simplified expression of the given Boolean function in sum-of-products form. To obtain a simplified expression of the complement of 𝐹, we just mark by 0’s all the squares not marked by 1’s and then combine them into valid adjacent squares, as shown in Figure 9-3. Because the minterms not included in the standard sum-of-products form of a function denote the complement of a function, we see that 𝐹′ is represented in the map by squares not marked by 1’s. That is, we have

𝐹′ 𝐴, 𝐵, 𝐷 = 𝐴𝐵 + 𝐷′ .

as the simplified expression of the complement of the given Boolean function in sum-of-products form.

A

BD

00 01 11 10

0

𝑚0

0 𝑚1

1 𝑚3

1 𝑚2

0

1 𝑚4

0 𝑚5

1 𝑚7

0 𝑚6

0

Figure 9-2. Obtaining the

simplified expression of 𝐹

A

BD

00 01 11 10

0 𝑚0

0 𝑚1

1 𝑚3

1 𝑚2

0

1 𝑚4

0 𝑚5

1 𝑚7

0 𝑚6

0

Figure 9-3. Obtaining the

simplified expression of 𝐹′

PROCEDURE

To verify that the obtained simplified expressions of 𝐹 and 𝐹′ are indeed complements of each other, the circuit of Figure 9-4 is constructed on the digital trainer, as illustrated in Figure 9-5. After double-checking all pin connections, power is supplied to the trainer. The logic outputs of 𝐹 and 𝐹′ in response to varying states of inputs are then observed and recorded in Figure 9-6 and Table 9-1.

𝑨′𝑫 𝑩′𝑫 𝑫′ 𝑨𝑩

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Figure 9-4 Circuit diagram—with pin assignments and logic gate analysis—for the

implementation of 𝐹 = 𝐴′𝐷 + 𝐵′𝐷 and 𝐹′ = 𝐴𝐵 + 𝐷′ using two 7400 quad two-

input NAND gates

Figure 9-5. Connection diagram of the circuit in Figure 9-4

7400 7400

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RESULTS AND DISCUSSION

Figure 9-6 below shows the logic output levels at LED4 and LED5 for all the eight possible states: 000, 001, 010, 011, 100, 101, 110, and 111. We see from this diagram that LED4 and LED5 are always opposite—that is, LED5 is OFF whenever LED4 is ON, and vice versa. In terms of 0’s and 1’s, Table 9-1 shows that for all eight possible states, the truth values for 𝐹 and 𝐹′ are always complements of each other. This proves that the obtained simplified expressions for both 𝐹 and 𝐹′ using the map method are valid.

SWITCHES DATA STATUS

INPUT INPUT OUTPUT

S1 S2 S3 LED1 LED2 LED3 LED4 LED5

0

1

2

3

4

5

6

7

D1 D2 D3 IN1 IN2 IN3 IN4 IN5

Figure 9-6. Output diagram of 𝐹 and 𝐹′

Table 9-1. Truth Table

Corresponding to Figure 9-6

INPUT OUTPUT

D1 D2 D3 IN4

𝑨 𝑩 𝑫 𝑭 𝑭′

0 0 0 0 1

0 0 1 1 0

0 1 0 0 1

0 1 1 1 0

1 0 0 0 1

1 0 1 1 0

1 1 0 0 1

1 1 1 0 1

We can further verify whether the experimentally determined truth values of 𝐹 and 𝐹′ depicted in Table 9-1 are correct or not by constructing the truth tables for the Boolean functions 𝐹 = 𝐴′𝐷 + 𝐵′𝐷 and 𝐹′ = 𝐴𝐵 + 𝐷′ . Indeed, if we compare the theoretical truth values of 𝐹 and 𝐹′ listed in Table 9-2, we see that the experimental and theoretical logic level outputs of 𝐹 and 𝐹′ coincide.

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Table 9-2. Combined Truth Tables for 𝐹 = 𝐴′𝐷 + 𝐵′𝐷 and 𝐹′ = 𝐴𝐵 + 𝐷′

𝑨 𝑩 𝑫 𝑨′𝑫 𝑩′𝑫 𝑭 = 𝑨′𝑫 + 𝑩′𝑫 𝑨𝑩 𝑫′ 𝑭′ = 𝑨𝑩 + 𝑫′

0 0 0 0 0 0 0 1 1

0 0 1 1 1 1 0 0 0

0 1 0 0 0 0 0 1 1

0 1 1 1 0 1 0 0 0

1 0 0 0 0 0 0 1 1

1 0 1 0 1 1 0 0 0

1 1 0 0 0 0 1 1 1

1 1 1 0 0 0 1 0 1

CONCLUSION

Based on the results of this experiment, it can be concluded that:

A Boolean function of 𝑛 variables can be simplified using an 𝑛-variable Karnaugh map, with the simplified function easily expressed in either sum-of-products of sum-of-minterms form.

By combining 0’s instead of 1’s in a K-map, a simplified expression of the function’s complement can be obtained.