Chapter 4 Homework · ELN-133 Chapter 4 Homework Spring 2015 Boolean Algebra AB Tech Page 2 of 15...

ELN-133 Chapter 4 Homework Spring 2015

Boolean Algebra

AB Tech of 15 McCormick

Name: _____________________________ Date:___________

MULTIPLE CHOICE. Circle or underline the one alternative that best completes the statement or

answers the question. USE A PENCIL!

1) Which of the following is a correct form of Boolean addition?

A) 0 ∙ 0 = 0 B) 0 + 1 = 1 C) 0 ∙ 1 = 0 D) 0 + 1 = 0 2) Which of the following is a correct form of Boolean multiplication?

A) 0 ∙ 1 = 1 B) 0 + 1 = 1 C) 0 ∙ 1 = 0 D) 0 + 1 = 0 3) In Boolean algebra, the expression "_ " means the complement of A.

A) negative A B) inverse A C) A negative D) not A

Figure 4-2

4) The symbol shown in Figure 4-2 is a(n) , and the output will be for the input shown.

A) buffer; 1 B) inverter; �̅� C) inverter; 1 D) buffer; not A


Boolean Algebra


Refer to the figure below to answer the following question(s).

Figure 4-3 5) The symbol shown in Figure 4-3 is a(n) gate.

A) AND B) OR C) NAND D) AND-OR 6) The Boolean expression for the symbol in Figure 4-3 is .

A) X = A + B + C + D B) X = ABCD C) X = AB + BC D) X = (A + B)(B + C)

Refer to the figure below to answer the following question(s).

Figure 4-4 7) The symbol shown in Figure 4-4 is a(n) gate. A) AND B) OR C) AND-OR D) Exclusive-OR 8) The Boolean expression for the symbol in Figure 4-4 is . A) X = A + B + C + D B) X = ABCD C) X = AB + BC D) X = (A + B)(B + C)


Boolean Algebra


9) The information shown below represents one line from the truth table for a two input NAND gate. A B X The value of X will be

1 0 ?

A) 0 B) B C) 1 D) not A 10) The expression for a 3-input NOR gate is .

A) A/B/C B) A ∙ B ∙ C C) �̅� + �̅� + 𝐶̅ D) 𝐴 + 𝐵 + 𝐶̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅ 11) Which of the examples below expresses the commutative law of multiplication? A) A + B = B + A B) AB = B + A C) AB = BA D) AB = AB 12) Which of the examples below expresses the associative law of addition? A) A + (B + C) = (A + B) + C B) A + (B + C) = A + (BC)

C) A(BC) = (AB) + C D) ABC = A + B + C 13) Which of the examples below expresses the distributive law? A) (A + B) + C = A + (B + C) B) A(B + C) = AB + AC

C) A + (B + C) = AB + AC D) A(BC) = (AB) + C

Figure 4-5

14) Which rule of Boolean algebra does the example shown in Figure 4-5 represent? A) A + 0 = A B) A + 1 = 1 C) A ∙ 0 = 0 D) A ∙ 1 = A 15) Which of the following is based on DeMorgan's theorems?

A) 𝑋 + 𝑌̅̅ ̅̅ ̅̅ ̅̅ = �̅� + �̅� B) X(1) = X C) 𝑋 ∙ 𝑌̅̅ ̅̅ ̅̅ = �̅� + �̅� D) X + 0 = 0


Boolean Algebra


Figure 4-6 16) Which of the figures is the DeMorgan equivalent of the Reference Circuit shown in Figure 4-6?

A) ( A ) B) ( B ) C) ( C ) D) ( D ) 17) Simplify the expression, using DeMorgan's theorems.

(𝐀𝐁)̅̅ ̅̅ ̅ + 𝐂̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅

A) 𝐴𝐵̅̅ ̅̅̅̅ ̅̅ + 𝐶 B) 𝐴 ∙ 𝐵 ∙ 𝐶̅ C) 𝐴 ∙ 𝐵 + 𝐶 D) (𝐴 + 𝐵) ∙ 𝐶

Figure 4-7

18) Solve the network in Figure 4-7 for X.

A) A + BC + D B) ((A + B)C) + D C) D(A + B + C) D) (AC + BC)D


Boolean Algebra


19) Which of the following expressions is in the sum-of-products form? A) (A + B)(C + D) B) (AB)(CD) C) AB(CD) D) AB + CD 20) Which of the following is an important feature of the simplest sum-of-products form of expressions? A) All logic circuits are reduced to nothing more than simple AND and OR gates.

B) The delay times are greatly reduced over other forms. C) No signal must pass through more than 2 gates, not including inverters. D) The maximum number of gates that any signal must pass through is reduced by a factor of two.

21) Which of the following expressions is in the product-of sums form? A) (A + B)(C + D) B) (AB)(CD) C) AB(CD) D) AB + CD

22) What is the primary motivation for using Boolean algebra to simplify logic expressions?

A) May make it easier to understand the circuit.

B) May reduce the number of gates. C) May reduce the number of inputs required. D) All of the above are correct.


Boolean Algebra


Figure 4-8 23) Which of the circuits in Figure 4-8 is equivalent to the Reference Circuit?

A) Figure (A) B) Figure (B) C) Figure (C) D) Figure (D)

24) Which statement below best describes a Karnaugh map?

A) A Karnaugh map can be used to replace Boolean rules.

B) The Karnaugh map eliminates the need for using NAND and NOR gates. C) Variable complements can be eliminated by using Karnaugh maps. D) Karnaugh maps provide a cookbook approach to simplifying Boolean expressions.


Boolean Algebra


Figure 4-9 25) Which of the Karnaugh maps in Figure 4-9 represents the expression,

X = AC + BC + B? A) Figure I. B) Figure II. C) Figure III. D) Figure IV.


Boolean Algebra


Figure 4-10 26) The simplest equation which implements the Karnaugh map in Figure 4-10 is .

A) X = AC + B B) X = AB C) ABC + ABC + ABC D) AB + AB


Boolean Algebra


SHORT ANSWER Questions. Write the word or phrase that best completes each statement or answers the question. 27) Simplify the following circuit using any simplification method. Draw the resulting circuit. Will work through in Class – BUT, do the Truth Table at the end as Homework! Answer: It’s important to get the logical equation correct to start with. Create interim terms or

equations to help. For example, the first NOT-AND gate is �̅� ∙ �̅� ∙ �̅� . Write that above the output of the NOT-AND gate. Keep working left to right until you reach the output. Notice the Output gate: A NOT-OR has its inputs from NOR gates, which means the two “bubbles” cancel each other. In fact, if you look from left to right, you really have four OR functions in the path, so you should end up with 5 equations from the outputs of the AND-type gates that are OR’d together. The author takes some liberties in not drawing individual inverters for all the signals, but this is not unusual when creating the logic diagram. There may even be some commercially available ICs that duplicate this kind of combination of some inverting and some non-inverting inputs.

So: 𝑋 = (�̅� ∙ 𝐵 ∙ 𝐶̅) + (�̅� ∙ �̅� ∙ �̅�) + (𝐶 ∙ �̅�) + (�̅� ∙ 𝐵 ∙ 𝐶) + (𝐴 ∙ �̅� ∙ 𝐶̅ ∙ �̅�)


Boolean Algebra


Nothing simple jumps out that would allow us to start factoring the equation, so we’re left with Karnaugh mapping. You could write down all 16 combinations of ABCD and then fill in the blanks in the map, or you could set up the map and just fill in starting with Row 1, Column 1.

C D A B

0 0 0 1 1 1 1 0

0 0

0 1

1 1

1 0

𝑋 = (�̅� ∙ 𝐵 ∙ 𝐶̅) + (�̅� ∙ �̅� ∙ �̅�) + (𝐶 ∙ �̅�) + (�̅� ∙ 𝐵 ∙ 𝐶) + (𝐴 ∙ �̅� ∙ 𝐶̅ ∙ �̅�)

We have to ask 16 questions. If the combinations of ABCD lead to one “1” result, then a 1 goes in

the box identified by the row and column.

1) Are any of the products true if ABCD are all zero? Yes, the second term, so the first box gets

a “1”.

C D A B

0 0 0 1 1 1 1 0

0 0 1

0 1

1 1

1 0

𝑋 = (�̅� ∙ 𝐵 ∙ 𝐶̅) + (�̅� ∙ �̅� ∙ �̅�) + (𝐶 ∙ �̅�) + (�̅� ∙ 𝐵 ∙ 𝐶) + (𝐴 ∙ �̅� ∙ 𝐶̅ ∙ �̅�)

2) Maybe we can get a little cleverer. For the second column, first row, are there any terms that are

true when only D is a 1? No. We can eliminate any terms that have �̅� and the remaining terms

need at least A, B, or C = 1.

C D A B

0 0 0 1 1 1 1 0

0 0 1 0

0 1

1 1

1 0

𝑋 = (�̅� ∙ 𝐵 ∙ 𝐶̅) + (�̅� ∙ �̅� ∙ �̅�) + (𝐶 ∙ �̅�) + (�̅� ∙ 𝐵 ∙ 𝐶) + (𝐴 ∙ �̅� ∙ 𝐶̅ ∙ �̅�)


Boolean Algebra


3) For the next column, ABCD = 0011. Any matches? No. How about ABCD = 0010? Yes. The

second term and the third term.

C D A B

0 0 0 1 1 1 1 0

0 0 1 0 0 1

0 1

1 1

1 0

𝑋 = (�̅� ∙ 𝐵 ∙ 𝐶̅) + (�̅� ∙ �̅� ∙ �̅�) + (𝐶 ∙ �̅�) + (�̅� ∙ 𝐵 ∙ 𝐶) + (𝐴 ∙ �̅� ∙ 𝐶̅ ∙ �̅�)

4) Repeat the process for the second row with ABCD = 0100, 0101, 0111, and 0110.

C D A B

0 0 0 1 1 1 1 0

0 0 1 0 0 1

0 1 1 1 1 1

1 1

1 0

𝑋 = (�̅� ∙ 𝐵 ∙ 𝐶̅) + (�̅� ∙ �̅� ∙ �̅�) + (𝐶 ∙ �̅�) + (�̅� ∙ 𝐵 ∙ 𝐶) + (𝐴 ∙ �̅� ∙ 𝐶̅ ∙ �̅�)

First column, “1” because of the first term,

Second column, “1” because of the first term,

Third column, “1” because of the fourth term,

Fourth column, “1” because of third and fourth terms. In fact, wherever CD = 10, we’ll have a “1”

because of the third term.

C D A B

0 0 0 1 1 1 1 0

0 0 1 0 0 1

0 1 1 1 1 1

1 1 1

1 0 1

𝑋 = (�̅� ∙ 𝐵 ∙ 𝐶̅) + (�̅� ∙ �̅� ∙ �̅�) + (𝐶 ∙ �̅�) + (�̅� ∙ 𝐵 ∙ 𝐶) + (𝐴 ∙ �̅� ∙ 𝐶̅ ∙ �̅�)


Boolean Algebra


5) Finish up the third and fourth rows.

C D A B

0 0 0 1 1 1 1 0

0 0 1 0 0 1

0 1 1 1 1 1

1 1 0 0 0 1

1 0 1 0 0 1

𝑿 = (�̅� ∙ 𝑩 ∙ �̅�) + (�̅� ∙ �̅� ∙ �̅�) + (𝑪 ∙ �̅�) + (�̅� ∙ 𝑩 ∙ 𝑪) + (𝑨 ∙ �̅� ∙ �̅� ∙ �̅�) 6) We have three combinations of four sets of “1”: Second Row, Fourth Column, and Corners.

7) The Second Row tells us we only care about A = 0 and B =1, the first term is (�̅� ∙ 𝐵).

8) The Fourth Column tells us we only care about C = 1 and D = 0, the second term is (𝐶 ∙ �̅�).

9) The Corners have B = 0 and D = 0 in common, the third term is (�̅� ∙ �̅�).

10) So the reduction is X = (�̅� ∙ 𝐵) + (𝐶 ∙ �̅�) + (�̅� ∙ �̅�).

11) Prove it! Fill in the Truth Table below! Use the space to do each term individually, then sum.

A B C D (�̅� ∙ 𝑩 ∙ �̅�) + (�̅� ∙ �̅� ∙ �̅�) + (𝑪 ∙ �̅�) + (�̅� ∙ 𝑩 ∙ 𝑪) + (𝑨 ∙ �̅� ∙ �̅� ∙ �̅�) (�̅� ∙ 𝑩) + (𝑪 ∙ �̅�) + (�̅� ∙ �̅�)

0 0 0 0

0 0 0 1

0 0 1 0

0 0 1 1

0 1 0 0

0 1 0 1

0 1 1 0 0 + 0 + 1 + 1 + 0 + 1 = 1 1 + 1 + 0 = 1

0 1 1 1

1 0 0 0

1 0 0 1

1 0 1 0

1 0 1 1

1 1 0 0

1 1 0 1

1 1 1 0

1 1 1 1


Boolean Algebra


28) Design the logic to control a seven segment display to generate the following characters:

B A a b c d e f g

0 0

0 1

1 0

1 1

29) Reduce the following equation using Boolean Algebra (you WILL need Rule 11).

𝐴 ∙ 𝐵 ∙ 𝐶 + �̅� ∙ 𝐵 ∙ 𝐶̅ + 𝐴 ∙ �̅� ∙ 𝐶 + �̅� ∙ �̅� ∙ 𝐶 + 𝐴 ∙ 𝐵 ∙ 𝐶̅ = 𝑋

a

b

c

d

e

f

g


Boolean Algebra


30) Generate a Boolean expression for the following circuit and reduce. Work through in Class. BUT, draw the reduced circuit at the end of this homework problem. This requires a little finesse and some brute force. Brute force is just write out the equation. We could note that the NOR gate for X and A is input into a NOT-NAND, so the “bubble” on the NOR gate gets cancelled.

( (𝑋 + 𝐴) ∙ (𝐴 ∙ �̅�))̅̅ ̅̅ ̅̅ ̅̅ ̅̅̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ∙ 𝑃 = 𝑂𝑈𝑇

We can temporarily ignore the ∙ 𝑃 and focus on the rest of the reduction. We can treat (𝑋 + 𝐴) as one

term and 𝐴 ∙ �̅� as another and apply DeMorgan’s rule. Then we have

( (𝑋 + 𝐴) ∙ (𝐴 ∙ �̅�))̅̅ ̅̅ ̅̅ ̅̅ ̅̅̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ = (𝑋 + 𝐴)̅̅ ̅̅ ̅̅ ̅̅ ̅̅ + (𝐴 ∙ �̅�)

We apply DeMorgan’s rule again to the (𝑋 + 𝐴)̅̅ ̅̅ ̅̅ ̅̅ ̅̅ term, which equals (�̅� ∙ �̅�)

Recombining everything, we have [(�̅� ∙ �̅�) + (𝐴 ∙ �̅�)] ∙ 𝑃 = 𝑂𝑈𝑇

The finesse occurs when we multiply through by P.

(�̅� ∙ �̅�) ∙ 𝑃 + (𝐴 ∙ �̅� ∙ 𝑃) = 𝑂𝑈𝑇 = (�̅� ∙ �̅�) ∙ 𝑃


Boolean Algebra


Another way is to redraw the circuit and find the only ways OUT = 1.

Now, just looking at the circuit, we know OUT can only be high when P = 1 AND either X AND A = 0, or A = 1 AND P = 0. But if P = 0, then OUT is guaranteed to be 0, so the only answer is

𝑂𝑈𝑇 = (�̅� ∙ �̅�) ∙ 𝑃

Now Draw the Reduced Circuit !

X

A

P

OUT

Chapter 4 Homework · ELN-133 Chapter 4 Homework Spring 2015 Boolean Algebra AB Tech Page 2 of 15...

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