Boolean Expression

Section 2

Examples of Solved Challenges

This section gives some common problems that trainees may encounter, and reveals how such problems could be solved. As well as the identities given in Section 2 . 5, these types of examples utilize an identification known as general opinion, defined under.



xВ·y+yВ·z+xВ·z =xВ·y+xВ·z

(x & y) В· (y + z) В· (x + z) sama dengan (x & y) В· (x + z)


Example installment payments on your 1

Trouble: Determine if this equation can be valid

x1 x3 + x2 x3 + x1 x2 = x1 x2 + x1 x3 & x2 x3

Solution: The equation can be valid in case the expressions around the left- and right-hand attributes represent the same function. To do the comparison, we could create a fact table for each and every side and see if the truth dining tables are the same. A great algebraic approach is to obtain a canonical sum-of-product form for each appearance.

Using the fact that x & x sama dengan 1 (Theorem 8b), we can manipulate the left-hand part as follows: LHS = x1 x3 + x2 x3 + x1 x2

= x1 (x2 + x2 )x3 + (x1 & x1 )x2 x3 & x1 x2 (x3 + x3 )

= x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3 These item terms stand for the minterms 2, zero, 7, several, 5, and 4, correspondingly. For the right-hand part we have

RHS = x1 x2 & x1 x3 + x2 x3

sama dengan x1 x2 (x3 + x3 ) + x1 (x2 + x2 )x3 + (x1 + x1 )x2 x3

= x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3 + x1 x2 x3 These product terms symbolize the minterms 3, a couple of, 7, 5, 4, and 0, correspondingly. Since equally expressions identify the same minterms, they signify the same function; therefore , the equation is usually valid. Other ways of addressing this function is by m(0, 2, a few, 4, five, 7). you

Example 2 . 2

Difficulty: Design the minimum-cost product-of-sums expression pertaining to the function f (x1, x2, x3, x4 ) = m(0, 2, 4, 5, 6th, 7, almost 8, 10, doze, 14, 15).

Solution: The function is usually defined when it comes to its minterms. To find a DETRAS expression we need to start with the definition regarding maxterms, which is f sama dengan ΠM (1, 3, on the lookout for, 11, 13). Thus, n

= M1 В· M3 В· M9 В· M11 В· M13

= (x1 + x2 + x3 + x4 )(x1 & x2 & x3 & x4 )(x1 + x2 + x3 + x4 )(x1 & x2 & x3 & x4 )(x1 + x2 + x3 + x4 )

We could rewrite the merchandise of the п¬Ѓrst two maxterms as

M1 В· M3 = (x1 + x2 + x4 + x3 )(x1 & x2 + x4 & x3 )

= x1 + x2 + x4 + x3 x3

sama dengan x1 & x2 & x4 + 0

sama dengan x1 & x2 + x4





commutative property 10b

distributive real estate 12b

theorem 8a

theorem 6b

In the same way, M9 В· M11 sama dengan x1 + x2 + x4. Today, we can use M11 once again, according to property 7a, to obtain M11 В· M13 sama dengan x1 + x3 & x4. Therefore

f = (x1 + x2 + x4 )(x1 + x2 + x4 )(x1 + x3 + x4 )

Applying 12b again, we have the п¬Ѓnal answer

n = (x2 + x4 )(x1 + x3 + x4 )

Example installment payments on your 3

Difficulty: A routine that handles a given digital system provides three inputs: x1, x2, and x3. It has to recognize three different conditions:

• Condition A is true if perhaps x3 applies and possibly x1 is valid or x2 is phony • Condition B is valid if x1 is true and either x2 or x3 is fake • Condition C holds true if x2 is true and either x1 is true or perhaps x3 can be false The control signal must develop an output of 1 in the event at least two of situations A, N, and C are authentic. Design the simplest circuit that can be used for this purpose. Remedy: Using you for accurate and zero for false, we can exhibit the three conditions as follows: A = x3 (x1 & x2 ) = x3 x1 + x3 x2

B sama dengan x1 (x2 + x3 ) = x1 x2 + x1 x3

C = x2 (x1 & x3 ) = x2 x1 + x2 x3


Then, the desired outcome of the signal can be stated as farreneheit = STOMACH + ALTERNATING CURRENT + BC. These item terms can be discovered as:

AB = (x3 x1 & x3 x2 )(x1 x2 + x1 x3 )

= x3 x1 x1 x2 + x3 x1 x1 x3 + x3 x2 x1 x2 + x3 x2 x1 x3

= x3 x1 x2 + 0 + x3 x2 x1 + zero

= x1 x2 x3

AC sama dengan (x3 x1 + x3 x2 )(x2 x1 & x2 x3 )

sama dengan x3 x1 x2 x1 + x3 x1 x2 x3 & x3 x2 x2 x1 + x3 x2 x2 x3

sama dengan x3 x1 x2 & 0 & 0 & 0

= x1 x2 x3

BC = (x1 x2 & x1 x3 )(x2 x1 + x2 x3 )

= x1 x2 x2 x1 & x1 x2 x2 x3 + x1 x3 x2 x1 & x1 x3 x2 x3

= zero + zero + x1 x3 x2 + x1 x3 x2

= x1 x2 x3

Therefore , f can be written as


= x1 x2 x3 + x1 x2 x3 + x1 x2 x3

= x1 (x2 & x2 )x3 + x1 x2 (x3 + x3 )

= x1 x3...



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