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13 Question 3 A Karnaugh map is nothing more than a special form of truth table. Fawad Khan 0321-6386013 www. according to the values found in the above truth table: Answer Notes: Showing students that Karnaugh maps are really nothing more than truth tables in disguise helps them to more readily learn this powerful
Electronics Club Handout # 2. Logic simplification with Karnaugh maps Now you can think about these problems. • Can you make a counter which count from 3 to 8 ? • Repeat the 7 segment BCD display example if (solution is given ,so verify it and try for more bit). • Truth table of 1 bit comparator. Logical Implementation
problems can be solved. Example 4.1. Problem: Determine the minimum-cost SOP and POS expressions for the function f(x1,x2,x3,x4) = ? m(4,6,8,10,11,12,15) + D(3,5,7,9). Solution: The function can be represented in the form of a Karnaugh map as shown in Figure. 4.1a. Note that the location of minterms in the map is as
Construct a K-map for F'. Since the truth values on the K-map are based on the minterms of F', in order to find the minimum sum of product terms for F, we need to group all zeros. AB. CD 00 01 11 10. 0. 1. 1. 1. 0. 0. 0. 0. 0. 1. 1. 0. 0. 1. 1. 1. F(A,B,C,D) = A'B' + C'D + B'D. 00. 01. 11. 10.
C-H 1. Lecture 5. Karnaugh Maps. • Algebraic procedures: • Difficult to apply in a systematic way. • Difficult to tell when you have arrived at a minimum solution. • Karnaugh map (K-map) can be used to minimize functions of up to 6 variables. – K-map is directly applied to two- level networks composed of AND and OR gates.
Homework Problem Set: K-Maps. We will collect these sheets from students at the start of class. Please write clearly and legibly. Answer all questions on this sheet. You may attach additional pages if necessary. Student Name: .. Student Number: .. Session ID (C1 or C2):
is particularly straightforward for most problems of three or four variables. Although there is no guarantee of finding a minimum solution, the meth- variable maps. The Karnaugh map consists of one square for each possible minterm in a function. Thus, a two-variable map has 4 squares, a three- variable map has 8
Step 1: Draw the Karnaugh map that represents the expression by placing a mark in each appropriate square. Step 2: “Cover" the marked boxes as best you can with ovals. Allowed coverings for this setup are limited to 2x1 rectangles. This example is covered by one 2x1 rectangle. Step 3: Expressions/rectangles that are
The minimization will result in reduction of the number of gates. (resulting from less number of terms) and the number of inputs per gate (resulting from less number of variables per term). • The minimization will reduce cost, efficiency and power consumption. • y(x+x`)=y.1=y. • y+xx`=y+0=y. • (x`y+xy`)=x?y. • (x`y`+xy)=(x?y)`
Therefore, the answer is 2. 3. = 8. 10. True or False: If done properly, there is exactly one possible arrangement for the rectangles of ones in a Karnaugh map. The answer is FALSE. To show this, look at the following example (note the red rectangle):. Problems 11, 12, and 13 use the state machine diagram to the right.
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