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## EE2310 Homework 2

by: Safwan Mazhar

12

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5

# EE2310 Homework 2 EE 2310

Safwan Mazhar
UTD

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Homework 2 Solutions
COURSE
Introduction to Digital Systems
PROF.
Dr. Nathan Dodge
TYPE
Class Notes
PAGES
5
WORDS
KARMA
25 ?

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This 5 page Class Notes was uploaded by Safwan Mazhar on Tuesday October 18, 2016. The Class Notes belongs to EE 2310 at University of Texas at Dallas taught by Dr. Nathan Dodge in Fall 2016. Since its upload, it has received 12 views.

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Date Created: 10/18/16
EE 2310 Homework #2 Solutions – Digital Logic Circuits and Simplifying Logic with Karnaugh Maps 1. Given the truth table below, write its Boolean expression in SOP form. Then draw the logic circuit that represents the Boolean expression. x y z f 0 0 0 1 f xyz xyz xyz 0 0 1 1 0 1 0 1 x 0 1 1 0 f 1 0 0 0 y 1 0 1 0 1 1 0 0 1 1 1 0 z 2. Draw the SOP circuit that represents the Boolean expression below. zyzyww   w x f y z 3. Given the Boolean SOP function shown, simplify using non-Karnaugh-map techniques and draw the simplified SOP circuit. f xyz wxyz wxyz wxyz Simplified: f  xz 4. For the truth table below, write its Boolean expression in SOP form. Then draw the circuit. Using the techniques learned in class, simplify the circuit (do NOT use a K- map) to a simpler SOP form and draw that SOP circuit. z yzxyzxfx x y z f 0 0 0 0 0 0 1 1 x 0 1 0 0 0 1 1 1 y 1 0 0 0 f 1 0 1 0 1 1 0 0 1 1 1 1 z Simplified:zfyx 5. Write the Boolean expression in SOP form for the truth tablew x y f to the right. Plot 1’s on the Karnaugh Map and simplify the 0 0 0 0 Boolean expression by determining prime implicants. Draw 0 0 1 0 both the original and the simplified circuit. 0 1 0 0 xy 0 1 1 1 1 0 0 0 00 01 11 10 1 0 1 0 w 0 1 1 1 0 1 1 11 1 1 1 1 Original:yxyxw  Simplified: f  wx  xy w w x f x f y y 2 EE 2310, Homework #2 6. A logic function can be expressed af   the K-map below, write the simplified SOP expression, and draw the simplified SOP circuit. yz 00 01 11 10 00 1 1 wx 01 11 10 1 1 Simpelfieession: f  xy SimpC 7. For the Karnaugh map at right, write the simplified expression, and draw the simplest possible circuit. The simplified expression is: zwx w x f y t i u c r i C d e i f i l p m i S z 8. For the Karnaugh map at right, write the simplified simplified circuit (only). Simplified expression: zfw  Simplified Circuit w y f z 3 9. Consider the K-mapped SOP Boolean function f   2,6, . The input variables m operate in a restricted space. Input combinations are limited such that x can never be 0 when w is 1. Map both the 1’s in the function and the special “don’t care” conditions listed, develop the simplified SOP expression, and draw ONLY the simplified circuit. yz Simplified expression:zfy 00 01 11 10 00 1 wx 01 1 11 1 10 X X X X Simplified Circuit 10. This problem shows that sometimes the POS Boolean expression is the more convenient. For the truth table on the right, fill in the K-map and write both the original and simplified POS expressions, and draw the simplified circuit. w x y z f y+z 0 0 0 0 1 00 01 11 10 0 0 0 1 1 00 1 1 1 1 0 0 1 0 1 1 1 1 1 0 0 1 1 1 w+x 01 0 1 0 0 1 11 0 1 1 1 0 1 0 1 1 10 0 1 1 1 0 1 1 0 1 1 0 0 0 0 1 0 0 1 1 1 0 1 0 1 Original expression: 1 0 1 1 1 1 1 0 0 0 ( ) f  w(yzwxyz . 1 1 0 1 1 1 1 1 0 1 1 1 1 1 1 Simplified expression: fwyz . w Simpclredit: y f z 4 EE 2310, Homework #2 11. It is desired to multiplex four different input data lines, a-d, onto one output. Five address lines, (“v” [MSB] through “z” [LSB]) control input-to-output selection. The five-bit address can be stated as a hexadecimal number ranging from 0x00 to 0x1f. Input a is MUXed out on address 0x06, b on address 0x0f, c on 0x1b, and d on 0x1e. Draw the MUX circuit below. a b MUX out c d v w x y z 12. The truth table for a 1-bit subtract only circuit is shown at the bi x y Dbo right. This circuit could be used in an n-bit subtractor by tying 0 0 0 0 0 0 0 1 1 1 n of these 1-bit subtractors together. This is a full subtractor, 0 1 0 1 0 which means that the inputs are x (the bit to be subtracted from), y (the bit that is subtracted from x), and borrow in (bi), 0 1 1 0 0 the number that results when the adjacent column to the right 1 0 0 1 1 1 0 1 0 1 has a y number bigger than x (or y and x are equal, but there is a borrow from the next-right column). Outputs are difference 1 1 0 0 0 (D) and borrow out (bo). Fill in the two Karnaugh maps below, 1 1 1 1 1 simplify, write the SOP expressions for D and bo, and draw the two SOP circuits. xy xy 00 01 11 10 00 01 11 10 bi 0 1 1 bi 0 1 1 1 1 1 1 1 1 Difference (D) (No simplification possible) Borrow Out (bo) D  bi i i ix y bx y bx y xO ii y xy bi bi x bo x D y y Borrow Out Difference 5 EE 2310, Homework #2

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