New version page

NAU EE 110 - Octal solution

Upgrade to remove ads

This preview shows page 1-2 out of 7 pages.

Save
View Full Document
Premium Document
Do you want full access? Go Premium and unlock all 7 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 7 pages.
Access to all documents
Download any document
Ad free experience

Upgrade to remove ads
Unformatted text preview:

EE 110 Spring 2017 CQUPT HW 7 Solution Mlsna EE 110 Spring 2017 CQUPT Homework 7 Solution 1. (10 points) In octal, perform the following calculation: 2404.38 + 156.28 Show all steps. Check your results by converting all numbers to decimal and making sure addition in decimal gives the same result. 2 4 0 4 . 3 + 0 1 5 6 . 2 (Carrying occurs by 8 since we’re working in octal) _________________ 2 5 6 2 . 5 Check: 2404.38 = 2x83 + 4x82 + 0x81 + 4x80 + 3x8–1 = 1024 + 256 + 4 + 3/8 = 1284.37510 156.28 = 1x82 + 5x81 + 6x80 + 2x8–1 = 64 + 40 + 6 + 2/8 = 110.2510 2562.58 = 2x83 + 5x82 + 6x81 + 2x80 + 5x8–1 = 1024 + 320 + 48 + 2 + 5/8 = 1394.62510 This checks, since in decimal: 1284.375 + 110.25 = 1394.625 2. (10 points) In octal, perform the following calculation: 2404.38 – 156.28 Show all steps. Check your results by converting all numbers to decimal and making sure subtraction in decimal gives the same result. 2 4 0 4 . 3 – 0 1 5 6 . 2 (Borrowing occurs by 8) _________________ 2 2 2 6 . 1 Check: 2226.18 = 2x83 + 2x82 + 2x81 + 6x80 + 1x8–1 = 1174.12510 This matches the decimal: 1284.375 – 110.25 = 1174.125EE 110 Spring 2017 CQUPT HW 7 Solution Mlsna 3. (10 points) In hexadecimal, perform the following calculation: A230C.7F16 – EDB.C16 Consider these to be positive, unsigned numbers. Show all steps. Show the decimal equivalent to all of the numbers involved. I’ll show this one column at a time A 2 3 0 C . 7 F – 0 0 E D B . C 0 (Borrowing occurs by 16 since we’re working in hex) ____________ F ! “F – 0” or 15 – 0 “7 – C” or 7 – 12. Need to borrow 16 from the left. B ! This produces 16+7–12 = 11 0 ! “B – B” (remember, we borrowed 16 from here) 3 ! “0 – D” Need to borrow 16 from left, producing 16+0–13 4 ! “2 – E” (we borrowed from here). Borrow: 16+2–14 1 ! “1 – 0” (we borrowed from here) A ! “A – 0” ____________ A 1 4 3 0 . B F Now let’s look at the decimal equivalent values. A230C.7F16 = 10 x 164 + 2 x 163 + 3 x 162 + 12 + 7 x 16-1 + 15 x 16-2 = 655360 + 8192 + 768 + 12 + 0.4375 + 0.05859375 = 664332.49609375 EDB.C16 = 14 x 162 + 13 x 16 + 11 + 12 x 16-1 = 3584 + 208 + 11 + 0.75 = 3803.75 A1430.BF16 = 10 x 164 + 163 + 4 x 162 + 3 x 16 + 11 x 16-1 + 15 x 16-2 = 655360 + 4096 + 1024 + 48 + 0.6875 + 0.05859375 = 660528.74609375EE 110 Spring 2017 CQUPT HW 7 Solution Mlsna 4. (10 points each) Perform the following calculations in 8-bit two’s complement binary and show all steps. Also check your results by showing the calculation in decimal and showing the answers are correct. a. 5510 + 11410 First convert numbers to 8-bit 2’s complement form: 5510 = 0 0 1 1 0 1 1 1 11410 = 0 1 1 1 0 0 1 0 ------------------------------------ now add 1 0 1 0 1 0 0 1 = -8710 This answer is negative! We added two 8-bit 2’s complement positive numbers and should have obtained a positive answer. The problem is sign-bit overflow because the correct answer cannot be represented in 8-bit two’s complement form. b. 5510 + (-11410) 5510 = 0 0 1 1 0 1 1 1 -11410 = 1 0 0 0 1 1 1 0 ------------------------------------ now add 1 1 0 0 0 1 0 1 = - 00111011 = -5910 correct! c. (-5510) + 11410 -5510 = 1 1 0 0 1 0 0 1 11410 = 0 1 1 1 0 0 1 0 ------------------------------------ now add 0 0 1 1 1 0 1 1 = 5910 correct!EE 110 Spring 2017 CQUPT HW 7 Solution Mlsna 5. (20 points each) Use the Quine-McCluskey method to fully minimize the following Boolean expressions, producing SOP expressions. Show all steps! You may check your results with a K-map, but don’t use the K-map as a substitute for doing Quine-McCluskey. a. € f (A, B,C, D) = m(1,2, 3, 6, 7,8,9,12,14)∑ First make a minterm list arranged by number of 1’s in the binary pattern Individual Minterms Number of 1’s in binary pattern Minterms covered Binary pattern Prime Implicants ( * = covered ) 1 1 2 8 0001 0010 1000 * * * 2 3 6 9 12 0011 0110 1001 1100 * * * * 3 7 14 0111 1110 * * Pairs of minterms Minterms covered Binary pattern Prime Implicants ( * = covered ) 1,3 1,9 2,3 2,6 8,9 8,12 00–1 –001 001– 0–10 100– 1–00 PI2 PI3 * * PI4 PI5 3,7 6,7 6,14 12,14 0–11 011– –110 11–0 * * PI6 PI7 Groups of 4 minterms (quads) Minterms covered Binary pattern Prime Implicants ( * = covered ) 2,3,6,7 0–1– PI1EE 110 Spring 2017 CQUPT HW 7 Solution Mlsna Prime Implicant Chart 1 2 3 6 7 8 9 12 14 PI1 X X X X PI2 X X PI3 X X PI4 X X PI5 X X PI6 X X PI7 X X Minterms 2 and 7 can only be covered by PI1, therefore PI1 is an essential prime implicant. That means PI1 must be included in the final answer. Reduced Prime Implicant Chart 1 8 9 12 14 PI2 X PI3 X X PI4 X X PI5 X X PI6 X PI7 X X Now we must choose a minimal (smallest) set of remaining prime implicants in order to cover all minterms in this chart. Possible solutions: PI1 + PI3 + PI5 + PI6 = € A C + B C D + C AD + BCD PI1 + PI3 + PI5 + PI7 = € A C + B C D + C AD + ABD PI1 + PI2 + PI4 + PI7 = € A C + A B D + AB C + ABDEE 110 Spring 2017 CQUPT HW 7 Solution Mlsna b. € f (A, B,C, D, E) = m( 5, 6, 7, 8,10,11,13,18,20,21,23,24,26)∑ First make a minterm list arranged by number of 1’s in the binary pattern Individual Minterms Number of 1’s in binary pattern Minterms covered Binary pattern Prime Implicants ( * = covered ) 1 8 01000 * 2 5 6 10 18 20 24 00101 00110 01010 10010 10100 11000 * * * * * * 3 7 11 13 21 26 00111 01011 01101 10101 11010 * * * * * 4 23 10111 * Pairs of minterms Minterms covered Binary pattern Prime Implicants ( * = covered ) 8,10 8,24 010–0 –1000 * * 5,7 5,13 5,21 6,7 10,11 10,26 18,26 20,21 24,26 001–1 0–101 –0101 0011– 0101– …


View Full Document
Download Octal solution
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Octal solution and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Octal solution 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?