ITI1100 Midterm 13 Solution 2 Article

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ITI1100 Digital Systems I

Mid-term Exam

Date: March a couple of, 2013

Time: 1h30 (10: 00-11: 30)

Professors: A. Karmouch and Hasan Ural

Instructions:

Response ALL questions.

This really is a close-book examination.

Utilize the provided space to answer the following questions. In the event that more space should be used, use the back of the webpage. Show your calculations to obtain full markings.

Calculators are certainly not allowed.

Examine all the questions carefully before you start.

Question

Points

Percentage

1

forty percent

2

60 per cent

Total

totally

Question 1 (40 details: 15 + 10+ 15):

(a) Convert the following binary into (i) decimal, (ii) octal et (iii) hexadecimal 10100001111. 1101

(i) to decimal

(10100001111)2

= 1 times 210 & 0 x 29 + 1 back button 28 + 0 x 27 + 0 back button 26 & 0 times 25 & 0 back button 24 + 1 times 23 + 1 x 22 + 1 by 21 + 1 x 20 = 1024 & 0 + 256 & 0 + 0 & 0 & 0 & 8 & 4 & 2 + 1 = 1295

(0. 1101)2

= 1 x 2-1 + 1 x 2-2 + zero x 2 - 3 + you x 2-4

= 0. 5 & 0. twenty-five + zero + 0. 0625

= 0. 8125

(10100001111. 1101)2 = 1295. 8125

(ii) to Octal

(10100001111)2

= 010 100 001 111. one hundred ten 100

= 2417. sixty four

(ii) to Hexadecimal

(10100001111)2

= 0101 0000 1111. 1101

= 50F. M

(b) Perform the following operation using 10's go with

(162)10 – (27)10

[N]r sama dengan rn – (N)r

and = several

(162)10 – (027)10

ciento tres – 027 = 973

10's Enhance of (27)10 = 973

(162)10 – (27)10 sama dengan (162)10 + 10's Go with of (27)10

= (162)10 + (973)10

= (135)10

(c) Convert the following figures into binary and carry out the math operations in (i) and (ii) applying signed binary numbers with 2's match. Use several bits to symbolize the integer part of decimal numbers and the sign little. Use 3 bits to symbolize the fractional part.

(i) (4. 5)10 – (9)10

(ii) (8. 5)10 + (9)10

(4. 5)10 = 0000100. 95

(9)10 = 0001001. 000

2's of (9)10 sama dengan 1110111. 500

(8. 5)10 = 0001000. 100

(i) (4. 5)10 – (9)10

sama dengan 0000100. 100 - 0001001. 000

= 0000100. 95 + 2's of (0001001. 000)

sama dengan 0000100. 95 + 1110111. 000

= 1111011. 90

= -- (0000100. 100)

sama dengan - (4. 5)10

(ii) (8. 5)10 + (9)10

= 0001000. 75 + 0001001. 000

sama dengan 0010001. 75

= (17. 5)10

Problem 2: (10 + 40 + your five + 12-15 = 70 points)

Design a 4- bit combinational circuit 2's complementer. The circuit creates at the outcome the 2's complement with the input binary numbers. Solution the following queries:

(i) Full the following fact table. A, B, C, D reveal the suggestions binary amount to be complemented using 2's complement and W, X, Y, Unces represent the outcome 2's match of the input binary number. The variable D may be the least significant bit and A is the most significant piece of the binary number.

A

N

C

D

w

x

y

unces

0

0

0

zero

0

0

0

0

0

zero

0

1

1

you

1

you

0

0

1

zero

1

you

1

zero

0

0

1

1

1

you

0

1

0

you

0

zero

1

you

0

0

0

you

0

one particular

1

zero

1

one particular

0

one particular

1

0

1

zero

1

zero

0

1

1

one particular

1

0

0

one particular

1

zero

0

zero

1

0

0

0

1

zero

0

1

0

one particular

1

1

1

zero

1

0

0

one particular

1

zero

1

0

1

1

0

1

0

you

1

one particular

0

0

0

you

0

0

1

one particular

0

you

0

zero

1

1

1

one particular

1

0

0

0

1

0

1

you

1

1

0

zero

0

one particular

(ii) Simplify the Boolean function T in its Quantity of Products type using K-Map

00

01

11

15

00

zero

1

you

1

01

1

1

1

1

11

zero

0

zero

0

10

1

0

0

0

W sama dengan A'B & A'D + A'C & AB'C'D'

(iii) Show the fact that Boolean function W can be constructed applying exclusive-OR entrances

W = A'B & A'D + A'C + AB'C'D'

sama dengan A' (B + M + C) + A B'C'D'

B'C'D' = (B + C + D)'

X sama dengan (B + C + D)

W = A' (B & D + C) + A (B + C + D)'

W = A' Times + A X'

W = A ⊕ X

W sama dengan A ⊕ (B & D + C)

(iv) Implement the Boolean function Z in its Product of Sums kind with a decoder constructed with NAND gates (see figure below) and external gate (s)...

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