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Exam 1 Solved Questions - Digital Electronics | EE 231, Exams of Digital Electronics

Material Type: Exam; Class: Digital Electronics; Subject: Electrical Engineering; University: New Mexico Institute of Mining and Technology; Term: Fall 2008;

Typology: Exams

Pre 2010

Uploaded on 08/08/2009

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EE 231 Fall 2008
EE 231
Exam 1
September 24, 2008
Name:
No calculators allowed. Show all work. Partial credit will be given. No credit will be given if an answer
appears with no supporting work.
1. You are tasked with designing a digital thermometer, which will display temperatures from -60oF to
+140oF. (The system displays the temperature to the nearest degree, so you will not need to display
fractions of a degree.) The numbers will be stored in a register in signed 2’s complement form. What
is the minimum number of bits your register will need to hold numbers from -60 to +140? Explain
your reasoning.
In 2’s compliment form, numbers range from 2N1to 2N11, where Nis the number of bits.
For N= 8, the range is -128 to +127. For N= 9 the range is -256 to +255. Thus, you need a 9 bit
register.
2. Convert the following decimal number to hexadecimal. You only need to keep two digits to the right
of the decimal point:
(73.28)10
Quot Rem Int Frac
73 / 16 = 4 R 9 0.28 x 16 = 4 + 0.48
4 / 16 = 0 R 4 0.48 x 16 = 7 + 0.68
To get the integer part, read the remainder column going up: (73)10 = (49)16 .
To get the fractional part, read the integer column going down: (0.28)10 = (0.47)16. (Actually, the
last fractional digit should be rounded up, since 0.68 is greater than 1/2.)
(73.28)10 = (49.48)16.
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EE 231

Exam 1 September 24, 2008

Name:

No calculators allowed. Show all work. Partial credit will be given. No credit will be given if an answer appears with no supporting work.

  1. You are tasked with designing a digital thermometer, which will display temperatures from -60o^ F to +140oF. (The system displays the temperature to the nearest degree, so you will not need to display fractions of a degree.) The numbers will be stored in a register in signed 2’s complement form. What is the minimum number of bits your register will need to hold numbers from -60 to +140? Explain your reasoning.

In 2’s compliment form, numbers range from − 2 N^ −^1 to 2 N^ −^1 − 1 , where N is the number of bits. For N = 8, the range is -128 to +127. For N = 9 the range is -256 to +255. Thus, you need a 9 bit register.

  1. Convert the following decimal number to hexadecimal. You only need to keep two digits to the right of the decimal point: (73.28) 10

Quot Rem Int Frac 73 / 16 = 4 R 9 0.28 x 16 = 4 + 0. 4 / 16 = 0 R 4 0.48 x 16 = 7 + 0.

To get the integer part, read the remainder column going up: (73) 10 = (49) 16. To get the fractional part, read the integer column going down: (0.28) 10 = (0.47) 16. (Actually, the last fractional digit should be rounded up, since 0.68 is greater than 1/2.) (73.28) 10 = (49.48) 16.

  1. For this problem, assume all numbers are held in 8-bit registers in a digital system, and the numbers are represented in 2’s complement form.

(a) Convert (+87) 10 to a 2’s complement 8-bit hex number.

Quot Rem 87 / 16 = 5 R 7 5 / 16 = 0 R 5 (+87) 10 = (57) 16

(b) Convert (−95) 10 to a 2’s complement 8-bit hex number. First, find the hex representation for +95: Quot Rem 95 / 16 = 5 R 15 5 / 16 = 0 R 5

(+95) 10 = (5F ) 16 Next, take the 2’s complement of 5F to make it negative: Ones’ complement of 5F is A0, add 1 to ones’ complement to get two’s complement: A0 + 1 = A1. (−95) 10 = (A1) 16

(c) Use the results from (a) and (b) to perform the following operation using 2’s complements: (+87) 10 − (−95) 10

(57) 16 − (A1) 16 = (57) 16 + (5F ) 16 = (B6) 16

(d) Convert the answer to of Part (c) to its decimal equivalent. Because (B6) 16 is negative, take its 2’s complement, convert that number to decimal, and put a minus sign in front: Ones’ complement of B6 is 49, add 1 to ones’ complement to get two’s complement: 49 + 1 = 4A. Convert 4A to decimal: 4 x 16 + 10 x 1 = 74. (B6) 16 = (−74) 10

  1. Use Boolean algebra to simplify the following expressions to a minimum number of literals:

(a) (xy + yz′^ + x′z)(x + z) (xy + yz′^ + x′z)(x + z) = xy + xyz′^ + x′z + xyz + yz′z + x′z (xy + yz′^ + x′z)(x + z) = xy + xyz′^ + x′z + xyz + x′z (xy + yz′^ + x′z)(x + z) = xy + xyz′^ + xyz + x′z (xy + yz′^ + x′z)(x + z) = xy(1 + z′^ + z) + x′z (xy + yz′^ + x′z)(x + z) = xy + x′z (xy + yz′^ + x′z)(x + z) = xy + x′z