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ECE 201 Spring 03 Quiz 1-1 Solution                  page 1 of 4

 

For each multiple choice question circle all correct answers. there may be one, zero or many correct answers. if none of the answers are correct, circle nothing.  each possible answer in the multiple choice will be graded separately-credit will be given for correctly circling or not circling each choice.

for all other questions please respond in the space provided.

 

1.)  (8 pts) Perform the indicated conversions:

  a.) Decimal 53 to 8-bit unsigned binary: __00110101____

 

32 + 16 + 4 + 1

 

b.) $(263)_8$ to unsigned binary: _______________

  

$\underbrace{010}_{2}\underbrace{110}_{6}\underbrace{011}_{3} $

 

c.) $(B7.A1)_{16}$ to unsigned binary: ____________

 

$\underbrace{1011}_{B}\underbrace{0111.}_{7}\underbrace{1010}_{.A}\underbrace{0001}_{1} $

  d.) $(D8.40)_{16}$ to octal: _______________

  

 
   D   8   .4   0  
   11011000.01000000
  3  3  0  .2  0

2.)(6 pts)   a.) the two's complement of 1011110 is _____0100010_______:

 

b.) the one's complement of 1011110 is ___0100001________: 

 

c.) What is the decimal value of 111001 given that is a 6-bit signed two's complement number?

Solution: This is a signed number whose most significant bit is 1. Therefore, the number is negative, and it's value is the two's complement of the number given

 

The two's complement of 111001 = 000111

ANSWER:-7

 

d.) For the numbers below, fill in the parity bit in the blank assigned assuming odd parity:

 

        _0_00110111

 

        _1_10000111

 

 

 

3.) (8 pts) Which of the following are true?

 

Answers: A, B, D , E, and H

a.) x+1 = 1

 

ANYTHING or'ed with 1 is 1.

 

b.)xy +0 = xy

 

ANYTHING or'ed with 0 is unchanged.

 

c.)y(0) = 1

 

ANYTHING and'ed with 0 is 0.

 

d.)xy*0 = 0

   

Anything AND'ed with 0 is 0

 

e.) a + 0 = x

 

ANYTHING or'ed with 0 is unchanged

 

f.) x+1=x'  

Anything or'ed with 1 is 1.

 

g.)y+y' = 0

 

Anything or'ed with its complement is 1. 

1+0 = 0+1 =  1

 

h.)xx'=0

 

Anything and'ed with its complement is 0. 

1*0 = 0*1 = 0

 

4.) (8 pts)Which of the following are NOT true?

Answers: D , E, and G  

a.) xy+y = y

(Absorption Theorem)

b.)x(x'+y) = xy

x(x'+y) = xx' + xy = 0+xy = xy

 

c.)(x+y+z)(x+y'+z)=(x+z)

 

(x+y+z)(x+y'+z) = (x+z)+yy' = x+z

 

d.) x + xy = x+y

 

x+xy = x by absorption

 

FALSE

e.)zx+z = x+z

 

zx + z = z (by absorption)

FALSE

 

f.)(xy'+z)'=(x'+y)z'     

 

This is a straightforword application of DeMorgan's theorem

 

g.) (x'+y')wz + w=w + zxy

False:

h.) (x')'= x + xy +xz

(x')' = x

x + xy + xz = x + xz = x

5.) (4 pts) Which of the following is the correct BCD interpretation of the word $\underbrace{0001}_{1} \underbrace{0011}_{3} $

a.) 19

\fbox{ b.) 13 }

c.) 31

d.) 023

6.)(8 pts) Which of the following expressions are true given the Boolean Function F(x,y,z)=\( \Sigma \)(0,1,2,3,5)?

Answers: A, B , C, and E

Note: If F(x,y,z)= \( \Sigma \)(0,1,2,3,5), F is also \( \Pi \)(4,6,7)

a.) F=x'yz' + x'y'z + x'y'z' + x'yz + xy'z

That's minterms 0,1,2,3,5 written algebraically

b.)F=\( \Pi \)(4,6,7)

c.)F=(x'+y+z)(x'+y') Expand into Maxterms:

$ F=(x'+y+z)(x'+y'+zz') = (x'+y+z)(x'+y'+z)(x'+y'+z')= \Pi(4,6,7) $

 

d.)F= \( \Pi \)(0,1,2,3,5)

 

e.)F=(xy'z'+xyz'+xyz)'

  Two possible approaches; this is the function that does NOT contain minterms 4,6, and 7 (therefore, it does contain those maxterms). Or, you could apply DeMorgan's theorem, and you will get an expression with maxterm 4,6, and 7.

 

f.)F=(x'+y'+z')(x'+y'+z)(x'+y+z')(x'+y+z)(x+y'+z)

 

That's Maxterms 7,6,5,4, and 2

 

g.)F=(x+y+z)(x+y+z')(x+y'+z)(x+y'+z')(x'+y+z')        

That's Maxterms 0,1,2,3,5

 

6.) (6 pts) Which of the following expressions are true given that:

F(w,x,y,z)=(w+x')' + x'yz

 

We need to expand this function into its canonical forms. 

F(w,x,y,z)=w'x + x'yz (DeMorgan's)
          =w'x(y+y')(z+z') + (w+w')x'yz
			 =( \Sigma )(3,11,12,13,14,15)

a.) F=\( \prod \)(3,4,5,6,7,11)

b.) F'=\( \Pi \)(0,1,2,8,9,10,12,13,14,15)

c.) F'=\( \Sigma \)(0,1,2,8,9,10,12,13,14,15)

d.) F=\( \Pi \)(0,1,3,4,5,7,8,9,12,13,15)

e.) F=\( \Sigma \)(3,4,5,6,7,11)

f.) F'=\( \Pi \)(3,4,5,6,7,11)

 

 

8.) (a) (12 pts) Convert the following eight-bit numbers to decimal given that they are in the following format:

 

Unsigned 11010011    __1 + 2 + 16 + + 64 + 128 = 211_____

 

Signed 2's complement 11010011     ___Negative - 00101101 = -45___

 

Signed 1's complement 11010011     ___{Negative - 00101100 = -44__




(b) Convert the following decimal numbers to binary in the format given:

 

42 as 7-bit signed 1's comp     __Positive: 32 + 8 + 2 = 0101010 _

 

-32 as 8-bit signed 2's comp     __32 = 0100000; take 2's comp: 1100000____

 

-35 as Signed 2's comp using the minimum # of bits     __35 = 0100011; take 2's comp: 1011101___

 

7 bits required: 6 bits can only represent +31 to -32

 

9.) (6 pts) Complete the following signed 2's complement additions, and circle any that produce overflow:

 
I II III
01010 01000 11001
+ 01110 + 11001 + 10011
<>

 

 

ANSWER: C. Overflow occurs when the carry into and out of the MSB are different. (Shortcut: this can only happen when adding two numbers of the same sign). Overflow is also the only situation when you will get the wrong answer!

 

 

 

10.) (12 pts) Add the following 2 eight-bit numbers given that they are in the following format:

 
Unsigned: Signed 1's Complement: Signed 2's Complement:
00010100 00010100 00010100
+ 11010110 + 11010110 + 11010110
 
 
_11101010_ ________ _______
Convert results
to decimal:
 
__234_____ __-21___ __-22__
<>

 

 

 

 

11.) (10 pts) Simplify the expressions below to a minimum number literals:

  a.) $ F= wxyz(wxyz'+wx'yz+w'xyz+wxy'z) $ 

	F = wwxxyyzz' + wwxx'yyzz + ww'xxyyzz + wxyy'zz = 0

  There are a number of possible ways through the algebra below. I have included the way that I think is the most straightforward. Many other paths to the solutions are possible, and are not necessarily either better or worse. Wherever possible, I tried to note the specific mistake you made on your test paper. If you got this problem wrong, try to figure out what you did wrong, don't just memorize this!

b.) $ F= (ab'+ab'c) + a(d'+ abd) $ 

 F = a(b'+b'c) + a(d'+d)(d'+ab)
   = ab'       + a(d'+ab)
   = ab'  + ad' + aab = ab' + ad' + ab
   = a(b' + b + d') = a(1+d')= a(1)
   = a

 


















12.) (12 pts) Express the equation below in Canonical sum-of-products form (sum of minterms):

 

F= xz + z'(y'x+x')

  

F = xz + xy'z' + x'z'
  = xz(y+y') + xy'z' +x'z'(y+y')
  = xy'z + xyz + xy'z' + x'y'z' + x'yz'

F= \( \Sigma \)(0,2,4,5,7)

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Dan Stanzione 2003-02-12