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Unformatted text preview: ECE 110 "1' . q V'Pii‘o‘feSSOrszrunet and Loui 1' ,, ‘_ November 12, 2007 HOUREXAMINATION #3
i ' 1) erte your: . ' ‘
' Last Name (use capital letters"): 50 LU. Ti 0 N S First Name (uSe capital letters)
Signature:. '2) Write your name and section at the hawk of the test. 3 Do NOT TURN THIS PAGE UNTIL YOUARE TOLD A. Write or print clearly. Answer each problem on the exam itself. If you 1 need extra paper, there 1s an extra sheet at the end of this eXam. Clearly
1dent1fy the problem number on any additional pages. The decimal/ binary/
hexadec1mal table, the Flip— ﬂop characteristic tables, the ASCII Code, the
Morse Code alphabet, the USPS Code, and numbers and properties for 10g
base 2 are also attached to the exam. B In order to receive partial or fun Credit, y0u must'sliow all your work,
'6. g., your Solution process, the equation(s) that you use, the values of the
variables used 1n the equatio‘n(s'), etc. You must also include the unit of
measurement in each answer. ‘ Studentscaught cheating on this exam will earn a grade of F for the
entire course. Other penalties may include SUSpension and/0r dismissal
frOm the university. Problem 1 (20 points) The sequential circuit below has positive—edgetriggered D and JK ﬂip
ﬂops and an XOR gate. The delay in the XOR gate is negligible. Initially q] = 1 and go = 0.
Complete the timing diagram by determining the waveforms for ql, go, and y. clock Problem 2 (20 points) 0 “P
counter down
counter + connected 1M connected Problem 3 (20 points) A digital picture is quantized using four different gray levels. The table
below gives the coding and color for various light intensity percentages. average light coding color
intensi ercenta e bina decimal
[0; 25 . 1
[50; 75) (2
[75; 100] 00
1 1 b‘E‘fo (a) [5 pts.] How many bytes are used to store a 2 x 4pixel digital image with the above
characteristics? Show work. ) (2 4' 'X&A)‘(2 6"“  ora 6:. es
x (96th agave) ’ St g u Byt Consider the 2 x ’4pixel digital image (below at left) transformed into a (smaller resolution)
1 x 2pixel image (below at right). L stands for Left, R stands for Right. ___> L R (b) [6 ptS. ] For both pixels (L, R) compute the average light intensity percentage using the 2 x 4
pixel digital image. Round results to two decimal digits. (c) [4 ptS.] For both pixels, use results of (b) to ﬁnd the binary coding corresponding to the
closest gray level, using two bits of information. pixel average light (b)
intensity % 33 binag (c) Show work:
codln  . ‘l ~
1,7“ §+é+é+®=ﬁ= 0.3331 R: gum 1&3.) {52% 205117 (d) [5 ptS.] Compare the overall light intensity percentages of the original 2 x 4 image (L4) and of
the transformed l x 2 image (L2). Show work. L4:A§(§..§t—;+mm§) =625Z, Cot rectiOn kProblem 4 (20 points) Here is a new method to encode a hexadecimal message that enables the f One bit error. First, a check digit is appended to the right end of the message so that
the sum of all digits is divisible by 16 (with zero remainder). For example, for the message 34E,
the check digit is B because 3 + 4 + 14 + 11 = 32. Then each digit is converted to its 4—bit binary
equivalent, and a parity bit is attached to the 1911 end to form a 5bit odd—parity word. The
message. 34B is encoded as 10011 00100 01110 01011
3 4 E B
(a) [4 pts.] Determine the 5bit oddparity words in this method for the digits 8 and 9. 8: OIOOO 9: llOO‘ (b) [8 pts.] Suppose this new method is used and the following is received. Assuming at most one
bit error occurred, determine the original message of three hexadecimal digits. Show your work.
even
10011 10100 10000 00001
V : v \_./‘
3 C. 0 1 3+O+\ :4, BinméCﬁ 1‘2 (:6 l\0\ :CCM)
Ll—&\’2::Ho 5’55 00w ‘ lewd? (A (H00
' Ow 6H:e.mr «.24: (0100 Original hexadecimal message (with no check digit): 3 C O (c) [8 pts.] In the US. Postal Service code, the check digit makes the sum of the digits divisible
by ten. Explain why in this new method, the check digit should not be chosen so that the
sum is divisible by ten. (Hint: use the message 34E above and suppose one bit error occurs.) ‘ SunsetWM oil at Mutts. M£vdiﬁ¢6szp
ﬁrttmwaumgz 34 , titcldaottgat «3 ‘6, 41%.
34.4, “44¢? :30. ak’trawi't I00“ OO\OO OHKO 7 1100‘ 3 '4 (5 ~ _
I5 0“? 6’56 carer 0:ch u'us th 6‘09— 6ft? laud OHOO bencﬂiuc’wkzcm‘ct to '
octooOt) or 0010 CE) km! 62% And beeme 23W aware W ti; 06.3% m amen 681a». Problem 5 (I 0 points) Design a counter that counts continuously 5, 4, 3, 5, 4, 3, 4bit
down
counter Problem 6 (IOpoints) The two parts below show separate coding trees for symbols P, Q, R, S, T
and V, W, X, Y, Z. The relative frequency of each symbol is shown. (a) [5 pts.] Compute the average code word length for the tree below. (0 5—
1/0 \ —3)c\ {'32’12 437)!» T(12/32) 3(10/32) 37’ 37’ 37 1 /°
0 2 . q
/ < \R(5/32) ' 2 0
Average length = P(3/32) Q(2/32)
(b) [5 pts.] Explain why the tree below is not the tree for a Huffman code. X uni ‘I are. Md
/\&Z(8/16) Wick" “kéwi’ )“tégrw
1/ \1 X" kc!» Pm“ (T‘ /\/\ $11me MM )
V(4/16) W(2/16)X(1/16) Y(1/16) p. MmJ mid W 62M
Math(7030 Maw ' _ r0 64—61}!th .' ...
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