This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: EECS 20N: Structure and Interpretation of Signals and Systems MIDTERM 2
Department of Electrical Engineering and Computer Sciences 11 October 2007
UNIVERSITY OF CALIFORNIA BERKELEY LAST Name Age—{L FIRST Name H I ”’1 0 Lab Time ~ [3 l o (10 Points) Print your name and lab time in legible, block lettering above
AND on the last page where the grading table appears. 0 This exam should take up to 70 minutes to complete. You will be given at
least 70 minutes, up to a maximum of 80 minutes, to work on the exam. 0 This exam is closed book. Collaboration is not permitted. You may not use
or access, or cause to be used or accessed, any reference in print or electronic
form at any time during the exam, except two doublesided 8.5” X 11” sheets
of handwritten notes having no appendage. Computing, communication,
and other electronic devices (except dedicated timekeepers) must be turned
off. Noncompliance with these or other instructions from the teaching staff——
including, for example, commencing work prematurely or continuing beyond the
announced stop time—is a serious violation of the Code of Student Conduct.
Scratch paper will be provided to you; ask for more if you run out. You may
not use your own scratch paper. 0 The exam printout consists of pages numbered 1 through 10. When you
are prompted by the teaching staff to begin work, verify that your copy of
the exam is free of printing anomalies and contains all of the ten numbered
pages. If you find a defect in your copy, notify the staff immediately. 0 Please write neatly and legibly, because if we can’t read it, we can’t grade it. o For each problem, limit your work to the space provided specifically for that
problem. No other work will be considered in grading your exam. No exceptions. 0 Unless explicitly waived by the specific wording of a problem, you must ex
plain your responses (and reasoning) succinctly, but clearly and convincingly. 0 We hope you do a fantastic job on this exam. MT2.1 (40 Points) The block diagram below shows an architecture for implement
ing an amplitude modulator using signal adders and squarelaw devices (SLDs).
The realvalued signals x1 and x2 are the inputs to the amplitude modulator and y
is its output. AMPLITUDE MODULATOR Each squarelaw device is characterized by the following parabolic inputoutput
graph, where xsu) denotes the input to the SLD and ySLD the output. The parame
ter K is a positive constant. ySLD K 2
YSLD(t)=K XSLD (t)
0 XSLD
Formulas and Facts of Potential Use or Interest:
cos(a + B) 9 = 00804 cosﬂ — sina sinﬁ
cos (1 cos B = %[cos(a + 3) + cos(a — 3)] (a) Show that if K = i, the amplitude modulator output y is characterized by
W) = 5610000205), Vt e R. 3: Klb‘ﬁxaf— @4311 :lKX‘X’Z 2 :5 d : X‘X‘Q‘ K: i
L\— (b) Select the strongest correct assertion from the following choices. Explain your
reasoning succinctl , but clearl and convincinl .
(I) . The am ulitude modulator above must be memoryless.
(II) The amplitude modulator above can be memoryles.
(III) The amplitude modulator above cannot be memoryless. Tilt SailEM \sz V\c> mamoﬁ‘ér Qlﬂmﬂﬁk’. T3 SLD ”‘3 QO00 \{gg lDQCMSQ ITS (“pr IS ‘lﬂs mm mngowgl Medal l3 \lS owvvk’x;l\olrlers 0w: memoralQSS,
A\Sq “CAVQ ﬁCCY Xt§1] l8 we, («\Swf oﬁﬁ SEEM 3(f\:£(xa’l):x(+\)lllf\ (c) Select the strongest correct assertion om the following choices. Explain your
reasoning succinctly, but clearly and convincingly. \ (I) The amplitude modulator above must be causal.
II The amplitude modulator above can be causal.
(III) The amplitude modulator above cannot be causal. Even memorale SS % 63—th \\ 3 waswl .
(D0 noTQ,\\0on\/Vcr) Tem'f ﬁg CWNU‘SQ is vxollfxzxt» (d) Select the strongest correct assertion from the following choices. Explain your
reasoning succinctly, but clearly and convincingly.
\ (I) The amplitude modulator above must be time invariant.
(II) The amplitude modulator above can be time invariant.
(III) The amplitude modulator above cannot be time invariant. EROﬁ "\QMota\C$S EAEM (ms out: AC§CAQ Tlptﬂ as) Q ox‘CSC) '\ '3 Moi \«woxcwwfl’. xlt) X
Y9K CR“ «\39 «Pakt “2&5 wad 5.5L): .1 >3€H=Nm ,zlﬂ LJ im = 5 m ljﬂl ) as 5m Amgm letﬂxatt—t.) : altt) \\
V, M
is m (e) Select the strongest correct assertion from the following choices Explain your X Q“)
reasoning succinctly, but clearly and convincingly X M‘ L: "U‘ ——> xx GIL ‘3 :zml 73~ «my
XL") (II) The amplitude modulator above can be linear. X :(Jfl'jsafsmgN a M“ “1M5 (9) III The am litude modulator above cannot be linear.
EZ:::::\ gum  .. x 1311a,. 31“ (f) Suppose the input signals x1 and 202 are sinusoids characterized by instanta
neous values (I) The amplitude modulator above must be linear. x1(t) = co‘s(w1t) and $205) 2 cos(w2t), Vt, 1
where O < wl < %' For simplicity, assume K = —. (i) Provide a welllabeled sketch of the spectrum of the output signal y. Be
sure to explain your reasoning succinctly, but clearly and convincingly. All/B: Cos «Ad'— cns w;T :— %{C03Km1+wﬂ‘ﬁ+ C03‘(W:~_w‘)f}i
X; y‘t yq—t 5.
~45 » Mw w w J~ a2  W141?) (ii) Prove that the output signal y is periodic if, and only if, the ratio of the
frequencies cal and tag is a rational number, 1. e , ?Q(‘\\ 9 A Yd. ?€m9c\1‘ _ E Q
W ' T‘Q SKM 9'? 1200 VKNMLQ Cl “Lb«4&8 \S QV’WQ‘X V9 «Kmi Oﬂglgr)
ﬂaw 9C3£>Q€lﬂyﬁ ”gaklcche “tack Fitmo‘ks Mt ”Emmi 1;le “ES 0’8”
{AAA 0% Y9; .... m M '74.,
l) _ g“ MAS—l. how“? if?“ I: )n EN >
0K. U’ \ I 4
{W\ X Wa—w“\ _ [VAw» ﬂW~QW:Mqun/\J
f _ 2“ ﬁrm ﬂ,—~B I n a \ ‘1 \
F“ urtJ\ 13?“: «x “Elﬁn
A WE’VKN W n—M 6Q
"—'=—\ l““"‘\ Wat” KMMW 1': 73"”? “T“ I MT2.2 (35 Points) The following diagram shows a combination lock. 4 The lock can be opened only if an ordered sequence of three numbers—selected in
strict adherence to the following steps—matches the lock’s unique ”combination”: Step I. Turn the dial clockwise two or more whole turns, and stop at the first num
ber of the combination. Step II. Turn the dial counterclockwise one whole turn past the number in Step 1,
and stop at the second number of the combination. Step 111. Turn the dial clockwise and stop at the third number of the combination. The combination for this lock (i.e., the only sequence of numbers that opens it) is
(1, 3, 5). This means that to open the lock, a user must stop at 1 at the end of Step I;
stop at 3 at the end of Step II;“and stop at 5 at the end of Step III. The thick vertical arrow at the top of the diagram is fixed, and it marks the number
selected by the user at the end of every step. The combination lock can be thought of as a mechanical system, where the ”input
signal” is the sequence of three numbers x 2 ($1,332,553) 6 {O,1,2,...,7}3 selected by the user who rotates the dial according to the steps and rules described
above; needless to say, the sequence of numbers selected by the user may or may
not match the lock’s combination (1, 3, 5). The ”output” signal y = (y1,y2,y3) shows the the sequence of states of the lock cor
responding to the input sequence (3:1, 332, $3). The state of the lock upon completion
of each step described above is either Locked (L) or Unlocked (U). By way of example, if the input signal is (1, 2, 4), the corresponding output signal is
(L, L, L), which means that the user has failed to open the lock. If, on the other hand,
the user applies the input signal (1, 3, 5), then the output signal will be (L, L, U),
which means that the user has succeeded in opening the lock. We can describe the combination lock by the function CombinationLock: {0,1,2,.. . ,7}3 —> S {(L,L,U) if 1:: (1,3,5) 3/ Z (L, L, L) otherwise. The set S is something you will determine below. In tackling this problem, consider only dial rotations that conform to the rules
described above. (a) Determine the size of the input signal space; that is, determine how many
valid input signals ($1, 3:2, 2:3) exist. 3 ‘ 3 . K  A
\gé)’\)"‘)7§ l A: 8 E4CL\ XL“ \f\a\$ 8 (DOSS\\0(\/tj(ES .
‘ (24,255) Also, determine 5 so that the description of the system is an onto function.
What is the size of 5? Tight are gﬁ\&_Two ()ossiL\Q 3Q \KivchS A; 3&3 1 LLJLIL) ““J‘ LL‘)L)\A>‘6
3—3,th law) (b) Select the strongest correct assertion from the following choices. Explain your
reasoning succinctly, but clearly and convincingly. (I) The combination lock must be a memoryless system. (II) The combination lock can be a memoryless system. ‘ (III) The combination lock cannot be a memoryless system. ! [ntadlﬁxml 1 \SQ knew) ﬁoj'ﬁﬁ \ché va's‘l— keg? Jﬁ‘aclg all
WC SQTWCWCQ cl" TRQ Co alts ew‘l/thol, 39 ll, mwsj how“: Mew/Leta. Vloft, CDAQPC—l’ﬂ\ ) CéoSlOle Flaﬁ {Bl/lowing l/NQ My“:w NRC/l
031st Si WQQRQQ ealFS‘. A \va’ /\
1 Q” 3:5,) w? (L)\~)U\\13‘ K31X$ H‘s—#65 X i9 t ‘H,
32: 032)?) “—3? LL)L)\—):3 50 tﬁqsgm Ck“ Q me mocﬁ e33 ~ (c) Select the strongest correct assertion from the following choices. Explain your
reasoning succinctly, but clearly and convincingly. l (I) The combination lock must be a causal system. l (11) The combination lock can be a causal system. l\ J— (III) The combinatlon lock cannot be a causal system. i’ A ‘ ‘ ‘
. mamas w wt . mwew ..T.,..i...\.ig lﬁ‘ilr Y‘QSFchlVE’RFSTinQsX‘ «we X, ) fmt’thCQ Q‘Nﬁ‘ﬁw OWE lt‘lﬂm‘lTCA
. . Mi . A
W? Ar") °"\°\ "\dwollma} «bf ms‘necl—Iwc hrlidﬂi El «Ml 3‘ 1
'\ \ alO‘l‘ﬂc‘m :31LL PK? *) y
3\ 3 .
a EV“ 0 in . N g“ n .. 03‘ “A A ﬁrst ox? \olechA m t «Allncl‘vllm,
Q a El VGA V F x rlx X ) Ycoclulce 0&1?le amp ’\
“ﬂair wesﬁe’d—(VQ SQLonoX Qr{\r(‘\‘€,§, * Xa mmA X92
6 «VAX/«Apt lkﬁwﬁi (Lax ugh) «WA lAC\U\l>l:v\ )TRQ;(‘ VQSQQLWQ $6CQQ1€IXV$QS
El “AA 1 3 L L W A .; Q. L
l J all?» 5.1V 3 . . .. a 3, ~ m x
“ No (MW Q'l’ Aisﬁacl'1wi3?s Cam lea {olewﬁéﬂx‘rx MAIEWQQ Qw—lﬂQS E;‘ "5:, 3:31 MT2.3 (30 Points) Consider an ideal interlocking pair of rotating mechanical gears
shown in the figure below. By ideal we mean that you can ignore friction and slip
page between the gears. —R2 0 Y“) R2 The respective radii of the two gears are shown in the diagram, and are related
according to 0 < R1 < R2. The smaller gear is the driver. The projection x can be thought of as the input
signal to this mechanical system. The larger gear responds to the rotation of the smaller gear. Accordingly, we can
think of the projection y as the output signal of this mechanical system. The smaller gear is shown rotating clockwise with angular velocity cal radians per
second; the larger gear rotates in the opposite direction. At the contact point P
where the two gears touch, they must have equal tangential velocities. Therefore, wlRl = (.ngg. An inputoutput depiction of the mechanical system is shown below. X Fay The instantaneous values 55(t) and y(t) of the input and output signals are shown
in the figure and are described by $0?) = R1 cos(w1t) and y(t) = R2 cos(w2t), Vt.
In answering the following questions, assume col = 7r / 3 and R1 = R2 / 2 (a) Select the strongest correct assertion from the list below. XU’) _. .—— c a 5 C21“) (i) F must be a timeinvariant system. 3“,.) 1 R ,1 co 3 Gill“)
(ii) F can be a timeinvariant system. \ (iii) F cannot be a timeinvariant system. 3
———\ xtﬂ "(03(L(+'é§) :P:C 043:6: X(+> Lél' Qbﬁ 1 x (l1 (a) 1’9 5853?“ wereﬁ,ﬁ&ﬂi\ a U2) “(1’43 R2 C95 (13%)): ~€l COSEV\ ‘\
Bud'siace 1‘ :5 \“ 363$“ W\S\\$\Q§§0m >9“:er gmws‘l'loe IJCvCchiJFé \hea‘t no . (b) Select the str gest correct assertion from the st below (i) F must be a memoryless system. (ii) F can be a memoryless system. A’ \s a new; 3 «6‘
(iii) F annot e a memoryless system. ”WT—X ’==B __‘ M
X(0\) : ley "z T; Gol’ TI) QQ’VMQMM less) at) = Ea + 4%; — 300) (c) Select the strongest correct assertion from the list below. «A3
(i) F must be a causal system. , EV 6'“ "8‘ 0 “Q “L {’ﬁe :6 'l’ 0
aﬂqw (‘C \Ook‘ \’\ G
(ii) F can be a causal system. Wfilcwi/s 565th Q’QJ’ (iii) F cannot be a causal system W& 0 T— $6§\/QW\ ‘ 5 CMWMCKEYWLQA bﬁﬁ 9&«r aft CC TEC Mr
S\&:: earmmwiiér ‘5‘ \S S QCJ \{JI TEE QNJYFQ 0 kai/ :3r 3 R0 \ch . \Té {hm “QQA nq‘l’ lcék Oslneocell vv 0 lee )“ “i7
l) a l 5? A o Golfer‘s/MAE Q 9W1? “ﬁr
G f A if Jﬁlbvw/DS lﬁaj’ a You Come look C :17 l‘Rls a; A5 we.“ we owﬁ 7:10 inf
ME (ital/(CA “~00 l/O 0:?wpfa::\?j:§ja ~lr/A/(e N”? cos 9J1“ AAA Cos («Kl/3 ﬁt FMJWCE ”(an/IL ;% LAST Name AL FIRST Name M ’ ”I 0
Lab Time ?! 10 ...
View
Full Document
 Spring '08
 EdwardA.Lee

Click to edit the document details