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 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: BALAKRISHNANI’RUNDELL
SEP 24, 2002 ECE 301 Midterm Examination #1 1. Enter your name, student ID number, email address, and signature in the space provided on
this page, NOW! Also circle your section. 2. This exam has two parts. Part I consists of questions for which no justiﬁcation is required. Enter the answers to Part I
in the spaces provided. Partial credit will not be provided for problems from Part I. Part II consists of three problems. Unless otherwise instructed, justify your answers to these
problems completely. Please note that answers provided without justiﬁcation to those prob
lems requiring a full justiﬁcation will be given zero credit. 3. This exam is worth 100 points. You have one hour to complete it. 4. There are 11 pages in the exam booklet. Use the back of each page for rough work, if
necessary. 5. Please work as neatly as you can.
6. No calculators or cribsheets are allowed. 7". You might want to read through all of the problems ﬁrst, to get a feel for how long each
one might take, but don’t worry—~several of the questions are easier than they might appear
when you just scan them. Good luck! IMPORTANT! Whenever a certain space is provided for the ﬁnal answer, be sure to
enter your answer there. Name:
StudentlD#: SO LUTl ON S
Email address: Section (circle one): Section 1 Section 2
Randell Balakn'shnan Signature: Questions for Part I Do not justify your answer. Partial credit is NOT available. 1. (30 points) A continuoustime signal x(t) is shown below. (The signal is zero over the time
intervals not shown in the ﬁgure.) Signal x0) 2 15 1 l .\ 2.
Em : riolxtuli'akt — FELHLM ﬁght M? (b) (2 points) What is the power in x(t) over the inﬁnite interval, that is, what is Pm? A5 Em<oo’ Pm=o Six signals labeled “Signal A” through “Signal F” are shown on the next page. (The signal
x0) is also shown at the top for your convenience.) Match these signals with the following
six signals, and enter your answers in the appropriate spaces in the following table. Each
entry must be a letter from “A" through “IF’ Each answer is worth four points. (c) The even part of x0), that is, xmnu).
(d) The odd part of x(t). that is, xoddﬂ). (e) x(1—t). (f) x(—1—t).
(g) X(1 r 23)
(h) x(2— 2:). Table I
(C) (e) “(D
xevenﬁ) lx(— —t} —1 t) x(l— )2!) (Easil‘j «revifi wk) ——'—————————————I—______._____—‘____ Signal x11) E‘ I I 1 I r
15 
I
0.5—
CI
.0 5' 
.In _
.I_5— _
.2 I I I I I __.___
_4 .3 .2 _1 o l 2 :3 4
Signal A Signal B
2 . I I —I— I I r— 2 I I 1"_“'I I l I
Is— . 15 _
1*  1 .
os . 95. _
0 o
as _ 4,5. _
.I . _ 1. _
.15   .15 _ _
.2 L.._. I I ..._.A .3 I I I .—rL———I——l._.._._*
4 3 —2 — O l 2 3 4 d 1! 2 l U  2 3 ‘
Signal 0 Signal D
2—Ié—II—r———.—_._.—,— 2m . . . .
15  15 
1  1 
05  ua 
D 0
—DE 1 45 ~
l—  —1— i 1
Is  15 _
.2 I I ".4. I I I _2—I___.._I_._. I I I I
4 4 2 — U l 2 3 4 —4 J —2 — ﬂ 1 2 3 Il
Signal E Signal F
2 . I I . I 2 I '—'—1_“"—l—l—l—l—
15  15— 
1 —‘  1 ~
95 I  os— 
o 1 o .
05—  05 
1.  _. _ 
.15 ~ 15
.2 I I I .._I_._..I I I _2_....___._.... I I I I I
—l —:i —2 l 0 l 2 i! I l —3 —2 — O l 2 3 4 2. (20 points) Determine if each of the following systems (with input x and output y) is memo ryless or withmemory; invertible or noninvertible; causal or noncausal; stable or unstable;
timeinvariant or timevarying; linear or nonlinear. (a) System I:
I) —— /1x(r  1):? Ta":
y( D ' (b) System II:
y(_t) = x(sin(t)). Enter your answers in the following table. In each entry in the table, write “Y” if you
can conclude that the property listed on the left holds for the system listed at the top of
the column. Enter “N" if it can be concluded that the property does not hold. (For each
case, there is sufﬁcient data to make this decision.) I l SystemI System [I I
I—_Il:I
I—IIIaI
l—_!
—_'I
I—Im“I
l—IKnl Every entry in the table is worth 1.5 points. You will get two additional points if all the
entries are correct. @ 2.
@ ﬂaw I ‘t
—t
«3H3 _—_, SIG"'08 0“: O MemogLess? 1N0" Beam AaLt) 6'4..th
an. {Item is Cowl} vawHLLa 3? thfle.
"Lt13)
\aL'H =3 Slit) 3 At (Chaﬁ‘
o Vinablcs)
t
QHKXLT) e At ..... (it)
0
(2:3 = ﬂ%Lt)+x_Lt) ; 0‘! CM 3d: 1%)
okt
*  Hz) 4— A Lt)
{m ‘jLt) ULS'Mj XLH 3 g xLJc)=uLt)
The“, usmj mam (51) 8mm anm
\"ﬂ‘" t
at“ _ a": 51006 M: ’ TM? *iKVaviawd' ? ‘NOI Take 1%): 8Ct+‘!2.), 1M «5(1) .— o
SkiH MP»): to ijM by 1, MA CWM 3. (10 points) Circle whether each of the statements is true or false, following these instructions: it Do not justify your answer. 0 A statement is true if it is always true, without further qualiﬁcations. It is false other
wise. (a) (2 points) If y(r) is the output of a linear timeinvariant system for an input x0), then
y(—t) is the output for the input x(—t). Gaga ‘l‘o ‘FWuA CWW‘l‘EYQXAMPLM. True
In f‘abi ) M‘s WM M We Lm ka'. (b) (2 points) For an unstable system, every bounded input x(t) yields an output that is not
bounded. Take, 0.“, LT I WW5 HELI True (False ) § SM) wil'k l“l"*""
L1()u‘l'puti' is bmd U“ ‘9‘“ “3“”) (c) (2 points) If x(r) is a periodic signal, then x(r) +x(at) is periodic for any real number
a. “a” waat be ka'wAﬂ. The (d) (2 points) If x[n] is a periodic signal, then x[n] +x[an] is periodic for any ' eer a.
6‘” False
TlmeL Period» 0} 1, [M Ma at (an) Y‘aHo 0i the. is Tmbmd . (e) (2 points) Let y(t) be the output of a linear timeinvariant system for a nonzero input
1:0). It is possible to deduce the impulse response h(t) from this information. Swﬂou lUcir—i ’6" “M l.“ “e
“m 3H?) .._._. A (6L Cmi'umi') ﬁv 0~M f) n .= A Mr. Cmt’r “km“
Win” A —{ .C‘C kL—lc) ij* Ms 1 Questions for Part II Justify your answer completely. No credit will be given for answers
without justiﬁcation. Partial credit is available. 4. (15 points) Consider a linear timeinvariant system with impulse response _—III_ e’, r<0,
h(t)—e "{e_‘,r20. (a) (10 points) Find the step response s0), i.e., the output of the system when the input is
the unit step function u(t). Lel— w.) um. Lam: gxttucyhtma'c
#063
Jaet) Cm £40“. / / Lm So (b) (2 points) Suppose the input to the system is x(t) = u(I)—u[r— 1). Express the output y(t) in terms of the step response s(t). gm ‘(meavifg l hmeviwaw‘auu, 5w: sLJciSUcI) (e) (3 points) Explicitly write out, as a function of time, the output y(t) that you obtained
in part (b) above. SLt—Ii: e ’
7,—9.0“); Ric030
SUﬂ— SUV!) 1: ('l:
.4)
e .... e , t 4 O
I l' * \
2—e—e”) o t<l
3“) ~17 5. (15 points) The impulse response of a discretetime linear timeinvariant system is given by
h[n] : 5M —5[n—— 1] +5[n—2] —5[n—3] +
= Z(—l)k5[n—k].
1:20 (a) (5 points) Find the step response s[n] of this system, that is, the output when the input
is the unit step function u[n]. LGLMVM 3m = 2 W3 “"3
kzﬁm Mir.) 6 I 1 3 q. 5 ”7K
JLDnL) 0 I 1. 3 ‘I‘ 5 . (“=4 SW L”) 1+ 3W 191 cLuw ﬁnd:
é \erKMLDALJ = 1 CAM “70, Ml, oM SAM: avammﬂ M BobM, Lu}: H’ SMLJ
lac. CW 1104' 2 kg; menL) —_. O
S; O , vx < 0
at“: 1 vx 7,0 MML w
0 I V\ 30 wot mitk 6“ (a) AHemak NATE51L
REM at kfvd = utm [m1 mm + Sva—zlSE‘An
+~ ] = [MEWS  “CMVJ) —r [kin1') _u[“ﬂnl .1. .. . 3va1+ Stu2.1+ EDA—43+ (Came W M Mn) 5 Kb) 100. 5m mun—.3
W1 = Jul dam13* sun2.3.— ans) + .. 3° xth km
2. {BOA} + EDA13h [5Lu1_g(_“_‘3+gc“_ﬂv m) Elm Stu‘3 ADA—2.3 .— SLus) ,f ...
+ ““43  5LWL3 + Stu—s.)   . 1
ﬂ : 6L“) COM “L60 SW Mg (b) (5 points) Find the output y([n] when the input is x[n] : 5[n]+5[n—l]. (c) (5 points) What is the impulse response of the inverse system? Rm» fax/«t Us) kabétEﬁLwl‘t'EUlnmﬁu) S; \waj} : 606+ 50“") 6. (10 points) The Fourier series representation (FSR) of a signal x0) is
x0?) = 2(0.5)keﬂ“.
k=0 (:1) (2 points) What is the fundamental period of x(t)? 030‘“ 1. (L3 iws‘acc’n‘m)
.s 2n: = 27C
T  730 (b) (4 points) Find the Fourier series representation of the even part of x(t), i.e., the FSR
of 1H.) = (c) (4 points) Find the Fourier series representation of the odd pan of x(t), i.e., the FSR of sam=§em—m—m. D \< (that °° t: jL’c
1mm = — 2W») e + 209 5) e
:—90 [(20
5 4:
o —i®6), k<o
(0M) '1 g:
H
O
W
\l
D ...
View
Full Document
 Spring '06
 V."Ragu"Balakrishnan

Click to edit the document details