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Unformatted text preview: ECSE2410 SIGNALS AND SYSTEMS FALL 2066
Rensselaer Polytechnic Institute EXAM #1 (1 hour and 50 minutes ) September 20, 2006 NAME: Section: 1 2 Do all work on these sheets.
A Table of Integrals and Identities will be handed out separately. One page of crib notes allowed. Calculators allowed.
Label and Scale axes on all sketches and indicate all key values. Show all work for full credit. Total Grades for Exam: Grades formg Points L Score J
J 31
10
5 T
7
10
7
6
T 6 10
8 F \DOOQC'AUlbDJNé—t 3—:
CD TOTAL 100 Exam #1. p.1 Label all key values! @ (b)(5). Express x(f)ﬂin terms of step functions. Any (vaiid) form of the expression is acceptabie. «my: “(2 maman) + gig— 3)<ML£~1)~u(£39 Exam #3. p.2 1‘ Continued. Given the signal shown, dxm . . . . lmﬁé’
(c)(8). Sketch (4 pts) y(r) m d: AND ﬁnd the equation (4 pts) of the derivative 1:: terms of stepﬁfunctions. (This problem can be solved in any order.) walllag a? M 36H =2: (Mm illii“l)  gates) ~ 2 SM) Q5 +31% wlv‘afilkél c2 29:!)
NH: 15 LLU—D" iﬁﬁlu‘ﬁéﬁj» 2u[£~[) $663)LLLﬁg) <3: 1
jalm 3% «r. (m) a... gutw  2 $6M) * : “(#3)
i . Scale vertical axis! Label all key values! Exam #1. 9.3 N '1# mm isanIeA [(9)1 1113 {9:121 '{WW m (DA 1401938 {911315 em HQAID (9)63) gsanim ((9:31 He {aqerl ‘(Ssd :2) [Isms CINV (>10 5? mm; mm K119) 2. (20(5) Find the equation for Mr) = Isin(r)5(r m @417. Justify answer. 7‘1} :2 Sta“ waxy?st Swiftﬁ Frnféfugﬂ
“Q” : ==~ g‘téttwwd’?‘ (31:2 Cpmgu 6’ 49% 5}?
a“ ewem 1172,) (b)(5) Find the value of ]si§_gr) 005$?er . Justify answer. Sammﬁj 0E4 W [MIJPS
27f ; {MM/£2: :0 U
«— 231’ value: @ Exam #1. p.6 3(5). Find c[n} : a{n] *b[n} , where a[n] m {1, g,1, 2} and b[n] = {1,1,;, 2}. Express c[n} in sequence form. “10am Exam #1‘ p.7 52(t) 60)
4(7). If a(r) *b(r) = C(t), where a(t) is and the convolution C(t) is ,
i G l «I 0 l sketch the sigma} b(r) . Justify your answer. We. 14%qu ab?) $3)
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Mm Scale vertical axis! f Label all key values! Exam #1. p.8 5(10). if am * b(r) = ca), where am is ﬁndthevalueofr,1<t< 5,30 that y(r) :0. H9?) Exam #1. p.9 6(7), When the input to a LTI system is x10), 1 Sketch the input signal, x20), that will produce the output, y2 (1‘), shown.
Justify your answer! «ch w) 0 , 2 3, I 1 M9 ﬁ
.4 m M», (
l f e J: Tam NLJFHMH =— 51M} Tb reﬁtﬁwﬁ
2959 a 2mm) Labs} all key values! Exam #1. p.10 7(6). An LTI system has an impulse response of Mn] 2 uEn}. Find the equation, in closed form (i.e., no
sequences) of the step response, ysrepm], ofthis system. (SHAPE; Q4 Mitt? 5M fofﬁiéﬁllé . ﬂ 3"! K Na K
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L» 4.: J mph] = <24“ (if) “5‘1 Exam #1. p.11 8(6). The LTI system shown 36(3) = e“,t 2 0 LTI W) is governed by the differential equation, 62:7?) ~+« 2 y(t) m 2x0), 2 2 0, with 31(0) 2 0. Find the output y(t) when the input is x(r) = e”, t 2 0. Use the ciassical approach for solving the differential equation. 43th mt. n+0 eta 22> “K +ZKg/m 2;?
Km; mat ...t: Mi: _ 1..
y(t)= ~26 +26: 2 “(:39 CK 359:2;(8 are? )RKU Exam #1.p.12 9(10). The input—output relationship of an LTE system is govemed by the differential equation
dyU) d:
input, with y(0) = O, the output gradually increases and at t m 0.5 see, it reaches a value of 1m 9“ = 0.6321, i.e., y(r) #05 a: I — 6". Find the value ofthe unknown constant, a. x(t) = u(t) y“) $40": gig die. if} fwd“? 3/3330. i... “E
gittts/iéla we ay(t) = axU) , Where the constant a is unknown. When an unit step function is applied to the Exam#1.p.13 10(8). A linear, time~invarianL discrete time system is described by the difference equation, y[n} mg y{n ml] = x91], with y[n} = 0 for n < 0, and speciﬁcally, yH] m 0. Use the classical solution approach lo ﬁnd the step response; ympﬁ’l]. Express yé,,ep[n} as a dosed form equation. No sequences. We Ag)“
#5:“: 3%: ‘> SM WW? «:5? his ==~l ar> krz.
i ..,., szﬂzgjuZ “l , , ?Ahﬁi End! Exam #14114 ...
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 Spring '07
 WOZNY
 LTI system theory, Impulse response, Step function, key values

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