solutions_test2_white_s08

solutions_test2_white_s08 - ECE 220 Exam #2 Spring 2008...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ECE 220 Exam #2 Spring 2008 Section 002 Kt], ( mm) LAST NAME (PRINT) Student Name: FIRST NAME (PRINT) (SIGNATURE) By signing I am stating that I have taken this exam in accordance with the NCSU honor code. Show all work — no credit for answer only PROBLEM 1 (20) Consider the differential equation and the initial condition given below. Find the total solution for t 2 0. dv(t) + 4v(t) = sin(4t)u (t) dt 1 11(0) ——Z‘ Useful identity: a cos a + [9 sin a 2 \la2 + b2 cos(a — arctan {2—) _L,4: \ JfFlfil’): B $in : C 6 ‘ a ~I - (o mpi‘ Aprilg COS (Wtfe " "(PM (> ‘ >< 2/) tr _ ,_ fl: - flu l2 :2 L— Sfiw A 2 a- {7 i l w {who “ 4‘” "” :3“ “é” W : " i (M «mm M- ig: 2 2 v A, PROBLEM 2 (20) Consider the differential equation and the initial conditions given below. d2v(t) + 2dv(t) a’tZ dt +10v(t)=4e‘3'u(t) v(0)=2 L(0)=0. a. Find the complementary solution, vC (t) , for t 2 0 . DO NOT SOLVE for constants. Your answer should not contain any complex terms. b. Find the particular solution. (X) 11+33+l020 73:2: —l%»j~» MM»): (C859, é{jg—(+33) Jc-Jr C636” €C~l~33l~)m(+) vow QC 6+ (03 ( 3+ +9} u M} b) We) ; % e’y‘um ”3»~>>“ 33 +2_(r:;‘:13+ “33 :17 13 3 L so A V’HSJFUCH» PROBLEM 3 (20) The relationship between the phasors of the input signal vs (t) , and the output signal v0 (t) is given by the fa) t 6 V5 (to) complex function Vo(a)) = . —a)+ 160 Find the output signal v0 (t) , when the input signal is vs (2‘) = 55in(21:10t —Jr/ 4) —10cos(2:rlOt + 15/ 4) r\/‘ .I _4LF -lC VS .— 58,011,€OL__ l0 (—337 {1317f _ “Legal.” n. no ,L i. ’5< (I. 3v?) (\{iflr 5%? “I, 53$ 4, = “if; __ L2. : '58 “7 V1 V; 1. 2:1: 1th -31 V a“ , lBE / 0 :- MW"WW"“MW"W m“ grflOC—l + j) ~$rr 3r- N - ~ ~,L' V : % Zflf 6 3 L! ” 3 6’0 2. O in: a}: C 33%? 177W; Z or VLH), 3 “942%” ~ '3: 1 PROBLEM 4 psi For the differential equation and the initial condition given below, find and graph the particular solution. dv(t) dt v(0) =1 +0.5v(t) =u(t) —u(t—2) PROBLEM 5 (10) The driving force and the total solution pair given below corresponds to a first order system. Give a differential equation representation of this system including the initial condition. vs (I) = 2e—4’u (t) —2t —41 v(t) =e u(t) —e u(t) OCH) :: 6” u (if) PROBLEM 6 (20) Consider the differential equation and the initial conditions given below. dv2 (t) + 2 dv(t) a’t2 dt v(0)=1, i2(O)=2 + v(t) = cos3(21rt)u(t) nd a) IdentifythetypeofDE: thxc’d‘f/ CaninA/‘l (Dc/{4, I 2 are)? 1'" fiD’l-L‘Wo “ b) Determine the value of the total solution v(t) for t = 0.2 using Euler’s method with a step size of h = 0.1 sec. 0 ' o 'X H xm- X‘H) - V“) if' H) : I I )i fi 32% m] " xza) ’ 3; 9} X2“) 4 v2. XLH) (05 q xHW) O l N”) l ifih 4% {want i, X2“) Xwii'l‘t‘l") «l ~2 MM) 0 1 L7. 0 1 1 fl x‘ (0“ -o'l + 04 a, l X (0 l) a /( ~Z Z (“wall-1 Z 1, E i Z ’ >< (0’2) :(OJHLQ) +<O.l)(o) + L 2 _—.- (-36 l X‘KGJ): '“*y*'[o,2) ...
View Full Document

Page1 / 7

solutions_test2_white_s08 - ECE 220 Exam #2 Spring 2008...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online