A#05_Solutions - Assignment #5 m Solutions « 9.1 ECSEu24I...

Info iconThis preview shows pages 1–5. 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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Assignment #5 m Solutions « 9.1 ECSEu24I 0 Signals & Systems -— Fall 2006 Due Fri 09/15/06 1(30). Find the equation and sketch the resulting convolution, w(r) = a(t) * b{r), where a t) 5(5) I} 2 I I 1 - l -l 0 at? (W 3 "TH ‘3” ~§<a+tw£2~14t<~l Fm Amati-<0 a: 6944:“ Assignment #5 ‘ Solutions — 13.2 BOSE-2410 Signals & Systems - Fall 2006 Due Fri 09/15/06 1 Cont. Find the equation and sketch the resulting convolution, w(t) = (2(1) * b0), where :20) 5(1) -i 0 l -l 0 i MATLAB 13101:. t:{u3:.01 :3]; E1 zi>:(w2)&t<m(- 1); g22t>(-l )&t<} ; g3 392} &t<=2; wm(0.5*(2+t)."2).*g1+(l~.5*t."2).*g2+(0.5*(2~t)."2).*g3; plot(t,w,‘LineWidth‘,2); grid Assignment #5 -— Solutions — 13.3 ECSE~2410 Signals & Systems - Fall 2006 Due Fri 09/15/06 2(35). The input-output relationship of an LTl system is govemed by the differential equation, 53:39 + 2w) m 2x(t) . x0) LTI W) dz Assume the system is causal and is in a condition of initial rest, so that y(0) = 0. (a) Solve this differential equation when the input is a unit step function. Recall that in classical solution of differential equations, the step input is treated as a constant of unity for z 2 0 . (b)Using the results from (a), find the system impulse response, [1(5), ((3) Find the step response of this system by convolution, y(t) = x(t) * 110) , using the Mt) found in (b). (d)Is the following true or false? Given the system impulse reSponse, the step response found by solving the differential equation is the same as the one solved by convolution. in other words, convolution is nothing more than the particular soiution of the differential equation! (a) w “3&1 M. +‘Zig “2.. 2..) “tea: at éfi:§g $§gqg flaking) (9+Zlam2, "A? [cal “flak if?" Sui-i:- Ae H W we M tweeter?” W ts __ ) i299 $3239 z: Chef} ‘ 4- aézfiérzi 4262,21“ l a M) d'i: (Cl $93? Mime. use 5min: idem “3 "I" 2: “gm 1%}. {2% t “if ‘ -2”? has“ “at i; i T 3543539: 23 irwe *- e a!) new} 6 r W L “l” _. ézéjusi “Sana/a £02 9 t: fittiqfi'); {5‘} aka“ [Q etc“) WW W; ‘i’éfia MSW l5 Assignment #5 - Solutions —- p.4 ECSE—2410 Signals & Systems - Fall 2006 Due Fri 09/15/06 6131(3) 3(35). Solve the differential equation, r +2y(r) = sin(z), t 2 0, with y(0) = 0, using classical techniques. Express your answer in the form, y(t) m A?“ + B sin(t + 6) ‘ You need to find the numbers for the unknown constants, A,a,8,6 . Express the phase shift, 9 , in degrees. -gfi léewgwm Sal‘s- ‘jl‘ts—a Aa taro J $45“;- :‘ KQSg'fl-{rdr k; Lafit— {gFaJ—alésfit lg @1935 Sal first“? WM ' s {a ole“ la; cost - KZ mist + zgisgst 4; zgawgf- LESFWR C03 Wag C;m{"&e{_'fis-:W)j a s.an gmfiweml; Kl+2¥L$O '2 4 “flip gamut? Semi-«s. Lie/126+ "6: V2.) Ki: 1.5.2 ll, 5 what-2K5: C his. gals: ME gag: Aézt+ 25.3134: «- 3§ 303% J __, 3.3..— ..atgs. yam Amfifgm m? A 5 O! ugh r— t “l, e: Kw \ WINES 259ml: gag—b Skfiwggwwttfimg Assignment #5 — Solutions ~— {3.5 ECSE-24l0 Signals & Systems - Fall 2006 Due Fri 09/15/06 3(35). Continued. Solve the differential equation, + 2 32(1) 3 sin(t), r 2 0, with y(0) m 0, using classical techniques. Express your answer in the form, y(t) m Aewo” + Bsin(r «t 49) . You need to find the numbers for the unknown constants, A, 0:, B, 6 , Express the phase shift, 6, in degrees. The solution is y(r) = 0.23""3‘ + 0.45 sin(r — 26.6”): 0262—Zl + 0.458in(z “— rad.) Need radians for piotting. steady — state term y(t) = 0.45 sin(I — 3i“— rad.) 57.3 transient term y(t) x: 0.2% complete solution y(t) m 0.29% + 0.45 sin(r w rad.) t={0:0.01:2*pi]; y1=0.2.*exp(—2_*t); y2=0.45.*sin(t—(26.6/57.3)); y=0.2.*exp(-2.*E)+0_45.*sin(t-(26.6/57.3)); plot(t,y1,':‘,t,y2,'-—’,t,y) grid ...
View Full Document

Page1 / 5

A#05_Solutions - Assignment #5 m Solutions « 9.1 ECSEu24I...

This preview shows document pages 1 - 5. Sign up to view the full document.

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