HW8-Key - 3-1 3-2)2(2)23(22(22 =ssssYand the original ode...

Info iconThis preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 3-1 3-2 )2(2)23(22)(22+++=ssssYand the original ode was tydtdydtyd2sin222322=++with )()(==′yyb)This is a unique result. c)The solution arguments can be found from )2()2)(1(222)(2++++=ssssYwhich in partial fraction form is 221)(22121++++α++α=sasasssYThus the solution will contain four functions of time e-t, e-2t, sin2t, cos2t3.3 a)Pulse width is obtained when x(t) = 0 Since x(t)= h – at tω: h -atω= 0 or tω= h/ab) x(t)= hS(t)– atS(t)+ a(t -tω)S(t-tω) x(t)x(t)hslope = aslope = -aslope = -a3-3 c) 2221)(seshsaesashsXstst-+=+-=ωω--d)Area under pulse = h tω/2 3.4 a) f(t) = 5 S(t) – 4 S(t-2) – S(t- 6) =)(sF= ( 296s-2s-e-4e-51sb) x(t)=x1(t) + x2(t) + x3(t) + x4(t) = at– a(t-tr)S(t -tr) -a(t-2tr)S(t-2tr) + a(t-3tr)S(t-3tr) following Eq. 3-101. Thus X(s) = [ ]stststrrreeesa3221---+--by utilizing the Real Translation Theorem Eq. 3-104. x(t)-a-ax2x3tr2tr3trx1x4aa3-4 3.5 T(t) = 20 S(t) + 3055tS(t) – 3055(t-30) S(t-30) ( 29ssessessssT302302211305520130551305520)(---+=-+=3.6 a) 432)4)(3)(2()1()(321+α++α++α=++++=sssssssssX1)4)(3()1(21=+++=α-=sssss6)4)(2()1(32-=+++=α-=sssss6)3)(2()1(43=+++=α-=sssss463621)(+++-+=ssssX234and( )66tttx teee---=-+b) )2)(2)(3)(2(1)4)(3)(2(1)(2jsjsssssssssX-++++=++++=jsjjsjsssX2232)(333321+β-α++β+α++α++α=81)4)(3(1221-=+++=α-=ssss132)4)(2(1322=+++=α-=ssss20811384021)2)(3)(2(1233jjjjssssjjs+-=---=-+++=β+α-=3-5 tteetxtt2sin2081122cos2083213281)(32+ -++-=--tteett2sin104112cos10431328132+-+-=--c) 2212)1(1)1(4)(+α++α=++=sssssX(1) 3)4(12=+=α-=ssIn Eq. 1, substitute any s≠-1 to determine α1. Arbitrarily using s=0, Eq. 1 gives 1or131141212=α+α=2)1(311)(+++=sssXand3ttx( t )ete--=+d) ( 29222214321111)(ω++=++=++=bsssssX13whereand22b=ω =tetetxtbt23sin32sin1)(2--=ωω=e) X(s) = sessss5.)3)(2(1-+++To invert, we first ignore the time delay term. Using the Heaviside expansion with the partial fraction expansion, 32)3)(2(1)(ˆ++++=+++=sCsBsAsssssXMultiply by sand let s→0 3-6 A= 61)3)(2(1=Multiply by (s+2) and let s→-2 B= 21)1)(2(1)32)(2(12=--=+--+-Multiply by (s+3) and let s→-3 C= 32)1)(3(2)23)(3(13-=---=+--+-Then 33222161)(ˆ+-+++=ssssXtteetx32322161)(ˆ---+=Imposing shift theorem )5.(3)5.(2322161)5.(ˆ)(-----+=-=tteetxtxfor t≥0.5 3.7 a) 221226)1()1(6)(sssssssYα+α==++=661222=α==α=sss26)(ssY=b)9)9()2(12)(23212+α+α+α=++=sssssssY3-7 Multiplying both sides by s(s2+9) ))(()9()2(123221ssssα+α++α=+or 132219)(2412α+α+α+α=+sssEquating coefficients of like powers of s, s2: α1+ α2= 0 s1: α3= 12 s: 9α1= 24 Solving simultaneously, 12,38,38321=α-=α=α91238138)(2++-+=ssssYc) 654)6)(5)(4()3)(2()(321+α++α++α=+++++=sssssssssY1)6)(5()3)(2(41=++++=α-=sssss6)6)(4()3)(2(52-=++++=α-=sssss6)5)(4()3)(2(63=++++=α-=sssss665641)(+++-+=ssssYd) [ ])2()22(1)2(1)1(1)(2222+++=+++=ssssssY= 2)22(2252243221+α+++α+α+++α+αsssssss3-8 Multiplying both sides by )2()22(22+++sssgives 1 = α1s4+ 4α1s3+ 6α1s2+4α1s+ α2s3+4α2s2+6α2s+4α...
View Full Document

Page1 / 26

HW8-Key - 3-1 3-2)2(2)23(22(22 =ssssYand the original ode...

This preview shows document pages 1 - 9. Sign up to view the full document.

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