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

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

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