信号与线性系统课后答案

—P2-1—第二章连续系统的时域分析习题解答2-1图题2-1所示各电路中，激励为f(t)，响应为i0(t)和u0(t)。试列写各响应关于激励微分算子方程。解：.1)p(;)1(1)p(,111,1111)()b(;105.7)625(3102;)(375)()6253(4)()()61002.041()a(0202200204006000fipfpupfpppuifpppppftupfippuitftuptftup2-2求图题2-1各电路中响应i0(t)和u0(t)对激励f(t)的传输算子H(p)。解：.1)()()(;11)()()()b(;6253105.7)()()(;6253375)()()()a(22020400pppptftipHppptftupHpptftipHptftupHfifufifu2-3给定如下传输算子H(p)，试写出它们对应的微分方程。.)2)(1()3()()4(;323)()3(;33)()2(;3)()1(pppppHpppHpppHpppH解：;3dd3dd)2(;dd3dd)1(ftfytytfyty.dd3dd2dd3dd)4(;3dd3dd2)3(2222tftfytytyftfyty2-4已知连续系统的输入输出算子方程及0–初始条件为：.4)(0y,0)(0y)y(0),()2(13)()3(;0)(0y,1)(0y,0)y(0),()84()12()()2(;1)(0y,2)y(0),()3)(1(42)()1(---2---2--tfppptytfpppptytfpppty1f(t)i0(t)+u0(t)-(b)1H11Ff(t)4k6k2Fi0(t)+u0(t)-(a)图题2-1—2—试求系统的零输入响应yx(t)(t0)。解：,ee)(,3,1)1(32121ttAAtypp.0,)e12(1)(121444200,)e()(,2,0)3(.0,2sine5.0)(905.00cos240)sin(cos21cos0,)2cos(e)(,2j2,0)2(;0,e5.1e5.3)(5.15.3312232132323123213,2123213233232132213,213212121tttyAAAAAAAAAAtAAtypptttyAAAAAAAAAAAAtAAtyppttyAAAAAAtttttt2-5已知图题2-5各电路零输入响应分别为：.0,Vsine6cose2)((b);0,Ve4e6)()a(3343xxttttuttutttt求u(0-)、i(0-)。解：;V246)0()0()a(xuu.0)66(1.0)0()0(V;202)0()0()b(A3512)1618(61)0()0(xxxxiiuuii2-6图题2-6所示各电路：(a)已知i(0-)=0，u(0-)=5V，求ux(t)；(b)已知u(0-)=4V，i(0-)=0，求ix(t)；(c)已知i(0-)=0，u(0-)=3V，求ux(t).解：0650650)()(2pppppaZ0.1F61H(b)i+u-图题2-511i1H(a)16F-u+5i1HF16(a)+u-(b)111F1H+u-i(c)1F+u-iH1415图题2-6—P2-3—.0,Ve10e15)(,10,1532050)0()0(',V5)0(ee)(3,232212121322121ttuAAAAAACiuuAAtuppttxxxttx.0,Vee4,4,1,15035)0(',V3)0(,ee,4,1,045045:)c(.0,Asine4)(,2,4sincos4cos04401)0(',0)0()cos(e)(11,11022011110)()b(4214212122121212121212tuAAuuAAupppppptttiAAAAAAAAiiAtAtijpjppppppttxxxttxtxxxtx同理/πY2-7已知三个连续系统的传输算子H(p)分别为：.)2(13)3(;)84()12()2(;)3)(1(42)1(22pppppppppp试求各系统的单位冲激响应h(t)。解：;)()ee()(3111)()1(3tthpppHtt.)()e41e2541()(241)2(5.241)()3(;)()2sine875.02cose8181()(2)2(2875.0)2(8181)(5.1,81)84(1)21()81(8481)()2(2222222222ttthppppHtttthppppHBAppppBpAppBApppHtttt2-8求图题2-8所示各电路中关于u(t)的冲激响应h(t)。解：(a)fupupuuiipuif480422111.V)(e5.0)(125.05.0184)(81tthppfupHtf2+2i1-i11F4+u-(a)—4—.V)()e4.0e4.2()(64.214.06723115.01111311)()c(.V)()e2e2()(221223235.015.01)()b(6222tthpppppppppHtthppppppppHtttt2-9求图题2-9所示各电路关于u(t)的冲激响应h(t)与阶跃响应g(t)。解：.)(2cos21)(02cos42)(21)()(),(2sin42)(21)(21)21(24121121)()a(t_0222tttttdhtgtttthppppppppH),(e41)(21)(2141211212111)()b(21ttthppppppHtεδ.)()ee()(0]ee[)()(),()ee2()(112211122)()c(.)()e211()(0e21)(21)()(22_022121_0tt][tttdhtgtthppppppHttttdhtgtttttεεττεεεετττττ2-10如图题2-10所示系统，已知两个子系统的冲激响应分别为h1(t)(t1)，h2(t)(t)，试求整个系统的冲激响应h(t)。解：求和号后的冲激响应为)1()(tt，于是整个系统的冲激响应为：1F0.5H+u-(b)+f-1310.5F1+u-(c)+f-F13图题2-8+f-1F1F1H+u-(a)+f-1F1+u-(b)1+f-1F1H(c)12+u-图题2-9图题2-10y(t)h2(t)h1(t)f(t)—P2-5—)1()()(ttth2-11各信号波形如题图2-11所示，试计算下列卷积，并画出其波形。.)(')()3(;)()()2(;)()()1(413121tftftftftftf***解：.)3()3(21)1()1(23)1()1(23)3()3(21)3()3(21)1()1()1()1(21)1()1(21)1()1()3()3(21)1()1()(')()3();6()6(21)5()5(21)4()4(21)3()3()2()2(21)1()1(21)(21)4()3()2()()()2();4()4(21)2()2()()2()2()4()4(21)2()2()()()1()2()2(21)()2()2(21)(11411113111211tttttttttttttttttttttftftftftttttttttttttttftftftftftttttttttttftftftftttttttf***2-12求下列各组信号的卷积积分。;)2()1()(,)]1()([sin)()5(;)(sin)(,)(e)()4(;)(e)(,)(e)()3(;)(e)(,)()()2(;)1()(,)()()1(21212212121tttfttttftttfttfttfttfttfttfttfttftttt.)(sin)(,)()()6(201ttTtfnTttfn解：;)()ee()()ee(121)()3(;)()e1()()de()()2(;)1()1()()1(22t0tttytttytttyt--tt--t-t-εεεε图题2-11f1(t)-2021t(a)f2(t)-202(1)t(b)(1)f3(t)024(1)t(c)(1)(-1)3f4(t)-101t(d)1f1(t)*f2(t)01t(1)-4-224f1(t)*f3(t)00.5t(2)312456(3)f1(t)*f4'(t)01t-3-113-1—6—.)()(sinsin)()(sin)()()6(;)]1()2([sin)]2()1()[1(sin)2()1()()5(;)()sincos(e21)(j4)ee()ee(jje2)()ee()1j(j21)()ee()1j(j21)()ee(j21)()4(00211jjjjjjjj2ttTtTnTtnTtTnTtftytttttttftftytttttttytfnn-tt-tt-t-tt--tt-tt-tεπεπεπεεπεεπεεεε2-13求图示各组波形的卷积积分y(t)=f1(t)*f2(t)。解：2-14已知)()1e()()(tttttft*，求f(t)。解：微分：)()e1()()1e0()()e1()()(0tttttftt*；再微分：)()(e)e1()(e)()(0tfttttftt*.2-15某LTI系统的激励f(t)和冲激响应h(t)如图题2-15图题2-130f1(t)t2410f2(t)t241e-t(t)(a)0f1(t)t2e2et1(b)0f2(t)t-1120f(t)t210y(t)tT2T13T4T0y(t)t-1-21210f1()2e2et1t+1f2(t-)2—P2-7—所示，试求系统的零状态响应yf(t)，并画出波形。解：][}{)2()(*)]2()([21)(0ttdtytf)4()]4()2([)2(41)2()]2()([41)2()()2()]2()([41222][}{tttttttttttttt