1、(1+2i(t)=+cost cos3t+1+(Q)21+()2Q2ATn=1i(t)=cosnt)sin(n/2)n/2书P200题4.2=T/2A 2A 2A 2A2 3 5cos5t+L10H(j)=1+jQ(0)0=3)|H(j)|=|101+jQ(0183|=|)|H(j5)|=|1501+jQ(0511615=2A 13 1002A2A52A 13 100Q 57Q 112.5|H(j3)|=1书P200题4.9a(t)=A(1+0.3cos1t+0.1cos2t)sinct3 3 6A=100V（1）部分调幅系数对调制频率1：m1=0.3对调制频率2：m2=0.2（2）调幅信号包
2、含的频率分量a(t)=(100+30cos1t+10cos2t)sinct=100sinct+30cos1tsinct+10cos2tsinct=100sinct+15sin(c+1)t+15sin(c 1)t+5sin(c+2)t+5sin(c 2)t所以该调幅波中包含有五个正弦分量振幅为100V的载频分量；振幅为15V的第一对上下边频分量，其频率分别为C 1 ；振幅为5V的第二对上下边频分量，其频率分别为C 2 。a(t)=A(1+0.3cos1t+0.1cos2t)sinctA=100V（2）绘出调制信号与调幅信号的频谱图；a(t)=A(1+0.3cos1t+0.1cos2t)sinct
3、A=100Va(t)=100sinct+15sin(c+1)t+15sin(c 1)t1 2100151555C+1CC+2+5sin(c+2)t+5sin(c 2)t幅度谱30102 1幅度谱频带宽度为1频带宽度为211 U m0=5W 1 Um1 m2 15 521 U1 Umax=9.8W载波功率1 100221000a(t)=A(1+0.3cos1t+0.1cos2t)sinctA=100V（3）此调幅信号加到1k电阻上产生的平均功率与峰值功率，载波功率与边频功率。a(t)=100sinct+15sin(c+1)t+15sin(c 1)t+5sin(c+2)t+5sin(c 2)t边频
4、功率平均功率峰值功率1(100+30+10)22 1000P max22 R=s2 2 2P=2 +=+=0.25W 2 R 2 R 1000 1000P=P c+P 旁=5.25W22 RPc =f(t)e dt F(s)e ds复习st+jj12jF(s)=f(t)=L 1dstF(s)=Ldf(t)=e (t)e t(t)积分(t)(t)t(t)tn(t)11S1S2 n!Sn+1-t-t n1S+n!(S+)n+1sint(t)cost(t)esint(t)-ts2+2ss2+2(s+)2+2ecost(t)ts+(s+)2+2f(t)=Kiesit(t)=(ss )F(s)e Res
5、k=(n1)!ds(s sk)F(s)e L =F()ds tstks=skResks=skn stn1 1 d n1dF(s)dsLt f(t)=ss f(t)Kis sini=1F(s)=ni=1Ki i s|=sikifi(t)k F(s)sF(s)f(0 )f ()d f()d+eF(s)i0线性微分积分时移ni=1df(t)dtt f(t t0)(t t0)ni=1F(s)s sst0s0tf(t)eF(ss0)频移拉氏变换的基本性质（1）a lim+f(t)=f(0 )=limsF(s)拉氏变换的基本性质（2）尺度变换f(at)1a s F +t0 slim f(t)=f()=li
6、msF(s)t s0卷积定理f1(t)*f2(t)F1(s).F2(s)f1(t).f2(t)F1(s)*F2(s)12j初值定理终值定理f2(t)=0 F 1(s)2、已知f1(t)0当f1(t)0当f1(t)0t f ()ds+0F(s)sr zi=Cie5-7 线性系统的LT分析法全响应零输入响应冲激响应D(p)=0,求出特征根 iNi=1itH(s)h(t)零状态响应rzs=e(t)h(t)Rzs(s)=H(s)E(s)dtddtddtd 22例1：已知一线性系统r(t)+2r(t)=e(t)+2e(t)r(t)+3若r(0)=1;r(0)=2;e(t)=2(t),求r(t)一、求零输
7、入响应rzi(t)=(4et 3e2t)(t)零输入响应，由初始储能引起，变化规律由系统微分方程的特征根。这样的分量叫自由分量s+2s2+3s+2H(s)=1=-1,2=-2rzi(t)=C1et+C2e2tdtH(s)=2Rzs(s)=2ddtddtd 22例1：已知一线性系统r(t)+2r(t)=e(t)+2e(t)r(t)+3若r(0)=1;r(0)=2;e(t)=2(t),求r(t)一、零输入响应二、求零状态响应L(2(t)=22s+4s +3s+2=2s+1trzi(t)=(4et 3e2t)(t)s+2s +3s+2dt零输入响应：零状态响应：t 2trzi(t)=(4et 3e2
8、t)(t)rzs(t)=2et(t)ddtddtd 22例1：已知一线性系统r(t)+2r(t)=e(t)+2e(t)r(t)+3若r(0)=1;r(0)=2;e(t)=2(t),求r(t)没有强迫分量没有稳态分量自由分量：暂态分量(6et 3e2t)(t)(6et 3e2t)(t)u(t)=L-u(t)=uC(0)+C或i(t)=CI(s)+U(0)一、对电路求响应i(t)+u(t)L-di(t)dtI(s)+LsU(s)Li(0)+U(s)=I(s)LsLi(0)1Cdu(t)dtt0i()d()1sc+U(0)-ss1sci(t)+u(t)-I(s)+U(s)-U(s)=1）电压、电流用
9、象函数表示，画出运算电路2）根据KCL或KVL方程列出方程组，求出响应R(s)3）通过L-1T，得到r(t)0.5F +0.1e-5t u(t)uc(t)2欧(s+5)(s+6)I(s)=(s+5)(s+6)=+复习对线性电路的分析例：已知电路如图所示，且iL(0_)=0,uC(0_)=4v，求电流i(t),t01+0.1/s+5-I(s)2U(s)0.1s2/s+4/s-0.1/(s+5)(1+0.1s)+21/(1+0.1s)+s/2+1/2解：U(s)=2(2s2+30s+101)2-iL(t)1欧 0.1H0.1ei(t)2s2+30s+10121 9 7(s+5)2 s+5 s+6i
10、(t)=(te-5t+9e-5t-7e-6t)(t)d f(t)dt =s F(s)s f(0 )L f(0 )二、由方程求响应D(p)r(t)=N(p)e(t)r(0-),r(0-),r(0-),r(n-1)(0-)nnLn n1 n1r(n)+an1r(n1)+L+a1r+a0r=bme(m)+bm1e(m1)+L+b1e+b0ed f(t)dt =s F(s)s f(0 )L f(0 )nnLn n1 n11、直接求全响应若e(t)为有始激励D(s)R(s)-p(s)=N(s)E(s)p(s)由ai及初态r(0-),r(0-),r(0-),r(n-1)(0-)决定D(p)r(t)=N(p
11、)e(t)1(sn+an1sn)+L+a 1s+a0)R(s)p(s)=(b msm+b m1sm1+L+bs+b 0)E(s)dts+b(s+a1)r(0 )+r(0 )(b1 0)E(s)+(s+a1)r(0 )+r(0 )=+s+b(b1 0)E(s)ddtddtd 22e(t)+b0e(t)r(t)+a 0r(t)=b1r(t)+a1 R(s)=s2+a1s+a0 s2+a1s+a0 s2+a1s+a0N(s)D(s)p(s)D(s)R(s)=E(s)+D(s)R(s)-p(s)=N(s)E(s)2 零状态零输入自由自由+受迫例:r zi=Cie2、分别求响应零输入响应零状态响应1）零
12、输入响应rzi(t)D(p)rzi(t)=0D(s)Rzi(s)-p(s)=0rzi(t)=L-1Rzi(s)D(p)=0,求出特征根 iNi=1it不如时域法方便p(s)D(s)Rzi(s)=2）零状态响应rzs(t)rzs=e(t)h(t)Rzs(s)=H(s)E(s)D(p)rzs(t)=N(p)e(t)D(s)Rzs(s)=N(s)E(s)N(s)D(s)E(s)Rzs(s)=N(s)D(s)H(s)=系统函数rzs(t)=L-1Rzs(s)2j j2j j=h()eh()e d三、LT法求零状态响应rzs(t)的含义e(t)零状态系统H(s)r(t)=?E(s)estds1 +je(
13、t)=1 +jr(t)=E(s)H(s)estds =L-1E(s)H(s)estH(s)est ,t(-，)est*h(t)=h(t)*eststs(t)s+d=e=H(s)estt(-，)e四、关于系统函数H(s)Rzs(s)E(s)1、H(s)=N(s)D(s)1）由I/O方程：D(p)r(t)=N(p)e(t)，得：H(s)=2）由电路：H(s)=H(p)|p=s3）由框图或流图求4）由状态方程求2、H(s)=Lh(t)(3、H(s)=est作用下的零状态响应st，t a2st0st+0+0 ptF(s)=Fa(s)+Fb(s)+Fa(s)=0a、将左边信号fb(t)反褶后形成的右边信
14、号fb(-t)；b、求右边信号fb(-t)的单边LT极其收敛区：c、将p=-s带入，得到Fb(s)及收敛区：Fb(s)=F-b(p)|p=-sRe(p)pRe(s)-p=bFb(s)=Fb(p)=fb(t)e dtF(s)=Fa(s)+Fb(s)aRe(s)b+0 ptp=sa a时，f(t)的双边LT存在，其收敛区为aRe(s)0例：求f(t)=e|t|的拉普拉斯变换j解：1pFb(p)=1s+Fb(s)=Fb(p)p=sRe(s)-0例：求f(t)=e|t的拉普拉斯变换1s jFa(s)=2）f a(t)=e t(t)3)f b(t)=e-t(-t)f b(-t)=e t(t)Re(p)0
15、j0j1 1s s+F(s)=Fa(s)+Fb(s)=4)a、当0时，双边LT不存在；收敛区：Re(s)-b、当0时，j例：求f(t)=e|t的拉普拉斯变换1s Fa(s)=2）f a(t)=e t(t)3)f b(t)=e-t(-t)Fb(s)=1s+Re(s)Re(s)-1例：求 F(s)=解：199e4900 t199f(t)=(t)+2）-200Re(s)-13）Re(s)0 ResL表示左侧极点的留数fb(t)=-Resrt 0=f(t)，t0则有L-1Fd(s)，0=-f(t)，t Re(s)-11Re(s)1R(s)=H(s)E(s)3 2s+1 s+2R(s)=1j-2-1-1
16、Re(s)16(s+1)(s+2)(s1)右边1s1左边函数+H(s)=H(2)=h(t)为此系统的冲激响应，满足h(t)+2h(t)=e-4t(t)+C(t)，C为常数。求（1）系统函数H(s)；（2）e(t)=e2t，t 0时对应的响应r(t)。例2：某零状态因果系统，激励e(t)=e2t，-t 时对应的响应为：r(t)=1e2t6-t。h(t)+2h(t)=e-4t(t)+C(t)解：(s+2)H(s)=e(t)=est1 Cs+4 sr(t)=H(s)est(1+C)s+4Cs(s+2)(s+4)16C=1H(s)=2s(s+4)解 （1）H(s)=E(s)=R(s)=1=+（2）e(t)=e2t，t 0H(s)；（2）e(t)=e2t，t 0时对应的响应r(t)。例2：某零状态因果系统，激励e(t)=e2t，-t01s2R(s)=H(s)E(s)2s(s+4)(s2)1 1 4t 1 2t4 12 62-400Re(s)21 14 12s s+4 s2零状态因果系统jRe(s)22s(s+4)
