1、1.+sin2tttdt=2 -+sint2ttdt=122.已知Fs=1s2+5s+4，则原函数f(t)=13e-t-e-4t(t) 拉普拉斯变换11+e-s=t-t-1+t-2-t-33.已知f(t)F(jw)则ej3tf(2t-4)的F1(jw)=12F(jw-32)e-2j(w-3)4.1-e-3s+2的反变换f(t)=e-2tt-e-2t-1(t-1)5.(k+1)(k+1)的Z变换F（Z）=z2(z-1)26.在1丨Z丨2下，zz2-3z+2的原函数fk=-2k-k-1-(k)7.按条件作图画出fk=-k+2+k+3的图形 20-1-3-2108.已知信号f(t)波形如图所示画出f
2、(1-2t)的波形211f(t)12-12112f(t)1-19.如图所示 10.已知f(t)F(jw)求f(3t-2)F(jw) f(t)F(s)=f(t)F(jw) f(3-2t)=f(-2(t-32)f(2t)12F(jw2)f(3-2t)=f(-2(t-32)12F(jw2)e-jw3211.(k-1)(k+1)的Z变换F(z)=(k-1)(k+1)=(k+1)(k+1)-2(k)k(k)z(z-1)2(k+1)(k+1)z2(z-1)2 (k-1)(k+1)z2(z-1)2-2zz-112.在-11下(3s2+6s-1)(s+1)(s+3)(s-1)的原函数f(t)F(s)=1s+1
3、+1s+3+1s-1 f(t)(e-t+e-3t-et(t)(t)112tf(t)13.画出f(k)=(-k+2)的图形0-2-112F(k) f(t)=(t)-2(t-1)+(t-2) F(s)1s-25e-s+15e-2s f(k)=(-k+2) -k+20 k22211014.已知f(t)的波形如图所示，画出f(-2-t)的波形-4-3-2-115.用图解法画卷积和f(k)的图形0021231-1k-1，f(k)=0 k=-1，f(k)=1 k=0，f(k)=11+1(-1)0 k=1，f(k)=11-11+11=1 k=2，f(k)=-11+11=0 k=3,f(k)=11=1 k=4
4、，f(k)=0216.差分方程y(k+2)-y(k+1)-2y(k)=e(k+2)+2e(k)已知yzi (0)=1 yzi (1)=5,e(k)=(k)（1）时域法或Z域法求全响应(E2-E-2)y(k)=（E2+2）e(k) y(k)=E2+2E2-E-2e(k)H(Z)=z2+2z2-z-2零输入：E2-E-2yk=0即(E-2)(E+1)y(k)=0 yzik=C1-1kk+C22k(k)yzi0=C1+C2=1yzi1=-C1+2C2=5C1=-1C2=2yzik=-1kk+22k(k)零状态：yzsz=HzEz=z2+2(z-2)(z+1)zz-1=12zz+1+-32zz-1+2
5、zz-2y（k）=yzik+yzsk=-12-1kk-32（k+42kk）q(k)(2)是否稳定：21，不稳定(3)画出系统模拟框图q(k+2)-q(k+1)-2q(k)=e(k) y(k)=q(k+2)+2q(k)q(k+1)e(k)DD22y(k) (k+2)17.已知某系统的初始条件相同，在e-t(t)作用下全响应为(t+1) e-t(t)在e-2t(t)作用下全响应为(2e-t-e-2t) (t)求电压e-3t(t)作用下的全响应y3(t) 解：yzit+e-tt*ht=(t+1)e-t(t)yzit+e-2tt*ht=(2e-t-e-2t)(t)同时进行拉式反变换yzit+H(s)s
6、+1=1(s+1)2+1s+1yzit+H(s)s+2=2s+1-1s+2解为Yzis与H(s)综上所述激励为e-3t(t)时全响应为E(s)=1s+3Y(s)=Yzis+E(s)H(s)解：yzis+1S+1Hs=1(s+1)2+1s+1yzis+1S+2Hs=2s+1-1s+2 yzis=1s+1Hs=1s+1 Y3s=yziS+Hs1s+3=1s+1+1s+11s+3=321s+1-121S+3 Y3t=32e-t-12e-3tq(t)18.根据零级图粗绘出该系统的幅频特性H0=11-1 H(s)=1s+219.如图所示，求系统的冲击响应和阶跃响应SS1e(t)y(t)解：(s2+s)Y
7、(s)s=E(s)Ys=1s+1E(s) f(t)1 (t)1sY1s=1s+1h(t)=e-t(t) Y2s=1s(s+1)=1s-1s+1 y(t)=(t)-e-t(t)20.一线性时不变系统在相同的初始条件下，当激励为f(t)时其全响应为y1t=2（e-2t+sin2t）(t)当激励为2f(t)时，其全响应为y2t=e-3t+2sin2t(t)求两信号的初始条件下当激励为f(t-2)时的全响应y3(t)解：yzit+ht=(2e-3t+sin2t)(t)yzit+2ht=(e-3t+2sin2t)(t)yzit=3e-3t(t)ht=(-e-3t+sin2t)(t) 当激励为f(t-2)
8、时的h(t-2)=-e-3t-2+sin2(t-2)(t-2) y(t)=3e-3tt+-e-3t-2+sin2(t-2)t-221.系统函数Hjw=-w2+4jw+5(jw+1)(jw+2)，求冲击响应h(t)，若e(t)= e-3t(t)求零状态相应解：H(jw)=1+21+jw-12+jwh(t)=(t)+(2e-t-e-2t)(t) rzst=ht*et=e-3t(t)+(e-t-e-3t) (t)+( e-3t-e-2t)(t)=（e-3t-e-2t+e-t）(t)22.系统函数H(jw)=-w2+4jw+5jw+1jw+2=1+2jw+1-1jw+2h(t)=(t)+ (2e-t t-e-2t tE(jw)=1jw+2Y(jw)=H(jw)E(jw)=2jw+1-1jw+2-1(jw+2)2y(t)=2e-tt-e-2t t-te-2t t七f(t)=costcos(1000t)=12cos1001t+cos999t Fjw=12w+1000+w+1000+j(w+999-(w-999)H(jw)=1 丨w丨10000 基波 y(jw)=F(jw)+H(jw)=12 八
