北理工信号与系统第一章作业参考答案

1、第二章作业参考答案第二章作业参考答案:P70.2.3已知系统的微分方程和初始状态如下，求其系统的全响应。 (a). (D2+5D+6)y(t)=u(t)，y(0)=0，y(1)(0)=1 (b). (D2+4D+3)y(t)=e-2tu(t)，y(0)=y(1)(0)=0 解: (a) 特征方程为 2+5+6=0 得 1= -2， 2= -3。则 yh(t)= c1e- 1 t+ c2e-2t= c1e- 2 t+ c2e-3t令 yp(t)= A,代入原方程，得A=1/6.全响应为 y(t)=yh(t)+yp(t)= c1e- 2 t + c2e-3t +1/6, 应用初始条件y(0)=0，

2、y(1)(0)=1，解得 c1=1/2，c2=-2/3， y(t)=（1/2)e- 2 t (2/3)e-3t +1/6 t0解: (b) 特征方程为 2+4+3=0 得 1= -3， 2= -1。则 yh(t)= c1e- 1 t+ c2e-2t= c1e- 3 t+ c2e-t令 yp(t)= c3e- 2 t,代入原方程，得c3=-1. yp(t)=-e- 2 t,全响应为 y(t)=yh(t)+yp(t)= c1e- 3 t + c2e-t - e- 2 t, 应用初始条件y(0)=0，y(1)(0)=0，解得 c1=c2=1/2， y(t)=（1/2)e- 3t +(1/2)e-t

3、- e- 2 t. t0 2.7已知描述系统的微分方程如下求其系统的单位冲击响应h(t). (a). (D2+3D+2)y(t)=x(t) (b). (D2+6D+8)y(t)=Dx(t) (c). (D2+4D+3)y(t)=(D+1)x(t) (d). (D2+4D+3)y(t)=x(t) (e). (D2+4D+4)y(t)=(D+3)x(t) (f). (D2+2D+2)y(t)=Dx(t) 解: (a) 特征方程为 2+3+2=0 得 1=-2， 2=-1。则 h(t)= (c1e1 t+ c2e2t)u(t)=( c1e- 2 t+ c2e-t)u(t)h(t)= (c1+ c2)

4、(t)+(-2c1e- 2 t-c2e-t)u(t)h(t)= (c1+ c2)(t)+(-2c1-c2) (t)+ (4c1e- 2 t+c2e-t)u(t) 将x(t)= (t), y(t)=h(t)代入原方程得： c1+ c2=0 c1+ 2c2= 1，得 c1=-1, c2=1， 所以 h(t)= (-e- 2 t+ e-t)u(t)解: (b) 特征方程为 2+6+8=0 得 1=-2， 2=-4。则 h(t)= (c1e1 t+ c2e2t)u(t)=( c1e- 2 t+ c2e-4t)u(t)h(t)= (c1+ c2)(t)+(-2c1e- 2 t-4c2e-4t)u(t)h

5、(t)= (c1+ c2)(t)+(-2c1-4c2) (t)+ (4c1e- 2 t+8c2e-4t)u(t) 将x(t)= (t), y(t)=h(t)代入原方程得： (c1+ c2)(t)+ (4c1+2c2) (t)= (t) c1+ c2=1 4c1+ 2c2= 0，得 c1= -1, c2=2， 所以 h(t)= (-e- 2 t+ 2e-4t)u(t)解: (c) 特征方程为 2+4+3=0 得 1=-3， 2=-1。则 h(t)= (c1e1 t+ c2e2t)u(t)=( c1e- 3 t+ c2e- t)u(t)h(t)= (c1+ c2)(t)+(-3c1e- 3 t-c

6、2e- t)u(t)h(t)= (c1+ c2)(t)+(-3c1-c2) (t)+ (9c1e- 3 t+c2e-t)u(t) 将x(t)=(t), y(t)=h(t)代入原方程得： (c1+ c2)(t)+ (c1+3c2) (t)= (t)+ (t) c1+ c2=1 c1+ 3c2= 1，得 c1=1, c2=0， 所以 h(t)= e- 3 tu(t)解: (d) 特征方程为 2+4+3=0 得 1=-3， 2=-1。则 h(t)= (c1e1 t+ c2e2t)u(t)=( c1e- 3 t+ c2e- t)u(t)h(t)= (c1+ c2)(t)+(-3c1e- 3 t-c2e

7、- t)u(t)h(t)= (c1+ c2)(t)+(-3c1-c2) (t)+ (9c1e- 3 t+c2e- t)u(t)将x(t)= (t), y(t)=h(t)代入原方程得： c1+ c2=0 c1+ 3c2= 1，得 c1=-1/2, c2=1/2， 所以 h(t)= (-e- 3 t+ e- t)u(t)/2解: (e) 特征方程为 2+4+4=0 得 1=-2， 2=-2。则 h(t)= (c1e1 t+ c2e2t)u(t)=( c1e- 3 t+ c2e-2 t)u(t)h(t)= (c1+ c2)(t)+(-3c1e- 3 t-2c2e- 2t)u(t) h(t)= (c1

8、+ c2)(t)+(-3c1-2c2) (t)+ (9c1e- 3 t+4c2e- 2t)u(t) 将x(t)= (t), y(t)=h(t)代入原方程得：c2(t)+ (c1+2c2) (t)= (t)+ (t) c11， c2= ， 所以 h(t)= (t+1)e- 2 tu(t)解: (f) 特征方程为 2+2+2=0 得 1=-1+j， 2=-1-j。则 h(t)= (c1e1 t+ c2e2t)u(t)=( c1e- tcost+ c2e- tsint)u(t)求出 h(t), h(t), 和x(t)= (t), y(t)=h(t)代入原方程得： c1=1, c1+ c2= 0，得

9、c1=1, c2=-1， 所以 h(t)= e- t(cost sint )u(t)= 21/2e- tcos(t+/4)u(t)2.18 见P376答案。2.20 图P2.20（a)所示电路图的输入信号如图(b)所示的矩形脉冲，其输出为i2(t).(a)求单位冲激响应h(t)(b)用卷积积分法求零状态响应i2(t).解: (a) )(323)(41)()()(41)(, 4/1)()(,4/1)0()(2)(3)(8)(2)(3)( 8)()()(0)()(, )()()(,838383221221322321tuetdtthdthtuethctucethhtththtetitiRRtite

10、tidttdiLRtitititiKVLKCLttt且可得由有得由0 T tE e(t)e(t) R1=1,R2=3 L=2H i2 (t) i1i3(b) 解一：利用卷积的微、积分性质解)()()()()(, 00,)(TttteTtututTtEte其他)(44)(44)(4)(4)(323)(4)(838383083)1(tueEEtueEEtueEdEdueEdEthttttttt)(4)(4)()()()()()()()()()()()()(8383)1()1()1()1()1()1(2TtueEtueETththTtthtthTttththtetiTtt解二：图解法)(4)(4)(

11、44)()(4,443230,4323)(40, 0)()(8383)(838383)(8383830838302TtueEtueETtueEeETtutueEtTeEeEdeETteEdeEdEttiTttTttttTtTttttttt0 t E(t-) h() Tt0E(t-) h() tT0 t-T t 2.22 某LTI系统的输入信号x(t)和其零状态响应yx(t)的波形如图P2.22所示。(a)求该系统的冲激响应h(t),(b)用积分器，加法器和延时器(T=1s)构成该系统。解: (a)0 1 2 3 4 t1 yx(t)0 1 2 tx(t)43,431, 110,)2() 1()

12、()()()2() 1()()()(43,431, 110,)()2() 1()()(tttttthththtythtttthtxttttttyttttxxxttttttthttththtthtttththththththtttthththththththttththththtth2 , 0, 021,210,)(0)2(2(04)2() 1(4)(,4(4).30)2()1(2(1)2() 1(1)(1)2() 1()(,3(3).22) 1(1) 1(1)(1) 1()()2() 1()(,21).2()()2() 1()(,10).1 (,0, 0)0(时时时时有1 h(t)0 1 2 t43,431, 110,)2() 1()(tttttththth有从而有2121,)1()()1()()() 1()()()()()(hhhhtututututhtututGthtGtGTTT又1 h(t)0 1 2 t 1 1 0 2 t (b) 1 1 * = GT(t) GT(t) h(t) 0 2 t (b) 1 0 2 t x(t)