第七章习题与答案.docx

1、第七章习题与答案第七章习题及解答7-1试求下列函数的z变换tT(1)e(t)a23t(2)()ettes1(3)E(s)2s(4)E(s)s(ss1)(3s2)解（1）nnE(z)azn011az1zza2Tz(z1)2（2）Zt3(z1)由移位定理：23T3T23T3TTze(ze1)Tze(ze)23tZte3T33T3(ze1)(ze)（3）E(s)s12s1s12szTzE(z)2z1(z1)（4）E(s)c0c1c2ss1s2c0lims0(ss1)(3s2)32c1lims1ss(s32)212c2lims2ss(s31)1232212ss1s2E(z)2(3zz21)zzezTz

2、e2(2T)7-2试分别用部分分式法、幂级数法和反演积分法求下列函数的z反变换。119(1)E(z)10z(z1)(z2)13z(2)E(z)1212zz解（1）E(z)(z101)(zz2)部分分式法E(z)101010z(z1)(z2)z1z210z10zE(z)(z1)(z2)nne(nT)10110210(21)幂级数法：用长除法可得E(z)(z101)(zz2)z210z3z210z12330z70z*e(t)10(tT)30(t2T)70(t3T)反演积分法ResE(z)zn1z1limz1n10zz210ResE(z)zn1z2n10zlim10z2z12nnne(nT)1011

3、0210(21)*en(t)10(21)(tnT)n013zz(3z1)z(3z1)（2）E(z)22212zzz2z1(z1)部分分式法E(z)13z2322z(z1)(z1)z1E(z)(z2z1)23zz12e(t)t31(t)T*e2(t)nT3(tnT)(2n3)(tnT)n0n0T120幂级数法：用长除法可得23zz123E(z)35z7z9z2z2z1*e(t)3(t)5(tT)7(t2T)9(t3T)反演积分法e(nT)ResE(z)z1dn1lim(3z2z)zz11!dzs1n1lims1nn13(n1)znz2n3*e(t)(2n3)(tnT)n07-3试确定下列函数的终

4、值(1)E(z)1Tz12(1z)20.792z(2)E(z)2(z1)(z0.416z0.208)解（1）ess1Tz1lim(1z)1z1(1z)2esslimz1(z1)E(z)（2）limz1z20.792z0.416z20.20810.7920.41610.2087-4已知差分方程为c(k)4c(k1)c(k2)0初始条件：c(0)=0,c(1)=1。试用迭代法求输出序列c(k),k=0，1，2，3，4。解依题有c(k2)4c(k1)c(k2)c(0)0,c(1)1c(2)4104c(3)44115c(4)4154567-5试用z变换法求解下列差分方程：(1)c(k2)6c(k1)8

5、c(k)r(k)r(k)1(k),c(k)0(k0)121(2)c(k2)2c(k1)c(k)r(k)c(0)c(T)0r(n)n,(n0,1,2,)(3)c(k3)6c(k2)11c(k1)6c(k)0c(0)c(1)1,c(2)0(4)c(k2)5c(k1)6c(k)cos(k/2)c(0)c(1)0解（1）令tT，代入原方程可得：c(T)0。对差分方程两端取z变换，整理得C(z)11R(z)2z(z2)(z4)z6z8z1C(z)111111z3z12z26z41z1z1zC(z)3z12z26z4111nnc(nT)124326n（2）对差分方程两端取z变换，整理得C(z)z212z1

6、(zz1)2(z1)2z(2z1)ResC(z)zn1z111!limz1ddzz(z1)2zn1limz1ndzdz(z1)2limz1n123nnz(z1)2(z1)z23n1n2224ResC(z)zn1z111!limz1ddzz(z1)2zn1limz1ddz(znz1)2limz1n123nz(z1)2(z1)nzn1n1(1)4n1n1c(nT)1(1)4*n1n1c(t)1(1)(tnT)4n0122(3)对差分方程两端取z变换得3zC(z)z3c(0)z2c(1)zc(2)6z2C(z)z2c(0)zc(1)11zC(z)zc(0)6C(z)0代入初条件整理得(3z6z232

7、11z6)C(z)z7z17zC(z)z3z367zz221711zz6C(zz)1111572z1z22213c(n)112(1)nn7(2)52(3)n(1)n1127n252n3（4）由原方程可得z(zcos)2z22(z5z6)C(z)2z12z2zcos12C(z)2(z5z2z6)(z21)(z2)(zz23)(2z1)C(z)z21311z122z(z2)(z3)(z1)5z210z310z1c(nT)25(2)n310(3)n110cosn2sinn2111n1n1n1(1)23cosnsinn51010227-6试由以下差分方程确定脉冲传递函数。c(n2)(1e0TcneTc

8、ne.rn.50.505T)(1)()(1)(1)解对上式实行z变换，并设所有初始条件为0得z2C(z)(1e0.5T0.5T0.5T)zC(z)eC(z)(1e)zR(z)根据定义有0.5TC(z)z(1e)G(z)20.5T0.5TR(z)z(1e)ze1237-7设开环离散系统如题6-7图所示，试求开环脉冲传递函数G(z)。22z解（a）Z2Ts2ze55zZ5Ts5ze25Z2s5(ze210z2Tze)(G(z)Z5Ts)252s5Z103s12103s15103(zz(ee2T2T)(ez5Te)（b）Z5Ts)2T5T2510z(ee)G(z)Z2T5Ts2s53(ze)(ze)

9、（c）Zs(1e)10110(1zs(s2)(s5)Zs(s12)(s5)10(z1)111111Zz10s6s215s5G(z)z1zzz153zze2T23zze5T153zze12T23zze15T52252T5T2T5T7T(1ee)zeee33332T5T(ze)(ze)1247-8试求下列闭环离散系统的脉冲传递函数(z)或输出z变换C(z)。题6-8图离散系统结构图解（a）将原系统结构图等效变换为图解6-8(a)所示图解7-8(a)G(z)G(z)E(z)1B(z1)B1(z)GG(z)E(z)B(121z)1GG(12z)B(1z)GG(z)E(z12)B(z)11GG12GG1

10、(2z)(z)E(z)G(z)111GG1G1(2G2z)(z)E(z)1G(z)21GG1(z)E(z)E(z)R(z)B(z)2B(z)G(z)C(z)23R(z)G(z)C(z)3C(z)1GG(212GG1z)(z)R(z)G3(z)C(z)1GG(z)C(z)G(z)R(z)G(z)C(z1213)1GG(z)G(z)G(z)C(1213z)G(z)R(1z)(z)C(z)R(z)1G(z)1GG(z)G(z)G(1213z)125（b）由系统结构图C(z)RG2G(z)E(z)G4hGG(34z)E(z)RG(z)C(z)1RGG(z)24GGG(z)RG(z)Ch341(z)C(

11、z)RGG(24z)1GGhGhG3G3G44(z)RG(z)1z)（c）由系统结构图C(z)NG2(z)R(z)D2(z)GhG1G2(z)E(z)D1(z)GhG1G2(z)E(z)R(z)C(z)NG2(z)R(z)D(z)GGG(z)D(z)GGG(z)R(z)C(z2h121h12)C(z)NG(z)R(z)D(z)G221D(z)G1GhG(12hz)D(1GG(z)12z)GGG(z)h12R(z)NG(Z)D2(z)D(z)12GGG(z)R(Zh12)1D(z)1GGG(h12Z)7-9设有单位反馈误差采样的离散系统，连续部分传递函数G(s)12s(s5)输入r(t)1(t)

12、，采样周期T1s。试求：（1）输出z变换C(z)；*（2）采样瞬时的输出响应()ct；（3）输出响应的终值c()。图解7-9解（1）依据题意画出系统结构图如图解6-9所示G(z)Z2s(1s5)51zz(1e)255(z1)5(z1)(ze)55(4e)z16ez2525(z1)(ze)(z)525G(z)(4e)z(16e)z1G(z)25(z1)255(ze)(4e)2z(16e5)z23.9933z0.9596z253z46.17472z26.2966z0.1684126zC(z)(z)R(z)(z)z12(0.1597z0.03838)zz42.8473z0.417899z21.058

13、6z0.00673612340.1597z0.4585z0.842z1.235z（2）*c(t)0.1597(tT)0.4585(t2T)0.842(t3T)1.235(t4T)（3）判断系统稳定性D(z)253z46.17472z26.2966z0.1684n3()奇数D(1)4.95330,D(1)97.63970列朱利表zz1z2z301-0.168426.2966-46.174725225-14.174726.2966-0.16843-624.971149.94-649.64a0.1684a2503b624.970b649.264(不稳定)闭环系统不稳定，求终值无意义。7-10试判断下

14、列系统的稳定性（1）已知离散系统的特征方程为D(z)(z1)(z0.5)(z2)0（2）已知闭环离散系统的特征方程为432D(z)z.zz.z.02036080（注：要求用朱利判据）（3）已知误差采样的单位反馈离散系统，采样周期T=1(s)，开环传递函数G(s)22.572s(s1)解（1）系统特征根幅值1，1120.50.53221有特征根落在单位圆之外，系统不稳定。4zzz32（2）()0.20.360.80Dzz用朱利稳定判据（n4）127z0z1z2z3z410.80.3610.21210.210.360.83-0.360.088-0.2-0.24-0.2-0.20.088-0.3650.0896-0.071680.0896D(z)3.360,D(1)2.240a0.8a1,b0.36040b03.2c00.896c02.0896所以，系统不稳定。（3）G(z)zs22.571)2(sz22.572s22.57s22s.571