自动控制原理（千博）习题答案第八章习题答案.docx
8-1(4)8-2 (1)(2)(4)8-3(1)(4)= EoNT6(t NT)= TE、°NMt_NT)f*(s)=£步(切二口心股一皿f*(t)=Etof(NT)6(t - NT)= lN=0NTe-aNTS(it- NT)=T£匕0 Ne-aNT8(t一 NT)f*G) = £ /*«)= T0Ne-aNTe-NTs/*Q) = %=o/WW-NT)=£/=o eaNTsin(ji)NTS(<t NT) f*(s) = £ /*(0 = T=oe-aNTsin(vNTe-NTs f*(t) = X=uf(NT)3(t - NT)=Sn=o e-aNTcoso)NT8(it - NT) f*(s) = £ /*«) = TN=aNTcoso)NTe-NTsf(t) = 1 eatF（Z）=z-l z-e-aTf(t) = t2e3t二 Zi) |Z二产八：占3：)' 7(z-1)3(ze3T-l)3f(t) = t3T3z3+4T3z2+t3z(Z-l)4f(t) = amt = amNTF(z)=zz-amF(s)=9f(t) = £/(£)=工露oNTb(t- NT)F（z）=合F(s)=()'(s+Q)(s+b) 、s+q s+匕，b-aF(z)=(z-eaT z-eF（S）=(s+a)2F(z)=KTze-aT(z-e-aT>)2F(s)=s+3(S+l)(S+2)s+1+2(s+l)(s+2)+S+2(s+l)(s+2)1221s+2 s+1 S+2_2 1s+1s+22zz-eT z-e2T8-4 F(z)=初值:lim e*(t) = lim F(z) = 1tT8Z->0终值:lim e*(t) = lim(z l)F(z) = 1 tT8Z->1(2) F(z) =ToZ-i T0z2初值:lim e*(t)= tT8lim F(z) = To zO终值:lim e*(t)=T8lim(z l)F(z) = ooZ-1 F(z)=z2(z0.5) (z0.1)初值:lim e*(t)= t-»oo期/=1终值:lim e*(t)= t->00lim(z l)F(z) = 0Z-18-5(l)F(z)(2)F(z)Sz5z 5z(z+l)(z+2)z+1z+2f(k) = 5(1/ - 5(-2)fcF&)=生f(k) = 102k - 10IOz lOzz_2z_ 1(4)F(z)=z 0.5z0.25z(z3)(z1)2z3(z-1)2z-1f(k) = 0.253k-k-0.25(5)F(z)=z3(ze-1)3(z0t)3f(k) = 0.5e-5fc(/c - l)e"fc+2 + 2e-4ke-fc+1 + e+3 = e-/c-3(0.5fc21.5/c + l)(6)F(z)=(z-e-aT)(z-e-/?T)z3e2aTe-2bTe-bT _ e-aTz-e-bTF(Z)e-aTe-bTz z eaTf(T) =p-2bT+ e-bTp-2aTp-aT e-aT _e-bTeG(s)=- = -(-)s(s+q) a、s s+ay"C C(z) K , zG(Z)=afczz-eaT8-8Gi =zG2(z) = z1S + 1- 1 -s + 1.2 -s + 2.1=2z,s + 2.zz eT1 -s + 2.2zzz= 2(z2z2z e2T (z e-T)(z e2T8-9(1) C(S)=X*(s)G2(s)X(s) = Gi(s)R(s) CG) = Gi(s)R(s) G1(s)X*(s)G2(s)X(z) = RGi(z) - GiG2(z)X(z)=X(z)=RGi(z)1+Gi G2 (2)C(z) = X(z)G2(z)=> C(z)=AG(z)G2(z)l+Ggz)(2) E(z) = R(z)- Gi”2(z)X(z) X(z) = E(z) - Gi"i(z)X(z)=X(z)=R(z)1+G1H2(z)+G1H1(z)C(z) = X(z)G1(z)=> C(z)=R(z)Gi(z)1+G1H2(z)+G1H1(z)(3) E(z) = R(z)- ”2(z)G2H3(z)X(z)X(z) = E(z)G1(z) - G2Hl (z)X(z)=X(z)=R(z)GMz)1+G2H1(z)+G2 H3(z) Gi (z) % 3)C(z) = X(z)G2(z)=> C(z)=R(z)Gi(z)G2(z)1+G2 Hl (z)+G2 H3 (z) G(z) % 8-10(l)G(z)=0.36z+0.26z2-1.36z+0.36将工=告代入上式:其闭环特征方程为：D(z) = z2- 1.36z + 0.36 + 0.26 = z2 - z + 0.62 = 0()2- -+ 0.62 = 0.62w2+ 0.76w + 2.62 = 0 w-l w-1列劳斯表:w2 0.622.62w1 0.76W0 2.62所以系统是稳定的。(2)G(z)=z+0.7(z-l)(z0.36)其闭环特征方程为：D(z) = (z- l)(z - 0.36) + z + 0.7 = 0即：=z2 0.36z + 1.06 = 0IV + 1 ow + 1Q(-)2 - 0.36-+ 1.06 = 1,7w2 - 0.121V + 2.42 = 0w 1w 1列劳斯表：w2 1.72.42w1 -0.12w° 2.42所以系统是不稳定的。8-11由结构图知：Gi(s)=-,G2(s)=i OIO I 工)o= Gq)=a-16(z)=六必 z)=六E(z) = R(z) G2(z)X(z), C(z) = G2(z)X(z),X(z) = E(z)G(z)=C(z)=R（Z）G1（Z）G2 （z）1 + Gi（Z）G2（Z）2z3(l-e-T)(z-l)3(z-e-r)+2z2(z-l)(l-e-，r)8-12由结构框图知：Gi（s） = K,G2（s）=7三O I O i 乙 J=6G2(z) = 2K(六-&)系统闭环特征方程为：D(z) = (z l)(z e-r) + 2Kz(1 e-1) = z2 + 2K(1 e-1) (1 + e-1 = 0将z =匕?代入上式： w-l/w + 1w + 1(-7)+2/<(l-e-1)-(l + e-1)-4-e-1 1/w 1=2K(1 e-1)w2 + 2(1 e-1)w 2K(1 e-1) + 2(1 + e-1) = 0 要使系统稳定需满足：2K(l-e-1) > 0-2K(1 -)+ 2(1 + J) > 0=0.8 V K V 2.168-13由系统结构图知系统开环传递函数为：K*2=(1-问而调z-l K( z z 一 (3z a z-lz-eai_ K(1 e")a(z eaT)所以闭环特征方程为：D(z) = a(z - e-aT) + K(1 - e-aT) = 0将工=言代入上式:IV + 1 nTLi+ K(1 e-QT)=a + K (K + a)e-aTw + a K + (K + a)eaT = 0系统临界稳定的条件为: a + K (K + a)eaT = 0l-e-aT对于 T=ls, Ki =a(l+ea)a(l+e-Q) 1+e-0.5a对于 T=o.5s, Ki =l-e-a i-e-0-5a八a(l+e-a)2(l + U-OSa) _ i+e-o.5a1£0.5。一？一0.5aKr<K2所以采样周期T减小会使得系统稳定性变好(或更容易稳定，或取值范围变宽)。 8-14 Gp(s) = l,G(s)=l-e-Ts Ks s+qGpG(z) = (1 z-1)XKs(s + a).a'_Kl-e-aT az - eaT、 R(z)E(z) = 1 ' “ 、1 + GpG(z)z zz-l z - eaT当r(t) = l(t), R(z) = 由终值定理得:z-lessv = lim(z l)E(z) = lim(z - 1) * z->iz->iR(z)1 + GpG(z) q + K当r(t) = t, R(z)= 三，由终值定理得:ooessv = lim(z l)E(z) = lim(z - 1)-勺，).、ssp zTi' ) zTi' + GpG(z) Gp(s) = L G(s) =ys(s+q)GpG(z) = XK.s(s + a).=3,一a Vz 1 z eaTK 1 - e-aTq(z 1) z eaT。,、 R(z)E(Z)=埒 GpG当r(t) = l(t), R(z) = 由终值定理得： / JLessv = lim(z l)E(z) = lim(z - 1)-、= 0ssp ztVZTl' 1 + GpG(z)当r(t) = t, R(z)=；,由终值定理得:(z-i)n1. z 1、”、/1、 R(z) aTen = hm(z - l)E(z) = lim(z - 1)- 、=ssp zTi' ' zTi' ，l + GpG(z) K若是连续系统：i p-Ts k Gp(s) = l,G(s)=V ys s+a,、1 la= limsF(s) = lims- =-s->o s->o 1 + Gp(s)G(s) s a + K1 1%皿=lim sE (s) = lim s 一 , 、 “ 、 = oossp2 sto - sto 1 + Gp(s)G(s)s2 Gp(s) = 1, G(s) =口s(s+a)1 1= limsE(s) = lims-，、“、一 = °sspi sto ' ' sto 1 + Gp(s)G(s) s,、11 ae$sp2 = limsE(s) = lim s=-“ stosto 1 + Gp(s)G(s) s' K结论：离散系统的误差在形式上与相同的连续系统传递函数相同，但对由于离散系统，稳态 误差除了与输入信号有关之外，还与采样周期T有关，T越小，其稳态误差越小。8-15Ts由框图知:Gi(s)=上二,G2G) = K,G3G)=jH(s) =0.5s ssC(s) = E* Gi(s)G2(s)GKs)E(s) = X(s) - C(s)H(S)Ei(s) = R(s) - C(s)=> E(s) = R(s)- E*(s)Gi(s)G2(s)G3(s) - E*(s)G1(s)G2(s)G3(s)H(s)E(z) = R(z)- E(z)G1G2G3(z) - E(z)G】G2G3”(z)E(z)=R(z)1 + GG2G3(z) + G1G2G3H (z)G1G2G3H(z) = 0.5/f-=- 1 (z - l)2 z-lG1G2G3(z) = (1-z-1)x = 7<z-1T2z(z+1)KT2(z+1)z 2(z-l)32(z-l)2z-1 Tz O.SKTK= 10, T = 0.2 =6"屹)二等絮 L)GiG2G3HQ) = 0.5Kz-l Tz 11 (z-l)2 z-lG(z) = G1G2G3(z) + G1G2G3/7(z)由终值定理得：essv= lim(z 一 l)E(z) = lim(z - 1) JR y ZTlZTl1+G(z),、0.2(z+ 1)1Kp = lim G(z) = lim+- = co口z->i z->i (z i)zz 1/0.2(z+ 1)1Kv = Hm(z - l)G(z) = lim(z - 1)-t-+ - = 8ZTlZT1(Z 1)/ Z 19/、9r0.2(z + 1)1 Ka = li吗(z l)2G(z) = lim(z - 1)2H+71= °-4ZTlZT1(Z 1JZ Z 1r(t) = l(t) + t + 0.5t2ITT2=FF = 0.1ssp 1 + Kp Kv Ka8-16G(z)=舒=Xl-eTs Ks s半Is2.z-lKTzz (z-l)2KTz-l=Wz)=悬=(3) T=l,系统闭环特征方程为：K+z-l=0将z = W代入上式：Kw+2-K=0 w-1=>0<K<28-17(l)G(z) = KXl-eTs 0.5-s s .0.571=K Xzz-1 O.STz(Zl)20.5KTz-1（Z）=G(z)0.5KT1+G(z)0.5KT+Z-1(2)K=2,4)(z)=TT+z-1系统闭环特征方程为：T+Z-1=O将z = H代入上式：Tw+2-T=0 iv-l=OVT V2(s)
