自動控制原理習題答案2

1、部分習題答案部分習題答案2-2(a)()()()(21txftxtxftxmooio - 即：即：)t (xf)t (x)ff()t (xmi1o21o (b)t (xfk)t (xkk)t (x)kk(fi2o21o21 (c)()()()()(121txktxftxkktxfiioo+=+2-3(a)t (u)t (u)CRCRCR()t (uCCRRooo+ 2122112121(b)t (xkk)t (x)kfkfkf()t (xffooo2112211121+ 2-4(a)t (uR)t (uCRR)t (u)RR()t (uCRRi2i21o21o21 (b)t (u)t (u)R

2、CRC()t (uCCRooo+ 122122（2-2題題2-4題）題）（僅供參考，不對之處敬請批評指正，謝謝！）（僅供參考，不對之處敬請批評指正，謝謝?。?()(2)(121tutuRCtuCRCiii )t (xkk)t (x)kfkf()t (xffiii21122121+= )t (u)CRCR()t (uCCRRii 22112121+=2-5(1)運動模態(tài)：運動模態(tài)：t5 . 0e tettx5 . 022)(- (2)tsine23t5 . 0 tsine)t (x23t5 . 0332 (3)(1+t)e-tte )t1(1)t(x 2-6PQoQ22k 2-7y11.12F

3、2-8)(sinEeooodd 2-9)2s)(1s (2s4s(s)2 tt2ee2) t (dt) t (dc) t (k 2-10零初態(tài)響應零初態(tài)響應t2ee21) t (ct1 零輸入響應零輸入響應tt22e2e) t (c 總輸出總輸出tteetctctc221241)()()( （2-5題題2-10題）題）運動模態(tài)：運動模態(tài)：運動模態(tài)：運動模態(tài)：（2-11題題2-15題）題）2-1125s23s12)1s4(100) s (R) s (C2 25s23s12)5s23s12(10) s (R) s (E22 2-12(a)sCR)10sCR)(1sCR() s (U) s (U1o

4、oo11io (b)1sCR(RR) s (U) s (Uooo1io (b)1sC)RR()1sCR(RR) s (U) s (U22122o1io 2-132123o22113o21ioRRsCRsCCRRRR) s (U) s (U 2-141sTK) s (U) s (m1am 1sTK) s (M) s (m2am 2-1511132k3)1sT( skmm 3skt111i o iuouu 1u2uautum3mt32miokk26.31s )kkk31(sT26.31) s () s ( （2-17題題2-21題）題）2-17(a)3221GG1GG) s (R) s (C (b

5、)21112121HHHG1)HH1(GG) s (R) s (C (c)22112231HGGHG1G)GG() s (R) s (C (d)3311332211321HGHGHGHGHG1GGG) s (R) s (C (e)232121123214HGGHGGHG1GGGG) s (R) s (C (f)121231HGG1G)GG() s (R) s (C 2-18(a)1212121HGGGG1GG) s (R) s (C 1212112123HGGGG1)HGG1(GG) s (N) s (C (b)434243421GGGG1GGGG)G1() s (R) s (C 43424G

6、GGG1G) s (N) s (C 2-19與與2-17同同 2-20與與2-18同同 2-21(a)23112132123111134321HGHGHHGGGHGHG1)HG1(GGGGG) s (R) s (C 2311213212311123423HGHGHHGGGHGHG1HHGG)HG1()s(R)s(E (b)21212121GG3GG1GG2GG) s (R) s (C 212121GG3GG1GG1) s (R) s (C （2-22題）題）2-22(a)23234313543216HGGHGGHG1GGGGGG) s (R) s (C (b)9個單獨回路：個單獨回路：5654

7、321545434363242121HGGGGGGL,HGGGL,HGL,HGL,HGL 1489568178568175654376HHGL,HGGHGL,HGGGL,HGGGGGL 6對兩兩互不接觸回路：對兩兩互不接觸回路：L1L2 L1L3 L2L3 L7L2 L8L2 L9L2三個互不接觸回路三個互不接觸回路1組：組：L1L2L34條前向通路及其余子式：條前向通路及其余子式： P1=G1G2G3G4G5G6 ,1=1 ; P2=G7G3G4G5G6 , 2=1 ;P3=-G7H1G8G6 ,3=1+G4H2 ; P4=G1G8G6 , 4=1+G4H2 ; 91a61321cba41k

8、kkLLLLLL1P) s (R) s (C(c)128.1539590) s (R) s (C (d)afchehgfchbgaf1)bg1(edabcd) s (R) s (C (e)adehcefgbcdehegcf1)eg1)(bca(adebcde) s (R) s (C1 adehcefgbcdehegcf1lehalehbc)cf1(le) s (R) s (C2 (f)fgdegf1)cicfdejcfdh() idegbbdejbdh(aegiaej)fg1(ah) s (R) s (C1 fgdegf1fjifdejfdh) s (R) s (C2 fgdegf1egiej

9、)fg1(h) s (R) s (C3 （3-1題題3-9題）題）3-1Tte1) t (hTT 3-2 (1)10)t (k t10)t (h 3-2 (2)t4sine425)t (kt3 )13.53t4sin(e451)t (hot3 3-3 (1)25. 1s0125. 0)s( (2)16s)4s(50s5)s(22 (3)1s3( s1 . 0)s( 3-4s917. 2ts96. 1t%478. 9%26 . 0spn 3-5686. 15 . 2z5 . 010066. 1r2dn %99.17%s0133. 6ts156. 3ts45. 1tspr 3-65 .2443.

10、1n 3-7311. 0k44. 1k21 3-8 (a)10n 系統(tǒng)臨界穩(wěn)定系統(tǒng)臨界穩(wěn)定(b)s51. 7t%8 .29%15 . 01ss1s)s(sn2 (c)s08. 8t%3 .16%15 . 01ss1)s(sn2 (b)比比(c)多一個零點多一個零點,附加零點有削弱阻尼的作用。附加零點有削弱阻尼的作用。3-9 (1)2 . 0es5 . 3t%09.35%101015k)1s5 . 0( s5)s(Gsssn (2)249. 11r10z1010110k)1s( s)1s1 . 0(10)s(Gdn 1 . 0es3t%06.37%2sss （3-11題題3-20題）題）3-11

11、勞斯表變號兩次，有兩個特征根在勞斯表變號兩次，有兩個特征根在s右半平面右半平面,系統(tǒng)不穩(wěn)定。系統(tǒng)不穩(wěn)定。3-12 (1)有一對純虛根：有一對純虛根：2js2, 1 系統(tǒng)不穩(wěn)定。系統(tǒng)不穩(wěn)定。(2)5s1s1s2js654,32, 1 系統(tǒng)不穩(wěn)定。系統(tǒng)不穩(wěn)定。(3)有一對純虛根：有一對純虛根：5js2,1 系統(tǒng)不穩(wěn)定。系統(tǒng)不穩(wěn)定。3-137 . 1k0 3-1400 3-15 (1) ssssee20k(2) sssse2 . 0e10k(3)20e0e1 . 0kssss 3-16 (1)(3)(2)0k0k50kavp 0k200kkkavp 1kkkavp 3-18 (1)(2)(3)0es

12、sr 0e1ssn 0e2ssn 3-20k1)1sT( sk12 1sT12 RuBCE(s)=R(s)-C(s)由題意得：由題意得：2122112o2TTkTTk)T(k1 （4-4題題4-5題）題）-0.8810j7=k-0.293-1.707-0.9-4.236op135= op0= （4-6題題4-10題）題）4-6 (1)11k=(2)30*k =63. 630199z=-0.4042 .73*k =036. 1= op73.92= 6j+-2-210j260*k =-3.2921j96*k =oop135,45 = 150kc=62. 9ko=-21.137 .70= （4-11

13、題題4-12題）題）s1= -9.98s2,3= -2.46354. 3j-4系統(tǒng)始終不穩(wěn)定！系統(tǒng)始終不穩(wěn)定！55. 2j= 75.22*k =75.22*k0時穩(wěn)定時穩(wěn)定（4-13題題4-14題）題）)3s2s ( s)2s (k) s (G2+=)2s)(1s ( s)10s (k) s (G+=-0.434j1.6976k=-8.47224)2s ()4s (b) s (G+=)40s ( sb30) s (G+=（4-15題題4-19題）題）2j2*k =-0.732536. 0*k =2.732464. 7*k = 整條實軸為根軌跡整條實軸為根軌跡2 12* k0k*12時系統(tǒng)穩(wěn)定時

14、系統(tǒng)穩(wěn)定系統(tǒng)始終不穩(wěn)定系統(tǒng)始終不穩(wěn)定（4-20題題4-21題）題） 主導極點離虛軸太近！主導極點離虛軸太近！主導極點離虛軸太近！主導極點離虛軸太近！6)24. 3)(3 .23()20(10)(221 ssssksGt6)24. 3)(3 .23()20(10)(2222 ssssksGt6)24. 3)(3 .23(10)(2223 sssksGt2 1k 0k1系統(tǒng)穩(wěn)定，系統(tǒng)穩(wěn)定，c(t)振蕩收斂，振蕩收斂，5 . 0)(cn （5-2題題5-8題）題）5-213j)36(36)j ()9s)(4s (36) s (2 5-3)57.26t2cos(79. 0)4 .48tsin(632.

15、 0) t (eooss 或：或：)90t2cos(707. 0)4 . 3tsin(447. 0) t (cooss ) t (c) t (r) t (essss 5-4848. 1653. 0n 5-50j0jT T5-60j0j0j0j1 v2 v3 v4 v5-7)1s20( s)1s05. 0(10) s (G 5-8o4 .1539 .17)5 . 0 j (G o53.327383. 0)2j (G 61 （5-9題題5-13題）題）j-10 2 20o90)0 j (G o4 .153)2j (G o4 .333)2j (G o3600)j (G 5-95-11 (1)02.

16、620 125. 025. 040 dB)(L dB05-11 (2)40 60 80 861 . 02611 . 2 dB0dB)(L 5-11 (3)06.3802.261 . 01243. 5 20 40 60 707. 0r dB25. 1Lm dB020 40 60 20 4006.78 dB0 11 . 020dB)(L 5-11 (4)5-135-5 ：T時系統(tǒng)穩(wěn)定，時系統(tǒng)穩(wěn)定，5-6 ：=1時系統(tǒng)穩(wěn)定，時系統(tǒng)穩(wěn)定，其余不穩(wěn)定。其余不穩(wěn)定。T時系統(tǒng)不穩(wěn)定。時系統(tǒng)不穩(wěn)定。（5-14題題5-23題）題）5-14(1) Z=0-2(-1)=2 不穩(wěn)定不穩(wěn)定 (2) z=0 穩(wěn)定穩(wěn)定 (

17、3) z=0-2(-1)=2 不穩(wěn)定不穩(wěn)定(4) z=0 穩(wěn)定穩(wěn)定 (5) Z=0-2(-1)=2 不穩(wěn)定不穩(wěn)定 (6) z=0-2(1-1)=0 穩(wěn)定穩(wěn)定 (7) z=0 穩(wěn)定穩(wěn)定 (8) z=0 穩(wěn)定穩(wěn)定 (9) Z=1 不穩(wěn)定不穩(wěn)定 (10) z=2 不穩(wěn)定不穩(wěn)定 5-15z=0-2(-1)=2 不穩(wěn)定不穩(wěn)定 5-16(1) k1.5 (2) T1/9 (3) k-11/T5-17 z=0-0=0 穩(wěn)定穩(wěn)定5-18（左圖）原系統(tǒng)穩(wěn)定，改變（左圖）原系統(tǒng)穩(wěn)定，改變k值。使值。使c 2 時系統(tǒng)穩(wěn)時系統(tǒng)穩(wěn)定定,（右圖）原系統(tǒng)（右圖）原系統(tǒng)z=0-2(-1)=2不穩(wěn)定。改變不穩(wěn)定。改變k值，使值

18、，使c 0時應有時應有 0k2.64 ; k0時應有時應有 -1k05-203686. 10 5-2184. 021a4 5-22oc156.6594455. 1 5-23on17.5324. 2517. 0 o21180)()( 其中其中其中其中o1180)( 5-24o4o3o2o156.33608 .15190 （6-1題題67題）題）6-1 (1)1667. 11623. 30521. 4924. 26 hkxoc1249. 383. 77673.298473. 3 hxoc(2)6-2 (1)57 k(2)11. 227572673. 2 vskst(3)%2 . 3%7 . 022

19、26.28)01. 0()091. 0)(3()( stkssssGsvc6-3 取取820 cksssGc04. 014 . 01)( 驗算得：驗算得：oc1 .62,93. 7 6-5 (1)sssGc004. 0119. 01)( 驗算得：驗算得：dBho7 .161 .30 (2)118016 . 2)( sssGc驗算得：驗算得：dBho3 .304 .50 6-6ococ5131.747962. 025.14969. 1 6-7 (1)取取5 . 05 cvk1200120)(1 sssGc驗算得：驗算得：oc4 .53,45. 0 (2)13 . 0161200120)( sss

20、ssGcoc3 .51,05. 2 驗算得：驗算得：(6-8題題6-15題題)6-8 (1)1s101s) s (Gc+=(a)(b)系統(tǒng)不穩(wěn)定系統(tǒng)不穩(wěn)定1s01. 01s1 . 0) s (Gc+=oc5 .346 .36 44.89 hx(c)1s025. 0)(1s10()1s5 . 0() s (G2c+=oc2.4810= = dB7.21h,8.58x 6-8 (2)采用采用(c)6-140k0.045時系統(tǒng)為過阻尼時系統(tǒng)為過阻尼;k10s) s (Gr=時時ess= 0 ,其中其中0k+=0k,ks1) s (G111c0k, 0TTkTk1k0,sk) s (G221oo2c+

21、=(6-16題題6-17題題)6-1695. 2466. 0)()3747. 1()(12 kssGsssGcn6-17100)2()12 . 0(505024 . 12 . 0)(12223 sssssssse04. 0028. 012= (Ts+1)寄生因子不影響系統(tǒng)穩(wěn)定性寄生因子不影響系統(tǒng)穩(wěn)定性,且因為它為非主導極點且因為它為非主導極點,所以也不太影響動態(tài)性能所以也不太影響動態(tài)性能,但附加極點有增大阻尼的作用。但附加極點有增大阻尼的作用。（7-3題題7-10題）題）7-3(1) e(nT)=10(2n-1) )T2t (7)Tt (5) t (3)nTt ()3n2() t (*e0n(

22、2)7-5 )( e0)( e (1)(2)7-756)4( c15)3( c4)2( c 7-8(1)nTt ()46122131() t (*c0)1( cTtTt0n 7-9T7T5T222ez )ee (zz10)z(G (a)(b)ezzezz(310)z(GT5T2 7-10(a)z(G)z(G)z(GG1)z(G)z(31211 (b)z(GGG1)z(GRG)z(GGG)z(RG)z(C43h4243h1 (c)z(GGG)z(D1)z(NG)z(GGG)z(D)z(D)z(R)z(C21h1221h21 7-4(1)1nnnn20nnn11a3a,z3a)z1(z31)z(E (2)25. 2)5 . 0)(1n3(5 . 1)n( enn (2)1(1)1n(4T)nT( cn (3)nnn)3(5)2(7)1(11)n( c （7-11題題7-21題）題）)nTt ()37. 047. 01()nT( cn 7-11 7-13 (1)0