自動控制原理黃堅課后習(xí)題答案

1、2-1試建立圖所示電路的動態(tài)微分方程(a)解：ii=i-i2ui二ui-uoui ui-u i ; _ i i ii=Ri= Riduii2=C-dt- =C. uoi=R2d(ui-uo)dtui-uouo cd(ui-uo)R1 =R2 c dt(b)解：(uWi)"i = i1+i2 i= R1i =上也 ii=R2 i2=C dtH H _ L duoI ui-uo=R2 dtR2(ui-uo)=Riu0-CRiR2(給-第)CRiR2duo+Riuo+R2u0=CRiR2 dLi+R2uiui-ui uodui-Rr=R2+Cdt I due ui”R2d"山 d

2、ol duo u。du%CLd2uoRi Ri-RiR2 dt = R2+c dt + R 2 dt2 出=瑤_a+9+外+1+)口 Ri= R2 dt2 +(C+RiR2) dt +(Ri+R2)uo2-2求下列函數(shù)的拉氏變換。(i) f(t尸sin4t+cos4t解：Lsin(ot= 0P+s2 Lcosa t= sLsin4t+cos4t = s2+4W + s2+si6=sS+46(2) f(t)=t 3+e4t解：Lt3+e4t = -+1 - 6s+24+s4 s-4- (s+4)f(t)=tneatf(t)=(t-1)2e2tc解：Ltneat=高解：L(t-1)2=aS)割2-

3、3求下列函數(shù)的拉氏反變換。s+1(1)F(s)=AA2(s+2)(s+3)=2+3解：A1=(s+2)(L_3ys=2=_1A2=(s+3s+(s+3)/s=-3=2F(s尸金一±f(t)=2e-3t-e-2tQ)F(s尸作+島s+2)=(i*+翁+翁解：A1=(s+1)2(s+1)s+2)/s=-1=-1人2=急品】屋=2A3=(s+2)(s+1)2(s+2)/s=-2=-2f二-2乳怕寸+2?F(s群平曾+令解：F(s)(s2+1)/s=+j=AlS+A4=+j2#-gS+%=j=A iS+A24=j“用A1+A2 -5j-1=-A 1+jA2A=1A2=-5 A3=F(s)s/

4、s=0=1F(s尸 + f(t)=1+cost-5sint F(s尸 s(s+需+3)右刀 A A2 A3 A4 斛：二由7+言+中+承 Ai= A3=- A4=12 A2=s=-1_ s(s+3A(s+2)(2s+3)y _-s(s+3)2/s=-1=4f(t)=儕3-金屋+會+自e®(2-4)求解下列微分方程。(1)耳段+5誓+6y(t)=6y(0)=y(0)=2解：s2Y(s)-sy(0)-y'(0)+5sY(s)-5y(0)+6>fs)=y=6+2金+患冬A2 A3F+S+3A1=1y(t)=1A+2=e52t-4e-3tA3=-42-5試畫題圖所示電路的動態(tài)結(jié)

5、構(gòu)圖，并求傳遞函數(shù)。ur-3i R2=+ucUr(S)(R1+sC)R2R2+R1R2sCUc(s)=1+(11+sC)R2=R1+R2+R1R2sCurR2L3=-1/sCR1 LiL3=R2/LCRiS2L i=-R2 /Ls L2=-/LCSP1=R2/LCR 1s2 A1=1R2U4sUc(s)= R1cLe+(R 1R2C+L)s+R+R2-8設(shè)有一個初始條件為零的系統(tǒng)，系統(tǒng)的輸入、輸出曲線如圖，求G(s)。占CK0Tc(t尸聿邛(t-T C(s尸表(1) C(s)=G(s)2-9若系統(tǒng)在單位階躍輸入作用時，已知初始 條件為零的條件下系統(tǒng)的輸出響應(yīng)，求系統(tǒng)的 傳遞函數(shù)和脈沖響應(yīng)。r(

6、t)=I(t)-2j -tc(t)=1-e +eR(s)= s1，一(s2+4s+2): C(s)= s - s+2+ s+1 =s(s+1)(s+2)G(s)=C(s)/R(s)=-(s2+4)s+2)C(s)=群解2) -tc(t戶 6 (t)+2e-eRG1Ric+G22G3(s)C(s)+G2(s)Gi(s)+H(s)R(s(c)R(s+1C(e)+3=1=1=1=1C(s) R(s)C(s) R(s)(a)R(s)(b)R(s)-1(sG1G1H1G11+636266+636465+61626364。7-G8)(d) R(s)L3G1G22-10已知系統(tǒng)的拉氏變換式，試畫出系 統(tǒng)的動

7、態(tài)結(jié)構(gòu)圖并求傳遞函數(shù)。解：X1(s)=R(s)G1(s)-G1(s)G7(s)-G8(s)C(s)=R(s)-C(s)G7(s)-G8(s)G1(s)X2(s)=G2(s)X1(s)-G6(s)X3(s)X3(s)=G3(s)X2(s)-C(s)G5(s)C(s)=G4(s)X3(s)H1H1C(s)G 7(s)-G8(s)C(s)G5(s)G61+G1G2H2Ms)G7A=1+G2H p1=g1g2 P2=G3G2眸：L-G162H L2=- =1+G1G4H+G 162H63G2|6162(1 GH1)l1=-g2h1L2=-G162H2L1=-G1G3 L24H P1=G162 Al =

8、1>2=G3G2 A2=1+G164HG3G4 1+G3G2G622 八2蹈二仁什巳）1+GL1=-G2H P=G1 A1G3 +n心 2 PC(s)- k=1R(s) Z14=6264 P1=G A 1=1 P2=G2 2=1C(s) _(G1+G2)R(s) 1+G1G3+G2G3 "6164-62G4:|G3(s) t -G1(s)'C(s)= 61G2+G2G3+G16263G4 H R(s)=1 + G162H+ G164HkkG2G1+G2G3；=1+G2Hl+G162H2解：Li=-GiG2L2=G2P1=G1A1=1-G=1+G1G2G-Gi(1-G2)

9、=1+G1G2C2L2=-G1G2H3P1=G1G2A1=1G2GlL1=G2H2=1-G2H2+G1G2H3P1=G2A1=1P2=-G1G2HlA2=1C(s)G20-G1H1)D(s)=1-G2H2+G1G2H3解：L1=-G1G2L2=-G1G2HP1=G1G2A1=1(b)C(s)GGR(s)=1+G1G2H+G1G2P1=GnG2A1=1P2=142=1+62HC(s)1+GnG2+G£2HD(s)=1+G1G2+G1G2H解：L1=-G2L2=-G1G2G3P1=G2G31=1P2=G1G2G3=1P1=-G2G3A1=1C(s)G2G3+G1G2G3R(s)=1+G2

10、+G1G2G3p2=1a2=1+g2C(s)_-G2G3+1+G2R(s)=1+G2+G1G2G3解：L1=-GG4L2=-GG3G5P產(chǎn)G1G5A1=1P2=G2G3G5A2=1Cs)_G1G2G5+GG521R(s)=1+G2G3G5+G3G4P產(chǎn)G1G5A1=1P2=1A2=1+3g4E(s)G1G5+(1+G1G5)R(s)=1+G2G3G5+G3G4EXs),1解：P1=G1G3a1=1p2=g2g3A2=1P3=G1G4A3=1P4=G2G4A4=1L1=-G1G3L2=-G2G3C(s)_(G1+G2)(G3+G4)113223R(s)=1+G3(G1+G2)一二XX(s)_c/

11、、C(s)R(s)=1+G3(G1+G2)E(s)=G2(s)D(s)=1C2(s)_G4G5G6(1-G1G2)麗T+G4+G1G4G5H1H2-G1G2-G1G2G4解：L產(chǎn)G1G2L2=-G1G4G5H1H2L3=-G4=1-G1G2+G1G4G35H1H2+G4-G1G2G4P1=G1G2G31=1+G4Ci.GiG2GKg)R1(S)=1+G4+GiG4G5H1H2-G1G2-G1G2G4C1(s)勺92G3G4G5H1R(S)=1+G4+G1G4G5H1H2-G1G2-G1G2G4C2(s)_G1G4G5G6H2RTS)=1+G4+G1G4G5H1H2-G1G2-G1G2G43-1

12、設(shè)溫度計需要在一分鐘內(nèi)指示出響應(yīng)值的98%,并且假設(shè)溫度計為一階系統(tǒng),求時間常數(shù)T。如果將溫度計放在澡盆內(nèi)，澡盆的溫度以10oC/min的速度線性變化,求溫度計的誤差。解:c(t)=c()98%t=4T=1minT=0.25r(t)=10tc(t)=10(t-T+e-t/T)=10(T-t/T)e(t)=r(t)-c(t)ess=lime(t)=10T=2.53-2電路如圖，設(shè)系統(tǒng)初始狀態(tài)為零。C=NFR0=20kQR1=200kQ(D求系統(tǒng)的單位階躍響應(yīng)及Uc(t1)=8時的t1值.UR0解:R1/R0G=R1Cs+1=Ts+1t2tUc(t)=K(1eT)=10(1-e2t)T=R1C=0

13、.5K=R1/R=100.8=1-e-2te-2t=0.28=10(1-e-2t)t=0.8解：t1=0.8R(s尸 s2 R(s)=3+T2-3)(2)求系統(tǒng)的單位脈沖響應(yīng)，單位斜坡響應(yīng)，及單位拋物響應(yīng)在t1時刻的值.R(s)=1g(t尸午e-t/T=4Uc(t)=K(t-T+Tet/T)=4Uc(s)=TsJ+7i3=K(i3-Juc(t)=1磴t2-0.5t+0.25-0.25e-2t)=1.23-3已知單位負(fù)反饋系統(tǒng)的開環(huán)傳遞函數(shù)，求系G(s)=T4K統(tǒng)的單位階躍響應(yīng)。s(s5)解:C(S)=S2+5S+4R(s)=1C(s尸乖+1扁)=1+睛-第1-4t4-tc(t)=1+3e-3e

14、3-4已知單位負(fù)反饋系統(tǒng)的開環(huán)傳遞函數(shù)，求系統(tǒng)的G(s尸s(g+i)升時間tr、峰值時間tp、超調(diào)量6。%和調(diào)整時間tso解：C(S)一1JZCi)n=1Jn=1d=Ci)n1-1'=0.866,R(S)=s+s+1ton2=1k=0.5B=tg-1受=60。一鄧一3.14-3.14/3fd,t一二一3.14,363tr=-d=0.866=2.42tp=wd=0.866=3.63(r%=e4'+100%6%ts=&=6ts=喙=8-1.8e-60t-10t3-6已知系統(tǒng)的單位階躍響應(yīng)：c(t)=1+0.2e-1.2e(1)求系統(tǒng)的閉環(huán)傳遞函數(shù)。(2)求系統(tǒng)的阻尼比和無阻

15、尼振蕩頻率。:小+_02_心600:C(s)=s+s+60s+10=s(s+60)(s+10)R(q)=1C(s)=,600-3n=70R(s)sR(s)s2+70s+600Ln2=6003-7設(shè)二階系統(tǒng)的單位階躍響應(yīng)曲線如圖，系統(tǒng)的為單位反饋，求系統(tǒng)的傳遞函數(shù)。解：jtp=-=0.1(兀g=ln3.31.19:產(chǎn)爸&九)2/1-12=1.42e=0.3e>MT2=339.86q2=1.42-1.4212(=0.35con=33.4G(s)=321115.6s(s+2cn)=s(s+22.7)3-1柒:，種M2移+9s+瞅妗W$+8s+Ws明6s15細(xì)穩(wěn)定性。勞斯表如下:s319

16、s220100s14s0100勞斯表如下：s41185s3816s2165q1216s16系統(tǒng)穩(wěn)定。s05系統(tǒng)穩(wěn)定。G(s2=網(wǎng)(0呼嬰饋系統(tǒng)的開環(huán)傳遞函9s(s+加瞥1就時K值范圍。解：s4+3s3+4s2+2s+Ks+2K=0s4s2s113b31b4142+K2K2K10二2Kb31=飛一K2+10K-20b41=10-K-(K-1.7)(K+11.7)>0K<103-13已知系統(tǒng)結(jié)構(gòu)如圖,統(tǒng)穩(wěn)定時T值范圍。試確定系解：G(s尸10(1+1).s2+s+10rs=s(s2+sms)s3s2不小-一10(s+1)(s)=s3+s2+10rs2+10s+106產(chǎn)也圖逑0ss,11

17、0(1+10)10b3110T>03-14已知系統(tǒng)結(jié)構(gòu)如圖，試確定系統(tǒng)穩(wěn)定時p值范圍。Rsk國用sS+MCS)解：G(s)=10(rS+1)s3s2S211b3110s+1)10T10(s)=10Gs+1)s3+s2+10rs+10b31=>0T>13-16已知單位反饋系統(tǒng)的開環(huán)傳遞函數(shù)，試求Kp、Kv和Ka.并求穩(wěn)態(tài)誤差ess.解：(1)G(s)=(0.1s+2%.2s+)u=0Kp=20Kt=0Ka=0ess=r(t)=I(t)+2t+t2R(s尸W+2+看R0_1ess1=1+K=21ess2=°°ess3=°°oo一一，、200

18、10.G(s)=s(s+2)(s+10)=s(0.5s+1)(0.1s+平)11Ku=10JKa=0ess1=0八_2._2ess2=K=10ess3=00ess=°°G=J0Qs+1)一2小偉+1)(3)G(s)=s2(s2+4s+10)=s2(0.1s2+0.4s+1)u=2Kp=00ess1=0Ko=00ess2=0Ka=1ess3=2ess=23-17已知系統(tǒng)結(jié)構(gòu)如圖。(1)單位階躍輸入：(r%=20%ts=1.8(5%)確定K1和p值。解：G=sfe (s尸 s2+K2s+K1p n2 =kt =0.453n=1.830.453.7K1= 3r2=13.71 I

19、ts= (3nr =0.24=0.2= 1.8 求系統(tǒng)的穩(wěn)態(tài)誤差：Kt)=i(t), t ,解:G=益莪s("s+1)u =1it21I R(s)=4R R(s)=4 s ，R(s)=?Kp=°°Ko =KKa=0ess1=0 ess2 =0.24ess3= 003-18已知系統(tǒng)結(jié)構(gòu)如圖。為使(二時，單 位斜坡輸入的穩(wěn)態(tài)誤差ess=定K和 值。K解: G(S)=S2+2s+Kr s =5(2+1)尸 s2+(2+Kr)S+K12 3n =2+Kr =2*0.7 JKesS=2+K=0.25K=31.6%n2 = Kr- 0 25K-2 r =0.186r = K3

20、-19系統(tǒng)結(jié)構(gòu)如圖D1(s)D2(s)r(t)=d(t)=d(t)=I(t)Rp炳升+ :囤 )>(1)求r(t)作下的穩(wěn)態(tài)誤差.H解: essr=sim0 s 1+G(ssF(s)=1+G(0)F(0)(2)求d1(t)和d2(t)同時作用下的穩(wěn)態(tài)誤差.Ed(s)=-G,(s)H(s)1+G 1(s)G2(s)H(s)D(s)e 產(chǎn) lim s F(s) +-1-1essdr "。s1+G(s)F(s) +1+G(s)F(s) s =-1+F(s)1+G(0)F(0)求di(t)作用下的穩(wěn)態(tài)誤差.1JSG(s)=Kp+-KF(s尸J1S-F-F(S)1i+(KP+-K)jSe

21、ssd=SimoS1+G(s)F(s)S=Simos4-1已知系統(tǒng)的零、極點分布如圖，大致繪制出系統(tǒng)的根軌跡。已知開環(huán)傳遞函數(shù)，試用解析法繪制出系統(tǒng)的根軌并判斷點(-2+j0),(0+j1),(-3+j2能否在根軌跡上。KrG=(s+1)4-2跡,解:K一(s尸s+K+Krs=-1-Kr-3+j2*j30+j1Kr=0s=-1Kr=foo s=-ooI M-2 -1p3zs=-2+j0s=0+j1s=-3+j24-3已知系統(tǒng)的開環(huán)傳遞函數(shù)，試?yán)L制出根軌跡圖。解：G©=Kr(s+1.5)(s+5.5):G(s)=s(s+1)(s+5)1)開環(huán)零、極點P1=0P2=-1P3=-5z1=-

22、1.5z2=-5.52)實軸上根軌跡段PiP2z1P3Z2-°°3)根軌跡的漸近線n-m=10=+180°4)分離點和會合點A(s)B'(s)=A'(s)B'A(s)=s3+6s2+5sB(s)=s2+7s+8.25A(s)'=3s2+12s+5B(s)'=2s+7s1=-0.63s2=-2.5s3=-3.6s4=-7.280G/7-Kr”5)、(2)G(s)=s(s+1)(s+4)1)開環(huán)零、極點P1=0P2=-1P3=-4z1=-1.52)實軸上根軌跡段P1P2P3Z13)根軌跡的漸近線n-m=29=+90°(

23、7=-1-4+1.5=-1.754)分離點和會合點JA(s)=s3+5s2+4sjB(s)=s+1.5U(s)'=3s2+10s+4Ib(s)'=1s=-0.62G(s)=s(S+iyi)開環(huán)零、極點Pl=0P2=1P3=12)實軸上根軌跡段P1P2P3-°°3)根軌跡的漸近線n-m=39=+60o,±180°kg1=-0.674)根軌跡與虛軸的交點s3+2s2+s+r0P2p3-0.6Ico1P1-1Kr=0Kr=2(4)G(s)=31=032,3=1Kr(s+8)s(s+3)(s+7)(s+15)5)分離點和會合點fA(s)=s3+2

24、s2+sIB(s)=1A(s)'=3s2+4s+1、B(s)'=0s=-0.331)開環(huán)零、極點P1=0P2=-3P3=-7P4=-15z1=-82)實軸上根軌跡段P1-P2P3Z1P4-x3)根軌跡的漸近線n-m=3ct=-3-7-15+8=-5.679=+6C0,+180°4)根軌跡與虛軸的交點s4+25s3+1712+323s+8Kr=0Kr=0co1=0Kr=638a2,3=土6.25)分離點和會合點A(s)=s4+25s3+171s2+315sA(s)'=4s3+75s2+342s+315B(s尸s+8B(s)'=2s+7s=-1.44-5已

25、知系統(tǒng)的開環(huán)傳遞函數(shù)。（1）試?yán)L制出根軌跡圖。（2）增益Kr為何值時,復(fù)數(shù)特征根的實部為-2。4-6G(s尸K(s+2)s(S+1)解：Pi=0P2=-1Zi=-2p1p2z1-°°分離點和會合點s2+4s+2=0Si=-3.41S2=-0.59閉環(huán)特征方程式s2+s+Krs+2Kr=0-s=-2+ja(-2+j。)2+(-2+jw)(1+Kr)+2Kr=0廠4川+(1+Kr筋=0l4-32-2(1+Kr)+2Kr=0卜=±1.41Kr=3已知系統(tǒng)的開環(huán)傳遞函數(shù)，試確定閉環(huán)極點L二時的Kr值(1) G(s)H(s)=s(s+1K)(s+3)解：P1=0P2=-1P

26、3=-3p1p2P38(y=33=-1.39=+60°,+180o根軌跡的分離點：A(s)B'(s)=A'(s)B(s)3s2+8s+3=0s1=-0.45s2=-2.2舍去與虛軸交點s3+4s2+3s+Kr=0f3+3w=0fKr=0a1=0、Kr-4a2=01Kr=12a2,3=±1.7(2) G(s)H(s)s+3)(K2r+2s+2)解：P1=0P2=-3p3.4=-1jp1p2(T=3-1-1=-1.259=+45°,+135°根軌跡的出射角03=+lt-01-02-04=±135090026.60=7.6°

27、與虛軸的交點s(s+3)(s2+2s+2)+Kr=0s4+5s3+8s2+6s+Kr=0(jw)4+5(j3)3+8(jco)2+j63+Kr=0%4-8a2+Kr=0fKr=031=0l533+6co=01Kr=8.16W2,3=±1.1s3P31M-3t=0.5得s1=-0.37+j0.8s3=-4+0.37x2=-3.26二惜3|23+1|口+3|=3.26x2.26x0.26=1.9分離點和會合點4s3+15s2+16s+6=0解得s=-2.3<=0.5得s1=-0.36+j0.75Kr=|s111sl+3|s1+1+j|s1+1-j|=2.92j co4-7已知系統(tǒng)的

28、開環(huán)傳遞函數(shù)，(1)試?yán)L制出根我跡圖。G(s)H(s尸s(s+2)(s+4)解：Pi=0P2=-2P3=-4PlP2P3-8L234=-29=±60°,+180°根軌跡的分離點：A(s)B'(s)=A'(s)B(s)33±8cjd=0 Kr=0 a 1=0:Kr-6a2=0 1 Kr=48 a 2,3= ±2.8(4) L =0.5 S=-O.7±j1.2s3=-6±0.7 X2=-4.6Kr=4.6X2.6X0.6=7.2si=-0.85s2=-3.15舍去(2)阻尼振蕩響應(yīng)的Kr值范圍s=-0.85Kr=

29、0.85x1.15X3.15=3.1s=±j2.8Kr=48(3)與虛軸交點s3±6s2±8s+Kr=05-1已知單位負(fù)反饋系統(tǒng)開環(huán)傳遞函數(shù)，當(dāng)輸入信號r(t)=sin(t±30°),試求系統(tǒng)的穩(wěn)態(tài)輸出。解：G(s尸部(s)=(s+°T)A(3)W+h?1襁=0.905。心)=也-阱=屯-1#=-5.2。cs(t)=0.9sin(t±24.8°)5-2已知單位負(fù)反饋系統(tǒng)開環(huán)傳遞函數(shù)，試?yán)L制系統(tǒng)開環(huán)幅相頻率特性曲線。解：G(s尸s(s+7jg±15)1型系統(tǒng)n-m=311m3=0A(co尸X。(3)=儂。G

30、0=°°顯3=00A(co)=0(|)(co)=-270°y解：(5) 6(5)=1)解：I型系統(tǒng)n-m=2=0 A(cd )= 00 (J)(cd )=-27(0°=00 A(w)=0(I)(co )=-H8©°3=0解:G(s)=(2s+1)(8s±1)II：_00型系統(tǒng)n-m=2一下eco=0A()=10小()=0°I二°°A()=0小仙)=180°解：G(s)=春箭毫5II型系統(tǒng)n-m=3產(chǎn)s(siW.I)(silw)3=0A(i)=00()(co)=-180°=0J

31、D=°°，0Re3=00A(a)=0()(w)=-270)°5-2已知單位負(fù)反饋系統(tǒng)開環(huán)傳遞函數(shù)，試?yán)L制系統(tǒng)開環(huán)對數(shù)頻率特性曲線。L(wydB10(1)G(s)(3)G(s)-20dB/dec(2s+1)(8s+1)-20dB/dec20lgKG(s)-40dB/decG(co)=01040dB/dec=01.33(5s+1)(7)G(s)20lgK=2.5dB1531=0.1=0-60dB/dec5-4L()dB(a),20dB/deccK=100(c)0-60dB/dei0100G(s)s(100s+1)(0.01s+1)40200-200-90-180-27

32、040200-20200-20-9018。3(co)=-180o(w)=-27Q°0.01-40dB/dec-60dB/dec0.10.2-60dB/decL(3),dB-20dB/decL(3)dB、-40dB/dec-60dB/dec102020dB/dec已知系統(tǒng)的開環(huán)幅頻率特性曲線，寫出傳遞函數(shù)并畫出對數(shù)相頻特性曲線。32=0.2w3=15G(s)=(s+1)(0.彳5+(0).01s+1)(c)K=100G(s尸s(100s+00)01s+1)10(s+0.2)s2(s+0.1)(s+15)s2(10s+1)(0.67s+1)解：20lgK=20dB3=0.12532=0.

33、5s(51s+1)(115s+1)20lgK10L(dB-20dB/dec()(a)=-1l80!)。以)dB-20dB/decL(ydB120lgK=20dB3=5m=153=03(D)=-900。3=°°j(co)=-270。-40dB/dec(3)-90-180-270,|處)0-90*-180''-270一"L(vdB丁-20dB/dec20lgK-40dB/dec110501001515-40dB/dec&3)-60dB/dec4020lgK=2020d(a)=-270)。-206心)=0)。j(co)=-11800。(d)20l

34、gK=4848K=2510(b)20lgK=-20K=0.10、01s-20G(s)=(0.05s+1)750s(s+5)(s+15)1020lgK=20K=1020G(s)=(011s+1)00.01-40dB/dec(e)由圖可得：20lgMr=4.58dBMr=1.7=。=±0.94上=±0.32X/1-t7得:1=0.3以3),dB0100-20dB/dec4.58dBJ100，CO3r=45.-60dB/dec3r=3nJT-23n=50K=W°=100T2=(n)2=0.0222TC=0.01G(s)=s(0.02s)2+0.01s+1)5-7已知奈氏曲線，p為不穩(wěn)定極點個數(shù)，u為積分環(huán)節(jié)個數(shù)，試判別系統(tǒng)穩(wěn)定性。(a)p=0i系統(tǒng)不穩(wěn)定系統(tǒng)穩(wěn)定(d)i系統(tǒng)穩(wěn)定ImIm0=0系統(tǒng)穩(wěn)定ImImp=1(h)oo