.目錄一、課設(shè)的任務(wù)與要求...........................................................................................................................................1I、課設(shè)的題目與問(wèn)題......................................................................................................................................1II課設(shè)的范化要求............................................................................................................................................1二、設(shè)計(jì)正文...........................................................................................................................................................2I、第一問(wèn)..........................................................................................................................................................31-1題目的處理........................................................................................................................................31-2題目的代碼........................................................................................................................................31-3運(yùn)行的的結(jié)果....................................................................................................................................31-4比較....................................................................................................................................................4II、第二問(wèn).......................................................................................................................................................42-1題目的處理........................................................................................................................................52-2題目的代碼........................................................................................................................................52-3運(yùn)行的結(jié)果........................................................................................................................................5K>0.............................................................................................................................................................6K<0.............................................................................................................................................................6III、第三問(wèn)........................................................................................................................................................63-1題目的處理........................................................................................................................................63-3運(yùn)行的結(jié)果........................................................................................................................................73-4結(jié)論....................................................................................................................................................8IV、第四問(wèn)........................................................................................................................................................8第四問(wèn)中的第一小題...............................................................................................................................84-1-1題目的分析..............................................................................................................................84-1-2題目的代碼..............................................................................................................................84-1-3運(yùn)行結(jié)果..................................................................................................................................94-1-4系統(tǒng)的校正.............................................................................................................................94-1-4-1超前校正......................................................................................................................94-1-4-2滯后校正....................................................................................................................10第四問(wèn)中的第二小題.............................................................................................................................114-2-1題目的分析............................................................................................................................114-2-2題目的代碼............................................................................................................................114-2-3運(yùn)行的結(jié)果...........................................................................................................................124-2-4系統(tǒng)校正...............................................................................................................................124-2-4-1超前校正....................................................................................................................124-2-4-2滯后校正..................................................................................................................13三、附錄.................................................................................................................................................................16第一問(wèn).............................................................................................................................................................16第二問(wèn).............................................................................................................................................................17K>0...........................................................................................................................................................17K<0...........................................................................................................................................................17;..第三問(wèn) 18第四問(wèn) 19第一題 19未校正的單位階躍響應(yīng)圖 19未校正前的 BODE圖 19超前校正 19滯后校正 21第二題 21未校正的單位階躍響應(yīng)圖 22可得未校正前的 BODE圖 22超前校正 22滯后校正 23滯后—超前校正 24四、 總結(jié) 26五、參考文獻(xiàn) 26;..一、課設(shè)的任務(wù)與要求、課設(shè)的題目與問(wèn)題題目：已知單位負(fù)反饋系統(tǒng)被控制對(duì)象的開(kāi)環(huán)傳遞函數(shù)Gk(s)

Ks(0.1s 1)問(wèn)題：分析系統(tǒng)單位階躍響應(yīng)的時(shí)域性能指標(biāo)2. 當(dāng)k , 時(shí)，繪制系統(tǒng)的根軌跡，分析系統(tǒng)的穩(wěn)定性對(duì)系統(tǒng)進(jìn)行頻域分析，繪制其N(xiāo)yquist圖及Bode圖，確定閉環(huán)系統(tǒng)的穩(wěn)定性用串聯(lián)校正的頻率域方法對(duì)系統(tǒng)進(jìn)行串聯(lián)校正設(shè)計(jì)，使系統(tǒng)滿足如下動(dòng)態(tài)及靜態(tài)性能指標(biāo)：4.1設(shè)計(jì)串聯(lián)校正滿足下列性能指標(biāo)（1）在單位斜坡信號(hào)r(t)t作用下，系統(tǒng)的穩(wěn)態(tài)誤差ess0.01；（2）系統(tǒng)校正后，相位裕量'('c)450。4.2設(shè)計(jì)串聯(lián)校正滿足下列性能指標(biāo)（1）在單位斜坡信號(hào)r(t)t作用下，系統(tǒng)的穩(wěn)態(tài)誤差ess0.005；（2）系統(tǒng)校正后，相位裕量'(c')450。（3）系統(tǒng)校正后，幅值穿越頻率'50。c課設(shè)的范化要求單位階躍響應(yīng)時(shí)域指標(biāo)的分析：繪制系統(tǒng)的單位階躍響應(yīng)曲線，利用曲線計(jì)算系統(tǒng)單位階躍響應(yīng)的時(shí)域性能指標(biāo)，包括上升時(shí)間，峰值時(shí)間，調(diào)節(jié)時(shí)間，超調(diào)，振蕩次數(shù)根據(jù)Matlab相關(guān)命令繪制系統(tǒng)的根軌跡，通過(guò)根軌跡分析系統(tǒng)的穩(wěn)定性根據(jù)Matlab相關(guān)命令繪制系統(tǒng)的Nyquist圖和Bode圖，由圖分析系統(tǒng)的穩(wěn)定性及穩(wěn)定裕度按如下步驟，利用頻率域串聯(lián)校正方法對(duì)系統(tǒng)進(jìn)行的串聯(lián)校正設(shè)計(jì)：（1）根據(jù)要求的穩(wěn)態(tài)品質(zhì)指標(biāo)，求系統(tǒng)的開(kāi)環(huán)增益值；;..（2）根據(jù)求得的值，畫(huà)出校正前系統(tǒng)的 Bode圖，并計(jì)算出幅值穿越頻率、相位裕量 (要求利用MATLAB軟件編程進(jìn)行輔助設(shè)計(jì)),以檢驗(yàn)性能指標(biāo)是否滿足要求。若不滿足要求，則執(zhí)行下一步；（3）畫(huà)出串聯(lián)校正結(jié)構(gòu)圖，分析并選擇串聯(lián)校正的類(lèi)型（超前、滯后和滯后-超前校正）。（若可以采用多種方法，分別設(shè)計(jì)并進(jìn)行比較）（4）確定校正裝置傳遞函數(shù)的參數(shù)；（5）畫(huà)出校正后的系統(tǒng)的Bode圖，并校驗(yàn)系統(tǒng)性能指標(biāo)(要求利用MATLAB軟件編程進(jìn)行輔助設(shè)計(jì))。若不滿足，跳到第（4）步。否則進(jìn)行下一步。（6）提出校正的實(shí)現(xiàn)方式及其參數(shù)。（要求實(shí)驗(yàn)實(shí)現(xiàn)校正前、后系統(tǒng)并得到的校正前后系統(tǒng)的階躍響應(yīng)）（7）若采用不同串聯(lián)校正方法進(jìn)行設(shè)計(jì)，比較不同串聯(lián)校正方法的特點(diǎn)。（穩(wěn)定性、穩(wěn)態(tài)性能、動(dòng)態(tài)性能和實(shí)現(xiàn)的方便性的比較）二、設(shè)計(jì)正文;..、第一問(wèn)1-1題目的處理設(shè)K=10 時(shí)K*=100GK=100/[s(s+1)]G=100/[s^2+10s+100]1-2題目的代碼在MATLAB中運(yùn)行以下代碼：GK=tf([100],[1100]);G=tf([100],[110100]);step(G)Wn=10;Kexi=10/(2*Wn);Beta=acos(Kexi);Wd=Wn*sqrt(1-Kexi^2);Tr=(pi-Beta)/Wd;Tp=pi/Wd;Ts=4.4/(Kexi*Wn);Ct=exp(-pi*Kexi/sqrt(1-Kexi^2))*100;1-3運(yùn)行的的結(jié)果Transferfunction ——G:100-----------------------s^2+10s+100Wn=10Kexi=0.5000Beta=;.1.0472Wd=8.6603Tr=0.2418Tp=0.3628Ts=0.8800Ct=16.30341.41.210.8edtlpmA0.60.40.200

.StepResponseSystem:GPeakamplitude:1.16Overshoot(%):16.3Attime(sec):0.36System:GSystem:GSystem:GFinalValue:1SettlingTime(sec):0.808Time(sec):0.216Amplitude:0.9130.2 0.4 0.6 0.8 1 1.2Time(sec)1-4比較以上數(shù)據(jù)與圖中數(shù)據(jù)相吻合、第二問(wèn);..2-1題目的處理GK=K*-------------s^2+10s分K*>0和K*<0兩項(xiàng)2-2題目的代碼在MATLAB中運(yùn)行以下代碼：當(dāng)K*>0時(shí)；num=[1];den=[1100];rlocus(num,den);[Kp]=rlocfind(num,den)title( ‘ROOTLOCUS’);K*<0 時(shí)num=[-1];den=[1100];rlocus(num,den);[K*p]=rlocfind(num,den) ；2-3運(yùn)行的結(jié)果K*=72.8796即K=7.28396p=-14.89344.8934;..K>0ROOT LOC US6isxAaringamI

0.890.810.70.560.380.240.95System:sysSystem:sysGain:0.207Gain:0.07120.988Pole:-9.98Pole:-0.00711Damping:1Damping:1Overshoot(%):0Overshoot(%):012Frequency(rad/sec):9.98Frequency(rad/sec):0.007112108640X:-5-20.988Y:-2.814-4 0.950.89 0.81 0.7 0.56 0.38 0.2-6-12 -10 -8 -6 -4 -2 0 2RealAxisK<0xisAynagimaI

RootLocus2.50.9960.9930.9860.9650.8620.9981.510.9990.51001252015105-0.51-10.999-1.50.998-2-2.50.9960.9930.9860.9650.86-30-25-20-15-10-505101520RealAxis2-4結(jié)論由圖可看出K*>0 即K>0可滿足系統(tǒng)穩(wěn)定；由圖可看出K*<0 即K<0 時(shí)，系統(tǒng)不穩(wěn)定、第三問(wèn)3-1題目的處理由以上可知K>0 時(shí),系統(tǒng)都穩(wěn)定故設(shè)K=10,K*=100;則GK=100/[s(s+10)];3-2題目的代碼;..在MATLAB中代碼為：num=[100];den=[1100];[magphasew]=bode(num,den)margin(mag,phase,w)nyquist(num,den)grid3-3運(yùn)行的結(jié)果)Bd(edtuingaM

BodeDiagramGm=92dB(at2e+003rad/sec),P m=51.8deg(at7.85rad/sec)500-50-100-90)ged(e-135sResponse:untitled1ahFrequency(rad/sec):8.22PPhase(deg):-129-180-101231010101010Frequency(rad/sec)NyquistDiagram200dB15105sx2dB-2dBAy4dB-4dBa0namI-5-10-15-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-0.10RealAxis;..3-4結(jié)論根據(jù)nyquist 穩(wěn)定判據(jù)，R=0，P=0，Z=0可知系統(tǒng)穩(wěn)定IV、第四問(wèn)第四問(wèn)中的第一小題4-1-1題目的分析由單位斜坡信號(hào)作用下ess<=0.01 得出K=100GK=100/[s(0.1s+1)];先做出未校正前的相關(guān)圖形；4-1-2題目的代碼Gk=tf([100],[0.110]);G=feedback(Gk,1);step(G)num1=[100];den1=[0.110];figure(1);[mag1,phase1,w]=bode(num1,den1);margin(mag1,phase1,w);..4-1-3運(yùn)行結(jié)果看圖可知不滿足相位裕度 >45etduilmA

StepR esponse1.81.61.41.210.80.60.40.2000.20.40.60.811.21.4Time(sec))(eudnitagM

BodeD iagramGm =72dB(at2e+003rad/sec),P m =18deg(at30.8rad/sec)100500-50-100-90)g(ashP

-135-1801010101010Frequency(rad/sec)（圖見(jiàn)附錄）4-1-4 系統(tǒng)的校正4-1-4-1超前校正4-1-4-1-1 問(wèn)題的計(jì)算由計(jì)算可得T=0.04s a=3.36;Gc=(0.s+1)/(0.04s+1);結(jié)構(gòu)圖;..原理圖R1U1 C R2 U24-1-4-1-2 題目的代碼代碼；Gk=tf(num0,den0);G=feedback(Gk,1);step(G)代碼；num2=[0.0311];den2=[0.0621];num0=conv(num2,num1);den0=conv(den2,den1);[mag0,phase0,w]=bode(num0,den0);figure;margin(mag0,phase0,w)4-1-4-1-3 運(yùn)行的結(jié)果（圖見(jiàn)附錄）4-1-4-2滯后校正4-1-4-2-1 問(wèn)題的計(jì)算計(jì)算可得T=11.87 b=0.104;Gc=(1+1.23s)/(1+11.87s)結(jié)構(gòu)圖;..原理圖R1R2U1 U2C4-1-4-2-2 問(wèn)題的代碼figurenum3=[1.231];den3=[11.871];num4=conv(num3,num1);den4=conv(den3,den1);[mag4,phase4,w]=bode(num4,den4);margin(mag4,phase4,w);由代碼Gk=tf(num4,den4);G=feedback(Gk,1);step(G)4-1-4-2-3 運(yùn)行的結(jié)果（圖見(jiàn)附錄）第四問(wèn)中的第二小題4-2-1題目的分析由題在單位斜坡信號(hào)作用下，系統(tǒng)的穩(wěn)態(tài)誤差 <=0.005 得出K=200;GK=200/[s(0.1s+1)];先做出未校正前的相關(guān)圖形；4-2-2題目的代碼由代碼Gk=tf([200],[0.110]);G=feedback(Gk,1);;..step(G)由代碼num1=[200];den1=[0.110];[mag1,phase1,w]=bode(num1,den1);margin(mag1,phase1,w)4-2-3 運(yùn)行的結(jié)果（圖見(jiàn)附錄）4-2-4 系統(tǒng)校正4-2-4-1超前校正4-2-4-1-1 題目的計(jì)算由計(jì)算可得T=0.016s a=1.63;Gc=(1+0.026s)/(1+0.016s);結(jié)構(gòu)圖原理圖R1U1 C R2 U24-2-4-1-2 題目的代碼;..由代碼num2=[0.0261];den2=[0.0161];num0=conv(num2,num1);den0=conv(den2,den1);[mag0,phase0,w]=bode(num0,den0);figure;margin(mag0,phase0,w);由代碼Gk=tf(num0,den0);G=feedback(Gk,1);step(G)4-2-4-1-3 運(yùn)行的結(jié)果（見(jiàn)附錄）4-2-4-2 滯后校正4-2-4-2-1 問(wèn)題的分析計(jì)算可得T=23.52 b=0.052;Gc=(1+1.22s)/(1+23.52s)結(jié)構(gòu)圖原理圖R1R2U1 U2C4-2-4-2-2 問(wèn)題的代碼由代碼Figure;..num3=[1.221];den3=[23.521];num4=conv(num3,num1);den4=conv(den3,den1);[mag4,phase4,w]=bode(num4,den4);margin(mag4,phase4,w);由代碼Gk=tf(num4,den4);G=feedback(Gk,1);step(G)4-2-4-2-3 運(yùn)行的結(jié)果（見(jiàn)附錄）4-2-4-3 滯后—超前校正4-2-4-3-1 問(wèn)題的分析計(jì)算可得 Tb=0.1 wC''=50 a=40 wa=0.51rad/sGc=[(1+1.28s)(1+0.1s)]/[(1+5.02s)(1+0.03s)];結(jié)構(gòu)圖原理圖R1R1R2UU1CRUU1222C4-2-4-3-2 問(wèn)題的代碼由代碼num5=conv([201],[0.11]);den5=conv([78.41],[0.00251]);num6=conv(num5,num1);den6=conv(den5,den1);[mag6,phase6,w]=bode(num6,den6);figuremargin(mag6,phase6,w);;..由代碼Gk=tf(num6,den6);G=feedback(Gk,1);step(G)4-2-4-3-3 運(yùn)行的結(jié)果（見(jiàn)附錄）;..三、附錄第一問(wèn)在MATLAB中運(yùn)行以下代碼：GK=tf([100],[1100]);G=tf([100],[110100]);step(G)Wn=10;Kexi=10/(2*Wn);Beta=acos(Kexi);Wd=Wn*sqrt(1-Kexi^2);Tr=(pi-Beta)/Wd;Tp=pi/Wd;Ts=4.4/(Kexi*Wn);Ct=exp(-pi*Kexi/sqrt(1-Kexi^2))*100;;..第二問(wèn)K>0ROOTLOCUS60.89 0.81 0.7 0.56 0.38 0.24 0.95sxiAyranigamI

System:sysSystem:sysGain:0.207Gain:0.07120.988Pole:-9.98Pole:-0.00711Damping:1Damping:1Overshoot(%):0Overshoot(%):012Frequency(rad/sec):9.986Frequency(rad/sec):0.00711108420-20.988X:-5Y:-2.814-40.95-60.890.810.70.560.380.2-12-10-8-6-4-202RealAxisK<0sxiAyranigaI

RootLocus2.50.9960.9930.9860.9650.8620.9981.510.9990.513025201510501-0.51-10.999-1.50.998-20.9960.9930.9860.9650.86-2.5-30-25-20-15-10-505101520RealAxis;..第三問(wèn))Bd(edtuingaM

BodeDiagramGm=92dB(at2e+003rad/sec),P m=51.8deg(at7.85rad/sec)500-50-100-90)ged(-135esResponse:untitled1ahFrequency(rad/sec):8.22PPhase(deg):-129-180-101231010101010Frequency(rad/sec)NyquistDiagram200dB15105sx2dB-2dBAy4dB-4dB0ngamI-5-10-15-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-0.10RealAxis;..第四問(wèn)第一題未校正的單位階躍響應(yīng)圖eudtilmA

StepR esponse1.81.61.41.210.80.60.40.2000.20.40.60.811.21.4Time(sec)未校正前的BODE圖)dB(deunitgaM

BodeD iagramGm =72dB(at2e+003rad/sec),P m =18deg(at30.8rad/sec)100500-50-100-90)ged(eshaP

-135-180-101231010101010Frequency(rad/sec)超前校正可得校正后的BODE圖;..)Bd(edutingaM)ged(esahP

BodeDiagramGm=102dB(at2.49e+004rad/sec),P m=50.6deg(at37.8rad/sec)100500-50-100-90-135-180-101234101010101010Frequency(rad/sec)可得超前校正后的單位階躍響應(yīng)圖StepResponse1.41.21e 0.8dtuilpm0.60.40.200 0.05 0.1 0.