机电系统动力学仿真matlab课后答案刘白雁

2020-10-06 来源：爱问旅游网

机电系统动力学仿真实验报告

机工 1004 班

201003130150宋康

习题一

1.3

Abe wu_2004两个是合法的。

1.4

(1)

x Command Window

|» (12+2* (7-4))/(3 2) axis =

(2)

;囚 How to Add ifl Wh^'s New

» A=llt2^A, 5,0 7, 8, 9]

Jk.

MSFS

holders

A =

2 5 8

3 6 9

沾h |

厂SI ert by L Show

snts actions thuml m

厂 Show

how functi 口

Dii^cto iv -

zes

? x I Com ma nd Window

(3)

clear;

x=-8:0.5:8;

y=x'；

X=ones(size(y))*x; Y=y*ones(size(x));

R=sqrt(X.A2+Y.A2)+eps; Z=sin(R)./R; mesh(X,Y,Z); colormap(hot) xlabel('x'),ylabel('y'),zlabel('z')

0.5

-0.5

1.7

(1)

]Whst's New

I .T ?~Command Window

» yl=2*sin(0. 3*pi)/(1+sqrt ⑸)

0.5000

r si

thumb

(2)

x I com mana window

>> yl=2*sinC0. 3*pi}/(1+sqrt (5))

0. 5000

>> y2=2*cos CO* 3*pi) / (l+sqrt (5))

：】y2 =

1 1

山 3633

5 a 5 a -

习题二

2.1

» K=1:0. 2:2

LOOOO

1.2000 1.4000 L600Q L 8000 2.0000

»尸2:乩2:1

Empty mat r is: i-by-0

2.2

» x=l in5p&ce(0^ 2=piF 50)

CdluHFiE L thrcAish. L6

Q.1282 0.2565

D. 33-17

0.5129

Celujuie

LT thioueh 蛙 2.0517

2.1799 2. 3OB1

2. <1363

乙 5646

CelminB

tlhnuA 4$

i.2315 4. 3融 4.1033

Ccliuiris <19 thioufh 5Q 6, 1550 乳 28 壮

ColuHne i L

D. IZ«Z D. 0. 3B47

Z54i5

0・51曲

Calwtris If throug3

2.0&K £ IT9S

2, 3031 2詞闭3

2.&S4S

Golwiris $3 t

hirough

呂

4. 1033 吐 359S 4■員囂

ColuAne

t^rou

钟

反

2532

&3 550

2.4

[L&11L

O. 7S94

2.6928

2. 8210

4. M0.. F6M

2. «92B

2F«

2L0

4 4.S72?

CL弱托 LQ2B§

2.9493

3-Q7T

6.0009

5.1291

0. S976 L.0Z5S 2. S4»3 3.0^5 E. 0009

E, 1291i4 0K/

D.3K/S

1.9231- L54. I

1.2823 L . 4105 1,5387 1.F9B2

3. 9751

5-25?^

3856

乩 613B 3.6420 5.7703 5.昭閃 6.user

区域

L IS4I L ZB2J ].4105i 】・阳前 9.6670 I.F9B@ L.9ZM

J,2®7 3. 3339 3^jfl22 3. &904

3.N86 区

或诙

3.B4SB 3. 9T&]

氐 2&74

5. 3B5^

乩瓯$

E ?7Q3

乩。沏

.?* x Command Window

B=【5」6 ; 8];

2.5

C=A*B c=A, *B C =

10 43

c 二

22 50

12 32

s„.事 x Command Window

2:1 :

0］； M

,5〕；

» 臥01;

arts 二

0 1

1 0

ans

—

1 1

2 3

3 5

4 0

ans

=\" 0

0 0

arts

；

= n

习题二

3.1

clf,clc,clear

t=(0:18/100:18)'; xi=0.2:0.2:0.8;

beita=sqrt(1-xi.A2); sita=atan(beita./xi);

y=1-exp(-t*xi).*sin(t*beita+ones(101,1)*sita)./(ones(101,1)*beita); figure(1) plot(t,y(:,1),'b:',t,y(:,2),'r-.',t,y(:,3),'g-',t,y(:,4),'ko')

Iegend('\\xi=0.2','\\xi=0.4','\\xi=0.6','\\xi=0.8')

II1 rIIr !=02_1.4 1.2

---- * =0.4 0.8

0.6

0.4

0.2

3.2

clf,clear x=-50:1:50; y=-40:1:60;

z=1./(sqrt((1-x).A2+y.A2)+sqrt((1+x).A2+y.A2)); figure⑴ plot3(x,y,z),box on [X,Y]=meshgrid(x,y);

Z=1./(sqrt((1-X).A2+Y.A2)+sqrt((1+X).A2+Y.A2)); figure(2) mesh(X,Y,Z) figure(3) surf(Z)

0.08

0.06

0.04

0.02

0 100

-50

0.5

0.4

0.3

0.2

0.1

0 100

0.5

0.4

0 150

100

150

3.3

clc; clear all; close all; t=0:pi/50:2*pi; x=s in (t); y=cos(t); z=t; plot3(x,y,z,'g-')

卜“ m辺 1 •B | ► X w k cr 1 o

IB =+1且二94 4

IIi i II 对 while i<=63 i i FJ+s

巴 ikH-厘Hf° 」 Command Window X lcomlnandwinQ.Qw

Command Window

» 1= (0:1:63); X=2. ?a K= sun (E!, 1) Kl=swn (X) K =

1.8447e+01ff

KI =

L8447B+019

习题五

5.1

» a=[l 35 291 1093 1700]; » b=[l 289 254 2541 4584 1700]: >> c=1：€ (為 b) Transfer Tunction:

s*4 + 35 s\"3 + 291 e*2 + 1093 s + 1700

srt5 十 289 s^4 + 254 s\"3 + 2541 s\"2 + 4684 s + 1700 » z=-3;

» p=[-l -5 -15]; » k=15;

>> sys=zpk (巧 k) Zer o/pole/g^in:

15 (s+3)

(s+1) ts+5) (s+155

» k=100； » 3=[0,-2.-2](

» P=M 1]；

>> sys 1 = zpk (zj p, k); » a=[l 3 2]； » b=[l 2 5 2]. >> sys2=tf (a, b);

» sys3=s er ies(sys1, sys 2) Zero/pole/gain:

100 s (s+2)*3 (s+1)

(s+l) (s+0-466?) (g-1)

+ 1.533s + 4.284)

5.4

» A=[3 2 1;0 4 6;0 -3 3]; » B=[l 2 3]'; » C=[l 2 5]: » D=0;

>> £ys=EE (扎 B,Cj D). » SYS=tf (sys) Transfer func

20 s 2 - 123 e + 234

s\"3 - 10 E*2 + 51 s-90

>> sys2-zpk(sys) Zero/pole/gain: 20 (s 2

- 6. 15s + 11.7)

(s-3) (S*2 - ?s + 3D) »

5.5

>> nun=[1 15 50 500]; >> den=[1 2 0];

>> sys—tf(den, nun)

Transfer fund ion:

s2 + 2 s

s\"3 + 15 s\"2 + 50 s + 500 » a=[l 3 6 4]； >> b= [4 0]; >> sys 1 =tf (bj a) Transfer fund ion:

4 s

5*3+3 5*2+6 s+4

5.6

a =

xl si 0x2

-2 x3

0b =

xL 0 s2

0 x3 0

x4 o.i?se

c =

zl x2 K3 K4

yl 0 0 1 0

d =

yl (J

Cent inuaus—t ine model.

Transfsr function;

0.1780 s\"2 4- □.02551 s + 0.3571 s*4 + 0- 2679『3 + 2. 005 s'2 + 0.25 s

x2 1 -0, 14290 0.08929

0 □0 0. 1429D 10

-0. 125

5.7

» ml=12;

» n2=38;k=1000;c=0. 1; >> sysl=tf ([c, k]3 [ml c k]) Transfer function:

0, 1 s + 1000 12 s*2 + 0. 1 s + 1000

» sys2=tf ([12 0. 1 1000]f [12**^ 0. 1*(*14*2)J 100D*Ul-Fs2). 0f 0]) Transfer function:

12 s\"2 + 0. 1 s + 1000 456 事4 + 5 s^3 + 50000 事2 >> sys3=eeries (sys sys2) Transfer function:

k 2 sA3 + L 2e004 事2 + 200 s + le006

5472 sA6 + 105.6 s^S + L 056e006 s\"4 + 10000 事3 + 5&007 s2

习题六

6.2

6.3

2 5

2 R I $ I I I i i I I I __________

'■ 0

2 4 6 8 10 12 14 16 18 20

Zero/pole/gain:

;(s+0. 5) 20 £ s (s+0. 1)

(s+2) (s+10)

6.4

M Figure 1 File Edit View Insert Tools Desktop Window Help D G

M Figure 1

□可 B O Time isec)

i回

File Edit View Insert Tools Desktop Window 出Ip u o

1 £ o. epnsdu^

0.4 -I 0 2 -I

0「1111

11111

.6 Step Response

・・

0 20 40 60 80 100 Time (sec)

120 140 160 180 200

习题七

7.1

(1)

Hie Edit View Insert Tools Desktop Window HelpFigure 1

Bode Dia^ranri

Gm > 0 dB i at 1 rad/sec), Pm D deg [at 1 rad/sec}

Nyquist Diagram

〔rmLlmro

BP)®p

agusdeffl

临界稳定

(2)

Frequency (rad/sec)

Real如总

10 1

Q Figure 1

0|超・・^-]

为

File Edit View [nsert Tools De&lctop Window Help

□ H e]

Open File Bode Diagram

Gm 二-19.2 dB fat Q.316 Md；aec), Pm = -34.3 de5 i at D.366 rad^secr

100 ---------- 1 ----------------- ■ ------------------ 2000

■叵

Nyquist Diagram

-1000

J 500

厂 10°

；I

-2000

-100 -50

■

Frequency (rad/sec?

Real Axis

-150

Q Figure 1

0|超・・^-]

不稳定 (3)

H Figure 1

File Edit Viev^ [nsert Tools Desktop Window Help □

[=I 回

fe以&紳®膜&匡I

a E

Bode Diagram

Gm = hf, Pm = -46.1 deg I at ZB& rad/^ec) 1CC ■■ ---------------------------------------------

(mpjIjpr4!LlmB Frequency (rad/sec

巨 (口 epj WGBLId Nyquist Diagram

15 ------------- 1 -----------------1 ---------------

Real Axis

.ssxytBUaraE-

不稳定 (4)

riyuie 丄

File Edit View Insert Tools Desktop Window Help

□

®I®i\"n目i ■叵

Bode Diagram

epripub®(

MP巨 )

■s ■o ■pJE) d

Fresuency

g畑口

Nyquist Diagram

Real Axis

x 10

不稳定

7.2 K=36

Q Figure 1

Edit View Insert File » ■ « □ G

0 m p

)

-50

3PNU6BS I □ 2

Tools Desktop MM MH Window Help Q S除釵QU®渥El匡］ B 0 Bode Diagram

-100

.Figure 1

File Edit View Insert Tools Desktop Window Help W ■ W [ o | 回 M

0 □ G Q S除題運 □ 0| H Q Lin ear Simulation Results Zero/pole/gain:

20 s (s+0.5)

s (s+0. 1) (s+2) (s+10)

5 2 2 5 1 5

15 Time (sec)

Figure 1

I u■叵1 0 J

File Edit View Insert Tools Desktop Window Help

Step Response

SPWQLUV玄 211. 5 3 5 2 5

□ U S h題Q斛® 隈口目7.3

5 3 5 CM 5 1.5O 玉2.1I.0. QP^QUUV

Pm =

-21. 6 345

Wcg =

O. 4083

Wop =

0. 4777

VP =

i.agga

k =

XP =

1.4994

l™

-e mcHFtKi

竹 w

MDO-m

8 c G 4.2 o.o.o. 5

10 15 20 25 30 35 40 45 50

Time (sec)

File Edit View Insert Tools Desktop Window Help □ o^Hei 硬|□目|■旦 (Bap) 8s£d

M nyure J

File Edit View Insert Tools Desktop Window Help □ Q S M Figure 6 致® D ® 运 □目 ■ 口 [=l ■回” Nyquist Diagram

File Edit View Insert Tools Desktop Window Help □Q e I

5000500050050005000 2 2 11 --1-1.2 0? AJeu-obeuj- 愆包的@|遲|□匡11■旦 Step Response5000

35030o250

200

5000

Real Axis

5 10 15 20 25 30 35 40 45 50

Time (sec)

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机电系统动力学仿真matlab课后答案刘白雁