Co-So-Ky-Thuat-Dien__Bai-Tap-Chuong-4 - [Cuuduongthancong.com].Pdf

Chương 4
4.1/

a)

co

ng

.c
om

Mạch từ tương đương:

du
o

ng

th

an

a/

L

di
dt

cu

v (t )  L

u

b/



1 1
 N 2W 2 0 (  )
i
g 2x

c/

i ( , x ) 


1 1
N 2W 2 i 0 (  )
g 2x

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

Wm   i( , x) d  

0

fe 

Wm ( , x )

x

1 1
2.N 2W 2i 0 (  )
g 2x
i


 2 x2

2.N W i o 
 x
 g

2

2

d) W 'm    (i, x)di  o2 (
0

1 1 2 i2
 ).N
2x g

2

 ' Wm
 2 Ni 2
 o 2
x
4x

ng

.c
om

f e (i, x) 

2

4.2/

co

a)

cu

Theo đề:

Ni

u

Ni

du
o

ng

th

an

Mạch từ tương đương:

L()  L1  L2cos(2) =>

 L(  0)  L1  L2

 L(  90)  L1  L2

N 2 o Ag o
N 21Ag1
 N N Ni
 N N Ni
L(  0)  
, L(  90)  





i
i
i 2g o
2g o
i
i
i 2g1
2g1
 o Ag o
o Ag1


N 2o Ag o
N 2o Ago Ag1
L1
L2
L1
(
)






2g o
4
go
g1





2
2
 L1  L2  N 1Ag1
 L2  N o ( Ago  Ag1 )


2g1
4
go
g1

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b)

  L()i  L1i  L 2cos(2)i
i

W 'm    (i, x)di  L1
0

Te 

i2
i2

 L 2 cos(2)
2
2

W 'm
 i 2 sin(2)L 2
x

4.3/

g
 o Wd

, Rx 

x
 o Wd

ng

b)

th

Rg 

an

co

ng

.c
om

a)

4.4/

cu

u

du
o

(N  N 2 )i

1  1

Rg
1Rg  N 2i  N1i



(1   2 )2Rx=N 2i   N 2i(Rg  2Rx)  N1i2Rx
 2
2RxRg

Mạch từ tương đương:

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Rx 




x
o w 2

Ni Ni o w 2

Rx
x

N 2 i o w 2
x
i

N 2i 2  o w 2
'
W m    (i, x)di 
2
x
0

.c

om

W 'm
N 2i 2  o w 2
 f (i, x) 

x
2
x2
e

ng

i
x
Wm  i  W 'm 
, i= 2
2
N o w 2


co

x 2
Wm   i( , x)d 
2N 2 o w 2
0

ias

u

du
o

ng

th

an

4.5/

i bs

ir

0

0

cu

W 'm   (L0  L1cos2)i 'as di 'as   (L0  L1cos2)i 'bs di 'bs   (L2cos4)i 'r di 'r
0

  N 

b)

Ni
 bD  o aD
 Ni( o

)
(2Rx / /2Ry)
2y
2x


1
1 di N 2i o bD 1 dy N 2i o aD 1 dx
2
v(t) 
N (

) 

t
2Rx 2Ry dt
2
y 2 dt
2
x 2 dt
i

N 2i 2 1
1
'
c) W m    (i, x)di 

(

)
2 2Rx 2Ry
0

W 'm
N 2i 2 o aD
 f (i, x) 

x
2 2x 2
e

4.7/

CuuDuongThanCong.com

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i1

i2

i3

0

0

0

a) W 'm   1 (i '1 ,0,0, )di '1    2 (i1 ,i '2 ,0, )di'2    3 (i1 ,i 2 ,i '3 , )di '3
i1

i2

i3

0

0

0

  (L11i '1 )di '1   (L 22i '2 )di '2   (Mcosi1  M sini 2  L33i '3 )di '3


1
(L11i12  L 22i 2 2  L33i32 )  Mcosi1i3  Mcosi 2i3
2

b) T e (i1 ,i 2 ,i3 , ) 

W 'm
  Msin i1i3  Mcosi 2i3
x

.c
om

4.8

1  (5  cos2)103 i1  0.1cosi 2
 2  0.1cosi1  (50  10cos2)i 2
i2

ng

i1

W 'm   (5  cos2)103 i '1 di '1   (0.1cosi1  (50  10cos2))i '2 di '2
0

co

0

i
i 22
 (5  cos2)10
 0.1cosi1i 2  (50  10cos2)
2
2

an

2
3 1

u

du
o

ng

th

W 'm
 103 sin 2i12  0.1sin i1i 2  10i 2 2 sin 2
x

cu

T e () 

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4.9/
i as

ibs

0

0

W 'm   (L 0  L1cos2)i 'as di 'as   (L 0  L1cos2)i 'bs di 'bs
ir

  ((Mcos)ias  (M sin )ias  L 2 cos4))i 'r di 'r
0

.c
om

i 2 as
i 2 bs
ir 2
 (L 0  L1cos2)
 (L 0  L1cos2)
 Mcosi asi r  M sin i bsi r  L 2 cos4
2
2
2
4.10/

a/

Wm ' 3i 2
i3


4x
4x
x

co

Wm ' 

ng

Ta có:

Wm'
i3
 2
x
4x

th

f e (i, x) 

an

b/

i3 3i3 i 3
i3



4x 4 x 4x 2 x

du

o

Wm  i  Wm'  i 

ng

c/

u

4.11/

cu

s  ( L cos 2 )i s  ( M sin  )ir
r  ( M sin  )is  ( L cos 2 )ir

a/
vs (t ) 

d s
d
 Mir cos
 MI cos
dt
dt

b/
is

ir

W 'm   s (is ' ,0, )dis '   r (is , ir ' ,  )dir ' 
0

0

Lis 2
Li 2
cos 2  Mis ir sin   r cos 2
2
2

c/
T e (is , ir ,  ) 

Wm '
  Lis 2 sin 2  Misir cos  Lir 2 sin 2


CuuDuongThanCong.com

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4.12 /
a/
d
vs  s  M .I ..cos 
dt
b/

is

ir

W ' m  i1 , i2 , x    s  i 's ,0,  di 's   r  is , i 'r ,   di 'r
0

0

2

is
i2
.cos 2  M .is .ir .sin   L. r .cos 2
2
2

'
W
m
 M .is .ir cos   (is2  ir2 ).L.sin 2
fe (is , ir ,  ) 


.c
om

 L.

co

ng

4.14 /

ve  0

th

du
o

ng

 dx
 dv
3
 dt  v
 dt  0,1.v.i  x  x

 2
 d x  dv
 di   1 .(2.v.i  i  5)
2
x
 dt
dt
 dt
b/

an

a/

cu

u

x  1

 x  x  0
i  5



  x  1
i  5  0

 i  5
3

CuuDuongThanCong.com

/>

4.15 /
a/
d 2 x dv
 ,
dt 2 dt

1
1
1
W ' m
  .L0 .I 0 2 .
.
fe 
2
x
2
 x a
1  
 a
(2)
v e  0  f e   M .g

(1)

.c
om

 L .I 2

0 0
 1  .a
Tu (1), (2)  x  
 2.M .g .a 


b/

 L .I 2

0 0
 1  .a  0
x0
 2.M .g .a 



ng

2.M .g .a
L0

co

 I0 

an

4.17 /
a/

b/

dW 'm (i, x)
C.i3
 2
dx
3.x

du
o

fe 

th

0

C.i 3
3.x

ng

i

W 'm (i, x)    (i ', x).di ' 

cu

u

Phương trình động lực học

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/>

M .g  f e  M .a  M .

d 2x
dv
 M.
2
dt
dt

dv
fe
 g
dt
M
1 d
I (t )  .
i
R dt
1 d  di
. i
 .
R di dt
1 2.C.i di
. i
 .
R x dt
di x.R.( I (t )  i )
 
dt
2.c
Wm (i, x)  i.  W 'm (i , x )

C.i 2 C.i 3 2.C.i 3


x
3.x
3.x

ng

 i.

.c
om



an

co

4.18 /
a/
t  0.1s
x1 (0,1)  x1 (0)  t.x2 (0)  1,05

th

x2 (0,1)  x2 (0)  t.  0,1.x2 (0).x3 (0)  x13 (0)  x1 (0)   0, 45

ng

x3 (0,1)  x3 (0)  t.  2.x2 (0).x3 (0)  x3 (0)  5   8,5
x1 (0, 2)  x1 (0,1)  t.x2 (0,1)  1, 095

du
o

x2 (0, 2)  x2 (0,1)  t.  0,1.x2 (0,1).x3 (0,1)  x13 (0,1)  x1 (0,1)   0, 4

cu

b/
x1  x2  x3  0

u

x3 (0, 2)  x3 (0,1)  t.  2.x2 (0,1).x3 (0,1)  x3 (0,1)  5   7,39

 x2  0

  x1  1, 1,0
x  5
 3

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4.21/



 .i 3
( x  a)
i

 W 'm (i, x)  
0

 .i3
( x  a)

.di 

 .i 4
4.( x  a )

 Wm ( , x)  i.  W 'm (i, x) 

3. .i 4
4.( x  a )

i1

i2

W 'm   1 (i1 ,0, ).di1   2 (i1 , i2 ,  )
0

0

Te 

i
i2
 M .i1.i2 .sin 2   L0  M .cos 2  . 2
2
2

W 'm
 2.M .i1.i2 .cos 2  M .i12 .sin 2  M .i2 2 .sin 2


an

W 'm
 M .I 2 .sin  2 s .t  2. 


4.23/
9.96.10 6
x2

du
o

f e ( x, i) 

ng

th

Te 

co

b/

ng

  L0  M .cos 2  .

2
1

.c
om

4.22 /
a/

v=0

cu

u

f e ( x, i )  K ( x  l )  Bv  0.

9.96.106

f ( x, i)  K ( x  l ) 
 K ( x  4.922.103 )
2
x
e

Vậy:
x1e  2.64.103
x1e  3.85 (mm)

4.24/
x1e (0,1)  x1 (0)  t * x2 (0)  1
x2e (0,1)  x2 (0)  t *( 5sin x1 (0)  x2 (0)) = 0+0.1(-5sin(1)-0)=-8.73* 103 .

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x1 (0, 2)  x1 (0,1) t ( x2 (0,1))  1  0,1( 8, 72.103 )  0,999
x2 (0,2)  x2 (0,1) t.(5sin1  8,73.103 )  16,58.103
x2 (0,3)  x1 (0, 2) t ( x2 (0, 2))  0,999  0,1.16,58.103  1

4.26/

ng

.c
om



x  x
2
 1

 x32
x

 40( x1  0,1)  80 x2 )
 2
2
x
1


xx
 x3  2 3  1000 x2  1000 x1  x3
x1


an

co

x1 (0, 001)  x1 (0) t .x2 (0)  0,1

ng

th

 x 2 (0)


x2 (0,001)  x2 (0) t   32
 40( x1 (0)  0,01)  80 x2 (0)   0
 x1 (0)


du
o

 x (0).x3 (0)

 1000 x1 (0)  1000 x1 (0).x3 (0)   0,1
x3 (0,001)  x3 (0) t  2
 x1 (0)


cu

u

x1 (0,002)  x1 (0,001) t.x2 (0,001)  0,1

 x 2 (0,001)

 40( x1 (0,001)  0,1)  80 x2 (0,001)   0,001
x2 (0,002)  x2 (0,001) t.   32
 x1 (0,001)


 x (0,001) x3 (0,001)


x3 (0,002)  x3 (0,001) t  2
 1000 x1 (0,001)  1000 x1 (0,001) x3 (0,001) 
x1 (0,001)


 0,1  0,001 0  1000.0,1  1000.0,1.0,1  0,19

4.27/
a/

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Mơ hình khơng gian trạng thái.

x  x2
x2  0,1.x2 .i  x3  x
i

2 x2
i
i 5
x
x
x

x(0,1)  x(0) t.x2 (0)  1  0,1.0,5  1,05

.c
om

x2 (0,1)  x2 (0) t  0,1.x2 (0).i (0)  x 3 (0)  x(0) 
 0,5  0,1  0,1.0,5.0  13  1  0,5

ng

5
 0,5
 0  0,1  2
 0    0, 4
1
1


co

 2 x2 (0)
i (0) 
i (0,1)  i (0) t 
i
x(0) 
 x (0)

an

x(0, 2)  x(0,1) t.x2 (0,1)  1,1

th

x2 (0, 2)  x2 (0,1) t  0,1x2 (0,1)i (0,1)  x 3 (0,1)  x(0,1) 

ng

 0,5  0,1( 0,1.0,5.0, 4  1, 053  1, 05)  0, 4872

cu

u

du
o

 2 x2 (0,1)
i (0,1)
5 
i (0, 2)  i (0,1) t 
i (0,1) 

x(0,1) x(0,1) 
 x (0,1)

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