Cascade Control of Dc machines - Newcastle University Control...num(s) den(s) Transfer Fcn1 Scope1 1...
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Cascade Control of DC machines Permanent magnet dc motor Armature circuit: () () ( ) () ⇒ + + = t E dt t di L R t i t v () () ( ) () t K dt t di L R t i t v E φω + + = (1) Mechanical dynamics: () () () i K T e T B T F L e T e L F dt t d J B T dt t d J T T T dt t d J T φ ω ω ω ω ω = = = ⇒ = − ⇒ = − − ⇒ = ∑ 0 () () () dt t d J t B t i K T ω ω φ = − (2) Laplace transform of (1) and (2): () () ( ) ( ) () () () s Js s B s I K s K s LsI R s I s V T E Ω = Ω − Ω + + = φ φ (3) Solve (3a) for I(s): () () ( )() () () () ( ) Ls R s K s V s I s I Ls R s K s V E E + Ω − = ⇒ + = Ω − φ φ and replace it in (3b):
Transcript of Cascade Control of Dc machines - Newcastle University Control...num(s) den(s) Transfer Fcn1 Scope1 1...
Cascade Control of DC machines Permanent magnet dc motor Armature circuit:
( ) ( ) ( ) ( )⇒++= tEdt
tdiLRtitv
( ) ( ) ( ) ( )tKdt
tdiLRtitv Eφω++= (1)
Mechanical dynamics:
( )
( )
( ) iKT
e
T
BTFLe
Te
L
F
dttdJBT
dttdJTTT
dttdJT
φ
ω
ωω
ω
ω
=
=
=
⇒=−
⇒=−−
⇒=∑0
( ) ( ) ( )dt
tdJtBtiKTωωφ =− (2)
Laplace transform of (1) and (2): ( ) ( ) ( ) ( )
( ) ( ) ( )sJssBsIKsKsLsIRsIsV
T
E
Ω=Ω−Ω++=
φφ
(3)
Solve (3a) for I(s): ( ) ( ) ( ) ( )
( ) ( ) ( )( )LsR
sKsVsI
sILsRsKsV
E
E
+Ω−
=
⇒+=Ω−φ
φ
and replace it in (3b):
( ) ( )( ) ( ) ( )
( )( )
( )( ) ( ) ( )
( )( ) ( ) ( )
( ) ( )( )( )( )( )
( )
( )( ) φφφ
φφφ
φφφ
φφφ
φφ
ET
T
ETT
ETT
ETT
ET
KKJsBLsRK
sVs
sJsBLsRKKsVK
sJsBLsR
KKLsRsVK
sJssBLsR
sKKLsRsVK
sJssBLsR
sKsVK
+++=
Ω⇒Ω+++=
⇒Ω⎟⎠
⎞⎜⎝
⎛++
+=
+
⇒Ω+Ω++Ω
=+
⇒Ω=Ω−+
Ω−
It is possible to simulate that using the following numerical values:
Nms/rad0001.0Kgm5
mH1052.0
rad/Vs65.3Nm/A65.3
2
==
=Ω=
==
BJ
LRKK
E
T
φφ
0 0.05 0.1 0.15 0.20
0.1
0.2
0.3
Step Response
Time (sec)
Ampl
itude
Also we can use a PI controller to improve the motor’s performance:
num(s)
den(s)T ransfer Fcn1 Scope1
1s
Integrator1
50
Gain1
30
Gain1
Constant1
0 0.05 0.1 0.15 0.20
0.5
1
1.5
But this system has a peculiar property which can be seen if we slightly change the model: ( ) ( ) ( )( )LsRsIsKsV E +=Ω− φ
( ) ( ) ( )sJsBsIKT Ω+=φ
1
w*
J.s+B
1
T ransfer Fcn3L.s+R
1
Transfer Fcn2
Scope3
Scope2
1s
Integrator1Kt
Gain5
Ke
Gain4
50
Gain3
30
Gain2 w
w
w
w
So let’s see the current:
0 0.05 0.1 0.15 0.2-200
-100
0
100
200
300
This is happening because the electrical time constant (L/R=0.0192) is much smaller (i.e. faster) than the mechanical one (i.e. J/B=50000). But we can use that difference to break the system into 2 smaller and separated subsystems, i.e. the electrical (A) and the mechanical (B) part and hence to use 2 PI controllers:
L.s+R
1
T ransfer Fcn4 Scope4
1s
Integrator21
I* 10
Gain7
5
Gain6 I
0 0.5 1 1.5 2x 10-3
0
0.2
0.4
0.6
0.8
1
J.s+B
1
Transfer Fcn6L.s+R
1
Transfer Fcn4
Scope7
Scope6
Scope4
1s
Integrator4
1s
Integrator2
10
Gain7
5
Gain6
40
Gain13
20
Gain12
Ke
Gain11
Kt
Gain101
Constant3
E_IE_II*
ITe
w
w
wV
E
w*
0 0.1 0.2 0.3 0.40
0.5
1
1.5
Current:
0 0.1 0.2 0.3 0.4-10
0
10
20
30
40
This type of control is called cascade control and is very popular in electric drives as we first tune the current loop and then we separately tune the speed loop.
