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PID Control (1) | 制御系CAD
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投稿日時: 2015/11/14
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PID Control (1)
1. Speed Control
Consider the above situation we we are driving a car with a constant speed. Then assume that we are
necessary to reduce the speed to stop somewhere. How to manipulate the driving force?
Let
,
and
be the mass, the velocity and the driving force at time
of the car respectively.
Its motion is governed by the following differential equation:
The problem is how to give the driving force
Control):
http://cacsd2.sakura.ne.jp/2015/11/14/whats-pid-control/[2015/11/24 7:55:19]
(1)
. Consider the following proportional control (P
PID Control (1) | 制御系CAD
(2)
Substituting (2) to (1), we have the closed-loop system:
(3)
That is, the driving force is given which is proportional to the velocity with negative sign. This means
that for the high speed the large regulating force is produced in the reverse direction, and for the low
speed the small regulating force is applied. This control strategy is called as feedback mechanism
and depicted as follows:
In general, the solution of a differential equation
is given by
(4)
Therefore the solution of (3) is given by
(5)
This means
That is, we can reduce the car speed gradually and stop it.
http://cacsd2.sakura.ne.jp/2015/11/14/whats-pid-control/[2015/11/24 7:55:19]
(6)
PID Control (1) | 制御系CAD
By the way, assume that we are required to change the current speed to the new speed
. How to
manipulate the driving force for the speed up or speed down?
Let the speed error from the current speed
to the required constant speed
be
(7)
Then consider the following proportional control:
(8)
i.e.
(9)
Substituting (8) into (1), we have
Noting
(10)
, this can be rewritten as
(11)
Therefore, we have
This means
http://cacsd2.sakura.ne.jp/2015/11/14/whats-pid-control/[2015/11/24 7:55:19]
(12)
PID Control (1) | 制御系CAD
(13)
The above control strategy is depicted as follows:
Exercise 1
Execute the following program under SCILAB. By changing the P gain
(vs), investigate the corresponding reaching distance
(kv) and the target speed
.
——————————————————————————————————————–
//cart1.sce
function dx=f(t,x),dx=A*x,endfunction
m=1; kv=1; A=[0 1;0 -kv/m];
v0=1; vs=0.5; x0=[0;vs-v0];
t0=0; t=0:0.1:10;
x=ode(x0,t0,t,f);
v=vs-x(2,:); r=-x(1,:);
scf(1);
subplot(211), plot(t,v), mtlb_grid, title(‘v(t)’)
subplot(212), plot(t,r), mtlb_grid, title(‘reaching distance’)
——————————————————————————————————————–
http://cacsd2.sakura.ne.jp/2015/11/14/whats-pid-control/[2015/11/24 7:55:19]
PID Control (1) | 制御系CAD
2. Position Control
Under the speed control described in the above, we can’t specify the stop position. For example, in
the case of encountering an obstacle at the distance
from the current position as shown in the
following, we will be required to stop in front of the obstacle. How to manipulate the driving force in
order to stop within the running distance of
Let
,
, and
.
be the mass, the position and the driving force at time
respectively. The velocity is
of the car
. Its motion is governed by the following differential
equation:
Let the position error from the current position
http://cacsd2.sakura.ne.jp/2015/11/14/whats-pid-control/[2015/11/24 7:55:19]
(14)
to the required target position
be
PID Control (1) | 制御系CAD
(15)
Then consider the following proportional and derivative control (PD Control):
(16)
i.e.
Here
(17)
is the speed error given by (7), in which
must be 0. Note
.
Substituting (16) into (1), we have
Noting
(18)
, this can be rewritten as
Furthermore, taking account of
(19)
, we have
As show here, the solution is given by
http://cacsd2.sakura.ne.jp/2015/11/14/whats-pid-control/[2015/11/24 7:55:19]
(20)
PID Control (1) | 制御系CAD
Aa
(21)
is a stable matrix, the following holds.
(22)
This means
(23)
The above control strategy is depicted as follows:
Exercise 2
Execute the following program under SCILAB. Determine the D gain
such that no undershoot occurs in the position
(kv) for collision avoidance,
.
——————————————————————————————————————–
//cart2.sce
function dx=f(t,x),dx=A*x,endfunction
m=1; kr=1; kv=2; A=[0 1;-kr/m -kv/m];
r0=-1; rs=0; v0=0; vs=0; x0=[rs-r0;vs-v0];
t0=0; t=0:0.1:10;
x=ode(x0,t0,t,f);
v=vs-x(2,:); r=rs-x(1,:);
scf(1);
subplot(211), plot(t,v), mtlb_grid, title(‘v(t)’)
subplot(212), plot(t,r), mtlb_grid, title(‘r(t)’)
http://cacsd2.sakura.ne.jp/2015/11/14/whats-pid-control/[2015/11/24 7:55:19]
PID Control (1) | 制御系CAD
——————————————————————————————————————–
Question:
Why the control law (2) is called as not D control but P control?
Answer:
In the situation of the speed control, we are assumed to measure the velocity, that is, we don’t have
to differentiate the position. So we need to feed back the value proportional to the measured velocity.
On the other hand, in the situation of the position control, we are assumed to measure the position,
that is, we have to differentiate the position to get the velocity.
カテゴリー: 未分類 作成者: admin パーマリンク
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