sistemas de control i
TRANSCRIPT
SISTEMAS DE CONTROL I umich+matlab www.drexel.edu
LABORATORIO DE CONTROL CON MATLAB. Suponemos que tenemos un modelo en el espacio de estados [A,B,C,D]
A= matriz de estado, multiplicaba las variables de estado.
MODELO DE FUNCIÓN DE TRANSFERENCIA
De función de transferencia a espacio de estados
[A,B,C,D] = tf2ss(num,dem)
[num,dem]=ss2tf(A,B,C,D,iµ)
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Ejercicio #1 A=[0,1;-2,-3];
B=[1,0;0,1];
C=[1,0];
D=[0,0];
[num,den]=ss2tf(A,B,C,D,1)
num =
0 1 3
den =
1 3 2
1 Uso de superposición en MATLAB
>> [num,den]=ss2tf(A,B,C,D,2)
num =
0 0 1
den =
1 3 2
>> num=[2 5 3 6];
>> den=[1 6 11 6];
>> [r,p,k]=residue(num,den)
r =
-6.0000
-4.0000
3.0000
p =
-3.0000
-2.0000
-1.0000
k =
2
Entonces:
ESCALÓN step(num,dem)
step(num,dem,t)
step(A,B,C,D,iµ,t)
[y,x,t]=step(num,den,t)
[y,x,t]=step(A,B,C,D,iµ,t)
>> num=[0 0 2 4];
>> den=[1 1.3 7 4];
>> step(num,den)
RESPUESTA DEL SISTEMA
MULTIPLICACIÓN DE VECTORES. help conv
Ejemplo (s+1)*(s+5)
conv([1 1],[1 5])
ans =
1 6 5
MULTIPLICACIÓN DE BLOQUES help series
Entonces el ejercicio:
Queda:
Bloque 2:
s(s+1)(s+5)
conv([0 1 0],[1 6 5])
ans =
0 1 6 5 0
num=[0 0 0 0 1]
dem=[0 1 6 5 0]
Su FT:
A=tf([0 0 0 0 1],[0 1 6 5 0])
A =
1
-----------------
s^3 + 6 s^2 + 5 s
Bloque 1:
6.3223(s+1.4235)2
>> conv([1 1.4235],[1 1.423])
ans =
1.0000 2.8465 2.0256
>> ans*6.3223
ans =
6.3223 17.9964 12.8067
B=tf([6.3223 17.9964 12.8067],[0 1 0])
B =
6.322 s^2 + 18 s + 12.81
------------------------
s
Continuous-time transfer function.
Y para tener un solo bloque:
M=series(B,A)
M =
6.322 s^2 + 18 s + 12.81
------------------------
s^4 + 6 s^3 + 5 s^2
Continuous-time transfer function.
Dándonos:
Para el sistema en retroalimentación:
help feedback2
F=feedback(M,1)
F =
6.322 s^2 + 18 s + 12.81
--------------------------------------
s^4 + 6 s^3 + 11.32 s^2 + 18 s + 12.81
Continuous-time transfer function.
Con respuesta al paso:
>> num=[0 0 6.322 18 12.81];
>> den=[1 6 11.32 18 12.81];
>> step(num,den)
En 10 segundos se estabiliza
2 Asume directamente retroalimentación negativa, para colocarla positiva se coloca F=feedback(M,1,+1)
Con tiempo estimado:
>> num=[0 0 6.322 18 12.81];
>> den=[1 6 11.32 18 12.81];
>> t=0:0.01:20;
>> step(num,den,t)
USO DE SCRIPT’S
Ejemplo en script:
num=[0 0 6.322 18 12.81]; den=[1 6 11.32 18 12.81]; [c,x,t]=step(num,den); plot(t,c,'0') grid on tittle('Respuesta al escalón') xlabel('t seg') ylabel('salida C')
Ejemplo:
Control de un de:
Caso 1: J=1 Kp=1 b=0.5
Caso 2: J=1 Kp=1 b=0.5
num1=[0 0 1]; den1=[1 0.5 1]; num2=[0 0 1]; den2=[0 0.5 4]; step(num1,den1,'-r') hold on step(num2,den2,'-g') grid on title('Respuesta al escalón de dos sistemas') text(9,0.9,'Sistema 1')
TAREA:3
1) [(
)] [
]
A=[0,1,0,0;0,0,1,0;0,0,0,1;-100,-80,0,-8]
B=[0;0;0;15]
C=[0,0,0,0]
D=[0]
A =
0 1 0 0
0 0 1 0
0 0 0 1
-100 -80 0 -8
B =
0
0
0
15
C =
0 0 0 0
D =
0
>> [num,den]=ss2tf(A,B,C,D,1)
num =
0 0 0 0 0
3 Leer Richard Dorf
den =
1.0000 8.0000 -0.0000 80.0000 100.0000
2)
Polos y ceros
Efectuamos en el denominador la función convolución y la multiplicamos por la constante 5:
conv([1 1],[1 10])
ans =
1 11 10
>> ans*5
ans =
5 55 50
Por lo que para hallar polos y ceros realizamos la operación:
num=[20 20 10]
den=[5 55 50]
[r,p,k]=residue(num,den)
num =
20 20 10
den =
5 55 50
r =
-40.2222
0.2222
p =
-10
-1
k =
4
3) Graficar la respuesta al paso
t=0:0.01:10
num=[0 0 25]
den=[1 4 25]
t=0:0.01:10;
step(num,den,t)
num =
0 0 25
den =
1 4 25
AYUDAS DE COMANDOS
HELP CONV conv Convolution and polynomial multiplication.
C = conv(A, B) convolves vectors A and B. The resulting vector is
length MAX([LENGTH(A)+LENGTH(B)-1,LENGTH(A),LENGTH(B)]). If A and B are
vectors of polynomial coefficients, convolving them is equivalent to
multiplying the two polynomials.
C = conv(A, B, SHAPE) returns a subsection of the convolution with size
specified by SHAPE:
'full' - (default) returns the full convolution,
'same' - returns the central part of the convolution
that is the same size as A.
'valid' - returns only those parts of the convolution
that are computed without the zero-padded edges.
LENGTH(C)is MAX(LENGTH(A)-MAX(0,LENGTH(B)-1),0).
Class support for inputs A,B:
float: double, single
See also deconv, conv2, convn, filter and,
in the signal Processing Toolbox, xcorr, convmtx.
Overloaded methods:
gf/conv
gpuArray/conv
Reference page in Help browser
doc conv
HELP SERIES series Series connection of two input/output models.
+------+
v2 --->| |
+------+ | M2 |-----> y2
| |------->| |
u1 ----->| |y1 u2 +------+
| M1 |
| |---> z1
+------+
M = series(M1,M2,OUTPUTS1,INPUTS2) connects the input/ouput models
M1 and M2 in series. The vectors of indices OUTPUTS1 and INPUTS2
specify which outputs of M1 and which inputs of M2 are connected
together. The resulting model M maps u1 to y2.
If OUTPUTS1 and INPUTS2 are omitted, series connects M1 and M2
in cascade and returns M = M2 * M1.
If M1 and M2 are arrays of models, series returns a model array M of
the same size where
M(:,:,k) = series(M1(:,:,k),M2(:,:,k),OUTPUTS1,INPUTS2) .
For dynamic systems SYS1 and SYS2,
SYS = series(SYS1,SYS2,'name')
connects SYS1 and SYS2 by matching their I/O names. The output names
of SYS1 and input names of SYS2 should be fully defined.
See also append, parallel, feedback, InputOutputModel, DynamicSystem.
Overloaded methods:
InputOutputModel/series
Reference page in Help browser
doc series
HELP FEEDBACK feedback Feedback connection of two input/output systems.
M = feedback(M1,M2) computes a closed-loop model M for the feedback loop:
u --->O---->[ M1 ]----+---> y
| | y = M * u
+-----[ M2 ]<---+
Negative feedback is assumed and the model M maps u to y. To apply
positive feedback, use the syntax M = feedback(M1,M2,+1).
M = feedback(M1,M2,FEEDIN,FEEDOUT,SIGN) builds the more general feedback
interconnection:
+------+
v --------->| |--------> z
| M1 |
u --->O---->| |----+---> y
| +------+ |
| |
+-----[ M2 ]<---+
The vector FEEDIN contains indices into the input vector of M1 and
specifies which inputs u are involved in the feedback loop. Similarly,
FEEDOUT specifies which outputs y of M1 are used for feedback. If SIGN=1
then positive feedback is used. If SIGN=-1 or SIGN is omitted, then
negative feedback is used. In all cases, the resulting model M has the
same inputs and outputs as M1 (with their order preserved).
If M1 and M2 are arrays of models, feedback returns a model array M of
the same dimensions where
M(:,:,k) = feedback(M1(:,:,k),M2(:,:,k)) .
For dynamic systems SYS1 and SYS2,
SYS = feedback(SYS1,SYS2,'name')
connects SYS1 and SYS2 by matching their I/O names. The I/O names of
SYS1 and SYS2 must be fully defined.
See also lft, parallel, series, connect, InputOutputModel, DynamicSystem.
Overloaded methods:
InputOutputModel/feedback
iddata/feedback
Reference page in Help browser
doc feedback