I am used to make my discrete time control systems simulations in Matlab and now I'm trying python and numpy.
So, my code bellow is working, but I would like to iterate over the numpy vector instead appending values into a list. Is it possible?
In other words, instead of using
xl.append(xt)
ul.append(uc)
I would like to use some Matlab equivalent like x[:, k 1] = np.dot(Ad, x[:, k]) Bd*uc, but it's not working on my code. If I do that, instead of obtaining a two line column vector that is the expected, I got a 2x2 matrix and an error.
Another question: Why it's neccessary to use plt.plot(tk, u[:, 0], label='u') instead plt.plot(tk, u, label='u') ?
from control.matlab import *
import math
import numpy as np
import matplotlib.pyplot as plt
Ts = 0.1
N = 50
#x = np.zeros((2, N 1))
tk = np.zeros(N)
u = np.zeros(N)
v = np.random.randn(N)/86.6 #% measurement noise
wn = 1.12
wn2 = pow(wn, 2)
A = [[0, 1], [-1.5, -1.4]]
B = [[0], [1.5]]
C = [[1, 0]]
D = 0
# Control gains
K = np.array([2.64, 3.41071429])
# Now build a feedback with control law u = -K*x
Ad = np.eye(2) np.multiply(A, Ts)
Bd = np.multiply(B, Ts)
Cd = C
xt = [[1.0], [0.12]] # initial states
xl = []
ul = []
for k in range(0, N):
tk[k] = k*Ts
uc = -K.dot(xt)
xt = np.dot(Ad, xt) Bd*uc
xt[1, 0] = v[k]
xl.append(xt)
ul.append(uc)
x = np.array(xl)
u = np.array(ul)
#x = np.delete(x, N, 1) # delete the last position of x
#s = TransferFunction.s
#Gs = wn2/(s**2 0*s wn2) # This is the KF solution
#yout, T = step(Gs)
plt.rcParams["figure.figsize"] = (10, 7)
plt.figure()
#plt.plot(T, yout, label='Open loop')
plt.plot(tk, x[:, 0], label='x_0')
plt.plot(tk, x[:, 1], label='x_1')
plt.plot(tk, u[:, 0], label='u')
plt.legend()
plt.title('Pendulum ex. 7.14 Franklin book')
plt.xlabel('Time')
plt.ylabel('amp.')
plt.show()
what I want is the code like this:
from control.matlab import *
import math
import numpy as np
import matplotlib.pyplot as plt
Ts = 0.1
N = 50
x = np.zeros((2, N 1))
tk = np.zeros(N)
u = np.zeros(N)
v = np.random.randn(N)/86.6 #% measurement noise
wn = 1.12
wn2 = pow(wn, 2)
A = [[0, 1], [-1.5, -1.4]]
B = [[0], [1.5]]
C = [[1, 0]]
D = 0
# Control gains
K = np.array([2.64, 3.41071429])
# Now build a feedback with control law u = -K*x
Ad = np.eye(2) np.multiply(A, Ts)
Bd = np.multiply(B, Ts)
Cd = C
for k in range(0, N):
tk[k] = k*Ts
u[k] = -K.dot(x[:, k])
x[1, k] = v[k]
x[:, k 1] = np.dot(Ad, x[:, k]) Bd*u[k]
x = np.delete(x, N, 1) # delete the last position of x
#s = TransferFunction.s
#Gs = wn2/(s**2 0*s wn2) # This is the KF solution
#yout, T = step(Gs)
plt.rcParams["figure.figsize"] = (10, 7)
plt.figure()
#plt.plot(T, yout, label='Open loop')
plt.plot(tk, x[:, 0], label='x_0')
plt.plot(tk, x[:, 1], label='x_1')
plt.plot(tk, u[:, 0], label='u')
plt.legend()
plt.title('Pendulum ex. 7.14 Franklin book')
plt.xlabel('Time')
plt.ylabel('amp.')
plt.show()
But it results in a following error:
Traceback (most recent call last):
File "C:\Users\ ... \np_matrices_v1.py", line 46, in <module>
x[:, k 1] = np.dot(Ad, x[:, k]) Bd*u[k]
ValueError: could not broadcast input array from shape (2,2) into shape (2,)
CodePudding user response:
In [248]: A = np.array([[1, 2], [2, 3]])
...: x = np.array([[0.5], [2.0]])
In [249]: A.shape, x.shape
Out[249]: ((2, 2), (2, 1))
In [250]: y = A.dot(x)
In [251]: y.shape
Out[251]: (2, 1)
Note the shapes. x
is (2,1), and as a result y
is too. y
can be assigned to a (2,1) slot, but not a (2,) shape.
In [252]: xa = np.zeros((2,5),int)
In [253]: xa
Out[253]:
array([[0, 0, 0, 0, 0],
[0, 0, 0, 0, 0]])
In [254]: xa[:,2]
Out[254]: array([0, 0]) # (2,) shape
In [255]: xa[:,[2]]
Out[255]:
array([[0], # (2,1) shape
[0]])
In contrast to MATLAB numpy
arrays can be 1d, e.g. (2,). Also leading dimensions are the outermost, as opposed to trailing. MATLAB readily reduces a (2,3,1) shape to (2,3), but a (2,1,1) only becomes (2,1).
broadcasting
the way numpy
uses arrays that can differ in shape. The two basic rules are that
- leading size 1 dimensions can added automatically to match
- size 1 dimensions can be adjusted to match
Thus a (2,) can become a (1,2).
If you remove the inner [] from x
, you get a 1d array:
In [256]: x = np.array([0.5, 2.0])
In [257]: x.shape
Out[257]: (2,)
In [258]: A.dot(x)
Out[258]: array([4.5, 7. ]) # (2,) shape
This can then be assigned to a row of xa
: xa[:,2] = A.dot(x)
reshape
and ravel
can be used to remove dimensions. Also indexing A.dot(x)[:,0]
CodePudding user response:
I don't know why, but if you try:
A = np.array([[1, 2], [2, 3]])
x = np.array([[0.5], [2.0]])
y = A.dot(x)
print(y)
xa = np.zeros((2, 10))
xa[:, 2] = A.dot(x)
You'll got:
Traceback (most recent call last):
File "C:\Users\eletr.spyder-py3\temp.py", line 19, in xa[:, 2] = A.dot(x)
ValueError: could not broadcast input array from shape (2,1) into shape (2,)
But if you do:
import numpy as np
A = np.array([[1, 2], [2, 3]])
x = np.array([[0.5], [2.0]])
y = A.dot(x)
print(y)
xa = np.zeros((2, 10))
#xa[:, 2] = A.dot(x)
xa[:, [2]] = A.dot(x)
print(xa)
You'll got the correct answer:
[[4.5]
[7. ]]
[[0. 0. 4.5 0. 0. 0. 0. 0. 0. 0. ]
[0. 0. 7. 0. 0. 0. 0. 0. 0. 0. ]]
Can anyone explain it ...?