Python linalg.companion函数代码示例

    def test_basic(self):
        c = companion([1, 2, 3])
        expected = array([[-2.0, -3.0], [1.0, 0.0]])
        assert_array_equal(c, expected)

        c = companion([2.0, 5.0, -10.0])
        expected = array([[-2.5, 5.0], [1.0, 0.0]])
        assert_array_equal(c, expected)

def test_fiedler_companion():
    fc = fiedler_companion([])
    assert_equal(fc.size, 0)
    fc = fiedler_companion([1.])
    assert_equal(fc.size, 0)
    fc = fiedler_companion([1., 2.])
    assert_array_equal(fc, np.array([[-2.]]))
    fc = fiedler_companion([1e-12, 2., 3.])
    assert_array_almost_equal(fc, companion([1e-12, 2., 3.]))
    with assert_raises(ValueError):
        fiedler_companion([0, 1, 2])
    fc = fiedler_companion([1., -16., 86., -176., 105.])
    assert_array_almost_equal(eigvals(fc),
                              np.array([7., 5., 3., 1.]))

 def test_from_state_space(self):
     # Ensure that freqresp works with a system that was created from the
     # state space representation matrices A, B, C, D.  In this case,
     # system.num will be a 2-D array with shape (1, n+1), where (n,n) is
     # the shape of A.
     # A Butterworth lowpass filter is used, so we know the exact
     # frequency response.
     a = np.array([1.0, 2.0, 2.0, 1.0])
     A = linalg.companion(a).T
     B = np.array([[0.0],[0.0],[1.0]])
     C = np.array([[1.0, 0.0, 0.0]])
     D = np.array([[0.0]])
     with warnings.catch_warnings():
         warnings.simplefilter("ignore", BadCoefficients)
         system = lti(A, B, C, D)
         w, H = freqresp(system, n=100)
     expected_magnitude = np.sqrt(1.0 / (1.0 + w**6))
     assert_almost_equal(np.abs(H), expected_magnitude)

    def test_from_state_space(self):
        # Ensure that bode works with a system that was created from the
        # state space representation matrices A, B, C, D.  In this case,
        # system.num will be a 2-D array with shape (1, n+1), where (n,n)
        # is the shape of A.
        # A Butterworth lowpass filter is used, so we know the exact
        # frequency response.
        a = np.array([1.0, 2.0, 2.0, 1.0])
        A = linalg.companion(a).T
        B = np.array([[0.0], [0.0], [1.0]])
        C = np.array([[1.0, 0.0, 0.0]])
        D = np.array([[0.0]])
        with suppress_warnings() as sup:
            sup.filter(BadCoefficients)
            system = lti(A, B, C, D)
            w, mag, phase = bode(system, n=100)

        expected_magnitude = 20 * np.log10(np.sqrt(1.0 / (1.0 + w**6)))
        assert_almost_equal(mag, expected_magnitude)

 def time_companion(self, size):
     sl.companion(self.x)

def lfilter_zi(b, a):
    """
    Compute an initial state `zi` for the lfilter function that corresponds
    to the steady state of the step response.

    A typical use of this function is to set the initial state so that the
    output of the filter starts at the same value as the first element of
    the signal to be filtered.

    Parameters
    ----------
    b, a : array_like (1-D)
        The IIR filter coefficients. See `scipy.signal.lfilter` for more
        information.

    Returns
    -------
    zi : 1-D ndarray
        The initial state for the filter.

    Notes
    -----
    A linear filter with order m has a state space representation (A, B, C, D),
    for which the output y of the filter can be expressed as::

        z(n+1) = A*z(n) + B*x(n)
        y(n)   = C*z(n) + D*x(n)

    where z(n) is a vector of length m, A has shape (m, m), B has shape
    (m, 1), C has shape (1, m) and D has shape (1, 1) (assuming x(n) is
    a scalar).  lfilter_zi solves::

        zi = A*zi + B

    In other words, it finds the initial condition for which the response
    to an input of all ones is a constant.

    Given the filter coefficients `a` and `b`, the state space matrices
    for the transposed direct form II implementation of the linear filter,
    which is the implementation used by scipy.signal.lfilter, are::

        A = scipy.linalg.companion(a).T
        B = b[1:] - a[1:]*b[0]

    assuming `a[0]` is 1.0; if `a[0]` is not 1, `a` and `b` are first
    divided by a[0].

    Examples
    --------
    The following code creates a lowpass Butterworth filter. Then it
    applies that filter to an array whose values are all 1.0; the
    output is also all 1.0, as expected for a lowpass filter.  If the
    `zi` argument of `lfilter` had not been given, the output would have
    shown the transient signal.

    >>> from numpy import array, ones
    >>> from scipy.signal import lfilter, lfilter_zi, butter
    >>> b, a = butter(5, 0.25)
    >>> zi = lfilter_zi(b, a)
    >>> y, zo = lfilter(b, a, ones(10), zi=zi)
    >>> y
    array([1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.])

    Another example:

    >>> x = array([0.5, 0.5, 0.5, 0.0, 0.0, 0.0, 0.0])
    >>> y, zf = lfilter(b, a, x, zi=zi*x[0])
    >>> y
    array([ 0.5       ,  0.5       ,  0.5       ,  0.49836039,  0.48610528,
        0.44399389,  0.35505241])

    Note that the `zi` argument to `lfilter` was computed using
    `lfilter_zi` and scaled by `x[0]`.  Then the output `y` has no
    transient until the input drops from 0.5 to 0.0.

    """

    # FIXME: Can this function be replaced with an appropriate
    # use of lfiltic?  For example, when b,a = butter(N,Wn),
    #    lfiltic(b, a, y=numpy.ones_like(a), x=numpy.ones_like(b)).
    #

    # We could use scipy.signal.normalize, but it uses warnings in
    # cases where a ValueError is more appropriate, and it allows
    # b to be 2D.
    b = np.atleast_1d(b)
    if b.ndim != 1:
        raise ValueError("Numerator b must be rank 1.")
    a = np.atleast_1d(a)
    if a.ndim != 1:
        raise ValueError("Denominator a must be rank 1.")

    while len(a) > 1 and a[0] == 0.0:
        a = a[1:]
    if a.size < 1:
        raise ValueError("There must be at least one nonzero `a` coefficient.")

    if a[0] != 1.0:
        # Normalize the coefficients so a[0] == 1.
        a = a / a[0]
#.........这里部分代码省略.........

import numpy as np
import scipy.linalg as la;
import matplotlib.pyplot as plt

P = [4,5,8,1]

c = la.companion(P)
eigenval,eigenvec=np.linalg.eig(c)

r = eigenval

print(eigenval)

x = []
y = []
for r1 in r:
	x.append(r1.real)
	y.append(r1.imag)	


plt.plot(x,y,'r.')
plt.axis("equal")
plt.show()

from scipy import linalg as la
import numpy as np
x = np.roots([10,5,-5,1])
print x
a = la.companion([10,5,-5,1])
print a
b=la.eig(a)
print b