numpy help

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  • Chris Smith

    #1

    numpy help

    Howdy,

    I'm a college student and for one of we are writing programs to
    numerically compute the parameters of antenna arrays. I decided to use
    Python to code up my programs. Up to now I haven't had a problem,
    however we have a problem set where we are creating a large matrix and
    finding it's inverse to solve the problem. To invert the matrix I've
    tried using numpy.numarray. linear_algebra. inverse and
    numpy.oldnumeri c.linear_algebr a.inverse which both give me the same
    error ( I was hoping they called different routines but I think they
    call the same one ).


    This is the error message I receive:

    Traceback (most recent call last):
    File "C:\Documen ts and Settings\Chris &
    Esther\Desktop\ 636_hw5_2\elen6 36_hw5_2.py", line 60, in <module>
    matrix_inverse =
    numpy.numarray. linear_algebra. generalized_inv erse(matrix)
    File
    "C:\Python25\li b\site-packages\numpy\ oldnumeric\line ar_algebra.py", line
    59, in generalized_inv erse
    return linalg.pinv(a, rcond)
    File "C:\Python25\li b\site-packages\numpy\ linalg\linalg.p y", line
    557, in pinv
    u, s, vt = svd(a, 0)
    File "C:\Python25\li b\site-packages\numpy\ linalg\linalg.p y", line
    485, in svd
    a = _fastCopyAndTra nspose(t, a)
    File "C:\Python25\li b\site-packages\numpy\ linalg\linalg.p y", line
    107, in _fastCopyAndTra nspose
    cast_arrays = cast_arrays + (_fastCT(a.asty pe(type)),)
    TypeError: can't convert complex to float; use abs(z)



    I've tried inverting small complex matrices and it worked fine. Does
    anyone know why it won't work for this larger matrix? Any ideas how I
    can work around this problem and get the correct inverse matrix?

    Chris

    P.S. elen636_math.py is my personal library of functions I've create to
    solve the problem while elen636_hw5_2.p y is the program that I'm
    actually running


    # Purpose:
    # This is a library of functions for ELEN 636 that
    # so far has the ability to calculate the Sine and
    # Cosine integrals as well as the mutual impedance
    # between two parallel antennas.
    #
    # Author: Christopher Smith
    # E-mail: mc_anjo@tamu.ed u
    # Date: 10/30/2006

    ###############
    ### NOTE: The functions below for the sine and cosine integrals are similar
    ### to the functions I turned in for homework assignment 4 problem 6
    ### except that I added the ability to check for convergence.
    ### I also added the factor into the mutual impedance formula so that the
    ### answer is given in terms of the terminal input impedance instead of
    ### the loop impedance as it was formerly giving.
    ###############

    # depends on the math library
    from math import *
    import numpy.numarray, numpy

    def factorial(n):
    """
    This function calculates the factorial of a number.
    """
    sum = 1.0
    for m in range(1, int(n)+1):
    sum = float(m)*sum
    return sum

    def Si(x):
    """
    This function computes the sine integral. It uses a power series
    expansion that can be found in Abramowitz and Stegun's math
    functions reference book.
    """
    start = 0.0
    stop = 10.0
    sine_int = 0.0
    convergence = 1.0*10**(-6) # want to have the difference between
    # the last run and this run below
    # this value
    while 1:
    for n in range(int(start ), int(stop)):
    n = float(n)
    sine_int += ((-1)**n)*x**(2*n +1)/((2*n+1)*factor ial(2*n+1))
    sine_int_new = sine_int + ((-1.)**stop)*x**( 2.*stop +1.)/((2.*stop+1.)*f actorial(2.*sto p+1.))
    converge_check = sine_int_new - sine_int
    if abs(converge_ch eck) < convergence:
    break
    else:
    start = stop
    stop += 5.0
    return sine_int_new

    def Ci(x):
    """
    This function computes the cosine integral. It uses a power series
    expansion that can be found in Abramowitz and Stegun's math
    functions reference book.
    """
    start = 1.0
    stop = 10.0
    convergence = 1.0*10.**(-6) # want to have the difference between
    # the last run and this run below
    # this value
    # The first number in the sum is Euler's constant to 10 digits
    cosine_int = 0.5772156649 + log(x)
    while 1:
    for n in range(int(start ), int(stop)):
    m = float(n)
    cosine_int = cosine_int +((-1)**m)*x**(2*m)/((2*m)*factoria l(2*m))
    cosine_int_new = cosine_int + ((-1)**stop)*x**(2 *stop)/((2*stop)*facto rial(2*stop))
    converge_check = cosine_int_new - cosine_int
    if abs(converge_ch eck) < convergence:
    break
    else:
    start = stop
    stop += 5.0
    #print stop
    return cosine_int_new


    def mutual_impedanc e(length1_tot, length2_tot, stagger, d):
    """
    This function computes the mutual impedance between two antennas
    for the Parallel in Echelon Configuration. The formulas are taken
    from a paper by Howard King, "Mutual Impedance of Unequal Length
    Antennas in Echelon"

    NOTE: all measurements should be entered in wavelengths
    """

    # stagger (this is the vertical separation between antenna centers)
    # d (this is the horizontal separation between the antennas)

    # length1 and length2 are the half length of the antennas, this is
    # to conform to King's formulas
    length1 = length1_tot/2.0
    length2 = length2_tot/2.0

    # vertical separation between center of antenna 1 and bottom of antenna 2
    h = stagger - length2

    # wave propagation constant
    beta = 2.0*pi

    # formulas to put into mutual impedance equation
    u0 = beta*(sqrt(d**2 +(h -length1)**2) +(h -length1))
    v0 = beta*(sqrt(d**2 +(h -length1)**2) -(h -length1))
    u0prime = beta*(sqrt(d**2 +(h +length1)**2) -(h +length1))
    v0prime = beta*(sqrt(d**2 +(h +length1)**2) +(h +length1))
    u1 = beta*(sqrt(d**2 +(h -length1 +length2)**2) +(h -length1 +length2))
    v1 = beta*(sqrt(d**2 +(h -length1 +length2)**2) -(h -length1 +length2))
    u2 = beta*(sqrt(d**2 +(h +length1 +length2)**2) -(h +length1 +length2))
    v2 = beta*(sqrt(d**2 +(h +length1 +length2)**2) +(h +length1 +length2))
    u3 = beta*(sqrt(d**2 +(h -length1 +2.0*length2)** 2) +(h -length1 +2.0*length2))
    v3 = beta*(sqrt(d**2 +(h -length1 +2.0*length2)** 2) -(h -length1 +2.0*length2))
    u4 = beta*(sqrt(d**2 +(h +length1 +2.0*length2)** 2) -(h +length1 +2.0*length2))
    v4 = beta*(sqrt(d**2 +(h +length1 +2.0*length2)** 2) +(h +length1 +2.0*length2))
    w1 = beta*(sqrt(d**2 +h**2) -h)
    y1 = beta*(sqrt(d**2 +h**2) +h)
    w2 = beta*(sqrt(d**2 +(h +length2)**2) -(h +length2))
    y2 = beta*(sqrt(d**2 +(h +length2)**2) +(h +length2))
    w3 = beta*(sqrt(d**2 +(h +2.0*length2)** 2) -(h +2.0*length2))
    y3 = beta*(sqrt(d**2 +(h +2.0*length2)** 2) +(h +2.0*length2))
    #print u0,v0,u0prime,v 0prime,u1,v1,u2 ,v2,u3,v3,u4,v4 ,w1,y1,w2,y2,w3 ,y3

    # real part of the mutual impedance between two antennas
    R12 = 15*(cos(beta*(l ength1 -h))*(Ci(u0) +Ci(v0) -Ci(u1) -Ci(v1)) \
    +sin(beta*(leng th1 -h))*(-Si(u0) +Si(v0) +Si(u1) -Si(v1)) \
    +cos(beta*(leng th1 +h))*(Ci(u0prim e) +Ci(v0prime) -Ci(u2) -Ci(v2)) \
    +sin(beta*(leng th1 +h))*(-Si(u0prime) +Si(v0prime) +Si(u2) -Si(v2)) \
    +cos(beta*(leng th1 -2.0*length2 -h))*(-Ci(u1) -Ci(v1) +Ci(u3) +Ci(v3)) \
    +sin(beta*(leng th1 -2.0*length2 -h))*(Si(u1) -Si(v1) -Si(u3) +Si(v3)) \
    +cos(beta*(leng th1 +2.0*length2 +h))*(-Ci(u2) -Ci(v2) +Ci(u4) +Ci(v4)) \
    +sin(beta*(leng th1 +2.0*length2 +h))*(Si(u2) -Si(v2) -Si(u4) +Si(v4)) \
    +2.0*cos(beta*l ength1)*cos(bet a*h)*(-Ci(w1) -Ci(y1) +Ci(w2) +Ci(y2)) \
    +2.0*cos(beta*l ength1)*sin(bet a*h)*(Si(w1) -Si(y1) -Si(w2) +Si(y2)) \
    +2.0*cos(beta*l ength1)*cos(bet a*(2.0*length2 +h))*(Ci(w2) +Ci(y2) -Ci(w3) -Ci(y3)) \
    +2.0*cos(beta*l ength1)*sin(bet a*(2.0*length2 +h))*(-Si(w2) +Si(y2) +Si(w3) -Si(y3)))

    # imaginary part of the mutual impedance between two antennas
    X12 = 15*(cos(beta*(l ength1 -h))*(-Si(u0) -Si(v0) +Si(u1) +Si(v1)) \
    +sin(beta*(leng th1 -h))*(-Ci(u0) +Ci(v0) +Ci(u1) -Ci(v1)) \
    +cos(beta*(leng th1 +h))*(-Si(u0prime) -Si(v0prime) +Si(u2) +Si(v2)) \
    +sin(beta*(leng th1 +h))*(-Ci(u0prime) +Ci(v0prime) +Ci(u2) -Ci(v2)) \
    +cos(beta*(leng th1 -2.0*length2 -h))*(Si(u1) +Si(v1) -Si(u3) -Si(v3)) \
    +sin(beta*(leng th1 -2.0*length2 -h))*(Ci(u1) -Ci(v1) -Ci(u3) +Ci(v3)) \
    +cos(beta*(leng th1 +2.0*length2 +h))*(Si(u2) +Si(v2) -Si(u4) -Si(v4)) \
    +sin(beta*(leng th1 +2.0*length2 +h))*(Ci(u2) -Ci(v2) -Ci(u4) +Ci(v4)) \
    +2.0*cos(beta*l ength1)*cos(bet a*h)*(Si(w1) +Si(y1) -Si(w2) -Si(y2)) \
    +2.0*cos(beta*l ength1)*sin(bet a*h)*(Ci(w1) -Ci(y1) -Ci(w2) +Ci(y2)) \
    +2.0*cos(beta*l ength1)*cos(bet a*(2.0*length2 +h))*(-Si(w2) -Si(y2) +Si(w3) +Si(y3)) \
    +2.0*cos(beta*l ength1)*sin(bet a*(2.0*length2 +h))*(-Ci(w2) +Ci(y2) +Ci(w3) -Ci(y3)))

    R12_in = R12/(sin(beta*lengt h1)*sin(beta*le ngth2))
    X12_in = X12/(sin(beta*lengt h1)*sin(beta*le ngth2))
    impedance = (R12_in, X12_in)
    return impedance


    def top_row_matrix( length1, length2, stagger, stagger_image, radius, m, n):
    """
    This function will find the top row of a mutual impdedance matrix
    over a ground plane. From the top row we can find the overall matrix
    since it is a block Toeplitz matrix
    """

    z = [] # list to store our impedance values

    # index to step over for the staggering and separation between antennas
    stagger_range = range(0, m)
    separation_rang e = range(0, n)

    # calculate the mutual impedance values for the real planar array
    # the first loop gives us the stagger between rows while the second
    # loop gives us the separation between dipoles on the same row
    for m in stagger_range:
    for n in separation_rang e:
    h = stagger*m
    d = stagger*n
    if d == 0: d = radius
    trans = mutual_impedanc e(length1, length2, h, d)
    z.append(comple x(trans[0],trans[1]))

    # Suppose the real antenna array is a plane in x-y at z = 0 and the
    # imaginary antenna array is a plane in x-y at z = some spacing
    # Since the mutual impedance would include spacing in the
    # x,y and z directions we need to condense it down to two spacings
    # so that we can input the spacings into our formulas. For the
    # spacing between dipoles we will use the hypotenuse formed by the
    # x and z coordinates. The y coordinate will be the staggering between
    # dipoles.
    for m in stagger_range:
    for n in separation_rang e:
    separation = sqrt((stagger_i mage)**2+(stagg er*n)**2)
    trans = mutual_impedanc e(length1, length2, stagger*m, separation)
    z.append(comple x(trans[0],trans[1]))

    return z


    def Toeplitz(a):
    """
    This function takes a list in. The list represents
    the top row in a Toeplitz matrix. From this row we
    can fill the rest of the matrix and then output it.

    Note: This function has a dependency on the numpy
    library.
    """

    # need the length of the list being input
    n = len(a)

    # fill a matrix the size we need with zeros
    matrix = numpy.zeros((n, n), dtype = type(a))

    a_new = []
    for i in range(-n+1,n):
    a_new.append( a[ abs(i) ] )

    for i in range(0,n):
    matrix[i,:] = a_new[ (n-1) -i : (2*n -1) -i ]

    return matrix


    def BlockToeplitz(a , M, N):
    """
    This function takes a list of N Toeplitz matrices in
    and creates a Block Toeplitz matrix from them.

    Note: This function has a dependency on the numpy
    library.
    This function is also specifically for an array
    over a groundplane problem. Which means that
    the matrix is twice the length and width of a
    normal array matrix. The code would need to be
    modified to apply to a normal Block-Toeplitz.
    """


    # fill a matrix the size we need with zeros
    matrix = numpy.zeros((2* M*N,2*M*N), dtype = type(a))

    a_new = []
    for i in range(-2*N+1,2*N):
    a_new.append( a[ abs(i) ] )

    for i in range(0, 2*N):
    i_new = i*M
    for j in range(0, 2*N):
    j_new = j*M
    matrix[ i_new:i_new+N, j_new:j_new+N ] = a_new[2*N-1-j+i]

    return matrix


    def h_array_over_gn dplane(M, N):
    """
    This function generates the voltage excitation matrix for an
    array over a ground plane. The antennas are parallel to the
    ground plane which means that according to image theory the
    current along the antenna images are in the opposite direction
    from those of the real array.

    Note: This function has a dependency on the numpy library.
    """

    h = []
    for n in range(0, 2*M*N):
    if n < M*N:
    h.append(1.0)
    else:
    h.append(-1.0)
    h = numpy.numarray. array((h), shape = (2*M*N,1))
    return h


    # Purpose:
    # This program will calculate the impedance matrix
    # for a planar antenna array over a ground plane.
    # This is for ELEN 636 homework 5 problem 2
    #
    # Author: Christopher Smith
    # E-mail: mc_anjo@tamu.ed u
    # Date: 10/30/06

    # library so arrays (matrices) can be used
    import numpy.numarray, numpy
    # library so we can perform linear algebra operations
    # like inverse( )
    import numpy.numarray. linear_algebra
    # import special functions
    from elen636_math import mutual_impedanc e, top_row_matrix, Toeplitz
    from elen636_math import BlockToeplitz, h_array_over_gn dplane
    # math library
    from math import *


    # length and radius of dipoles (wvlgths)
    # horizontal (same in x and y) stagger of matrix (wvlgths)
    # vertical separation between array and it's image (wvlgths)
    # or twice the vertical separation between array and ground plane
    # and the number of elements in the y and x directions m, n
    length1 = 0.5
    length2 = length1
    radius = 0.0025
    stagger = 0.55
    stag_image = 0.5
    M = 7
    N = 7



    # function to give us back the top row of the m x n matrix
    z = top_row_matrix( length1, length2, stagger, stag_image, radius, M, N)

    # We now have a list with 2*M*N elements. Using symmetry we can create
    # N NxN arrays which correspond to N Toeplitz matrices for our N x N
    # Block Toeplitz matrix of 2*M*N x 2*M*N elements. So from those N
    # matrices we can completely fill the matrix.
    toeplitz_matric es = [] # a list of N Toeplitz matrices
    for i in range(0, 2*M*N, N):
    toeplitz_matric es.append( Toeplitz( z[i:i +N] ) )

    # Now we fill our Block-Toeplitz matrix
    matrix = BlockToeplitz(t oeplitz_matrice s, M, N)

    # generate the voltage excitations of all the dipoles
    h = h_array_over_gn dplane(M, N)





    # now we compute Cn by multiply the inverse mutual impedance matrix
    # and multiplying it by the h matrix
    matrix_inverse = numpy.numarray. linear_algebra. inverse(matrix)
    Cn = numpy.numarray. matrixmultiply( matrix_inverse, h)

    input_impedance _matrix = h/Cn

    #input_impedanc e_element25 = input_impedance _matrix[24]

    # creates and opens files to output results
    try:
    file = open( 'mutual_impedan ce_results.txt' , "w" )
    except IOError, message:
    print >sys.stderr, "File could not be opened:" , message
    sys.exit(1)

    print >file, input_impedance _matrix
    file.close()









    _______________ _______________ _______________ __
    Tutor maillist - Tutor@python.or g



  • robert

    #2
    Re: numpy help

    Chris Smith wrote:
    Howdy,
    >
    I'm a college student and for one of we are writing programs to
    numerically compute the parameters of antenna arrays. I decided to use
    Python to code up my programs. Up to now I haven't had a problem,
    however we have a problem set where we are creating a large matrix and
    finding it's inverse to solve the problem. To invert the matrix I've
    tried using numpy.numarray. linear_algebra. inverse and
    numpy.oldnumeri c.linear_algebr a.inverse which both give me the same
    error ( I was hoping they called different routines but I think they
    call the same one ).
    >
    try scipy.linalg.in v etc.

    (you are using outdated old NumPy modules with recent array type which sometimes creates type conflicts.

    robert

    Comment

    • Chris Smith

      #3
      Re: numpy help

      robert wrote:
      Chris Smith wrote:
      >
      >>Howdy,
      >>
      >>I'm a college student and for one of we are writing programs to
      >>numerically compute the parameters of antenna arrays. I decided to use
      >>Python to code up my programs. Up to now I haven't had a problem,
      >>however we have a problem set where we are creating a large matrix and
      >>finding it's inverse to solve the problem. To invert the matrix I've
      >>tried using numpy.numarray. linear_algebra. inverse and
      >>numpy.oldnume ric.linear_alge bra.inverse which both give me the same
      >>error ( I was hoping they called different routines but I think they
      >>call the same one ).
      >>
      >
      >
      try scipy.linalg.in v etc.
      >
      (you are using outdated old NumPy modules with recent array type which sometimes creates type conflicts.
      >
      robert
      >
      I tried that one also, and it didn't work.

      Chris

      Comment

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