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Solution to an eigenvalue problem ?

  1. Jun 29, 2010 #1
    "solution to an eigenvalue problem" ?

    I am trying to reproduce the results from a paper. The essence of the paper is hidden in just one equation (eq. 11) and some lines of text. For me this is going somewhat too fast.
    Below are the essential parts of the paper, describing the problem (and solution?)

    It may be something trivial, but I cannot get how to solve the equations. According to the text it is some kind of eigenvalue problem. I can follow the text up to the point of "Not only ...", from there I am lost.

    I am aiming for a way to solve this problem numericly.

    I hope someone can help.

    Given are:
    Yi: generator admittance (complex)
    Ti: noise temperature
    Wi: Output power spectrum
    Pi: Output power
    k: Boltzmann constant

    Wu: voltage noise power spectrum
    Wi: current noise power spectrum
    Wui: Wu/Wi correlation (complex)
    Yin: input admittance (complex)

    Stolle, Reinhard; Schiek, Burkhard; , "The Complete and Accurate Determination of Two-Port Noise Parameters without Seperate Measurements of the Two-Port Input Impedance," Microwave Conference, 1998. 28th European , vol.1, no., pp.241-246, Oct. 1998


    Each of these methods [cut] requires the knowledge of the input admittance Yin
    of the device under test (DUT). This paper presents a method to overcome this problem. [cut] The input admittance Yin of the device under test is found as the solution to an eigenvalue problem. The determinant of a matrix which is composed of the noise measurement values and of the input admittance Yin can be set to zero and solved for Yin.


    According to linear circuit theory, the output power spectrum W1 [cut] is given by

    Wi = mu / |Yin+Yi|^2 * (
    4 k Ti Re{Yi}
    + Wu |Yi|^2
    + Wi
    + Re{Wui} 2 Re{Yi}
    + Im{Wui} 2 Im{Yi}
    ) . . . (3)

    where mu denotes a real conversion factor and is assumed to be a constant and k is Boltzmann's
    constant. The elimination of mu is achieved with the introduction of

    wij = Pi/Pj |Yin+Yi|^2 / |Yin+Yj|^2 . . . (4)

    and yields

    4k(wij Tj Re{Yj} - Ti Re{Yi}) =
    Wu (|Yi|^2 - wij |Yj|^2)
    + Wi (1 - wij)
    + Re{Wui} 2*(Re{Yi} - wij Re{Yj})
    + Im{Wui} 2*(Im{Yi} - wij Im{Yj}) . . . (5)

    which constitutes a linear equation in the four unknowns Wu, Wi, Re{Wui} and Im{Wui}.
    Hence, there are five admittances Yi needed to make up a system of four equations to solve for the four unknowns:

    A * [Wu, Wi, Re{Wui}, Im{Wui}]^T = b . . . (6)

    where the 4 x 4-matrix A and the 4 x 1-vector b consist of the measurement data Wi, Yi for
    i = 1, ..., 5 and of Yin.


    At least one of the five source temperatures Ti corresponding to each Yi has to be different
    from the others, otherwise the 4 equations will be dependent. This can be seen from the fact, that if Ti = To for every i, then the left side of Eq. (5) becomes proportional to the coefficient of Re{Wuj} on the right side. In this case the right-hand vector b of the corresponding matrix equation (6) becomes proportional to the third column of A, such that the matrix equation (6) can be transformed into a homogeneous matrix equation with the same solution as (6). As this equation is homogeneous, it is not solvable and its determinant becomes zero.

    Not only does this indicate the necessity of at least one measurement at an excess temperature, but it implies a way to determine the input admittance Yin of the device under test with the same noise measurements that are needed to solve for the four noise parameters. For Yi, unknown, four equations of the type (5) yield a matrix A, the coefficients of which are parameterized by Yin. If Ti = To for every i, its determinant det(A(Yin)) is zero. This determinant is a surface in the Yin plane which cuts the zero plane. By means of one more equation of the type (5) one obtains another solution matrix B which cuts the zero plane as well. The intersection point of both cuts is Yin and is the solution to

    det(A(Yin))^2 + det(B(Yin))^2 = 0. . . . (11)

    For numerical benefits the matrices A and B should be normalized to their norms.

    Finally, at least seven measurements of noise power have to be performed, six of them at
    the temperature To and one of them at a different temperature, in order to obtain the noise
    parameters without seperate measurements of the input impedance.

    Last edited: Jun 29, 2010
  2. jcsd
  3. Jun 30, 2010 #2
    Re: "solution to an eigenvalue problem" ?

    I now get what they intend to do.

    I was tricked by "By means of one more equation of the type (5)", Which equation? It turns out that you just have to pick another set of data points.

    I could write the determinants of A and B out symbolicly and solve for wij, from there the Yin can be solved using eq 4.

    Furthermore, the matrix A is overdetermined. I would like to use this fact as well. I could solve eq(11) in pairs of four data points and then average over all pairs.

    Surely there is a more elegant way to do this.
    Last edited: Jun 30, 2010
  4. Jul 1, 2010 #3
    Re: "solution to an eigenvalue problem" ?

    Don't call me Shirley
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