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    Diagonalization of Impedance Matrix

    Generally an impedance matrix [Z] is a complex symmetrical matrix.
    It is neither a real symmetrical matrix nor a Hermitian matrix.

    However a complex impedance matrix [Z] can be diagonalized.
    I don't think any impedance matrix [Z] can be diagonalized.
    How can I interpret whether diagonalization is possible or not physically >

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    Re: Diagonalization of Impedance Matrix

    Diagonalization is only possible iff there exists a complete set of eigenvectors.

    Is [Z] a Hermitian matrix?

    See Horn and Johnson's Matrix Analysis.



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    Re: Diagonalization of Impedance Matrix

    Quote Originally Posted by PlanarMetamaterials View Post
    Diagonalization is only possible iff there exists a complete set of eigenvectors.
    I know such very basic thing.

    Quote Originally Posted by PlanarMetamaterials View Post
    Is [Z] a Hermitian matrix?
    No. [Z] is not a Hermitian matrix.

    Here [Z]T means transpose of [Z], [Z]+ means conjugate transpose of [Z].

    For lossless reciprocal circuit, [Z] is a pure imaginary symmetrical matrix which is a skew Hermitian Matrix at same time.
    [Z]T==[Z], [Z]+==-[Z]
    [Z]T*[Z]==[Z]*[Z]T, [Z]+*[Z]==[Z]*[Z]+
    In this case, eigenvectors are real, eigenvalues are pure imaginary.
    We can always diagonalize [Z] by Orthogonal Matrix. It is not by Unitary Matrix.

    On the other hand, for reciprocal circuit with loss, [Z] is neither a real symmetrical matrix nor Hermitian Matrix. It is complex symmetrical matrix.
    [Z]T==[Z], [Z]+!=[Z]
    [Z]T*[Z]==[Z]*[Z]T, [Z]+*[Z]!=[Z]*[Z]+
    However we can still diagonalize [Z] by Orthogonal Matrx. It is not by Unitary Matrix.

    Of course, if eigenvectors are not linearly independent, diagonalization is impossible.

    How can I interpret whether diagonalization is possible or not physically ?
    Last edited by pancho_hideboo; 14th October 2019 at 20:04.



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    Re: Diagonalization of Impedance Matrix

    I think it may be instructive to look at diagonalization of the matrix in the first place.

    If we have such a diagonalization, then: [Z]I = Iλ, with λ being the eigenvalues and I being the currents. This means that, since the impedance matrix definition is V = [Z]I, that:

    V = Iλ.

    I.e., there exists a certain set of currents for which all of the corresponding voltages are related by the same value (λ, or if you prefer, a modal impedance Zm).

    As to what this means physically.... good question. A series of impedances connected to all ports might do it; but I can't think what purpose it might serve.



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    Re: Diagonalization of Impedance Matrix

    Quote Originally Posted by PlanarMetamaterials View Post
    I think it may be instructive to look at diagonalization of the matrix in the first place.
    Not useful at all.



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    Re: Diagonalization of Impedance Matrix

    Code:
    Z = [2,   j,   0;
         j,   2, j*2;
         0, j*2,   2]
    This [Z] is neither Hermitian, Skew-Hermitian nor Unitary, yet it is Symmetric and Normal.
    This [Z] can be diagonalized by Orthogonal Matrix, [P].
    Code:
    [P, D] = eig(Z)
    
    P =
    
       0.3162 + 0.0000i   0.8944            -0.3162          
       0.7071             0.0000 - 0.0000i   0.7071          
       0.6325 - 0.0000i  -0.4472 + 0.0000i  -0.6325 - 0.0000i
    
    
    D =
    
       2.0000 + 2.2361i        0                  0          
            0             2.0000 - 0.0000i        0          
            0                  0             2.0000 - 2.2361i
    Code:
    Z = [2,   j,   0;
         j,   2, j*2;
         0, j*2,   2.1]
    This [Z] is also neither Hermitian, Skew-Hermitian nor Unitary, yet it is Symmetric.
    However [Z] is not Normal.

    This [Z] still can be diagonalized by [P]
    However [P] is neither Unitary nor Orthogonal Matrix.
    Code:
    [P, D] = eig(Z)
    
    P =
    
       0.3162 + 0.0057i   0.8942            -0.3162 + 0.0057i
       0.7071             0.0000 - 0.0179i   0.7071          
       0.6322 - 0.0170i  -0.4473 - 0.0000i  -0.6322 - 0.0170i
    
    
    D =
    
       2.0400 + 2.2354i        0                  0          
            0             2.0200 + 0.0000i        0          
            0                  0             2.0400 - 2.2354i
    How can I interpret ?
    Last edited by pancho_hideboo; 17th October 2019 at 19:43.



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    Re: Diagonalization of Impedance Matrix

    [P, D] = eig(Z)


    (Case-1)
    Z = [1, j; j, 1]
    Z : Symmetric=1
    Z : Hermitian=0
    Z : Skew-Hermitian=0
    Z : Orthogonal=0
    Z : Unitary=0
    Z : Normal=1

    P = [0.7071,0.7071; 0.7071,-0.7071]
    P : Orthogonal=1
    P : Unitary=1
    rank(P)=2
    det(P)=-1+j*-5.88785e-017
    cond(P)=1

    D = [1.0000+1.0000*j,0; 0,1.0000-1.0000*j]


    (Case-2)
    Z = [1, j; j, 2]
    Z : Symmetric=1
    Z : Hermitian=0
    Z : Skew-Hermitian=0
    Z : Orthogonal=0
    Z : Unitary=0
    Z : Normal=0

    P = [0.7071,-0.6124+0.3536*j; 0.6124-0.3536*j,0.7071]
    P : Orthogonal=0
    P : Unitary=0
    rank(P)=2
    det(P)=0.75+j*-0.433013
    cond(P)=1.73205

    D = [1.5000+0.8660*j,0; 0,1.5000-0.8660*j]


    (Case-3)
    Z = [1, j; j, 3]
    Z : Symmetric=1
    Z : Hermitian=0
    Z : Skew-Hermitian=0
    Z : Orthogonal=0
    Z : Unitary=0
    Z : Normal=0

    P = [0.7071,0.7071*j; -0.7071*j,0.7071]
    P : Orthogonal=0
    P : Unitary=0
    rank(P)=1
    det(P)=2.10734e-008+j*0
    cond(P)=9.49063e+007

    D = [2.0000,0; 0,2.0000]

    Eigenvalues are degenerated and Eigenvectors are not linearly independent in Case-3.



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    Re: Diagonalization of Impedance Matrix

    Just for example.

    For series RC circuit, [Z] can not be diagonalized at omega=2/(R*C).



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    Re: Diagonalization of Impedance Matrix

    Quote Originally Posted by LMA View Post
    For series RC circuit, [Z] can not be diagonalized at omega=2/(R*C).
    Thanks for very good example.

    Why is a diagonalization impossible in this frequency ?
    What does this frequency mean?
    This frequency is twice of 3-dB cutoff frequency.



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