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10th October 2019, 17:37 #1
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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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11th October 2019, 15:23 #2
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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.

14th October 2019, 19:49 #3
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Re: Diagonalization of Impedance Matrix
I know such very basic thing.
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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15th October 2019, 15:31 #4
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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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15th October 2019, 16:39 #5

17th October 2019, 20:13 #6
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Re: Diagonalization of Impedance Matrix
Code:Z = [2, j, 0; j, 2, j*2; 0, j*2, 2]
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]
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
Last edited by pancho_hideboo; 17th October 2019 at 19:43.

20th October 2019, 06:34 #7
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Re: Diagonalization of Impedance Matrix
[P, D] = eig(Z)
(Case1)
Z = [1, j; j, 1]
Z : Symmetric=1
Z : Hermitian=0
Z : SkewHermitian=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*5.88785e017
cond(P)=1
D = [1.0000+1.0000*j,0; 0,1.00001.0000*j]
(Case2)
Z = [1, j; j, 2]
Z : Symmetric=1
Z : Hermitian=0
Z : SkewHermitian=0
Z : Orthogonal=0
Z : Unitary=0
Z : Normal=0
P = [0.7071,0.6124+0.3536*j; 0.61240.3536*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; 0,1.50000.8660*j]
(Case3)
Z = [1, j; j, 3]
Z : Symmetric=1
Z : Hermitian=0
Z : SkewHermitian=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)=2.10734e008+j*0
cond(P)=9.49063e+007
D = [2.0000,0; 0,2.0000]
Eigenvalues are degenerated and Eigenvectors are not linearly independent in Case3.

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Yesterday, 18:33 #8
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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).

Today, 02:33 #9
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