User:Netshine2/sandbox

Tridiagonal example

Some applications may require multiple zero entries in a similarity matrix, possibly in the form of a tridiagonal matrix. Since Jacobian rotations may remove zeros from other cells that were previously zeroed, it is usually not possible to achieve tridiagonalization by simply zeroing each off-tridiagonal cell individually in a medium to large matrix. However, if Jacobian rotations are repeatedly performed on the above-tridiagonal cell with the highest absolute value using an adjacent cell just below or to the left to rotate on, then all of the off-triangular cells are expected to converge on zero after several iterations. In the example below, $A$ is a 5X5 matrix that is to be tridiagonalized into a similar matrix, $T$.

${\begin{aligned}&A={\begin{bmatrix}2&5&9&11&7\\5&3&6&-2&5\\9&6&7&3&1\\11&-2&3&1&3\\7&5&1&3&4\\\end{bmatrix}}\\&{\text{eigenvalues of }}A={\begin{bmatrix}4.1714549&5.9308002&-5.3682585&24.054762&-11.788758\end{bmatrix}}\\&|A|=37662\end{aligned}}$

To tridiagonalize matrix $A$ into matrix $T$, the off-tridiagonal cells [1,3], [1,4], [1,5], [2,4], [2,5], and [3,5], must continue to be iteratively zeroed until the maximum absolute value of those cells is below an acceptable convergence threshold. This example will use 1.e-14.. The cells below the diagonal will be zeroed automatically, due to the symmetric nature of the matrix. The first Jacobian rotation will be on the off-tridiagonal cell with the highest absolute value, which by inspection is [1,4] with a value of 11. To make this entry zero, the condition specified in the above equations must be met for the cell coordinates to be zeroed ($h=1,l=4$) and for the selected rotational coordinates of $S$ ($h=1,k=3$), and are reproduced below for the first iteration.

To force cell[1,4] and [4,1] to be 0 by rotating on cell[1][3]:

${\begin{aligned}&{\text{setting }}a'_{h\ell }=a'_{\ell h}=0{\text{ in above equation, }}a'_{h\ell }=a'_{\ell h}=ca_{h\ell }+sa_{hk}\,\!\\&9\sin(\theta )+11\cos(\theta )=0\\&\theta =\tan ^{-1}(-11/9)=-0.8850668\\&\sin(\theta )=-0.77395730\\&\cos(\theta )=0.66323890\\&\\&{\text{More conveniountly:}}\\&r={\sqrt {11^{2}+9^{2}}}=14.2126704\\&\sin(\theta )=-11/r=-0.77395730\\&\cos(\theta )=9/r=0.66323890\\&\\&S={\begin{bmatrix}&1&0&0&0&0\\&0&0.66323890&0&-0.77395730&0\\&0&0&1&0&0\\&0&0.77395730&0&0.66323890&0\\&0&0&0&0&1\\\end{bmatrix}}\\&\\&T_{1}=S^{T}AS={\begin{bmatrix}2&12.083046&9&0&7\\12.083046&-0.16438356&5.2139171&0.56164384&4.8001142\\9&5.2139171&7&-4.22079&1\\0&0.56164384&-4.22079&4.1643836&-3.3104236\\7&4.8001142&1&-3.3104236&4\\\end{bmatrix}}\\&{\text{eigenvalues of }}T_{1}={\begin{bmatrix}4.1714549&5.9308002&-5.3682585&24.054762&-11.788758\end{bmatrix}}\\&|T_{1}|=37662{\text{, and cell [1,4] = [4,1]=0}}\end{aligned}}$

The first rotation iteration, $T_{1}$, produces a matrix with cells [1,4] and [4,1] zeroed, as expected. Furthermore, the eigenvalues and determinant of $T_{1}$are identical to those of $A$ and T1 is also symmetric, confirming that the Jacobian rotation was performed correctly. The next iteration for $T_{2}$ will select cell [2,5] which contains the highest absolute value, 4.8001142, of all the cells to be zeroed..

After 10 iterations of zeroing the cell with the maximum absolute value using Jacobian rotations on the cell just below it, the maximum absolute value of all off-tridiagonal cells is 2.6e-15. Assuming this convergence criteria is acceptably low for the application it is being performed for, the similar triangularized $T$ matrix is shown below.

${\begin{aligned}&T={\begin{bmatrix}2&16.613248&0&0&0\\16.613248&10.184783&5.1109346&0&0\\0&5.1109346&4.0304188&4.5541372&0\\0&0&4.5541372&-3.3901672&0.43515335\\0&0&0&0.43515335&4.1749658\\\end{bmatrix}}\\&{\text{eigenvalues of }}T={\begin{bmatrix}4.1714549&5.9308002&-5.3682585&24.054762&-11.788758\end{bmatrix}}\\&|T|=37662\end{aligned}}$

Since $A$ and $T$ have identical eigenvalues and determinants and $T$ is also symmetric, $A$ and $T$ are similar matrices with $T$ being tridiagonalized.

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Eigenvalues example

Jacobian rotation can be used to extract the eigenvalues in a similar manner as the triangulation example above, but by zeroing all of the cells above the diagonal, instead of the tridiagonal, and performing the Jacobian rotation directly in the cells to be zeroed, instead of an adjacent cell.

Starting with the same matrix $A$ as the tridiagonal example,

${\begin{aligned}&A={\begin{bmatrix}2&5&9&11&7\\5&3&6&-2&5\\9&6&7&3&1\\11&-2&3&1&3\\7&5&1&3&4\\\end{bmatrix}}\\&{\text{eigenvalues of }}A={\begin{bmatrix}4.1714549&5.9308002&-5.3682585&24.054762&-11.788758\end{bmatrix}}\\&|A|=37662\end{aligned}}$

The first Jacobian rotation will be on the off-diagonal cell with the with the highest absolute value, which by inspection is [1,4] with a value of 11, and the rotation cell will also be [1,4], $l=4,k=1$ in the equations above. The rotation angle is the result of a quadratic solution, but it can be seen in the equation that if the matrix is symmetric, then a real solution is assured.

To force cell[1,4] and [4,1] to be 0 by rotating on cell[1][4]:

${\begin{aligned}&{\text{setting }}a'_{k\ell }=a'_{\ell k}=0{\text{ in above equation: }}a'_{k\ell }=a'_{\ell k}=(c^{2}-s^{2})a_{k\ell }+sc(a_{kk}-a_{\ell \ell })\,\!\\&11(cos^{2}(\theta )-sin^{2}(\theta ))+(2-1)cos(\theta )sin(\theta )=0\\&{\text{dividing by -}}cos^{2}(\theta ){\text{:}}\\&tan^{2}(\theta )-tan(\theta )/11-1=0\\&tan(\theta )={\bigg [}1/11+{\sqrt {1/11^{2}+4}}{\bigg ]}/2=1.0464871\\&\theta =tan^{-1}(1.0464871)=-0.80810980\\&sin(\theta )=0.72298259\\&cos(\theta )=0.69086625\\&\\&{\text{More conveniountly:}}\\&r={\sqrt {1.0464871^{2}+1}}=1.4474582\\&sin(\theta )=1.0464871/r=0.72298259\\&cos(\theta )=1/r=0.69086625\\&\\&S={\begin{bmatrix}&0.69086625&0&0&0.72298259&0\\&0&1&0&0&0\\&0&0&1&0&0\\&-0.72298259&0&0&0.69086625&0\\&0&0&0&0&1\\\end{bmatrix}}\\&\\&T_{1}=S^{T}AS={\begin{bmatrix}-9.5113578&4.9002964&4.0488484&0&2.6671159\\4.9002964&3&6&2.2331805&5\\4.0488484&6&7&8.5794421&1\\0&2.2331805&8.5794421&12.511358&7.1334769\\2.6671159&5&1&7.1334769&4\\\end{bmatrix}}\\&{\text{eigenvalues of }}T_{1}={\begin{bmatrix}4.1714549&5.9308002&-5.3682585&24.054762&-11.788758\end{bmatrix}}\\&|T_{1}|=37662\end{aligned}}$

The first rotation iteration, $T_{1}$, produces a matrix with cells [1,4] and [4,1] zeroed, as expected. Furthermore, the eigenvalues and determinant of $T_{1}$are identical to those of $A$ and T1 is also symmetric, confirming that the Jacobian rotation was performed correctly. The next iteration for $T_{2}$ will select cell [3,4] which contains the highest absolute value, 8.5794421, of all the cells to be zeroed..

After 25 iterations of zeroing the cell with the maximum absolute value using Jacobian rotations on the cell just below it, the maximum absolute value of all off-diagonal cells is 9.0233029E-11. Assuming this convergence criteria is acceptably low for the application it is being performed for, the similar diagonalized $T$ matrix is shown below.

${\begin{aligned}&T={\begin{bmatrix}24.054762&0&0&0&0\\0&5.9308002&0&0&0\\0&0&-11.788758&0&0\\0&0&0&4.1714549&0\\0&0&0&0&-5.3682585\\\end{bmatrix}}\\&{\text{eigenvalues of }}T={\begin{bmatrix}4.1714549&5.9308002&-5.3682585&24.054762&-11.788758\end{bmatrix}}\\&|T|=37662\end{aligned}}$

The eigenvalues are now displayed across the diagonal, and may be directly extracted for use elsewhere.

Since $A$ and $T$ have identical eigenvalues and determinants and $T$ is also symmetric, $A$ and $T$ are similar matrices with $T$ being successfully diagonalized.

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Coupled transmission lines

Transmission lines may be placed in proximity to each other such that they electrically interact, such as two microstrip lines in close proximity. Such transmission lines lines are said to be coupled transmission lines. Coupled transmission lines be characterized by an even and odd mode analysis. The even mode is characterized by excitation of the two conductors with a signal of equal amplitude and phase. The odd mode is characterized by excitation with signals of equal and opposite magnitude. The even and odd modes each have their own characteristic impedances (Zoe, Zoo) and phase constants ($\beta _{e}{\text{, }}\beta _{o}$). Lossy coupled transmission lines have their own even and odd mode attenuation constants ($\alpha _{e}{\text{, }}\alpha _{o}$), which in turn leads to even and odd mode propagation constants ($\gamma _{e}{\text{, }}\gamma _{o}$) .

Coupled matrix parameters

Coupled transmission lines may be modeled using even and odd mode transmission line parameters defined in the prior paragraph as shown with ports 1 and 2 on the input and ports 3 and 4 on the output,

${\begin{aligned}Y&={\begin{bmatrix}y11&y12&y13&y14\\y21&y22&y23&y24\\y31&y32&y33&y34\\y41&y42&y43&y44\\\end{bmatrix}}\\Z&=[Y]^{-1}\\&\\{\text{Where:}}&\\{\text{For lossless coupled lines:}}&\\y11&=y22=y33=y44={\frac {-j}{2}}{\bigg (}{\frac {cot(\beta _{e}l)}{Z_{oe}}}+{\frac {cot(\beta _{o}l)}{Z_{oo}}}{\bigg )}\\y12&=y22=y34=y43={\frac {-j}{2}}{\bigg (}{\frac {cot(\beta _{e}l)}{Z_{oe}}}-{\frac {cot(\beta _{o}l)}{Z_{oo}}}{\bigg )}\\y13&=y31=y24=y42={\frac {j}{2}}{\bigg (}{\frac {csc(\beta _{e}l)}{Z_{oe}}}+{\frac {csc(\beta _{o}l)}{Z_{oo}}}{\bigg )}\\y14&=y41=y23=y32={\frac {j}{2}}{\bigg (}{\frac {csc(\beta _{e}l)}{Z_{oe}}}-{\frac {csc(\beta _{o}l)}{Z_{oo}}}{\bigg )}\\{\text{For lossy coupled lines:}}&\\y11&=y22=y33=y44={\frac {1}{2}}{\bigg (}{\frac {coth(\gamma _{e}l)}{Z_{oe}}}+{\frac {coth(\gamma _{o}l)}{Z_{oo}}}{\bigg )}\\y12&=y22=y34=y43={\frac {1}{2}}{\bigg (}{\frac {coth(\gamma _{e}l)}{Z_{oe}}}-{\frac {coth(\gamma _{o}l)}{Z_{oo}}}{\bigg )}\\y13&=y31=y24=y42={\frac {-1}{2}}{\bigg (}{\frac {csch(\gamma _{e}l)}{Z_{oe}}}+{\frac {csch(\gamma _{o}l)}{Z_{oo}}}{\bigg )}\\y14&=y41=y23=y32={\frac {-1}{2}}{\bigg (}{\frac {csch(\gamma _{e}l)}{Z_{oe}}}-{\frac {csch(\gamma _{o}l)}{Z_{oo}}}{\bigg )}\\\end{aligned}}$..

Losses

Attenuation due to losses from the conductor and dielectric are generally considered when simulating a microstrip. Total losses are a function of microstrip length, so attenuation is generally calculated in units of attenuation per unit length, with total losses calculated by attenuation * length, with attenuation units of Nepers, although some applications may use attenuation in units dB. When the microstrip characteristic impedance (Zo), effective dielectric constant (Ere), and total losses ($\alpha l$) are all known the microstrip may be modeled as a standard transmission line.

Conductor losses

Conductor losses are defined by the "specific resistance" or "resistivity" of the conductor material, and generally expressed as $\rho$ in the literature. Each conductor material generally has a published resistivity associated with it. For example, the common conductor material of copper has a published resistivity of 1.68E-8 Ohm-m . E. Hammerstad and Ø. Jensen proposed the following expressions for attenuation due to conductor losses:

${\begin{aligned}\alpha _{c}&={\frac {R^{s}}{Z_{o}W}}K_{r}K_{i}\\{\text{Where :}}&\\R_{s}&={\sqrt {\frac {\rho \omega u_{o}}{2}}}\\K_{i}&=exp{\bigg [}-1.2{\bigg (}{\frac {Z_{o}}{n_{o}}}{\bigg )}^{0.7}{\bigg ]}\\K_{r}&=1+{\frac {2tan^{-1}(1.4({\frac {\Delta }{\delta }})^{2})}{\pi }}\end{aligned}}$.

And:

$R^{s}$ = sheet resistance of the conductor

$K_{i}$ = current distribution factor

$K_{r}$ = correction term due to surface roughness

$u_{o}$ = Vacuum permeability ($4\pi \times 10^{-7}H/m$)

$\rho$ = Specific resistance, or resistivity, of the conductor

$\Delta$ = effective (rms) surface roughness of the substrate

$\delta$ = skin depth

$n_{o}$ = wave impedance in a vacuum (376.730313412 Ohms)

Note that if surface roughness is neglected, the $K_{r}$ disappears from the expression, and it frequently is.

Dielectric losses

Dielectric losses are defined by the "loss tangent" of the dielectric material, and generally expressed as $tan\delta _{d}$ in the literature. Each dielectric material generally has a published loss tangent associated with it. For example, the common dielectric material is alumina has a published loss tangent of 0.0002 to .0003 depending on the frequency. Welch and Pratt, and Schneider proposed the following expressions for attenuation due to dielectric losses.:

${\begin{aligned}\alpha _{d}&={\frac {tan\delta _{d}{\text{ }}\omega }{2}}{\frac {\varepsilon _{r}}{\sqrt {\varepsilon _{re}}}}{\frac {\varepsilon _{re}-1}{\varepsilon _{r}-1}}{\text{Np per unit length}}\\\alpha _{d}&={\frac {27.3{\text{ }}tan\delta _{d}{\text{ }}\omega }{2\pi }}{\frac {\varepsilon _{r}}{\sqrt {\varepsilon _{re}}}}{\frac {\varepsilon _{re}-1}{\varepsilon _{r}-1}}{\text{dB per unit length}}\\\end{aligned}}$.

Dielectric losses are in general much less that conductor losses and are frequently neglected in some applications.

Coupled microstrip losses

Coupled microstrip losses may be estimated using the same even and odd mode analysis as is used for characteristic impedance, dielectric constant. and effective dielectric constant for single line microstrips. Coupled line even mode and odd mode each have their independently calculated conductor and dielectric loss values calculated from the corresponding single line parameter, with the exception that even and odd current distribution factor are the same value, and is calculated from a Zo consisting of the average of Zoe and Zoo. .

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Inverse solution

Some applications require that k be calculated from a known $K({\sqrt {1-k^{2}}})/K(k)$, which we will refer to as $F$ for the remainder of this section, and the modulus of k, ${\sqrt {1-k^{2}}}$, will be referred to as k'. Although it is possible to obtain a value for k that solves the equation from table interpolations, Newton's method provides for a more efficient solution that is both rapid and precise for a wide variety of $F$. Since the range of $K(k')/K(k)$ for 0.1<k<.9 is not too far off from linear, as observed from the table below, the standard equation for Newton's method is sufficient to solve the problem for this range. Limits to the states may be applied as needed to keep k within the range 0<k<1,

$k_{n+1}=k_{n}-{\frac {F-{\frac {K(k')}{K(k)}}}{d{\Bigl [}{\frac {K(k')}{K(k)}}{\Bigl ]}/dk}}$.

The extremely nonlinear nature of $K(k')/K(k)$ for k<0.1 and k>0.9 prevents the normal convergence of the equation in these ranges. However, when k is replaced by $lgk=ln(k)$, and k' is replace by $lgk'=ln(k)'$, then $K(lgk')/K(lgk)$ is significantly more linear in nature for high values, which in turn allows Newton's method to rapidly converge for F>.9. For F<.1, 1/F may be substituted for F, and the final k solution substituted with k'. The modified equations is

$lgk_{n+1}=lgk_{n}-{\frac {F-{\frac {K((\varepsilon ^{lgk})')}{K(\varepsilon ^{lgk})}}}{d{\Bigl [}{\frac {K((\varepsilon ^{(lgk)})')}{K(\varepsilon ^{lgk})}}{\Bigl ]}/dlgk}}$

then

$k=\varepsilon ^{lgk}$.

The derivatives in the denominators may be determined from the K(k) derivative definitions above and the rules from the derivative page. For high values of F, singularities occur due to finite 64 bit arithmetic precision. If it is not desired to use high precision arithmetic, approximations based on the limits as $lgk\rightarrow -\infty$ remove the singularities.

${\begin{aligned}d{\Biggl [}{\frac {K(k')}{K(k)}}{\Biggl ]}/dk&={\frac {K(k){\frac {dK(k')}{dk}}-K(k'){\frac {dK(k)}{dk}}}{K^{2}(k)}}\\{\text{where: }}{\frac {dK(k')}{dk}}&={\Biggl [}{\frac {dK(k')}{dk'}}{\Biggl ]}{\frac {-k}{\sqrt {1-k^{2}}}}\\\end{aligned}}$

Then:

${\begin{aligned}d{\Biggl [}{\frac {K((\varepsilon ^{(lgk)})')}{K(\varepsilon ^{(lgk)})}}{\Biggl ]}/d(lgk)&{\begin{cases}=\varepsilon ^{lgk}d{\Biggl [}{\frac {K(\varepsilon ^{(lgk)})'}{K(\varepsilon ^{(lgk)})}}{\Biggl ]}/d(\varepsilon ^{(lgk)}),&{\text{if }}lgk>-14\\\\\approx -\pi /2,&{\text{otherwise }}\end{cases}}\\\end{aligned}}$

The initial value for lgk may be -F for the ln(k) case and k may be $\varepsilon ^{-F}$ for the straight case, based on values observed in the table below. For smaller F, ln(1-tanh²(F)) and (1-tanh²(F)) have been observed to be a crude approximation and may serve as a suitable starting point.

Below is a table of $k$, $ln(k)$, ${\frac {K(k')}{K(k)}}$, and the computed ${\Bigg [}{\frac {K(k')}{K(k)}}{\Bigg ]}^{-1}$ using Newton's method as described above.

${\begin{aligned}&{\frac {K(k')}{K(k)}}{\text{ as a function of }}k{\text{ and }}ln(k),\\&{\text{ and }}{\Biggl [}{\frac {K(k')}{K(k)}}{\Biggl ]}^{-1}{\text{ computed from }}{\frac {K(k')}{K(k)}}\end{aligned}}$

$k$	$lgk{\text{, }}ln(k)$	${\frac {K(k')}{K(k)}}$	$k={\Bigg [}{\frac {K(k')}{K(k)}}{\Bigg ]}^{-1}$
1.e-20	-46.051702	30.199966	1.e-20
1.e-09	-20.723266	14.075383	1.e-09
1.e-05	-11.512925	8.2118984	1.e-05
0.001	-6.9077553	5.2801558	0.001
0.01	-4.6051702	3.8142689	0.01
0.1	-2.3025851	2.3468155	0.1
0.2	-1.6094379	1.9006702	0.2
0.5	-0.69314718	1.2792616	0.5
${\sqrt {2}}/2$	-0.34657359	1	0.70710678
0.8	-0.22314355	0.87743766	0.8
0.9	-0.10536052	0.72553432	0.9
0.99	-0.010050336	0.47032697	0.99
0.999	-0.0010005003	0.34958259	0.999
0.999999	-1.0000005E-06	0.1976472	0.999999
1-1.0e-09	-9.9999997E-10	0.1377727	1-1.0e-09

Complete elliptic integral of the second kind

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Multivariate interpolation

${\begin{aligned}V_{D}&=INTERP(Mn,V_{In})\\V_{D}n&=INTERP(M_{n-1},V_{I})\\\end{aligned}}$