BER vs confidence level

1. General telegraph equation
From telegraph equation

P'(\epsilon)=\frac{\epsilon}{n},when ~ n->\infty , P'(\epsilon)->P(\epsilon)

Transmission line equations

1. General telegraph equation
From telegraph equation

\begin{align*}
- & \frac{\partial \bf{v}(z,\omega) }{\partial z}=[\bf{R}(\omega )+j\omega \bf{L}(\omega)]\bf{i}(z,\omega )=\bf{Z}(\omega) \bf{i}(z,\omega) \\
- & \frac{\partial \bf{i}(z,\omega) }{\partial z}=[\bf{G}(\omega )+j\omega \bf{C}(\omega)]\bf{v}(z,\omega )= \bf{Y}(\omega ) \bf{v} (z, \omega )\\
\end{align*}

we have the following wave equation

\begin{align*}
- \frac{\partial ^2 \bf{v}(z)}{\partial z} = [Z][Y]\bf{v}(z) \\
- \frac{\partial ^2 \bf{i}(z)}{\partial z} = [Y][Z]\bf{i}(z)
\end{align*}

The modal solution to the wave equation will be of the form

\begin{align*}
 \bf{v}(z) & = \bf{V}^{\pm}e ^{\mp \gamma _0 z} \\
 \bf{i}(z) & = \bf{I}^{\pm}e ^{\mp \gamma _1 z }\\
\end{align*}

Substitute above equation into wave equation, we have

\begin{align*}
( \gamma _0 ^2 \bf{U} &  -[Z][Y]) \bf{V} ^{\pm} =0 \\
( \gamma _1 ^2 \bf{U} &  -[Y][Z]) \bf{I} ^{\pm} =0 ....(4)\\
\end{align*}

For non-trivial solution, the determinant of

\begin{align*}
| \gamma _0 ^2 \bf{U} &  -[Z][Y] | =0 \\
( \gamma _1 ^2 \bf{U} &  -[Y][Z]| =0 ....(5)\\
\end{align*}

It can be shown r0=r1=r, due to symmetric of [Z],[Y] and ([Z][Y])T=[YT][ZT]=[Y][Z].
The solution to (4) will be called modes, while the eigenvalues will be called spectrum.
Assume(5) (4) has n distinct eigenvalues and eigenvectors, combine all eigenvectors to form matrix, it will be

\begin{align*}
[\bf{V}_1 e ^ {- \gamma _1 z}, \bf{V}_2 e ^ {- \gamma _1 z},...\bf{V}_N e ^ {- \gamma _N z} ] = \bf{M} _V diag(e ^ {- \gamma _1 z},e ^ {- \gamma _2 z},...,e ^ {- \gamma _N z})=\bf{M}_v \bf{D} ....(6)\\
\end{align*}

where Mv=[V1,V2,...VN] are eigenvectors of voltage wave, Mv is non-singular and span a N-dimensional space. Similarity, we have

\begin{align*}
[\bf{I}_1 e ^ {- \gamma _1 z}, \bf{I}_2 e ^ {- \gamma _1 z},...\bf{I}_N e ^ {- \gamma _N z} ] = \bf{M} _I diag(e ^ {- \gamma _1 z},e ^ {- \gamma _2 z},...,e ^ {- \gamma _N z})=\bf{M}_I \bf{D} 
\end{align*}

Substitute (6) into (1), we have

\begin{align*}
\bf{M}_V =[Z]\bf{M}_I diag(\gamma_1 ^ {-1},\gamma_2 ^ {-1},...\gamma_N^ {-1})=\bf{Z}\bf{M}_I \bf{S} 
\end{align*}

Similarity, we have

\begin{align*}
\bf{M}_I =[Y]\bf{M}_V diag(\gamma_1 ^ {-1},\gamma_2 ^ {-1},...\gamma_N^ {-1})=\bf{Y}\bf{M}_V \bf{S} 
\end{align*}

Now, we have the N engenvectors spanning N-dimensional space, any voltage V(z) can be represented as linear combination of these eighenvectors.

\bf{V}(z)=\bf{V}_i(z)+\bf{V}_r(z)=\bf{M_v(DV_0^++D^{-1}V_0^-)}

where

\begin{align*}
\bf{V_0^+}= &(V_{01}^+,V_{02}^+,....,V_{0N}^+)\\
\bf{V_0^-}= & (V_{01}^-,V_{02}^-,....,V_{0N}^-)
\end{align*}

Same thing for I(z),where

\bf{I}(z)=\bf{I}_i(z)+\bf{I}_r(z)=\bf{M_I(DI_0^+-D^{-1}I_0^-)}

Note: minus (-1) sign is due to the directions of current Ir is opposite of Ii.
The characteristic matrix Zc is defined as

\begin{align*}
\bf{V_i}(z) & =\bf{Z}_C \bf{I}_i(z) \\
\bf{V_r}(z) & =-\bf{Z}_C \bf{I}_r(z) \\
w&here\\
  \bf{Z}_C & =\bf{M}_V \bf{M}_I ^{-1}
\end{align*}

S-parameter vs impedance matrix

1.Generalized scattering parameter

\begin{align*}
a_n &=\frac{V_n ^+}{\sqrt {Z_{0n}}}\\
b_n &=\frac{V_n ^-}{\sqrt {Z_{0n}}}\\
w&here~ a_n,b_n ~are~incident~wave~and~reflected~wave,respectively\\
V_n &=V_n^+ + V_n^- =(a_n + b_n) \times \sqrt{Z_{0n}}\\
I_n &=I_n^+ + I_n^- =(V_n^+ -V_n^-)/ Z_{0n}=(a_n-b_n)/\sqrt{Z_{0n}}\\
P_n & =\frac{1}{2}Re[V_nI_n^*]=\frac{1}{2}[a^2-b^2]\\
\bf{b} & =\bf{S}\bf{a}\\
w&here~ S_{ij}=\frac{b_i}{a_j}|a_k=0~ for~ k\ne j
\end{align*}

For Z0n=Z0, ie, all ports have same characteristic impedance, we have

\begin{bmatrix} V_1 ^- \\ V_2 ^-\\.\\.\\.\\V_n ^- \end{bmatrix}=
\begin{bmatrix}S_{11}&S_{12}&... &S_{1n}\\S_{21}&S_{22}&...&S_{2n}\\.\\.\\.\\S_{n1} &S_{n2}&...&S_{nn} \end{bmatrix}
\begin{bmatrix} V_1 ^+ \\ V_2 ^+\\.\\.\\.\\V_n ^+ \end{bmatrix} \\
where \\
  S_{ij}=\frac{V_i^-}{V_j^+} |_{V_k^+ =0,~ for~ k \ne j} \\
V_k ^+=0~ --~ port~k~terminated~to~characteristic~impedance\\

For impedance matrix
[V]=[Z][I]
where

\begin{align*}
V_{ij} &=V_{ij}^+ + V_{ij}^- \\
I_{ij} &=I_{ij}^+ - I_{ij}^- \\
\end{align*}

The - sign is due to inverted direction of current \inline I^- . See below for the illustration of V+/V- and I+/I-.
I+
---->--------+-----------+
| | | |
V+ V- | port k |
| | | |
-----<-------+-----------+
I-

inductance calculation

1.two transmission lines ( one transmission line over ground plane)
============ t
h
============
<--- W ---->

Assume W>>h, h>>t, the partial inductance of the two-pair trace is

L_{trace_pair}\approx \frac{\mu_0 \mu_r h len}{W}

where u0= 4*pi nH/meter=32pH/mil (permeability of air), ur=relative permeability of the medium. len=transmission line length.
when len=W, L_trace=32pH*h (independent of the size of the square.)

Frequency domain(time-harmonic) maxwell equations

Frequency domain maxwell equation (vs. regular time-domain maxwell equations)
That's most frequently used maxwell equations. Every time domain function a(t) can be obtained using inverse fourier transform, A(w)

\begin{align*}
\pmb{e} & (\pmb{r},t)=\frac{1}{2\pi }\int \pmb{E}(\pmb{r},\omega)e^{j \omega t}d\omega,
\pmb{h}  (\pmb{r},t)=\frac{1}{2\pi }\int \pmb{H}(\pmb{r},\omega)e^{j \omega t}d\omega \\
F & or ~time ~ harmonic ~ components,\\
\mathcal{E} &(r,t)=\pmb{E}(r,\omega)e^{j\omega t},
\mathcal{H}(r,t)=\pmb{H}(r,\omega)e^{j\omega t}\\
\end{align*}

Substitute time harmonic E/H into maxwell equations, with d/dt=jw, we have frequency domain maxwell equations

\begin{align*}
\nabla & \times \pmb{E}= - j\omega \pmb{B} \\
\nabla & \times \pmb{H}= \pmb{J}+  j\omega \pmb{D}\\
\nabla & \cdot \pmb{D}= \rho\\
\nabla & \cdot \pmb{B}= 0
\end{align*}

Note: for the above equations, E=E(r,w). Often, we drop w for simplicity.
Once we know the solution of time harmonic E(r,w) and H(r,w) , we can obtain E(r,t) and H(r,t) in time domain by doing inverse fourier transform.

Signal phasor representation and time average
In real word, the signal are real. For a real value sinusodal signal,

\begin{align*}
 \mathcal{A} & (t)=|A|cos(\omega t+\phi) ~ \Longleftrightarrow  A(t)=Ae^{j\omega t}\\
o & r ~ \mathcal{A}(t)=Re \{A(t)\}\\
w&here ~ A=|A|e^{j\phi}~ is~ a ~complex ~phasor.
\end{align*}

Time average of two harmonic signals,

\begin{align*}
\mathcal{A} &(t)=Re [A(t)] = Re [Ae^{j\omega t} ]\\
\mathcal{B} &(t)=Re [B(t)] = Re [Be^{j\omega t} ]\\
<\mathcal{A} &(t) \mathcal{B}(t)>=<\frac{1}{2}(Ae^{j\omega t}+A^*e^{-j\omega t})*\frac{1}{2}(Be^{j\omega t}+B^*e^{-j\omega t})\\
= & \frac{1}{4}<(AB^*+BA^*)+ABe^{-j2\omega t}+A^*B^*e^{-j2\omega t}> \\
= & \frac{1}{2} ( < Re[AB^*] > + < Re[AB*e^{-j2\omega t} ] > ) \\
= & \frac{1}{2}  Re[AB^*]  
\end{align*}

energy conservation and charge conservation

1. Charge conservation
Do the divergence of ampere law(2nd maxwell equation). Since curl has no divergence(another one-gradient has no curl), we have

\begin{align*}
0 &=\nabla \cdot (\nabla \times \pmb{H})=\nabla \cdot (J+\frac{\partial \pmb{D}}{\partial t}) ~\Rightarrow \\
\nabla & \cdot J +\frac{\partial \rho}{\partial t}=0 \\
o &r, in~ integral~ form \\
I &=\int_V\nabla \cdot Jdv=\oint_SJda=\frac{\partial} {\partial t}\int_V \rho dv=\frac{\partial Q} {\partial t}
\end{align*}

Current flow out of the volume V equals to the charge depletion per unit time.
For conductors, J=sigmaE,

\begin{align*}
\nabla & \cdot J= \nabla  \cdot \sigma E=\frac{\sigma \rho }{\epsilon} ~ \Rightarrow \\
\nabla & \cdot J+\frac{\partial \rho}{\partial t}=\frac{\sigma \rho }{\epsilon}+\frac{\partial \rho}{\partial t}=0  ~ \Rightarrow \\
\rho &(r,t)  = \rho_0  e^{-\sigma t / \epsilon } \\
d &ue ~ to ~ \sigma >> \epsilon ~ in ~ conductor, \rho -> 0 ~ in~ a ~short~ time. 
\end{align*}

e.g. Copper, sigma=5.7x10^7, epsilon=8.85x10^-12, epsilon/sigma=1.6x10^-19 sec.


2. Energy conservation.
* Energy per unit volume, \inline w=\frac{1}{2}\epsilon |E|^2+\frac{1}{2}\mu |H|^2
* Energy flux (Poynting vector): \inline \pmb{P}=\pmb{E} \times \pmb{H}

\begin{align*}
 \nabla  &\cdot (\pmb{E} \times \pmb{H})=\pmb{H}\cdot (\nabla \times \pmb{E})-\pmb{E}\cdot (\nabla \times \pmb{H})= \\
-\mu & \pmb{H}\cdot \frac{\partial \pmb{H}}{\partial t} -\mu \pmb{E}\cdot \frac{\partial \pmb{E}}{\partial t}-\pmb{E}\cdot \pmb{J}  ~\Rightarrow\\
\nabla  &\cdot \pmb{P}=  -\frac{\partial w}{\partial t}- \pmb{E}\cdot \pmb{J}\\
N&ote: \epsilon \pmb{E} \cdot \frac{\partial \pmb{E} }{\partial t}=\frac{1}{2}\epsilon \frac{\partial |E|^2}{\partial t}, \mu \pmb{H} \cdot \frac{\partial \pmb{H} }{\partial t}=\frac{1}{2}\mu \frac{\partial |H|^2}{\partial t}
\end{align*}

the rate of energy flow out of volume equals to the depletion rate of stored energy in the volume and ohmic loss in the volume.

curvilinear coordinates

A point can be represented as (x,y,z).Consider 3 independent,smooth and unique invertible(and hopefully orthogonal) functions: u1=f1(x,y,z), u2=f2(x,y,z),u3=f3(x,y,z).The intersection of constant u1,u2,u3 defines a point in a space. The coordinate (u1,u2,u3) is called curvilinear coordinates.Base on inversion, we have x=f1'(u1,u2,u3),y=f2'(u1,u2,u3),z=f3'(u1,u2,u3)

\begin{align*}
\pmb{r} & (u_1,u_2,u_3) =x(u_1,u_2,u_3)\hat{x}+y(u_1,u_2,u_3)\hat{y}+z(u_1,u_2,u_3)\hat{z}\\
d &\pmb{r}=\frac{\partial \pmb{r}}{\partial u_1}du_1+\frac{\partial \pmb{r}}{\partial u_2}du_2+\frac{\partial \pmb{r}}{\partial u_3}du_3\\
=& \pmb{a}_1'du_1+\pmb{a}_2'du_2+\pmb{a}_3'du_3\\
i&f ~ \pmb{a}_1' \perp \pmb{a}_2' \perp \pmb{a}_3', this ~ is ~ orthogonal ~ curvilinear ~ coordinates ~ system\\
d \pmb{r} & = h_1\pmb{a}_1du_1+h_2\pmb{a}_2du_2+h_3\pmb{a}_3du_3\\
w&here ~ \pmb{a}_1,\pmb{a}_2,\pmb{a}_3 ~ are ~ unit ~ vectors
\end{align*}

Similarly,for small surface and volume

\begin{align*}
d\pmb{S}_i &  =  d\pmb{s}_j \times d\pmb{s}_k \\
wh&ere ~ d\pmb{s}_j=h_jdu_j\pmb{a}_j,d\pmb{s}_k=h_kdu_k\pmb{a}_k,\pmb{a}_j \times \pmb{a}_k=\pmb{a}_i\\
dV  = & d\pmb{s}_i \cdot d\pmb{S}_i =(h_idu_i\pmb{a}_i)\cdot (h_jh_kdu_jdu_k\pmb{a}_i)\\
= &(h_1h_2h_3)du_1du_2du_3
\end{align*}

For Cartesian coordinate,

\begin{align*}
u_1 & =x,u_2=y,u_3=z \\
\pmb{a}_1 & =\hat{x},\pmb{a}_2=\hat{y},\pmb{a}_z=\hat{z}\\
d\pmb{r} & =d\pmb{s}=\hat{x}dx+\hat{y}dy+\hat{z}dz \\
h_1 & =h_2=h_3=1\\
d\pmb{S} & =\hat{x}dS_x+\hat{y}dS_y+\hat{z}dS_z=dydz\hat{x}+dxdz\hat{y}+dxdy\hat{z} \\
dV & =dxdydz
\end{align*}

For cylindrical coordinates,

\begin{align*}
u_1 & =\rho, u_2=\psi, u_3=z \\
\pmb{a}_1 & =\hat{u}_{\rho},\pmb{a}_2=\hat{u}_{\psi},\pmb{a}_3=\hat{u}_{z}\\
d\pmb{r} & =\hat{u}_{\rho}d\rho+\hat{u}_{\psi}\rho d\psi+\hat{u}_{\z}dz\\
h_1 &=1, h_2=\rho, h_3=1 \\
d\pmb{S} & =\rho d\psi dz\hat{u}_{\rho} +d\rho dz \hat{u}_{\psi}+\rho d\rho d\psi \hat{u}_{z}\\
dV & =\rho d\rho d\psi dz
\end{align*}

For spherical coordinates,

\begin{align*}
u_1 & =\rho, u_2=\psi, u_3=\theta \\
\pmb{a}_1 & =\hat{u}_{\rho},\pmb{a}_2=\hat{u}_{\psi},\pmb{a}_3=\hat{u}_{\theta}\\
d\pmb{r} & =\hat{u}_{\rho}d\rho+\hat{u}_{\psi}\rho sin\theta d\psi+\hat{u}_{\theta}\rho d\theta \\ 
h_1 & =1, h_2=\rho, h_3=\rho sin \theta \\
d\pmb{S} &=dS_r\hat{u}_r+dS_{psi}\hat{u}_\psi+dS_{\theta}\hat{u}_{\theta} \\
d\pmb{S} &=\rho^2sin\theta d\psi d \theta\hat{u}_r+\rho sin\theta d\rho d\theta \hat{u}_\psi+\rho d\rho d\psi \hat{u}_{\theta}\\
dV &=\rho^2sin\theta d\rho d\psi d\theta
\end{align*}

For gradient,

\begin{align*}
d\phi & (u_1,u_2,u_3)=\nabla \phi (u_1,u_2,u_3) \cdot d\pmb{s} \\
d\phi & =\frac{\partial \phi}{\partial u_1}du_1+\frac{\partial \phi}{\partial u_2}du_2+\frac{\partial \phi}{\partial u_3}du_3\\
d\pmb{s} & =h_1du_1\pmb{a}_1+h_2du_2\pmb{a}_2+h_3du_3\pmb{a}_3 ~~\Rightarrow \\
\nabla & \phi (u_1,u_2,u_3)=\frac{\partial \phi}{h_1\partial u_1}\pmb{a}_1+\frac{\partial \phi}{h_2\partial u_2}\pmb{a}_2+\frac{\partial \phi}{h_3\partial u_3}\pmb{a}_3
\end{align*}

For divergence,

\begin{align*}
\nabla \cdot \pmb{A}=\frac{1}{h_1h_2h_3}[\frac{\partial(A_1h_2h_3)}{\partial u_1}+\frac{\partial(A_2h_1h_3)}{\partial u_2}+\frac{\partial(A_3h_1h_2)}{\partial u_3}]
\end{align*}

For curl,

\begin{align*}
\nabla \times \pmb{A}=\frac{1}{h_1h_2h_3}
\begin{vmatrix}
h_1\pmb{a}_1 & h_2\pmb{a}_2 & h_3\pmb{a}_3 \\
\frac{\partial}{\partial u_1} & \frac{\partial }{\partial u_2} & \frac{\partial}{\partial u_3} \\
h_1A_1 & h_2A_2 & h_3A_3
\end{vmatrix}
\end{align*}

For laplacian,

\begin{align*}
\nabla ^2 \phi  & =\nabla \cdot (\nabla \phi)
=\frac{1}{h_1h_2h_3}[\frac{\partial(A_1h_2h_3)}{\partial u_1}+\frac{\partial(A_2h_1h_3)}{\partial u_2}+\frac{\partial(A_3h_1h_2)}{\partial u_3}] \\
wh & ere ~ A_1=\frac{\partial \phi}{h_1\partial u_1},A_2=\frac{\partial \phi}{h_2\partial u_2},A_3=\frac{\partial \phi}{h_3\partial u_3} \Rightarrow \\
\nabla ^2 \phi & =\frac{1}{h_1h_2h_3}[\frac{\partial}{\partial u_1}(\frac{h_2h_3}{h_1}\frac{\partial \phi}{\partial u_1})+\frac{\partial}{\partial u_2}(\frac{h_1h_3}{h_2}\frac{\partial \phi}{\partial u_2})+\frac{\partial}{\partial u_3}(\frac{h_1h_2}{h_2}\frac{\partial \phi}{\partial u_3})

\end{align*}