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Three phase

As compared to a single-phase AC power supply that uses two current-carrying conductors with no neutral, a three-phase supply with no neutral and the same phase-to-phase voltage can transmit the same power by using just 0.75 times as much conductor material.^[a]^[1]

Blacklisting

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External inductance of a round wire

$L_{ext}=$

Funky display

${\begin{aligned}{\frac {{\text{d}}U}{{\text{d}}t}}&=L(i)\,i\,{\frac {{\text{d}}i}{{\text{d}}t}}\\[3pt]{\text{d}}U&=L(i)\,i\,{\text{d}}i\end{aligned}}$

Dipole antenna fields

I feel like we can bring some insight to the near field and the far field. The is stuff that "I know is true," and is "obvious to me," but maybe not to the other editors here. Please let me have your comments, suggestions, improvements.

The upshot is:

1. The electric potential, φ, is a function of only the charge distribution. There are many WP:RS for this.
2. The magnetic vector potential, A, is a function of only the current distribution. There are many RS for this.
3. The charge distribution is a dipole. (Obvious?)
4. The current distribution is a monopole (a current monopole, not a magnetic monopole).
5. Dipole fields die off rapidly with distance compared to monopoles. (Common knowledge)
6. At a distance, only A remains.
7. The E and B fields are given by $E=-\nabla \phi -{\frac {dA}{dt}}$ and $B=\nabla \times A$ . Many RS.
8. At a distance, the E and B fields reduce to $E=-{\frac {dA}{dt}}$ and $B=\nabla \times A$ . Obviously follows from #7.
9. ${\frac {dA}{dt}}$ and $\nabla \times A$ . are orthogonal. Basic property of curl (I think).

Frequency domain

The preceding time domain equations can be expressed in the frequency domain.^[2]^: 139

Lorenz gauge $\nabla \cdot \mathbf {A} +{\frac {j\omega }{c^{2}}}\phi =0\qquad$ or $\qquad \phi ={\frac {j\omega }{k^{2}}}\nabla \cdot \mathbf {A}$
Solutions $\mathbf {A} \!\left(\mathbf {r} ,\omega \right)={\frac {\mu _{0}}{\ 4\pi \ }}\int _{\Omega }{\frac {\mathbf {J} \left(\mathbf {r} ',\omega \right)}{R}}\ e^{-jkR}d^{3}\mathbf {r} '\qquad \phi \!\left(\mathbf {r} ,\omega \right)={\frac {1}{4\pi \epsilon _{0}}}\int _{\Omega }{\frac {\rho \left(\mathbf {r} ',\omega \right)}{R}}\ e^{-jkR}d^{3}\mathbf {r} '$
Wave equations $\nabla ^{2}\phi +k^{2}\phi =-{\frac {\rho }{\epsilon _{0}}}\qquad \nabla ^{2}\mathbf {A} +k^{2}\mathbf {A} =-\mu _{0}\ \mathbf {J} .$
Electromagnetic field equations $\mathbf {B} =\nabla \times \mathbf {A} \ \qquad \mathbf {E} =-\nabla \phi -j\omega \mathbf {A} =-j\omega \mathbf {A} -j{\frac {\omega }{k^{2}}}\nabla (\nabla \cdot \mathbf {A} )$

where

\phi

and

\rho

are scalar phasors.

\mathbf {A} ,\mathbf {B} ,\mathbf {E} ,

and

\mathbf {J}

are vector phasors.

k={\frac {\omega }{c}}

Frequency domain notes

There are a few notable things about $\mathbf {A}$ and $\phi$ calculated in this way:

The Lorenz gauge condition is satisfied: $\textstyle \phi =-{\frac {c^{2}}{j\omega }}\nabla \cdot \mathbf {A} .$ This implies that the frequency domain electric potential, $\phi$ , can be computed entirely from the current density distribution, $\mathbf {J}$ .
The position of $\mathbf {r} ,$ the point at which values for $\phi$ and $\mathbf {A}$ are found, only enters the equation as part of the scalar distance from $\mathbf {r} '$ to $\ \mathbf {r} .$ The direction from $\mathbf {r} '$ to $\mathbf {r}$ does not enter into the equation. The only thing that matters about a source point is how far away it is.
The integrand uses the phase shift term $e^{-jkR}$ which plays a role equivalent to retarded time. This reflects the fact that changes in the sources propagate at the speed of light; propagation delay in the time domain is equivalent to a phase shift in the frequency domain.
The equation for $\mathbf {A}$ is a vector equation. In Cartesian coordinates, the equation separates into three scalar equations:^[3] ${\begin{aligned}\mathbf {A} _{x}\!\left(\mathbf {r} ,\omega \right)&={\frac {\mu _{0}}{4\pi }}\int _{\Omega }{\frac {\mathbf {J} _{x}\left(\mathbf {r} ',\omega \right)}{R}}\ e^{-jkR}\ d^{3}\mathbf {r} ',\qquad \mathbf {A} _{y}\!\left(\mathbf {r} ,\omega \right)&={\frac {\mu _{0}}{4\pi }}\int _{\Omega }{\frac {\mathbf {J} _{y}\left(\mathbf {r} ',\omega \right)}{R}}\ e^{-jkR}\ d^{3}\mathbf {r} ',\qquad \mathbf {A} _{z}\!\left(\mathbf {r} ,\omega \right)&={\frac {\mu _{0}}{4\pi }}\int _{\Omega }{\frac {\mathbf {J} _{z}\left(\mathbf {r} ',\omega \right)}{R}}\ e^{-jkR}\ d^{3}\mathbf {r} '\end{aligned}}$ In this form it is apparent that the component of $\mathbf {A}$ in a given direction depends only on the components of $\ \mathbf {J} \$ that are in the same direction. If the current is carried in a straight wire, $\mathbf {A}$ points in the same direction as the wire.

Noise

Memory aid

The noise voltage of a 1kΩ resister at room temperature is 4 nV per root hertz.

$v_{n}=4.07{\sqrt {\tfrac {T}{300K}}}{\sqrt {\tfrac {R}{1k\Omega }}}{\tfrac {nV}{\sqrt {Hz}}}$

Lorentz transformation of the potentials

${\begin{aligned}\phi '&=\gamma \left(\phi -v\mathbf {n} \cdot \mathbf {A} \right)\,,\\\mathbf {A} '&=\mathbf {A} +(\gamma -1)(\mathbf {A} \cdot \mathbf {n} )\mathbf {n} -{\frac {\gamma \phi v\mathbf {n} }{c^{2}}}\,.\end{aligned}}$

where:

$\mathbf {A}$ = magnetic vector potential
$\phi$ = scalar electric potential
c = speed of light
v = magnitude of velocity
$\mathbf {n}$ = unit normal in the direction of motion
γ = Lorentz factor

ESR

Equivalent series resistance or ESR is the value of a resister posited in series with an ideal reactance used to account for the power losses of an actual reactance. The value is defined for a capacitor by $R_{esr}=({\text{Disapation factor}})\times {|X_{c}|}={\frac {\text{Disapation factor}}{\omega C}}=({\text{Disapation factor}})\times \omega L$ where $C$ is the capacitance and $\omega$ is the frequency. ^[4]

Wave Vector Notation

${\tilde {\nu }}$

${\tilde {\boldsymbol {\nu }}}$

$\mathbf {F} =q\mathbf {E} +q\mathbf {v} \times \mathbf {B}$

${\overrightarrow {F}}=q{\overrightarrow {E}}+q{\overrightarrow {v}}\times {\overrightarrow {B}}$

${\overrightarrow {\mathbf {F} }}=q{\overrightarrow {\mathbf {E} }}+q{\overrightarrow {\mathbf {v} }}\times {\overrightarrow {\mathbf {B} }}$

$\mathbf {F} =q\mathbf {E} +q\mathbf {v} \times \mathbf {B}$

Aharonov–Bohm effect apparatus

Real field

Every time I read the intro of electric field, I cringe. It is pretty much 19th century physics. Now, 19th century physics is useful and widely taught, especially to non-physicists. It is great for engineering. Yet it contains misconceptions that hinder and cause frustration for folks trying to “get it” on a deeper level. Let me quote Feynman, which you can read here The Vector Potential. In section 15-4, Feynman says.

"A real field is a mathematical function we use for avoiding the idea of action at a distance."

And

"A “real” field is then a set of numbers we specify in such a way that what happens at a point depends only on the numbers at that point."

The chapter is about the vector potential, but Feynman isn't limiting himself to only the vector potential. He is addressing all real fields, including the electric field.

So, this is my elaboration. The field is made of nothing but numbers. The numbers are not unique. Your numbers may be different from my numbers. Thankfully we have the theory of relativity that allows us to understand each other's numbers. The numbers at each point in the field are useful for computing the forces on particles at that point. The field exists only because we humans find it useful. The field is not physical. It is not fundamental. It doesn't move. It doesn't do anything. It is not attached to charge particles. There is only one field of a given type, therefore the proper article is "the", as in "the electric field." A charged particle does not have "an electric field." Electric fields do not interact because there is only one electric field. Electric fields do not propagate. However, we do say, write, and repeat those things. We can find plenty of examples in reliable sources. It is not wrong; it is a type of jargon. It allows to say things using fewer words. If we were writing carefully what we would say is that a charged particle influences the value of the field in its vicinity. The values of the field change dynamically over space and time in accordance with a wave equation. The electric field is such a useful and reliable artifice for computing outcomes, that we sometimes tend to think of it as a physical thing. It is not. It is nothing but imagination. The electric force is real. It does things. The electric field is a purely human construct. Once you embrace that, you can stop wasting time by asking unanswerable questions.

I am not proposing to rewrite the entire article, but only the first few sentences.

Power factor

The general expression for power factor is given by

{\mbox{power factor}}=P/P_{a}

P_{a}=I_{rms}V_{rms}

where $P$ is the real power measured by an ideal wattmeter, $I_{rms}$ is the rms current measured by an ideal ammeter, and $V_{rms}$ is the rms voltage measured by an ideal voltmeter. Apparent power, $P_{a}$ , is the product of the rms current and the rms voltage.

Periodic waveforms

If the waveforms are periodic with a period that is much shorter than the averaging time of the physical meters, then the power factor can be computed by the following

{\mbox{power factor}}=P/P_{a}

P_{a}=I_{rms}V_{rms}

P={\frac {1}{T}}\int _{t'}^{t'+T}i(t)v(t)dt

I_{rms}^{2}={\frac {1}{T}}\int _{t'}^{t'+T}{i(t)}^{2}dt

V_{rms}^{2}={\frac {1}{T}}\int _{t'}^{t'+T}{v(t)}^{2}dt

where $i(t)$ is the instantaneous current, $v(t)$ is the instantaneous voltage, $t'$ is an arbitrary starting time, and $T$ is the period of the waveforms.

Nonperiodic waveforms

If the waveforms are not periodic and the physical meters have the same averaging time, then the equations for the periodic case can be used with the exception that $T$ is the averaging time of the meters instead of the waveform period.

References for reflection coefficient

The reflection coefficient (RC) is a widely used concept that appears in many reliable sources across many subject areas that involve waves. It applies to circuit quantities (voltage and current), electromagnetic quantities (E and B), sonic quantities (velocity and displacement). Typically, the symbol Γ is used for reflection coefficient, although ρ and r also appear in reliable sources.

In circuits, the voltage RC has the opposite sign to the current RC. $\Gamma _{current}=-\Gamma _{voltage}$ When not specified, voltage RC is usually assumed.

In electromagnetics, the E-field RC has the opposite sign to the B-field RC. $\Gamma _{B}=-\Gamma _{E}$ When not specified, E-field RC is usually assumed.

Here are some reliable sources that can be accessed from the internet.

Harrington, Time-Harmonic Electromagnetic Fields, p. 55, eq. 2-45 $\Gamma ={\frac {\eta _{2}-\eta _{1}}{\eta _{2}+\eta _{1}}}$ for a wave propagating from media 1 to media 2. η is the wave impedance of the media.

Hayt, Engineering Electromagnetics, 8th ed, p. 321, eq. 73 $\Gamma ={\frac {Z_{L}-Z_{0}}{Z_{L}+Z_{0}}}$

Wadell, Transmission Line Design Handbook, p. 501, eq C.2 $\Gamma =\rho ={\frac {Z_{L}-Z_{0}}{Z_{L}+Z_{0}}}$

Steer, Microwave And Rf Design: Transmission Lines, p. 68, eq. 2.59 $\Gamma _{L}={\frac {Z_{L}-Z_{0}}{Z_{L}+Z_{0}}}$

Rosenstark, Transmission Lines in Computer Engineering, p. 23, eq. 2.6 $\rho _{L}={\frac {R_{L}-Z_{0}}{R_{L}+Z_{0}}}$

Harrington is a widely cited graduate level text book used in the study of wave guiding structures.

Hayt is a widely cited under-graduate level text book used in electrical engineer schools.

First, I agree that you got the correct result. However, there are three problems with your derivation.

Wadell doesn’t give a derivation. He gives the result of the derivation. It is a reliable source for the result, but it is not a reliable source for the correctness of your derivation.

You define Γ as the relative difference between the actual current and the optimum current. That definition does not appear in any reliable source. It may be coincidentally correct, but it is an observation and not a definition. Γ is defined as the ratio of the reflected signal to the incident signal. Its purpose is to let you calculate the amplitude and phase of the reflection. Its purpose is not for calculating the relative difference between the actual current and the optimum current. No one cares about that. They care about the reflected signal.

You start with the knowledge of the optimum current, which is the current that you get when there is no reflection. But you don’t know that until you have derived it. You cannot start with that.

The usual approach is to assume that there are 2 coefficients, Γ and T, such that the reflected and transmitted signals are given by $v_{r}=\Gamma v_{i}$ and $v_{t}=Tv_{i}$ then you apply the continuity requirements which are:

$v_{r}+v_{i}=v_{t}$ This is the voltage continuity requirement. In words: incident voltage + reflected voltage = transmitted voltage.
${\frac {v_{i}}{Z_{0}}}-{\frac {v_{r}}{Z_{0}}}={\frac {v_{t}}{Z_{L}}}$ This is the current continuity requirement. In words: incident current - reflected current = transmitted current.

From that you derive $1+\Gamma =T$ and

${\frac {1}{Z_{0}}}-{\frac {\Gamma }{Z_{0}}}={\frac {1+\Gamma }{Z_{L}}}$

which reduces to $(1-\Gamma )Z_{L}=(1+\Gamma ){Z_{0}}$ which can be manipulated into $\Gamma ={\frac {Z_{L}-Z_{0}}{Z_{L}+Z_{0}}}$

The flow of results goes like this:

Fundament physical requirements → reflection coef → optimum load → optimum current.

What you have done is this:

optimum load + optimum current + convenient definition of Gamma → Gamma = reflection coef

All that you have shown is that the algebra at the tail end of derivation is reversable.

The only thing important and notable is the fundament physical requirements that start the entire deductive chain.

Caeseum beam resonator

Loaded cable - Heaviside condition

Example of an external link

Sign convention in Fourier transform

I use the same convention that you are advocating. I have no problem with using that convention. If you want to state a reason for that convention in the article you need a reliable secondary source that states that reason. No amount of WP:OR will change that. However, I do not mind dabbling in OR here on the talk page.

Let

{\begin{aligned}&{\hat {f}}_{-}(\omega )=\int _{-\infty }^{\infty }f(t)e^{-i\omega x}\,dt\end{aligned}}

This is the conventional forward transform.

{\begin{aligned}&{\hat {f}}_{+}(\omega )=\int _{-\infty }^{\infty }f(t)e^{+i\omega x}\,dt\end{aligned}}

This is the other convention. It is mathematically equal to the conventional reverse transform.

I hope it is obvious that

{\begin{aligned}&{\hat {f}}_{-}={\hat {f}}_{+}^{*}\end{aligned}}

Thus the results of these two conventions are simply conjugates of each other.

This has no physical effect because physical effects are caused by energy or power. The power of a Fourier transform is computed by multiplying the transform by its conjugate.

Again, I hope it is obvious that

{\begin{aligned}&{\hat {f}}_{-}{\hat {f}}_{-}^{*}=&{\hat {f}}_{+}{\hat {f}}_{+}^{*}\end{aligned}}

So, lets look at a couple of examples. I will suppress multiplicative constants that clutter up the results.

First, consider the Fourier transform of $cos(\omega t)$ .

The Fourier transform under the usual convention is

\delta (\omega -a)+\delta (\omega +a)

. It has Fourier components at both

+a

and

-a

.

The Fourier transform under the other convention is

\delta (\omega +a)+\delta (\omega -a)

. The result is exactly the same result.

Next, consider the Fourier transform of $sin(\omega t)$ .

The Fourier transform under the usual convention is

-i\delta (\omega -a)+i\delta (\omega +a)

. It has Fourier components at both

+a

and

-a

.

The Fourier transform under the other convention is

-i\delta (\omega +a)+i\delta (\omega -a)

. It has Fourier components at both

+a

and

-a

. The result is the conjugate of the result using the usual convention.

Now let me go way off into OR la-la land to speculate why engineers prefer the usual convention. Consider the Fourier transform of cos(ωt) + sin(ωt). It is $\delta (\omega -a)+\delta (\omega +a)-i\delta (\omega -a)+i\delta (\omega +a)$ . The component at the positive frequency of $+a$ is $\delta (\omega -a)-i\delta (\omega -a)$ . Notice in particular that the sign of the imaginary part is negative. Engineers prefer this because $cos(\omega t)+sin(\omega t)$ lags $cos(\omega t)$ by 45°. When an engineer plots this in Cartesian space, it is [1,-1]. The principal argument is negative. Engineers prefer that because the phase of cos(ωt) + sin(ωt) relative to cos(ωt) is negative. Mathematicians consider cos(ωt) and sin(ωt) as basis vectors and they plot cos(ωt) + sin(ωt) as [1,1]. That is all there is to it. Engineers prefer that the Fourier component of sin(ωt) should be negative at positive frequency.

Wolfram Mathworld

The following quoted material comes from Wolfram Mathworld at . It is copyrighted material, but the terms of use permit this use.

" (1) ${\begin{aligned}f(x)=\int _{-\infty }^{\infty }F(k)e^{+i2\pi kx}\,dk\end{aligned}}$
(2) ${\begin{aligned}F(k)=\int _{-\infty }^{\infty }f(x)e^{-i2\pi kx}\,dx\end{aligned}}$ "

Here x and k are continuous, unitless domains with no particular meaning. k comes from the idea of harmonic number with the spacing between harmonics reduced to zero.

and

" In general, the Fourier transform pair may be defined using two arbitrary constants, a and b as

(15) ${\begin{aligned}F(\omega )={\sqrt {\frac {|b|}{(2\pi )^{1-a}}}}\int _{-\infty }^{\infty }f(t)e^{ib\omega t}\,dt\end{aligned}}$

(16) ${\begin{aligned}f(t)={\sqrt {\frac {|b|}{(2\pi )^{1+a}}}}\int _{-\infty }^{\infty }F(\omega )e^{-ib\omega t}\,d\omega \end{aligned}}$

… By default, the Wolfram Language takes FourierParameters as (0,1). Unfortunately, a number of other conventions are in widespread use. For example, (0,1) is used in modern physics, (1,-1) is used in pure mathematics and systems engineering, (1,1) is used in probability theory for the computation of the characteristic function, (-1,1) is used in classical physics, and (0,-2pi) is used in signal processing. "

Unwinding that yeilds

modern physics and Wolfram default (0,1)

{\begin{aligned}F(\omega )={\sqrt {\frac {1}{2\pi }}}\int _{-\infty }^{\infty }f(t)e^{+i\omega t}\,dt\qquad f(t)={\sqrt {\frac {1}{2\pi }}}\int _{-\infty }^{\infty }F(\omega )e^{-i\omega t}\,d\omega \end{aligned}}

classical physics (-1,1)

{\begin{aligned}F(\omega )={\frac {1}{2\pi }}\int _{-\infty }^{\infty }f(t)e^{+i\omega t}\,dt\qquad f(t)=\int _{-\infty }^{\infty }F(\omega )e^{-i\omega t}\,d\omega \end{aligned}}

probability (1,1)

{\begin{aligned}F(\omega )=\int _{-\infty }^{\infty }f(t)e^{+i\omega t}\,dt\qquad f(t)={\frac {1}{2\pi }}\int _{-\infty }^{\infty }F(\omega )e^{-i\omega t}\,d\omega \end{aligned}}

pure mathematics and systems engineering (1,-1)

{\begin{aligned}F(\omega )=\int _{-\infty }^{\infty }f(t)e^{-i\omega t}\,dt\qquad f(t)={\frac {1}{2\pi }}\int _{-\infty }^{\infty }F(\omega )e^{+i\omega t}\,d\omega \end{aligned}}

signal processing (0,-2π)

F(\omega )={\sqrt {\frac {1}{2\pi }}}\int _{-\infty }^{\infty }f(t)e^{-i2\pi \omega t}\,dt\qquad f(t)={\sqrt {\frac {1}{2\pi }}}\int _{-\infty }^{\infty }F(\omega )e^{+i2\pi \omega t}\,d\omega

Doodles

Q=\epsilon _{0}\iint \limits _{S}\mathbf {E} \cdot d\mathbf {A}

5

where $H$ and $E$ are the magnitudes of $H$ and $E$ . Multiplying the last two equations gives

n=c\,{\sqrt {\mu \epsilon }}\,.

4

Dividing (or cross-multiplying) the same two equations gives $H = YE ‍,$ ‍ where

Y={\sqrt {\epsilon /\mu }}\,

.

5

\mu '\mu ^{'}\mu ^{'}

From (4) we obtain

From (5

Mutual inductance

Don't do this: $4.35*10^{-17}$

Instead, do this: $4.35\times 10^{-17}$

Or better still, within text, do this: 4.35×10⁻¹⁷

Propagating Plane Wave

Plane waves in linear media

The propagation factor of a sinusoidal plane wave propagating in the x direction in a linear material is given by

P=e^{-jkx}

^[1]^: 51

where

k=k'-jk''={\sqrt {-j\omega \mu (\sigma +j\omega \epsilon )}}={\sqrt {-(\omega \mu ''+j\omega \mu ')(\sigma +\omega \epsilon ''+j\omega \epsilon ')}}\;

= wavenumber

k'=

phase constant in the units of radians/meter

k''=

attenuation constant in the units of nepers/meter

\omega =

frequency in the units of radians/meter

x=

distance traveled in the x direction

\sigma =

conductivity in S/meter

\epsilon =\epsilon '-j\epsilon ''=

complex permitivity

\mu =\mu '-j\mu ''=

complex permeability

j={\sqrt {-1}}

The sign convention is chosen for consistency with propagation in lossy media. If the attenuation constant is positive, then the wave amplitude decreases as the wave propagates in the x direction.

manipulation

By a straightforward, if lengthy, algebraic calculation

k=k'-jk''={\sqrt {-j\omega \mu ^{'}(1-j\chi )j\omega \epsilon ^{'}(1-j(\kappa +\xi ))}}=\omega {\sqrt {\mu ^{'}\epsilon ^{'}}}{\sqrt {(1-j\chi )(1-j(\kappa +\xi )}}=\omega {\sqrt {\mu ^{'}\epsilon ^{'}}}{\sqrt {a-jb}}

\xi ={\frac {\epsilon ^{''}}{\epsilon ^{'}}}

dielectric loss tangent

\chi ={\frac {\mu ^{''}}{\mu ^{'}}}

magnetic loss tangent

\kappa ={\frac {\sigma }{\omega \epsilon ^{'}}}={\frac {1}{\omega \tau }}

\tau ={\frac {\epsilon ^{'}}{\sigma }}=\rho \epsilon ^{'}

dielectric relaxation time

Typical values for

\tau

: copper: 150 X 10^-9 ps ( 10⁶ THz), silicon: 244 ns (650 kHz), polyethylene: 150,000 (1 μHz), soil: 300 ps (500 MHz) to 20 ns (8 MHz)

a=(1-\kappa \chi -\xi \chi )

b=(\kappa +\chi +\xi )

Using the formula for the square root of a complex number a + jb

k^{'}=\omega {\sqrt {\frac {\mu ^{'}\epsilon ^{'}}{2}}}{\sqrt {{\sqrt {a^{2}+b^{2}}}+a}}

k^{''}=\omega {\sqrt {\frac {\mu ^{'}\epsilon ^{'}}{2}}}{\frac {b}{\sqrt {{\sqrt {a^{2}+b^{2}}}+a}}}

If b=0 and a<0 then the previous equation has a divide by zero problem. However, it is mathematically impossible for both b=0 and a<0 at the same time. The second order terms $\xi ^{2},\chi ^{2},\xi \chi ,\xi ^{2}\chi ^{2}$ are shown shown in red in the two following equations.

k^{'}=\omega {\sqrt {\frac {\mu ^{'}\epsilon ^{'}}{2}}}{\sqrt {{\sqrt {1\color {Red}+\xi ^{2}+\chi ^{2}+\xi ^{2}\chi ^{2}\color {Black}+(1\color {Red}+\chi ^{2}\color {Black})\kappa ^{2}+2\xi (1\color {Red}+\chi ^{2}\color {Black})\kappa }}+(1-\kappa \chi \color {Red}-\xi \chi \color {Black})}}

k^{''}=\omega {\sqrt {\frac {\mu ^{'}\epsilon ^{'}}{2}}}{\frac {(\kappa +\chi +\xi )}{\sqrt {{\sqrt {1\color {Red}+\xi ^{2}+\chi ^{2}+\xi ^{2}\chi ^{2}\color {Black}+(1\color {Red}+\chi ^{2}\color {Black})\kappa ^{2}+2\xi (1\color {Red}+\chi ^{2}\color {Black})\kappa }}+(1-\kappa \chi \color {Red}-\xi \chi \color {Black})}}}

low magnetic and dielectric loss

When magnetic and dielectric losses are low, then the second order terms $\xi ^{2},\chi ^{2},\;\xi \chi ,\;\xi ^{2}\chi ^{2}$ may be neglected.

k^{'}=\omega {\sqrt {\frac {\mu ^{'}\epsilon ^{'}}{2}}}{\sqrt {{\sqrt {1+\kappa ^{2}+2\xi \kappa }}+(1-\kappa \chi )}}

k^{''}=\omega {\sqrt {\frac {\mu ^{'}\epsilon ^{'}}{2}}}{\frac {(\kappa +\chi +\xi )}{\sqrt {{\sqrt {1+\kappa ^{2}+2\xi \kappa }}+(1-\kappa \chi )}}}

no magnetic loss

k^{'}=\omega {\sqrt {\frac {\mu ^{'}\epsilon ^{'}}{2}}}{\sqrt {{\sqrt {1+\kappa ^{2}+2\xi \kappa }}+1}}

k^{''}=\omega {\sqrt {\frac {\mu ^{'}\epsilon ^{'}}{2}}}{\frac {(\kappa +\xi )}{\sqrt {{\sqrt {1+\kappa ^{2}+2\xi \kappa }}+1}}}

Substituting $\kappa =1/(\rho \omega \epsilon ^{'})$

k^{'}=\omega {\sqrt {\frac {\mu ^{'}\epsilon ^{'}}{2}}}{\frac {{\sqrt {\rho \omega \epsilon ^{'}}}{\sqrt {{\sqrt {1+\kappa ^{2}+2\xi \kappa }}+1}}}{\sqrt {\rho \omega \epsilon ^{'}}}}

k^{''}=\omega {\sqrt {\frac {\mu ^{'}\epsilon ^{'}}{2}}}{\frac {(\kappa +\xi ){\sqrt {\rho \omega \epsilon ^{'}}}}{{\sqrt {\rho \omega \epsilon ^{'}}}{\sqrt {{\sqrt {1+\kappa ^{2}+2\xi \kappa }}+1}}}}

k^{'}={\sqrt {\frac {\omega \mu ^{'}}{2\rho }}}{\sqrt {{\sqrt {1+2\xi \rho \omega \epsilon ^{'}+{(\rho \omega \epsilon ^{'})}^{2}}}+\rho \omega \epsilon ^{'}}}

k^{''}={\sqrt {\frac {\omega \mu ^{'}}{2\rho }}}{\frac {1+\xi \rho \omega \epsilon ^{'}}{\sqrt {{\sqrt {1+2\xi \rho \omega \epsilon ^{'}+{(\rho \omega \epsilon ^{'})}^{2}}}+\rho \omega \epsilon ^{'}}}}

\delta ={\sqrt {\frac {2\rho }{\omega \mu ^{'}}}}{\frac {\sqrt {{\sqrt {1+2\xi \rho \omega \epsilon ^{'}+{(\rho \omega \epsilon ^{'})}^{2}}}+\rho \omega \epsilon ^{'}}}{1+\xi \rho \omega \epsilon ^{'}}}

\delta ={\sqrt {\frac {2\rho }{\omega \mu ^{'}}}}{\frac {\sqrt {{\sqrt {1+2\rho \omega \epsilon ^{'}\tan \delta _{e}+{(\rho \omega \epsilon ^{'})}^{2}}}+\rho \omega \epsilon ^{'}}}{1+\rho \omega \epsilon ^{'}\tan \delta _{e}}}

At high frequency

\delta ={\frac {2\rho }{1+\rho \omega \epsilon ^{'}\tan \delta _{e}}}{\sqrt {\frac {\epsilon ^{'}}{\mu ^{'}}}}

Limiting skin depth in silicon

There is a statement about the skin depth in silicon not being less than about 11m. This statement does not have a reliable source and I believe that it is incorrect, so I will remove it. There does not need to be any justification other than there is no reliable source cited, but I will give a more elaborate justification. The formula, as given, is correct if you can neglect dielectric loss. Any type of loss will decrease skin depth. Using these equations from wavenumber:

\delta ={\frac {1}{k''}}

k=k'-jk''={\sqrt {-j\omega \mu (\sigma +j\omega \epsilon )}}={\sqrt {-(\omega \mu ''+j\omega \mu ')(\sigma +\omega \epsilon ''+j\omega \epsilon ')}}\;

,

In the equation for wavenumber, you see $\sigma +\omega \epsilon ''$ . The dielectric loss term ands to the conductivity term. This effectively reduces resistivity as frequency increases. It is straight forward, but tedious, to carry out all the multiplications, gather terms, and apply the formula for the square root of a complex numbers. The result shows that if there is any dielectric (or magnetic) loss, then there is no non-zero lower bound. If someone would like to check the math, I would be grateful.

I plugged numbers in, assuming a dielectric loss tangent of 0.3% at 10GHz I got 0.9m at 10GHz. The conclusion that skin depth in silicon is deep enough to ignore is still correct. Of course, I could have made a mistake. If anyone would like to run the numbers, I used ${\epsilon ''}/{\epsilon '}=0.003,\epsilon _{r}=12,\sigma =435\mu S/m$

[1]
Harrington, Roger F. (1961), Time-Harmonic Electromagnetic Fields (1st ed.), McGraw-Hill, ISBN 0-07-026745-6 {{citation}}: ISBN / Date incompatibility (help)

Feynman Lectures

dBm

P_{\mathrm {dBm} }=10\ \log _{10}(k_{\text{B}}T\Delta f/1\,{\textrm {mW}})\ {\textrm {dBm}}=10\times \log _{10}(1.38\times 10^{-23}\times 300\times 1/10^{-3})=10\times \log _{10}(4.14\times 10^{-18})=-173.8

Googled "thermal noise power formula"

I found the sengpielaudio calculator. They calculate dBu and dBv, neither of which is dBm.

I also found

Thermal noise in a 50 Ω system at room temperature is -174 dBm / Hz.

calculator produces -173.8277942 dBm

Eq. 2.121 verifies formula for power P = kTB

Verifies Thermal noise power equation

-173.9 dBm⁄Hz

and some printed sources

"Noise power at 300 K = -173.83 dBm/Hz"^[1]^: 260

"Noise power = kTB" ^[2]^: 621

"P = kTB" ^[3]^: 203

[1]
Vizmuller, Peter (1995), RF Design Guide, Artech House, ISBN 0-89006-754-6
[2]
Ludwig, Reinhold; Bretchko, Pavel (2000), RF Circuit Design, Prentice-Hall, ISBN 0-13-095323-7
[3]
Hayward, Wes (1994), Introduction to Radio Frequency Design, ARRL, ISBN 0-87259-492-0

Telegrapher's equations

$\gamma ={\sqrt {(G+j\omega C)(R+j\omega L)}}$

Since $\omega C\gg G\;$ and $\omega L\ll R\;$

$\gamma \approx {\sqrt {j\omega CR}}=(1+j){\sqrt {\frac {RC\omega }{2}}}$

$v={\frac {\omega }{Im(\gamma )}}={\frac {\omega }{\sqrt {\frac {RC\omega }{2}}}}={\sqrt {\frac {2\omega }{RC}}}$

Dynamical variables

These references appear in the Gyrator–capacitor model article:

Hamill^[1],

Mohammad^[2],

González^[3],

Lambert^[4]

Magnetic voltage

Magnetic voltage, $v_{m}$ , is an alternate name for magnetomotive force (mmf), ${\mathcal {F}}$ , which is analogous to electrical voltage in an electric circuit.^[2]^: 42^[3]^: 5 The SI unit of mmf is the ampre or amp-turn. Not all authors use the term magnetic voltage. The magnetomotive force applied to an element between point A and point B is equal to the line integral through the component of the magnetic field strength, $\mathbf {H}$ .

v_{m}={\mathcal {F}}=-\int _{A}^{B}\mathbf {H} \cdot \operatorname {d} \mathbf {l}

The resistance–reluctance model uses the same equivalence between magnetic voltage and magnetomotive force.

Magnetic current

Magnetic current, $i_{m}$ , is an alternate name for the time rate of change of flux, ${\dot {\Phi }}$ , which is analogous to electrical current in an electric circuit.^[4]^: 2429^[2]^: 37 In the physical circuit, ${\dot {\Phi }}$ , is magnetic displacement current.^[2]^: 37 The SI unit of ${\dot {\Phi }}$ is webers/second or volts. The magnetic current flowing through an element of cross section, $S$ , is the area integral of the magnetic flux density $\mathbf {B}$ .

i_{m}={\dot {\Phi }}={\frac {d}{dt}}\int _{S}\mathbf {B} \cdot \operatorname {d} \mathbf {S}

The resistance–reluctance model uses a different equivalence, taking magnetic current to be an alternate name for flux, $\Phi$ . This difference in the definition of magnetic current is the fundamental difference between the gyrator-capacitor model and the resistance–reluctance model. The definition of magnetic current and magnetic voltage imply the definitions of the other magnetic elements.^[2]^: 35

Magnetic charge

Magnetic flux, $\Phi$ , in the physical circuit is the analog of charge, $Q$ , in the model circuit.^[1]^: 99

Magnetic flux

Charge, $Q$ , in the physical circuit is the analog of flux in the model circuit. The units of charge (coulombs or ampere x seconds) and flux (weber or volts x seconds) are the duals of each other.

Summary of analogy between magnetic circuits and electrical circuits

The following table summarizes the mathematical analogy between electrical circuit theory and magnetic circuit theory.

More information

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...

Analogy between magnetic circuits and electrical circuits used in the gyrator-capacitor approach
Physical magnetic circuit			Model magnetic circuit
Name	Symbol	Units	Name	Symbol	Units
Magnetomotive force (MMF)	${\mathcal {F}}=\int \mathbf {H} \cdot \operatorname {d} \mathbf {l}$	ampere-turn	Electromotive force (EMF)	${\mathcal {E}}=\int \mathbf {E} \cdot \operatorname {d} \mathbf {l}$	volt
Magnetic field	H	ampere/meter = newton/weber	Electric field	E	volt/meter = newton/coulomb
Flux	$\Phi$	weber^[b]	Electric charge	Q	Coulomb
Flux rate of change	${\dot {\Phi }}$	weber/second = volt	Electric current	${\mathcal {}}I$	coulomb/second =ampere
Magnetic admittance	${\mathcal {}}Y(\omega )={\frac {{\dot {\Phi }}(\omega )}{{\mathcal {F}}(\omega )}}$	ohm	Admittance	${\mathcal {}}Y_{E}(\omega )={\frac {I(\omega )}{V(\omega )}}$	1/ohm = mho = siemens
Magnetic conductance	$G_{M}=\operatorname {Re} (Y(\omega ))$	ohm	Electric conductance	$G_{E}=\operatorname {Re} (Y_{E}(\omega ))$	1/ohm = mho = siemens
Permeance	${\mathcal {P}}={\frac {\operatorname {Im} (Y(\omega ))}{\omega }}$	Henry	Capacitance	$C={\frac {\operatorname {Im} (Y_{E}(\omega ))}{\omega }}$	Farad

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More information

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...

Equivalence between physical magnetic circuits and their electrical models in the gyrator–capacitor model
Physical				Model
Name	Symbol	Formula	Units	Name^[c]	Symbol	Formula	Units
Magnetomotive force (MMF)	${\mathcal {F}}$	$\oint \mathbf {\vec {H}} \cdot \operatorname {d\ell }$	A⋅t	Electromotive force^[2]^: p38	$v_{m}$	$\oint \mathbf {\vec {E}} \cdot \operatorname {d\ell }$	V
Flux	$\Phi$	$\iint \limits _{S}\mathbf {\vec {B}} \cdot \operatorname {d\mathbf {\vec {S}} }$ ^[2]^: p41	Wb^[d]	Charge	$q_{m}$	$\iint \limits _{S}\mathbf {\vec {D}} \cdot \operatorname {d\mathbf {\vec {S}} }$	C
Flux rate of change	${\dot {\Phi }}$	${\frac {d}{dt}}\Phi$ ^[2]^: p41	Wb/s = V	Current^[3]^: p5	$i_{m}$	${\frac {d}{dt}}q_{m}$	A
Charge	$q$	$\iint \limits _{S}\mathbf {\vec {D}} \cdot \operatorname {d\mathbf {\vec {S}} }$	C	Flux	$\Phi _{m}$	$\iint \limits _{S}\mathbf {\vec {B}} \cdot \operatorname {d\mathbf {\vec {S}} }$	Wb
Permeance	$P$	${\frac {\Phi }{\mathcal {F}}}$	H	Capacitance	$C_{m}$	${\frac {q_{m}}{v_{m}}}$	F
Capacitance	$C$	${\frac {q}{v}}$	F	Inductance	$L_{m}$	${\frac {\Phi _{m}}{i_{m}}}$	H
Conductance	$G$	${\frac {i}{v}}$	S	Resistance	$R_{m}$	${\frac {v_{m}}{i_{m}}}$	Ω
Admittance	$Y$	$G+j\omega C-{\frac {j}{\omega L}}$	S	Impedance	$Z_{m}$	$R_{m}+j\omega L_{m}-{\frac {j}{\omega C_{m}}}$	Ω
Impedance	$Z$	$R+j\omega L-{\frac {j}{\omega C}}$	Ω	Admittance	$Y_{m}$	$G_{m}+j\omega C_{m}-{\frac {j}{\omega L_{m}}}$	S

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notes

[a]
In a high-pressure system it is the difficulty of devising an insulation that will not break down which practically limits the voltage. Therefore, in comparing polyphase and single-phase we must take cases that are on a par from this point of view of insulation. It has been common in the case of single-phase (and also in that of continuous currents) to keep one line at the potential of the earth, and to insulate the other sufficiently, having regard to the pressure between the two lines. In this case it is clearly the maximum pressure between the lines that forms what we here are calling the voltage of the system. If, however, both the lines are independently insulated from earth so as to withstand safely the maximum pressure occurring between line and earth, then the lines may have between themselves a voltage equal to double that maximum pressure without fear of a breakdown, provided always the lines, and the respective circuits into which they lead, are so well insulated from one another as to obviate risks in this respect. The question then arises whether, in comparing the advantages of various systems, we shall take as the basis of comparison the pressure between any two lines or the pressure between line and earth. If we take the maximum pressure between any point of the line and earth as the basis of comparison, then there is no saving in copper by the employment of polyphase currents; for each line of any system, carrying a certain current at the maximum allowable pressure above the earth may be considered as dependently transmitting a certain amount of power; and therefore the total power is simply proportional to the number of line-wires, to which the total weight of copper is also proportional. For instance, a 3-phase system joined up in star fashion with the common junction to earth, and having a pressure of 1000 volts between each line and earth (and therefore 1732 volts between line and line), has no advantage so far as the insulation of the line is concerned, over a single-phase system having a pressure of 1000 volts between each line and earth (and therefore 2000 volts between line and line). To transmit equal power, with equal loss in the lines, each of the two wires of the single-phase system must be times as heavy as each of the three wires of the 3-phase system. The two systems will require equal total weights of copper. If, on the other hand, we take the maximum pressure between any two lines as the basis of comparison, we are now equating not the risks of breakdown of the lines, but the risks of breakdown of the apparatus, machines, transformers, &c., in which the goodness of the insulation must be considered equal. And on this basis of comparison there is a decided economy of copper by the employment of 3-phase currents, as can be seen by the following considerations. ... Or, to put it in another way: the single-phase system requires two wires, each of double cross-section, as against the three wires of the 3-phase system. The weight of copper required on the 3-phase system is only three-fourths of that required by any single-phase system.
[b]
Hamill parenthetically includes "(per turn)" on page 97. ^[1]
[c]
Often preceded by the term "magnetic"
[d]
Hamill parenthetically includes "(per turn)" on page 97. ^[1]

Frequency domain

Frequency domain notes

Periodic waveforms

Nonperiodic waveforms

Wolfram Mathworld

manipulation

low magnetic and dielectric loss

no magnetic loss

Limiting skin depth in silicon

Magnetic voltage

Magnetic current

Magnetic charge

Magnetic flux

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