Average power for circuit elements

February 9, 2016 ece1236 , , , , , , ,

In [2] section 2.2 is a comparison of field energy expressions with their circuit equivalents. It’s clearly been too long since I’ve worked with circuits, because I’d forgotten all the circuit energy expressions:

\label{eqn:averagePowerCircuitElements:20}
\begin{aligned}
W_{\textrm{R}} &= \frac{R}{2} \Abs{I}^2 \\
W_{\textrm{C}} &= \frac{C}{4} \Abs{V}^2 \\
W_{\textrm{L}} &= \frac{L}{4} \Abs{I}^2 \\
W_{\textrm{G}} &= \frac{G}{2} \Abs{V}^2 \\
\end{aligned}

Here’s a recap of where these come from

Energy lost to resistance

Given
\label{eqn:averagePowerCircuitElements:40}
v(t) = R i(t)

the average power lost to a resistor is

\label{eqn:averagePowerCircuitElements:60}
\begin{aligned}
p_{\textrm{R}}
&= \inv{T} \int_0^T v(t) i(t) dt \\
&= \inv{T} \int_0^T \textrm{Re}( V e^{j \omega t} ) \Real( I e^{j \omega t} ) dt \\
&= \inv{4 T} \int_0^T
\lr{V e^{j \omega t} + V^\conj e^{-j \omega t} }
\lr{I e^{j \omega t} + I^\conj e^{-j \omega t} }
dt \\
&= \inv{4 T} \int_0^T
\lr{
V I e^{2 j \omega t} + V^\conj I^\conj e^{-2 j \omega t}
+ V I^\conj + V^\conj I
}
dt \\
&= \inv{2} \textrm{Re}( V I^\conj ) \\
&= \inv{2} \textrm{Re}( I R I^\conj ) \\
&= \frac{R}{2} \Abs{I}^2.
\end{aligned}

Here it is assumed that the averaging is done over some integer multiple of the period, which kills off all the exponentials.

Energy stored in a capacitor

I tried the same sort of analysis for a capacitor in phasor form, but everything cancelled out. Referring to [1], the approach used to figure this out is to operate first strictly in the time domain. Specifically, for the capacitor where $$i = C dv/dt$$ the power supplied up to a time $$t$$ is

\label{eqn:averagePowerCircuitElements:80}
\begin{aligned}
p_{\textrm{C}}(t)
&= \int_{-\infty}^t C \frac{dv}{dt} v(t) dt \\
&= \inv{2} C v^2(t).
\end{aligned}

The $$v^2(t)$$ term can now be expanded in terms of phasors and averaged for

\label{eqn:averagePowerCircuitElements:100}
\begin{aligned}
\overline{{p}}_{\textrm{C}}
&= \frac{C}{2T} \int_0^T \inv{4}
\lr{ V e^{j \omega t} + V^\conj e^{-j \omega t} }
\lr{ V e^{j \omega t} + V^\conj e^{-j \omega t} } dt \\
&= \frac{C}{2T} \int_0^T \inv{4}
2 \Abs{V}^2 dt \\
&= \frac{C}{4} \Abs{V}^2.
\end{aligned}

Energy stored in an inductor

The inductor energy is found the same way, with

\label{eqn:averagePowerCircuitElements:120}
\begin{aligned}
p_{\textrm{L}}(t)
&= \int_{-\infty}^t L \frac{di}{dt} i(t) dt \\
&= \inv{2} L i^2(t),
\end{aligned}

\label{eqn:averagePowerCircuitElements:140}
\overline{{p}}_{\textrm{L}}
= \frac{L}{4} \Abs{I}^2.

Energy lost due to conductance

Finally, we have conductance. In phasor space that is defined by

\label{eqn:averagePowerCircuitElements:160}
G = \frac{I}{V} = \inv{R},

so power lost due to conductance follows from power lost due to resistance. In the average we have

\label{eqn:averagePowerCircuitElements:180}
\begin{aligned}
p_{\textrm{G}}
&= \inv{2 G} \Abs{I}^2 \\
&= \inv{2 G} \Abs{V G}^2 \\
&= \frac{G}{2} \Abs{V}^2
\end{aligned}

References

[1] J.D. Irwin. Basic Engineering Circuit Analysis. MacMillian, 1993.

[2] David M Pozar. Microwave engineering. John Wiley & Sons, 2009.

ECE1236H Microwave and Millimeter-Wave Techniques: Transmission lines. Taught by Prof. G.V. Eleftheriades

Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course ECE1236H, Microwave and Millimeter-Wave
Techniques, taught by Prof. G.V. Eleftheriades, covering [1] chap. 2 content.

Requirements

A transmission line requires two conductors as sketched in fig. 1, which shows a 2-wire line such a telephone line, a coaxial cable as found in cable TV distribution, and a microstrip line as found in cell phone RF interconnects.

../../figures/ece1236/deck4TxlineFig1: fig. 1. Transmission line examples.

A two-wire line becomes a transmission line when the wavelength of operation becomes comparable to the size of the line (or higher spectral component for pulses). In general a transmission line much support (TEM) transverse electromagnetic modes.

Time harmonic solutions on transmission lines

In fig. 2, an electronic representation of a transmission line circuit is sketched.

../../figures/ece1236/deck4TxlineFig2: fig. 2. Transmission line equivalent circuit.

In this circuit all the elements have per-unit length units. With $$I = C dV/dt \sim j \omega C V$$, $$v = I R$$, and $$V = L dI/dt \sim j \omega L I$$, the KVL equation is

\label{eqn:uwaves4TransmissionLines:20}
V(z) – V(z + \Delta z) = I(z) \Delta z \lr{ R + j \omega L },

or in the $$\Delta z \rightarrow 0$$ limit

\label{eqn:uwaves4TransmissionLines:40}
\PD{z}{V} = -I(z) \lr{ R + j \omega L }.

The KCL equation at the interior node is

\label{eqn:uwaves4TransmissionLines:60}
-I(z) + I(z + \Delta z) + \lr{ j \omega C + G} V(z + \Delta z) = 0,

or
\label{eqn:uwaves4TransmissionLines:80}
\PD{z}{I} = -V(z) \lr{ j \omega C + G}.

This pair of equations is known as the telegrapher’s equations

\label{eqn:uwaves4TransmissionLines:100}
\boxed{
\begin{aligned}
\PD{z}{V} &= -I(z) \lr{ R + j \omega L } \\
\PD{z}{I} &= -V(z) \lr{ j \omega C + G}.
\end{aligned}
}

The second derivatives are

\label{eqn:uwaves4TransmissionLines:120}
\begin{aligned}
\PDSq{z}{V} &= -\PD{z}{I} \lr{ R + j \omega L } \\
\PDSq{z}{I} &= -\PD{z}{V} \lr{ j \omega C + G},
\end{aligned}

which allow the $$V, I$$ to be decoupled
\label{eqn:uwaves4TransmissionLines:140}
\boxed{
\begin{aligned}
\PDSq{z}{V} &= V(z) \lr{ j \omega C + G} \lr{ R + j \omega L } \\
\PDSq{z}{I} &= I(z) \lr{ R + j \omega L } \lr{ j \omega C + G},
\end{aligned}
}

With a complex propagation constant

\label{eqn:uwaves4TransmissionLines:160}
\begin{aligned}
\gamma
&= \alpha + j \beta \\
&= \sqrt{ \lr{ j \omega C + G} \lr{ R + j \omega L } } \\
&=
\sqrt{ R G – \omega^2 L C + j \omega ( L G + R C ) },
\end{aligned}

the decouple equations have the structure of a wave equation for a lossy line in the frequency domain

\label{eqn:uwaves4TransmissionLines:180}
\boxed{
\begin{aligned}
\PDSq{z}{V} – \gamma^2 V &= 0 \\
\PDSq{z}{I} – \gamma^2 I &= 0.
\end{aligned}
}

We write the solutions to these equations as

\label{eqn:uwaves4TransmissionLines:200}
\begin{aligned}
V(z) &= V_0^{+} e^{-\gamma z} + V_0^{-} e^{+\gamma z} \\
I(z) &= I_0^{+} e^{-\gamma z} – I_0^{-} e^{+\gamma z} \\
\end{aligned}

Only one of $$V$$ or $$I$$ is required since they are dependent through \ref{eqn:uwaves4TransmissionLines:100}, as can be seen by taking derivatives

\label{eqn:uwaves4TransmissionLines:220}
\begin{aligned}
\PD{z}{V}
&= \gamma \lr{ -V_0^{+} e^{-\gamma z} + V_0^{-} e^{+\gamma z} } \\
&=
-I(z) \lr{ R + j \omega L },
\end{aligned}

so
\label{eqn:uwaves4TransmissionLines:240}
I(z)
=
\frac{\gamma}{ R + j \omega L } \lr{ V_0^{+} e^{-\gamma z} – V_0^{-} e^{+\gamma z} }.

Introducing the characteristic impedance $$Z_0$$ of the line

\label{eqn:uwaves4TransmissionLines:260}
\begin{aligned}
Z_0
&= \frac{R + j \omega L}{\gamma} \\
&= \sqrt{ \frac{R + j \omega L}{G + j \omega C} },
\end{aligned}

we have

\label{eqn:uwaves4TransmissionLines:280}
\begin{aligned}
I(z)
&=
\inv{Z_0} \lr{ V_0^{+} e^{-\gamma z} – V_0^{-} e^{+\gamma z} } \\
&=
I_0^{+} e^{-\gamma z} – I_0^{-} e^{+\gamma z},
\end{aligned}

where

\label{eqn:uwaves4TransmissionLines:300}
\begin{aligned}
I_0^{+} &= \frac{V_0^{+}}{Z_0} \\
I_0^{-} &= \frac{V_0^{-}}{Z_0}.
\end{aligned}

Mapping TL geometry to per unit length $$C$$ and $$L$$ elements

From electrostatics and magnetostatics the per unit length induction and capacitance constants for a co-axial cable can be calculated. For the cylindrical configuration sketched in fig. 3

../../figures/ece1236/deck4TxlineFig3: fig. 3. Coaxial cable.

From Gauss’ law the total charge can be calculated assuming that the ends of the cable can be neglected

\label{eqn:uwaves4TransmissionLines:520}
\begin{aligned}
Q
&= \int \spacegrad \cdot \BD dV \\
&= \oint \BD \cdot d\BA \\
&= \epsilon_0 \epsilon_r E ( 2 \pi r ) l,
\end{aligned}

This provides the radial electric field magnitude, in terms of the total charge

\label{eqn:uwaves4TransmissionLines:320}
E =
\frac{Q/l}{\epsilon_0 \epsilon_r ( 2 \pi r ) },

which must be a radial field as sketched in fig. 4.

../../figures/ece1236/deck4TxlineFig4: fig. 4. Radial electric field for coaxial cable.

The potential difference from the inner transmission surface to the outer is

\label{eqn:uwaves4TransmissionLines:340}
\begin{aligned}
V
&= \int_a^b E dr \\
&=
\frac{Q/l}{2 \pi \epsilon_0 \epsilon_r }
\int_a^b \frac{dr}{r} \\
&=
\frac{Q/l}{2 \pi \epsilon_0 \epsilon_r } \ln \frac{b}{a}.
\end{aligned}

Therefore the capacitance per unit length is

\label{eqn:uwaves4TransmissionLines:360}
C = \frac{Q/l}{V} = \frac{2 \pi \epsilon_0 \epsilon_r }{ \ln \frac{b}{a} } .

The inductance per unit length can be calculated form Ampere’s law

\label{eqn:uwaves4TransmissionLines:380}
\begin{aligned}
\int \lr{ \spacegrad \cross \BH } \cdot d\BS
&=
\int \BJ \cdot d\BS + \PD{t}{} \int \BD \cdot d\Bl \\
&=
\int \BJ \cdot d\BS \\
&=
I \\
&=
\oint \BH \cdot d\Bl \\
&=
H ( 2 \pi r ) \\
&=
\frac{B}{\mu_0} ( 2 \pi r )
\end{aligned}

The flux is

\label{eqn:uwaves4TransmissionLines:400}
\begin{aligned}
\Phi
&= \int \BB \cdot d\BA \\
&= \frac{\mu_0 I}{ 2 \pi } \int_A \inv{r} d dr \\
&= \frac{\mu_0 I}{ 2 \pi } \int_a^b \inv{r} l d dr \\
&= \frac{\mu_0 I l}{ 2 \pi } \ln \frac{b}{a}.
\end{aligned}

The inductance per unit length is

\label{eqn:uwaves4TransmissionLines:420}
L = \frac{\Phi/l}{I} = \frac{\mu_0}{ 2 \pi } \ln \frac{b}{a}.

For a lossless line where $$R = G = 0$$, we have $$\gamma = \sqrt{ (j \omega L)(j \omega C)} = j \omega \sqrt{L C}$$,
so the phase velocity for a (lossless) coaxial cable is

\label{eqn:uwaves4TransmissionLines:440}
\begin{aligned}
v_\phi
&= \frac{\omega}{\beta} \\
&= \frac{\omega}{\textrm{Im}(\gamma)} \\
&= \frac{\omega}{\omega \sqrt{LC})} \\
&= \frac{1}{\sqrt{LC})}.
\end{aligned}

This gives

\label{eqn:uwaves4TransmissionLines:460}
\begin{aligned}
v_\phi^2
&= \inv{ L } \inv{C} \\
&=
\frac{ 2 \pi }{ \mu_0 \ln \frac{b}{a} }
\frac
{\ln \frac{b}{a}}
{2 \pi \epsilon_0 \epsilon_r } \\
&=
\frac{1 }{ \mu_0 \epsilon_0 \epsilon_r } \\
&=
\frac{1 }{ \mu_0 \epsilon }.
\end{aligned}

So

\label{eqn:uwaves4TransmissionLines:480}
v_\phi = \inv{\sqrt{\epsilon \mu_0}},

which is the speed of light in the medium ($$\epsilon_r$$) that fills the co-axial cable.

This is \underline{not} a coincidence. In any two-wire homogeneously filled transmission line, the phase velocity is equal to the speed of light in the unbounded medium that fills the line.

The characteristic impedance (again assuming the lossless $$R = G = 0$$ case) is

\label{eqn:uwaves4TransmissionLines:500}
\begin{aligned}
Z_0
&= \sqrt{ \frac{R + j \omega L}{G + j \omega C} } \\
&= \sqrt{ \frac{j \omega L}{j \omega C} } \\
&= \sqrt{ \frac{L}{C} } \\
&= \sqrt{
\frac{\mu_0}{ 2 \pi } \ln \frac{b}{a}
\frac{ \ln \frac{b}{a} }{2 \pi \epsilon_0 \epsilon_r }
} \\
&=
\sqrt{ \frac{\mu_0}{\epsilon} } \frac{ \ln \frac{b}{a} }{ 2 \pi }.
\end{aligned}

Note that $$\eta = \sqrt{\mu_0/\epsilon_0} = 120 \pi \Omega$$ is the intrinsic impedance of free space. The values $$a, b$$ in \ref{eqn:uwaves4TransmissionLines:500} can be used to tune the characteristic impedance of the transmission line.

Lossless line.

The lossless lossless case where $$R = G = 0$$ was considered above. The results were

\label{eqn:uwaves4TransmissionLines:540}
\gamma = j \omega \sqrt{ L C },

so $$\alpha = 0$$ and $$\beta = \omega \sqrt{LC}$$, and the phase velocity was

\label{eqn:uwaves4TransmissionLines:560}
v_\phi = \inv{\sqrt{LC}},

the characteristic impedance is

\label{eqn:uwaves4TransmissionLines:580}
Z_0 = \sqrt{\frac{L}{C}},

and the signals are
\label{eqn:uwaves4TransmissionLines:600}
\begin{aligned}
V(z) &= V_0^{+} e^{-j \beta z} + V_0^{-} e^{j \beta z} \\
I(z) &= \inv{Z_0} \lr{ V_0^{+} e^{-j \beta z} – V_0^{-} e^{j \beta z} }
\end{aligned}

In the time domain for an infinite line, we have

\label{eqn:uwaves4TransmissionLines:620}
\begin{aligned}
v(z, t)
&= \textrm{Re}\lr{ V(z) e^{j \omega t} } \\
&= V_0^{+} \textrm{Re}\lr{ e^{-j \beta z} e^{j \omega t} } \\
&= V_0^{+} \cos( \omega t – \beta z ).
\end{aligned}

In this case the shape and amplitude of the waveform are preserved as sketched in fig. 7.

../../figures/ece1236/deck4TxlineFig7: fig. 7. Lossless line signal preservation.

Low loss line.

Assume $$R \ll \omega L$$ and $$G \ll \omega C$$. In this case we have

\label{eqn:uwaves4TransmissionLines:640}
\begin{aligned}
\gamma
&= \sqrt{ (R + j \omega L) ( G + j \omega C ) } \\
&=
j \omega \sqrt{L C} \sqrt{
\lr{ 1 + \frac{R}{j\omega L} }
\lr{ 1 + \frac{G}{j\omega C} }
} \\
&\approx
j \omega \sqrt{L C}
\lr{ 1 + \frac{R}{2 j\omega L} }
\lr{ 1 + \frac{G}{2 j\omega C} } \\
&\approx
j \omega \sqrt{L C}
\lr{ 1 + \frac{R}{2 j\omega L} + \frac{G}{2 j\omega C} } \\
&=
j \omega \sqrt{L C}
+ j \omega \frac{R \sqrt{C/L}}{2 j\omega}
+ j \omega \frac{G \sqrt{L/C}}{2 j\omega} \\
&=
j \omega \sqrt{L C}
+
\inv{2} \lr{
R \sqrt{\frac{C}{L}}
+
G \sqrt{\frac{L}{C}}
},
\end{aligned}

so
\label{eqn:uwaves4TransmissionLines:660}
\begin{aligned}
\alpha &=
\inv{2} \lr{
R \sqrt{\frac{C}{L}}
+
G \sqrt{\frac{L}{C}}
} \\
\beta &= \omega \sqrt{L C}.
\end{aligned}

Observe that this value for $$\beta$$ is the same as the lossless case to first order. We also have

\label{eqn:uwaves4TransmissionLines:680}
Z_0
= \sqrt{ \frac{R + j \omega L}{G + j \omega C} }
\approx
\sqrt{ \frac{L}{C} },

also the same as the lossless case. We must also have $$v_\phi = 1/\sqrt{L C}$$. To consider a time domain signal note that

\label{eqn:uwaves4TransmissionLines:700}
\begin{aligned}
V(z)
&= V_0^{+} e^{-\gamma z} \\
&= V_0^{+} e^{-\alpha z} e^{-j \beta z},
\end{aligned}

so
\label{eqn:uwaves4TransmissionLines:720}
\begin{aligned}
v(z, t)
&= \textrm{Re} \lr{ V(z) e^{j \omega t} } \\
&= \textrm{Re} \lr{ V_0^{+} e^{-\alpha z} e^{-j \beta z} e^{j \omega t} } \\
&= V_0^{+} e^{-\alpha z} \cos( \omega t – \beta z ).
\end{aligned}

The phase factor can be written

\label{eqn:uwaves4TransmissionLines:740}
\omega t – \beta z
=
\omega \lr{ t – \frac{\beta}{\omega} z }
\omega \lr{ t – z/v_\phi },

so the signal still moves with the phase velocity $$v_\phi = 1/\sqrt{LC}$$, but in a diminishing envelope as sketched in fig. 8.

../../figures/ece1236/deck4TxlineFig8: fig. 8. Time domain envelope for loss loss line.

Notes

• The shape is preserved but the amplitude has an exponential attenuation along the line.
• In this case, since $$\beta(\omega)$$ is a linear function to first order, we have no dispersion. All of the Fourier components of a pulse travel with the same phase velocity since $$v_\phi = \omega/\beta$$ is constant. i.e. $$v(z, t) = e^{-\alpha z} f( t – z/v_\phi )$$. We should expect dispersion when the $$R/\omega L$$ and $$G/\omega C$$ start becoming more significant.

Distortionless line.

Motivated by the early telegraphy days, when low loss materials were not available. Therefore lines with a constant attenuation and constant phase velocity (i.e. no dispersion) were required in order to eliminate distortion of the signals. This can be achieved by setting

\label{eqn:uwaves4TransmissionLines:760}
\frac{R}{L} = \frac{G}{C}.

When that is done we have
\label{eqn:uwaves4TransmissionLines:780}
\begin{aligned}
\gamma
&= \sqrt{ (R + j \omega L) ( G + j \omega C ) } \\
&= j \omega \sqrt{L C} \sqrt{
\lr{ 1 + \frac{R}{j \omega L} }
\lr{ 1 + \frac{G}{j \omega C} }
} \\
&= j \omega \sqrt{L C} \sqrt{
\lr{ 1 + \frac{R}{j \omega L} }
\lr{ 1 + \frac{R}{j \omega L} }
} \\
&= j \omega \sqrt{L C}
\lr{ 1 + \frac{R}{j \omega L} } \\
&= R \sqrt{\frac{C}{L} }
+ j \omega \sqrt{L C} \\
&= \sqrt{R G }
+ j \omega \sqrt{L C}.
\end{aligned}

We have

\label{eqn:uwaves4TransmissionLines:800}
\begin{aligned}
\alpha &= \sqrt{R G } \\
\beta &= \omega \sqrt{L C}.
\end{aligned}

The phase velocity is the same as that of the lossless and low-loss lines

\label{eqn:uwaves4TransmissionLines:820}
v_\phi = \frac{\omega}{\beta} = \inv{\sqrt{L C}}.

Terminated lossless line.

Consider the load configuration sketched in fig. 9.

../../figures/ece1236/deck4TxlineFig9: fig. 9. Terminated line.

Recall that

\label{eqn:uwaves4TransmissionLines:840}
\begin{aligned}
V(z) &= V_0^{+} e^{-j \beta z} + V_0^{-} e^{+j \beta z} \\
I(z) &= \frac{V_0^{+}}{Z_0} e^{-j \beta z} – \frac{V_0^{-}}{Z_0} e^{+j \beta z} \\
\end{aligned}

At the load ($$z = 0$$), we have

\label{eqn:uwaves4TransmissionLines:860}
\begin{aligned}
V(0) &= V_0^{+} + V_0^{-} \\
I(0) &= \inv{Z_0} \lr{ V_0^{+} – V_0^{-} }
\end{aligned}

So

\label{eqn:uwaves4TransmissionLines:880}
\begin{aligned}
Z_{\textrm{L}}
&= \frac{V(0)}{I(0)} \\
&= Z_0 \frac{ V_0^{+} + V_0^{-} }{ V_0^{+} – V_0^{-} } \\
&= Z_0 \frac{ 1 + \Gamma_{\textrm{L}} }{1 – \Gamma_{\textrm{L}} },
\end{aligned}

where

\label{eqn:uwaves4TransmissionLines:900}
\Gamma_{\textrm{L}} \equiv \frac{V_0^{-} }{V_0^{+}},

is the reflection coefficient at the load.

The phasors for the signals take the form

\label{eqn:uwaves4TransmissionLines:920}
\begin{aligned}
V(z) &= V_0^{+} \lr{ e^{-j \beta z} + \Gamma_{\textrm{L}} e^{+j \beta z} } \\
I(z) &= \frac{V_0^{+}}{Z_0} \lr{ e^{-j \beta z} – \Gamma_{\textrm{L}} e^{+j \beta z} }.
\end{aligned}

Observe that we can rearranging for $$\Gamma_{\textrm{L}}$$ in terms of the impedances

\label{eqn:uwaves4TransmissionLines:940}
\lr{ 1 – \Gamma_{\textrm{L}} } Z_{\textrm{L}} = Z_0 \frac{ 1 + \Gamma_{\textrm{L}} },

or
\label{eqn:uwaves4TransmissionLines:960}
\Gamma_{\textrm{L}} \lr{ Z_0 + Z_{\textrm{L}} } = Z_{\textrm{L}} – Z_0,

or
\label{eqn:uwaves4TransmissionLines:980}
\Gamma_{\textrm{L}}
= \frac{Z_{\textrm{L}} – Z_0}
{ Z_0 + Z_{\textrm{L}} } .

Power

The average (time) power on the line is

\label{eqn:uwaves4TransmissionLines:1000}
\begin{aligned}
P_{ \textrm{av}}
&= \inv{2} \textrm{Re}\lr{ V(Z) I^\conj(z) } \\
&=
\inv{2} \textrm{Re}
\lr{
V_0^{+} \lr{ e^{-j \beta z} + \Gamma_{\textrm{L}} e^{+j \beta z} }
\lr{\frac{V_0^{+}}{Z_0}}^\conj \lr{ e^{j \beta z} – \Gamma_{\textrm{L}}^\conj e^{-j \beta z} }
} \\
&= \frac{ \Abs{V_0^{+}}^2 }{2 Z_0 } \textrm{Re}\lr{
1 + \Gamma_{\textrm{L}} e^{2 j \beta z} – \Gamma_{\textrm{L}}^\conj e^{-2 j \beta z} – \Abs{\Gamma_{\textrm{L}}}^2
} \\
&= \frac{ \Abs{V_0^{+}}^2 }{2 Z_0 } \lr{
1 – \Abs{\Gamma_{\textrm{L}}}^2
}.
\end{aligned}

where we’ve made use of the fact that $$Z_0 = \sqrt{L/C}$$ is real for the lossless line, and the fact that a conjugate difference $$A – A^\conj = 2 j \textrm{Im}(A)$$ is purely imaginary.

This can be written as

\label{eqn:uwaves4TransmissionLines:1020}
P_{ \textrm{av}} = P^{+} – P^{-},

where

\label{eqn:uwaves4TransmissionLines:1040}
\begin{aligned}
P^{+} &= \frac{ \Abs{V_0^{+}}^2 }{2 Z_0 } \\
P^{+} &= \frac{ \Abs{V_0^{+}}^2 }{2 Z_0 } \Abs{\Gamma_{\textrm{L}}}^2.
\end{aligned}

This difference is the power delivered to the load. This is not z-dependent because we are considering the lossless case. Maximum power is delivered to the load when $$\Gamma_{\textrm{L}} = 0$$, which occurs when the impedances are matched.

Return loss and insertion loss. Defined.

Return loss (dB) is defined as

\label{eqn:uwaves4TransmissionLines:1060}
\begin{aligned}
\textrm{RL}
&= 10 \log_{10} \frac{P_{\textrm{inc}}}{P_{\textrm{refl}}} \\
&= 10 \log_{10} \inv{\Abs{\Gamma}^2} \\
&= -20 \log_{10} \Abs{\Gamma}.
\end{aligned}

Insertion loss (dB) is defined as

\label{eqn:uwaves4TransmissionLines:1080}
\begin{aligned}
\textrm{IL}
&= 10 \log_{10} \frac{P_{\textrm{inc}}}{P_{\textrm{trans}}} \\
&= 10 \log_{10} \frac{P^{+}}{P^{+} – P^{-}} \\
&= 10 \log_{10} \inv{1 – \Abs{\Gamma}^2} \\
&= -10 \log_{10} \lr{ 1 – \Abs{\Gamma}^2 }.
\end{aligned}

Standing wave ratio

Consider again the lossless loaded configuration of fig. 9. Now let $$z = – l$$, where $$l$$ is the distance from the load. The phasors at this point on the line are

\label{eqn:uwaves4TransmissionLines:1100}
\begin{aligned}
V(-l) &= V_0^{+} \lr{ e^{j \beta l} + \Gamma_{\textrm{L}} e^{-j \beta l} } \\
I(-l) &= \frac{V_0^{+}}{Z_0} \lr{ e^{j \beta l} – \Gamma_{\textrm{L}} e^{-j \beta l} } \\
\end{aligned}

The absolute voltage at this point is
\label{eqn:uwaves4TransmissionLines:1120}
\begin{aligned}
\Abs{V(-l)}
&= \Abs{V_0^{+}} \Abs{ e^{j \beta l} + \Gamma_{\textrm{L}} e^{-j \beta l} } \\
&= \Abs{V_0^{+}} \Abs{ 1 + \Gamma_{\textrm{L}} e^{-2 j \beta l} } \\
&= \Abs{V_0^{+}} \Abs{ 1 + \Abs{\Gamma_{\textrm{L}}} e^{j \Theta_{\textrm{L}}} e^{-2 j \beta l} },
\end{aligned}

where the complex valued $$\Gamma_{\textrm{L}}$$ is given by $$\Gamma_{\textrm{L}} = \Abs{\Gamma_{\textrm{L}}} e^{j \Theta_{\textrm{L}}}$$.

This gives
\label{eqn:uwaves4TransmissionLines:1140}
\Abs{V(-l)}
= \Abs{V_0^{+}} \Abs{ 1 + \Abs{\Gamma_{\textrm{L}}} e^{j(\Theta_{\textrm{L}} -2 \beta l)} }.

The voltage magnitude oscillates as one moves along the line. The maximum occurs when $$e^{j (\Theta_{\textrm{L}} -2 \beta l)} = 1$$

\label{eqn:uwaves4TransmissionLines:1160}
V_{\mathrm{max}} = \Abs{V_0^{+}} \Abs{ 1 + \Abs{\Gamma_{\textrm{L}}} }.

This occurs when $$\Theta_{\textrm{L}} – 2 \beta l = 2 k \pi$$ for $$k = 0, 1, 2, \cdots$$. The minimum occurs when $$e^{j (\Theta_{\textrm{L}} -2 \beta l)} = -1$$

\label{eqn:uwaves4TransmissionLines:1180}
V_{\mathrm{min}} = \Abs{V_0^{+}} \Abs{ 1 – \Abs{\Gamma_{\textrm{L}}} },

which occurs when $$\Theta_{\textrm{L}} – 2 \beta l = (2 k – 1 )\pi$$ for $$k = 1, 2, \cdots$$. The standing wave ratio is defined as

\label{eqn:uwaves4TransmissionLines:1200}
\textrm{SWR} = \frac{V_{\mathrm{max}}}{V_{\mathrm{min}}} = \frac{ 1 + \Abs{\Gamma_{\textrm{L}}} }{ 1 – \Abs{\Gamma_{\textrm{L}}} }.

This is a measure of the mismatch of a line. This is sketched in fig. 10.

../../figures/ece1236/deck4TxlineFig10: fig. 10. SWR extremes.

Notes:

• Since $$0 \le \Abs{\Gamma_{\textrm{L}}} \le 1$$, we have $$1 \le \textrm{SWR} \le \infty$$. The lower bound is for a matched line, and open, short, or purely reactive termination leads to the infinities.
• The distance between two successive maxima (or minima) can be determined by setting $$\Theta_{\textrm{L}} – 2 \beta l = 2 k \pi$$ for two consecutive values of $$k$$. For $$k = 0$$, suppose that $$V_{\mathrm{max}}$$ occurs at $$d_1$$

\label{eqn:uwaves4TransmissionLines:1220}
\Theta_{\textrm{L}} – 2 \beta d_1 = 2 (0) \pi,

or
\label{eqn:uwaves4TransmissionLines:1240}
d_1 = \frac{\Theta_{\textrm{L}}}{2 \beta}.

For $$k = 1$$, let the max occur at $$d_2$$

\label{eqn:uwaves4TransmissionLines:1260}
\Theta_{\textrm{L}} – 2 \beta d_2 = 2 (1) \pi,

or
\label{eqn:uwaves4TransmissionLines:1280}
d_2 = \frac{\Theta_{\textrm{L}} – 2 \pi}{2 \beta}.

The difference is

\label{eqn:uwaves4TransmissionLines:1300}
\begin{aligned}
d_1 – d_2
&= \frac{\Theta_{\textrm{L}}}{2 \beta} – \frac{\Theta_{\textrm{L}} – 2 \pi}{2 \beta} \\
&= \frac{\pi}{\beta} \\
&= \frac{\pi}{2 \pi/\lambda} \\
&= \frac{\lambda}{2}.
\end{aligned}

The distance between two consecutive maxima (or minima) of the SWR is $$\lambda/2$$.

Impedance Transformation.

Referring to fig. 11, let’s solve for the impedance at the load where $$z = 0$$ and at $$z = -l$$.

../../figures/ece1236/deck4TxlineFig11: fig. 11. Configuration for impedance transformation.

At any point on the line we have

\label{eqn:uwaves4TransmissionLinesCore:1320}
V(z) = V_0^{+} e^{-j \beta z} \lr{ 1 + \Gamma_{\textrm{L}} e^{2 j \beta z} },

so at the load and input we have

\label{eqn:uwaves4TransmissionLinesCore:1340}
\begin{aligned}
V_{\textrm{L}} &= V_0^{+} \lr{ 1 + \Gamma_{\textrm{L}} } \\
V(-l) &= V^{+} \lr{ 1 + \Gamma_{\textrm{L}}(-1) },
\end{aligned}

where

\label{eqn:uwaves4TransmissionLinesCore:1360}
\begin{aligned}
V^{+} &= V_0^{+} e^{ j \beta l } \\
\Gamma_{\textrm{L}}(-1) &= \Gamma_{\textrm{L}} e^{-2 j \beta l}
\end{aligned}

Similarly

\label{eqn:uwaves4TransmissionLinesCore:1380}
I(-l) = \frac{V^{+}}{Z_0} \lr{ 1 – \Gamma_{\textrm{L}}(-1) }.

Define an input impedance as
\label{eqn:uwaves4TransmissionLinesCore:1400}
\begin{aligned}
Z_{\textrm{in}}
&= \frac{V(-l)}{I(-l)} \\
&= Z_0 \frac{1 + \Gamma_{\textrm{L}}(-1)}{1 – \Gamma_{\textrm{L}}(-1)}
\end{aligned}

This is analogous to

\label{eqn:uwaves4TransmissionLinesCore:1420}
Z_{\textrm{L}}
= Z_0 \frac{1 + \Gamma_{\textrm{L}}}{1 – \Gamma_{\textrm{L}}}

From \ref{eqn:uwaves4TransmissionLines:980}, we have

\label{eqn:uwaves4TransmissionLinesCore:1440}
\begin{aligned}
Z_{\textrm{in}}
&= Z_0 \frac{Z_0 + Z_{\textrm{L}} + \lr{Z_{\textrm{L}} – Z_0} e^{-2 j \beta l}}{Z_0 + Z_{\textrm{L}} – \lr{Z_{\textrm{L}} – Z_0} e^{-2 j \beta l}} \\
&= Z_0 \frac{\lr{Z_0 + Z_{\textrm{L}}} e^{j\beta l} + \lr{Z_{\textrm{L}} –
Z_0} e^{- j \beta l}}{\lr{Z_0 + Z_{\textrm{L}}} e^{j\beta l} – \lr{Z_{\textrm{L}} – Z_0} e^{- j \beta l}} \\
&= Z_0
\frac
{Z_{\textrm{L}} \cos( \beta l ) + j Z_0 \sin(\beta l ) }
{Z_0 \cos( \beta l ) + j Z_{\textrm{L}} \sin(\beta l ) },
\end{aligned}

or
\label{eqn:uwaves4TransmissionLinesCore:1460}
\boxed{
Z_{\textrm{in}} =
\frac
{Z_{\textrm{L}} + j Z_0 \tan(\beta l ) }
{Z_0 + j Z_{\textrm{L}} \tan(\beta l ) }.
}

This can be thought of as providing a reflection coefficient function along the line to the load at any point as sketched in fig. 12.

../../figures/ece1236/deck4TxlineFig12: fig. 12. Impedance transformation reflection on the line.

References

[1] David M Pozar. Microwave engineering. John Wiley \& Sons, 2009.

Updated notes for ece1229 antenna theory

I’ve now posted a first update of my notesÂ for theÂ antenna theory courseÂ that I am taking this term at UofT.

Unlike most of the other classes I have taken, I am not attempting to take comprehensive notes for this class. The class is taught on slides which go by faster than I can easily take notes for (and some of which match the textbook closely). In class I have annotated my copy of textbook with little details instead. This set of notes contains musings of details that were unclear, or in some cases, details that were provided in class, but are not in the text (and too long to pencil into my book), as well as some notes Geometric Algebra formalism for Maxwell’s equations with magnetic sources (something I’ve encountered for the first time in any real detail in this class).

The notes compilation linked above includes all of the following separate notes, some of which have been posted separately on this blog:

Notes for ece1229 antenna theory

I’ve now posted a first set of notes for the antenna theory course that I am taking this term at UofT.

Unlike most of the other classes I have taken, I am not attempting to take comprehensive notes for this class. The class is taught on slides that match the textbook so closely, there is little value to me taking notes that just replicate the text. Instead, I am annotating my copy of textbook with little details instead. My usual notes collection for the class will contain musings of details that were unclear, or in some cases, details that were provided in class, but are not in the text (and too long to pencil into my book.)

• Reading notes for chapter 2 (Fundamental Parameters of Antennas) and chapter 3 (Radiation Integrals and Auxiliary Potential Functions) of the class text.
• Geometric Algebra musings.Â  How to do formulate Maxwell’s equations when magnetic sources are also included (those modeling magnetic dipoles).
• Some problems for chapter 2 content.

Fundamental parameters of antennas

This is my first set of notes for the UofT course ECE1229, Advanced Antenna Theory, taught by Prof. Eleftheriades, covering ch. 2 [1] content.

Unlike most of the other classes I have taken, I am not attempting to take comprehensive notes for this class. The class is taught on slides that match the textbook so closely, there is little value to me taking notes that just replicate the text. Instead, I am annotating my copy of textbook with little details instead. My usual notes collection for the class will contain musings of details that were unclear, or in some cases, details that were provided in class, but are not in the text (and too long to pencil into my book.)

Poynting vector

The Poynting vector was written in an unfamiliar form

\label{eqn:chapter2Notes:560}
\boldsymbol{\mathcal{W}} = \boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{H}}.

I can roll with the use of a different symbol (i.e. not $$\BS$$) for the Poynting vector, but I’m used to seeing a $$\frac{c}{4\pi}$$ factor ([6] and [5]). I remembered something like that in SI units too, so was slightly confused not to see it here.

Per [3] that something is a $$\mu_0$$, as in

\label{eqn:chapter2Notes:580}
\boldsymbol{\mathcal{W}} = \inv{\mu_0} \boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{B}}.

Note that the use of $$\boldsymbol{\mathcal{H}}$$ instead of $$\boldsymbol{\mathcal{B}}$$ is what wipes out the requirement for the $$\frac{1}{\mu_0}$$ term since $$\boldsymbol{\mathcal{H}} = \boldsymbol{\mathcal{B}}/\mu_0$$, assuming linear media, and no magnetization.

It was mentioned that

U(\theta, \phi)
=
\frac{r^2}{2 \eta_0} \Abs{ \BE( r, \theta, \phi) }^2
=
\frac{1}{2 \eta_0} \lr{ \Abs{ E_\theta(\theta, \phi) }^2 + \Abs{ E_\phi(\theta, \phi) }^2},

where the intrinsic impedance of free space is

\eta_0 = \sqrt{\frac{\mu_0}{\epsilon_0}} = 377 \Omega.

(this is also eq. 2-19 in the text.)

To get an understanding where this comes from, consider the far field radial solutions to the electric and magnetic dipole problems, which have the respective forms (from [3]) of

\label{eqn:chapter2Notes:740}
\begin{aligned}
\boldsymbol{\mathcal{E}} &= -\frac{\mu_0 p_0 \omega^2 }{4 \pi } \frac{\sin\theta}{r} \cos\lr{w t – k r} \thetacap \\
\boldsymbol{\mathcal{B}} &= -\frac{\mu_0 p_0 \omega^2 }{4 \pi c} \frac{\sin\theta}{r} \cos\lr{w t – k r} \phicap \\
\end{aligned}

\label{eqn:chapter2Notes:760}
\begin{aligned}
\boldsymbol{\mathcal{E}} &= \frac{\mu_0 m_0 \omega^2 }{4 \pi c} \frac{\sin\theta}{r} \cos\lr{w t – k r} \phicap \\
\boldsymbol{\mathcal{B}} &= -\frac{\mu_0 m_0 \omega^2 }{4 \pi c^2} \frac{\sin\theta}{r} \cos\lr{w t – k r} \thetacap \\
\end{aligned}

In neither case is there a component in the direction of propagation, and in both cases (using $$\mu_0 \epsilon_0 = 1/c^2$$)

\label{eqn:chapter2Notes:780}
\Abs{\boldsymbol{\mathcal{H}}}
= \frac{\Abs{\boldsymbol{\mathcal{E}}}}{\mu_0 c}
= \Abs{\boldsymbol{\mathcal{E}}} \sqrt{\frac{\epsilon_0}{\mu_0}}
= \inv{\eta_0}\Abs{\boldsymbol{\mathcal{E}}} .

A superposition of the phasors for such dipole fields, in the far field, will have the form

\label{eqn:chapter2Notes:800}
\begin{aligned}
\BE &= \inv{r} \lr{ E_\theta(\theta, \phi) \thetacap + E_\phi(\theta, \phi) \phicap } \\
\BB &= \inv{r c} \lr{ E_\theta(\theta, \phi) \thetacap – E_\phi(\theta, \phi) \phicap },
\end{aligned}

with a corresponding time averaged Poynting vector

\label{eqn:chapter2Notes:820}
\begin{aligned}
\BW_{\textrm{av}}
&= \inv{2 \mu_0} \BE \cross \BB^\conj \\
&=
\inv{2 \mu_0 c r^2}
\lr{ E_\theta \thetacap + E_\phi \phicap } \cross
\lr{ E_\theta^\conj \thetacap – E_\phi^\conj \phicap } \\
&=
\frac{\thetacap \cross \phicap}{2 \mu_0 c r^2}
\lr{ \Abs{E_\theta}^2 + \Abs{E_\phi}^2 } \\
&=
\frac{\rcap}{2 \eta_0 r^2}
\lr{ \Abs{E_\theta}^2 + \Abs{E_\phi}^2 },
\end{aligned}

verifying \ref{eqn:advancedantennaL1:20} for a superposition of electric and magnetic dipole fields. This can likely be shown for more general fields too.

Field plots

We can plot the fields, or intensity (or log plots in dB of these).
It is pointed out in [3] that when there is $$r$$ dependence these plots are done by considering the values of at fixed $$r$$.

The field plots are conceptually the simplest, since that vector parameterizes
a surface. Any such radial field with magnitude $$f(r, \theta, \phi)$$ can
be plotted in Mathematica in the $$\phi = 0$$ plane at $$r = r_0$$, or in
3D (respectively, but also at $$r = r_0$$) with code like that of the
following listing

Intensity plots can use the same code, with the only difference being the interpretation. The surface doesn’t represent the value of a vector valued radial function, but is the magnitude of a scalar valued function evaluated at $$f( r_0, \theta, \phi)$$.

The surfaces for $$U = \sin\theta, \sin^2\theta$$ in the plane are parametrically plotted in fig. 2, and for cosines in fig. 1 to compare with textbook figures.

Visualizations of $$U = \sin^2 \theta$$ and $$U = \cos^2 \theta$$ can be found in fig. 3 and fig. 4 respectively. Even for such simple functions these look pretty cool.

fig 3. Square sinusoidal radiation intensity

fig 4. Square cosinusoidal radiation intensity

dB vs dBi

Note that dBi is used to indicate that the gain is with respect to an “isotropic” radiator.
This is detailed more in [2].

Trig integrals

Tables 1.1 and 1.2 produced with tableOfTrigIntegrals.nb have some of the sine and cosine integrals that are pervasive in this chapter.

Polarization vectors

The text introduces polarization vectors $$\rhocap$$ , but doesn’t spell out their form. Consider a plane wave field of the form

\label{eqn:chapter2Notes:840}
\BE
=
E_x e^{j \phi_x} e^{j \lr{ \omega t – k z }} \xcap
+
E_y e^{j \phi_y} e^{j \lr{ \omega t – k z }} \ycap.

The $$x, y$$ plane directionality of this phasor can be written

\label{eqn:chapter2Notes:860}
\Brho =
E_x e^{j \phi_x} \xcap
+
E_y e^{j \phi_y} \ycap,

so that

\label{eqn:chapter2Notes:880}
\BE = \Brho e^{j \lr{ \omega t – k z }}.

Separating this direction and magnitude into factors

\label{eqn:chapter2Notes:900}
\Brho = \Abs{\BE} \rhocap,

allows the phasor to be expressed as

\label{eqn:chapter2Notes:920}
\BE = \rhocap \Abs{\BE} e^{j \lr{ \omega t – k z }}.

As an example, suppose that $$E_x = E_y$$, and set $$\phi_x = 0$$. Then

\label{eqn:chapter2Notes:940}
\rhocap = \xcap + \ycap e^{j \phi_y}.

Phasor power

In section 2.13 the phasor power is written as

\label{eqn:chapter2Notes:620}
I^2 R/2,

where $$I, R$$ are the magnitudes of phasors in the circuit.

I vaguely recall this relation, but had to refer back to [4] for the details.
This relation expresses average power over a period associated with the frequency of the phasor

\label{eqn:chapter2Notes:640}
\begin{aligned}
P
&= \inv{T} \int_{t_0}^{t_0 + T} p(t) dt \\
&= \inv{T} \int_{t_0}^{t_0 + T} \Abs{\BV} \cos\lr{ \omega t + \phi_V }
\Abs{\BI} \cos\lr{ \omega t + \phi_I} dt \\
&= \inv{T} \int_{t_0}^{t_0 + T} \Abs{\BV} \Abs{\BI}
\lr{
\cos\lr{ \phi_V – \phi_I } + \cos\lr{ 2 \omega t + \phi_V + \phi_I}
}
dt \\
&= \inv{2} \Abs{\BV} \Abs{\BI} \cos\lr{ \phi_V – \phi_I }.
\end{aligned}

Introducing the impedance for this circuit element

\label{eqn:chapter2Notes:660}
\BZ = \frac{ \Abs{\BV} e^{j\phi_V} }{ \Abs{\BI} e^{j\phi_I} } = \frac{\Abs{\BV}}{\Abs{\BI}} e^{j\lr{\phi_V – \phi_I}},

this average power can be written in phasor form

\label{eqn:chapter2Notes:680}
\BP = \inv{2} \Abs{\BI}^2 \BZ,

with
\label{eqn:chapter2Notes:700}
P = \textrm{Re} \BP.

Observe that we have to be careful to use the absolute value of the current phasor $$\BI$$, since $$\BI^2$$ differs in phase from $$\Abs{\BI}^2$$. This explains the conjugation in the [4] definition of complex power, which had the form

\label{eqn:chapter2Notes:720}
\BS = \BV_{\textrm{rms}} \BI^\conj_{\textrm{rms}}.

Flat plate.

\label{eqn:chapter2Notes:960}
\sigma_{\textrm{max}} = \frac{4 \pi \lr{L W}^2}{\lambda^2}

fig. 6. Square geometry for RCS example.

Sphere.

In the optical limit the radar cross section for a sphere

fig. 7. Sphere geometry for RCS example.

\label{eqn:chapter2Notes:980}
\sigma_{\textrm{max}} = \pi r^2

Note that this is smaller than the physical area $$4 \pi r^2$$.

Cylinder.

fig. 8. Cylinder geometry for RCS example.

\label{eqn:chapter2Notes:1000}
\sigma_{\textrm{max}} = \frac{ 2 \pi r h^2}{\lambda}

Tridedral corner reflector

fig. 9. Trihedral corner reflector geometry for RCS example.

\label{eqn:chapter2Notes:1020}
\sigma_{\textrm{max}} = \frac{ 4 \pi L^4}{3 \lambda^2}

Scattering from a sphere vs frequency

Frequency dependence of spherical scattering is sketched in fig. 10.

• Low frequency (or small particles): Rayleigh\label{eqn:chapter2Notes:1040}
\sigma = \lr{\pi r^2} 7.11 \lr{\kappa r}^4, \qquad \kappa = 2 \pi/\lambda.
• Mie scattering (resonance),\label{eqn:chapter2Notes:1060}
\sigma_{\textrm{max}}(A) = 4 \pi r^2

\label{eqn:chapter2Notes:1080}
\sigma_{\textrm{max}}(B) = 0.26 \pi r^2.
• optical limit ( $$r \gg \lambda$$ )\label{eqn:chapter2Notes:1100}
\sigma = \pi r^2.

fig 10. Scattering from a sphere vs frequency (from Prof. Eleftheriades’ class notes).

FIXME: Do I have a derivation of this in my optics notes?

Notation

• Time average.
and the text [1] use square brackets $$[\cdots]$$ for time averages, not $$<\cdots>$$. Was that an engineering convention?
writes $$\Omega$$ as a circle floating above a face up square bracket, as in fig. 1, and $$\sigma$$ like a number 6, as in fig. 1.
• Bold vectors are usually phasors, with (bold) calligraphic script used for the time domain fields. Example: $$\BE(x,y,z,t) = \ecap E(x,y) e^{j \lr{\omega t – k z}}, \boldsymbol{\mathcal{E}}(x, y, z, t) = \textrm{Re} \BE$$.

fig. 11. Prof. handwriting decoder ring: Omega

fig 12. Prof. handwriting decoder ring: sigma

References

[1] Constantine A Balanis. Antenna theory: analysis and design. John Wiley \& Sons, 3rd edition, 2005.

[3] David Jeffrey Griffiths and Reed College. Introduction to electrodynamics. Prentice hall Upper Saddle River, NJ, 3rd edition, 1999.

[4] J.D. Irwin. Basic Engineering Circuit Analysis. MacMillian, 1993.

[5] JD Jackson. Classical Electrodynamics. John Wiley and Sons, 2nd edition, 1975.

[6] L.D. Landau and E.M. Lifshitz. The classical theory of fields. Butterworth-Heinemann, 1980. ISBN 0750627689.