## 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.

## Plane wave solution directly from Maxwell’s equations

Here’s a problem that I thought was fun, an exercise for the reader to show that the plane wave solution to Maxwell’s equations can be found with ease directly from Maxwell’s equations. This is in contrast to the what seems like the usual method of first showing that Maxwell’s equations imply wave equations for the fields, and then solving those wave equations.

## Problem. $$\xcap$$ oriented plane wave electric field ([1] ex. 4.1)

A uniform plane wave having only an $$x$$ component of the electric field is traveling in the $$+ z$$ direction in an unbounded lossless, source-0free region. Using Maxwell’s equations write expressions for the electric and corresponding magnetic field intensities.

The phasor form of Maxwell’s equations for a source free region are

\label{eqn:ExPlaneWave:40}
\spacegrad \cross \BE = -j \omega \BB

\label{eqn:ExPlaneWave:60}
\spacegrad \cross \BH = j \omega \BD

\label{eqn:ExPlaneWave:80}

\label{eqn:ExPlaneWave:100}

Since $$\BE = \xcap E(z)$$, the magnetic field follows from \ref{eqn:ExPlaneWave:40}

\label{eqn:ExPlaneWave:120}
-j \omega \BB
=
\begin{vmatrix}
\xcap & \ycap & \zcap \\
\partial_x & \partial_y & \partial_z \\
E & 0 & 0
\end{vmatrix}
=
\ycap \partial_z E(z)
– \zcap \partial_y E(z),

or

\label{eqn:ExPlaneWave:140}
\BB =
-\inv{j \omega} \partial_z E.

This is constrained by \ref{eqn:ExPlaneWave:60}

\label{eqn:ExPlaneWave:160}
j \omega \epsilon \xcap E
=
=
-\inv{\mu j \omega}
\begin{vmatrix}
\xcap & \ycap & \zcap \\
\partial_x & \partial_y & \partial_z \\
0 & \partial_z E & 0
\end{vmatrix}
=
-\inv{\mu j \omega}
\lr{
-\xcap \partial_{z z} E
+ \zcap \partial_x \partial_z E
}

Since $$\partial_x \partial_z E = \partial_z \lr{ \partial_x E } = \partial_z \inv{\epsilon} \spacegrad \cdot \BD = \partial_z 0$$, this means

\label{eqn:ExPlaneWave:180}
\partial_{zz} E = -\omega^2 \epsilon\mu E = -k^2 E.

This is the usual starting place that we use to show that the plane wave has an exponential form

\label{eqn:ExPlaneWave:200}
\BE(z) =
\xcap
\lr{
E_{+} e^{-j k z}
+
E_{-} e^{j k z}
}.

The magnetic field from \ref{eqn:ExPlaneWave:140} is

\label{eqn:ExPlaneWave:220}
\BB
= \frac{j}{\omega} \lr{ -j k E_{+} e^{-j k z} + j k E_{-} e^{j k z} }
= \inv{c} \lr{ E_{+} e^{-j k z} – E_{-} e^{j k z} },

or

\label{eqn:ExPlaneWave:240}
\BH
= \inv{\mu c} \lr{ E_{+} e^{-j k z} – E_{-} e^{j k z} }
= \inv{\eta} \lr{ E_{+} e^{-j k z} – E_{-} e^{j k z} }.

A solution requires zero divergence for the magnetic field, but that can be seen to be the case by inspection.

# References

[1] Constantine A Balanis. Advanced engineering electromagnetics. Wiley New York, 1989.

## Updated notes for ece1229 antenna theory

I’ve now posted a first update of my 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 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:

## Reciprocity theorem: background

The class slides presented a derivation of the reciprocity theorem, a theorem that contained the integral of

\label{eqn:reciprocityTheorem:360}
\int \lr{ \BE^{(a)} \cross \BH^{(b)} – \BE^{(b)} \cross \BH^{(a)} } \cdot d\BS = \cdots

over a surface, where the RHS was a volume integral involving the fields and (electric and magnetic) current sources.
The idea was to consider two different source loading configurations of the same system, and to show that the fields and sources in the two configurations can be related.

To derive the result in question, a simple way to start is to look at the divergence of the difference of cross products above. This will require the phasor form of the two cross product Maxwell’s equations

\label{eqn:reciprocityTheorem:100}
\spacegrad \cross \BE = – (\BM + j \omega \mu_0 \BH) % \BM^{(a)} + j \omega \mu_0 \BH^{(a)}

\label{eqn:reciprocityTheorem:120}
\spacegrad \cross \BH = \BJ + j \omega \epsilon_0 \BE, % \BJ^{(a)} + j \omega \epsilon_0 \BE^{(a)}

so the divergence is

\label{eqn:reciprocityTheorem:380}
\begin{aligned}
\lr{ \BE^{(a)} \cross \BH^{(b)} – \BE^{(b)} \cross \BH^{(a)} }
&=
\BH^{(b)} \cdot \lr{ \spacegrad \cross \BE^{(a)} } -\BE^{(a)} \cdot \lr{ \spacegrad \cross \BH^{(b)} } \\
&-\BH^{(a)} \cdot \lr{ \spacegrad \cross \BE^{(b)} } +\BE^{(b)} \cdot \lr{ \spacegrad \cross \BH^{(a)} } \\
&=
-\BH^{(b)} \cdot \lr{ \BM^{(a)} + j \omega \mu_0 \BH^{(a)} } -\BE^{(a)} \cdot \lr{ \BJ^{(b)} + j \omega \epsilon_0 \BE^{(b)} } \\
&+\BH^{(a)} \cdot \lr{ \BM^{(b)} + j \omega \mu_0 \BH^{(b)} } +\BE^{(b)} \cdot \lr{ \BJ^{(a)} + j \omega \epsilon_0 \BE^{(a)} }.
\end{aligned}

The non-source terms cancel, leaving

\label{eqn:reciprocityTheorem:440}
\boxed{
\lr{ \BE^{(a)} \cross \BH^{(b)} – \BE^{(b)} \cross \BH^{(a)} }
=
-\BH^{(b)} \cdot \BM^{(a)} -\BE^{(a)} \cdot \BJ^{(b)}
+\BH^{(a)} \cdot \BM^{(b)} +\BE^{(b)} \cdot \BJ^{(a)}
}

Should we be suprised to have a relation of this form? Probably not, given that the energy momentum relationship between the fields and currents of a single source has the form

\label{eqn:reciprocityTheorem:n}
\PD{t}{}\frac{\epsilon_0}{2} \left(\BE^2 + c^2 \BB^2\right) + \spacegrad \cdot \inv{\mu_0}(\BE \cross \BB) = -\BE \cdot \BJ.

(this is without magnetic sources).

This suggests that the reciprocity theorem can be expressed more generally in terms of the energy-momentum tensor.

## far field integral form

Employing the divergence theorem over a sphere the identity above takes the form

\label{eqn:reciprocityTheorem:480}
\int_S
\lr{ \BE^{(a)} \cross \BH^{(b)} – \BE^{(b)} \cross \BH^{(a)} } \cdot \rcap dS
=
\int_V \lr{
-\BH^{(b)} \cdot \BM^{(a)} -\BE^{(a)} \cdot \BJ^{(b)}
+\BH^{(a)} \cdot \BM^{(b)} +\BE^{(b)} \cdot \BJ^{(a)}
}
dV

In the far field, the cross products are strictly radial. That surface integral can be written as

\label{eqn:reciprocityTheorem:500}
\begin{aligned}
\int_S
\lr{ \BE^{(a)} \cross \BH^{(b)} – \BE^{(b)} \cross \BH^{(a)} } \cdot \rcap dS
&=
\inv{\mu_0}
\int_S
\lr{ \BE^{(a)} \cross \lr{ \rcap \cross \BE^{(b)}} – \BE^{(b)} \cross \lr{ \rcap \cross \BE^{(a)}} } \cdot \rcap dS \\
&=
\inv{\mu_0}
\int_S
\lr{ \BE^{(a)} \cdot \BE^{(b)} – \BE^{(b)} \cdot \BE^{(a)}
}
dS \\
&= 0
\end{aligned}

The above expansions used \ref{eqn:reciprocityTheorem:540} to expand the terms of the form

\label{eqn:reciprocityTheorem:560}
\lr{ \BA \cross \lr{ \rcap \cross \BC } } \cdot \rcap
= \BA \cdot \BC -\lr{ \BA \cdot \rcap } \lr{ \BC \cdot \rcap },

in which only the first dot product survives due to the transverse nature of the fields.

So in the far field we have a direct relation between the fields and sources of two source configurations of the same system of the form

\label{eqn:reciprocityTheorem:580}
\boxed{
\int_V \lr{
\BH^{(a)} \cdot \BM^{(b)} +\BE^{(b)} \cdot \BJ^{(a)}
}
dV
=
\int_V \lr{
\BH^{(b)} \cdot \BM^{(a)} +\BE^{(a)} \cdot \BJ^{(b)}
}
dV
}

## Application to antenna

This is the underlying reason that we are able to pose the problem of what an antenna can recieve, in terms of what the antenna can transmit.

More on that to come.

## Identities

Lemma: Divergence of a cross product.

\label{thm:polarizationReview:400}
\spacegrad \cdot \lr{ \BA \cross \BB } =

Proof.

\label{eqn:reciprocityTheorem:420}
\begin{aligned}
\spacegrad \cdot \lr{ \BA \cross \BB }
&=
\partial_a \epsilon_{a b c} A_b B_c \\
&=
\epsilon_{a b c} (\partial_a A_b )B_c

\epsilon_{b a c} A_b (\partial_a B_c) \\
&=
\end{aligned}

Lemma: Triple cross product dotted
\label{thm:polarizationReview:520}
\begin{aligned}
\lr{ \BA \cross \lr{ \BB \cross \BC } } \cdot \BD
=
\lr{ \BA \cdot \BC } \lr{ \BB \cdot \BD }
-\lr{ \BA \cdot \BB } \lr{ \BC \cdot \BD }
\end{aligned}

Proof.

\label{eqn:reciprocityTheorem:540}
\begin{aligned}
\lr{ \BA \cross \lr{ \BB \cross \BC } } \cdot \BD
&=
\epsilon_{a b c} A_b \epsilon_{r s c } B_r C_s D_a \\
&=
\delta_{[a b]}^{r s}
A_b B_r C_s D_a \\
&=
A_s B_r C_s D_r
-A_r B_r C_s D_s \\
&=
\lr{ \BA \cdot \BC } \lr{ \BB \cdot \BD }
-\lr{ \BA \cdot \BB } \lr{ \BC \cdot \BD }.
\end{aligned}