phasor

Bandwidth of some bandpass filters

October 18, 2025 math and physics play , , , , , , , , , , , , ,

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Karl had a couple of fun filter problems that he chatted with me about. I didn’t remember how do find the bandwidth, nor the resonant frequency of such a circuit, and after he solved the problems, I tried them for myself.

The circuits, as seen in fig. 1, were RC and RL respectively, and both ended up having the same transfer function.

fig 1a. RC bandpass filter.

fig 1b. RL bandpass filter.

 

For my attempt to solve these circuits, I used the loop current method that Karl recently taught me. Perhaps I once knew that method, but if I had, I’d forgotten. This loop method can be nicer than a standard nodal analysis, since it can automatically eliminate some current variables. Here’s the equations for the RC circuit
\begin{equation}\label{eqn:RLCbandwidth:20}
\begin{aligned}
V_a – V_s &= I_1 (1) \\
V_a &= (I_1 – I_2) \inv{s} \\
V_o – V_a &= I_2 \inv{s} \\
V_o &= I_2 (1).
\end{aligned}
\end{equation}
The equations for the RL circuit are
\begin{equation}\label{eqn:RLCbandwidth:40}
\begin{aligned}
V_a – V_s &= I_1 s \\
V_a &= (I_1 – I_2) (1) \\
V_o – V_a &= I_2 (1) \\
V_o &= I_2 s.
\end{aligned}
\end{equation}
The transfer functions for both is
\begin{equation}\label{eqn:RLCbandwidth:60}
H(s) = \frac{V_o}{V_s} = \frac{1}{1/s + 3 + s}.
\end{equation}
A plot of \( \Abs{H(j\omega)} \) can be found in fig. 2, and peaks at \( \omega = 1 \).

fig. 2. Transfer function.

 

Observe that the denominator of this transfer function looks just like a series RLC impedance. For example, for the circuit of fig. 3, we have
\begin{equation}\label{eqn:RLCbandwidth:80}
\frac{I}{V} = \inv{Z} = \inv{ R + L s + \inv{s C} }.
\end{equation}

fig. 3. RLC circuit.

 

Written out in the frequency domain, that impedance is
\begin{equation}\label{eqn:RLCbandwidth:100}
Z(\omega) = R + j \lr{ \omega L – \inv{\omega C} }.
\end{equation}
Max power transfer through this circuit will be for the specific frequency where the impedance is purely real. In this case, that is the frequency \(\omega_0\) that satisfies
\begin{equation}\label{eqn:RLCbandwidth:120}
\omega_0^2 = \inv{L C}.
\end{equation}
Karl’s textbook didn’t define bandwidth in any general sense, but did do so for an RLC circuit of this form, stating that the bandwidth was \( \omega_2 – \omega_1 \) where these are the frequencies of the half (average) power points.

We need to remind ourselves what the formula for average power is
\begin{equation}\label{eqn:RLCbandwidth:140}
\begin{aligned}
P
&= \inv{T} \int_0^T v(t) i(t) dt \\
&= \inv{T} \int_0^T \textrm{Re} \lr{ V e^{j\omega t} } \Real \lr{ I e^{j\omega t} } dt \\
&= \inv{4 T} \int_0^T
\lr{ V e^{j \omega t} + \bar{V} e^{-j\omega t} }
\lr{ I e^{j \omega t} + \bar{I} e^{-j\omega t} } dt \\
&= \inv{4} \lr{ V \bar{I} + \bar{V} I } \\
&= \inv{2} \textrm{Re} \lr{ V \bar{I} } \\
&= \inv{2} \textrm{Re} \lr{ I Z \bar{I} } \\
&= \inv{2} \Abs{I}^2 R \\
&= \inv{2} \Abs{V}^2 \frac{R}{\Abs{Z}^2}.
\end{aligned}
\end{equation}
The maximum average power is for purely real impedance
\begin{equation}\label{eqn:RLCbandwidth:160}
P = \inv{2 R} \Abs{V}^2,
\end{equation}
so the half power points are when \( \Abs{Z} = \sqrt{2} R \). For the RLC circuit that is when
\begin{equation}\label{eqn:RLCbandwidth:180}
\Abs{R + j \lr{ \omega L – \inv{\omega C} }}^2 = 2 R^2,
\end{equation}
or
\begin{equation}\label{eqn:RLCbandwidth:200}
R^2 = \lr{ \omega L – \inv{\omega C} }^2.
\end{equation}
We seek solutions for
\begin{equation}\label{eqn:RLCbandwidth:220}
\begin{aligned}
R &= \omega L – \inv{\omega C} \\
-R &= \omega L – \inv{\omega C}.
\end{aligned}
\end{equation}
We find the same solutions for either, both leading to
\begin{equation}\label{eqn:RLCbandwidth:240}
B = \Abs{\omega_2 – \omega_1} = \frac{R}{L}.
\end{equation}
This is the (half-power) bandwidth for the RLC circuit. We may now re-express the transfer functions for the filters in terms of the resonant frequency and bandwidth
\begin{equation}\label{eqn:RLCbandwidth:260}
H(s) = \frac{s}{\omega_0^2 + B s + s^2},
\end{equation}
and see by inspection that \( \omega_0 = 1 \) for our circuit (also seen in the plot) and \( B = 3 \,\textrm{rad/s} \).

Transverse electric and magnetic field relations.

August 10, 2025 math and physics play , , , , , , , , , , , , , , ,

[Click here for a PDF version of this post]

I found a sign error in my book. Here’s I’ll re-derive all the results for myself here in a standalone fashion, also verifying signs as I go.

Setup

Suppose that a field is propagating in a medium along the z-axis. We may represent that field as the real part of
\begin{equation}\label{eqn:transverseField:20}
F = F(x,y) e^{j(\omega t – k z)}.
\end{equation}
This is a doubly complex relationship, as we have a scalar complex imaginary \( j \), as well as the spatial imaginary \(I = \Be_1 \Be_2 \Be_3 \) that is part of the multivector field itself
\begin{equation}\label{eqn:transverseField:40}
F = \BE + I \eta \BH.
\end{equation}

Let’s call
\begin{equation}\label{eqn:transverseField:60}
F_z = \lr{ \BE \cdot \Be_3} \Be_3 + I \eta \lr{ \BH \cdot \Be_3 } \Be_3,
\end{equation}
the propagation component of the field and \( F_t = F – F_z \) the transverse component of the field. We can write these in a more symmetric fashion by expanding the dot products and regrouping
\begin{equation}\label{eqn:transverseField:80}
\begin{aligned}
F_z
&= \lr{ \BE \cdot \Be_3} \Be_3 + I \eta \lr{ \BH \cdot \Be_3 } \Be_3 \\
&= \inv{2} \lr{ \BE \Be_3 + \Be_3 \BE } \Be_3 + \frac{I \eta}{2} \lr{ \BH \Be_3 + \Be_3 \BH} \Be_3 \\
&= \inv{2} \lr{ \BE + \Be_3 \BE \Be_3 } + \frac{I \eta}{2} \lr{ \BH + \Be_3 \BH \Be_3} \Be_3 \\
&= \inv{2} \lr{ F + \Be_3 F \Be_3 }.
\end{aligned}
\end{equation}
By subtraction, we also have
\begin{equation}\label{eqn:transverseField:100}
F_t = \inv{2} \lr{ F – \Be_3 F \Be_3 }.
\end{equation}

Relating the transverse and propagation direction fields

The multivector form of Maxwell’s equation, for source free conditions, is
\begin{equation}\label{eqn:transverseField:120}
0 = \lr{ \spacegrad + \inv{c} \partial_t } F.
\end{equation}
We split the gradient into a propagation direction component and a transverse component \( \spacegrad_t \)
\begin{equation}\label{eqn:transverseField:140}
\spacegrad = \spacegrad_t + \Be_3 \partial_z,
\end{equation}
so
\begin{equation}\label{eqn:transverseField:160}
\begin{aligned}
0
&= \lr{ \spacegrad_t + \Be_3 \partial_z + \inv{c} \partial_t } F \\
&= \lr{ \spacegrad_t + \Be_3 \partial_z + \inv{c} \partial_t } F(x,y) e^{j(\omega t – k z) } \\
&= \lr{ \spacegrad_t – j\Be_3 k + j\frac{\omega}{c} } F(x,y) e^{j(\omega t – k z) },
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:transverseField:180}
-j \lr{ \frac{\omega}{c} – k \Be_3 } F = \spacegrad_t F.
\end{equation}

Observe that
\begin{equation}\label{eqn:transverseField:200}
-j \lr{ \frac{\omega}{c} – k \Be_3 } \Be_3 F \Be_3 = -\spacegrad_t \Be_3 F \Be_3,
\end{equation}
which means that
\begin{equation}\label{eqn:transverseField:220}
-j \lr{ \frac{\omega}{c} – k \Be_3 } \inv{2} \lr{ F \pm \Be_3 F \Be_3 } = \spacegrad_t \inv{2} \lr{ F \mp \Be_3 F \Be_3 },
\end{equation}
or
\begin{equation}\label{eqn:transverseField:240}
\begin{aligned}
-j \lr{ \frac{\omega}{c} – k \Be_3 } F_z &= \spacegrad_t F_t \\
-j \lr{ \frac{\omega}{c} – k \Be_3 } F_t &= \spacegrad_t F_z.
\end{aligned}
\end{equation}

Provided \( \omega^2 \ne k^2 c^2 \), this can be inverted, meaning that \( F_t \) fully specifies \( F_z \) if known, as well as the opposite.

That inversion provides the propagation direction field in terms of the transverse
\begin{equation}\label{eqn:transverseField:260a}
F_z = j \frac{ \frac{\omega}{c} + k \Be_3 }{ \omega^2 \mu \epsilon – k^2 } \spacegrad_t F_t,
\end{equation}
and the transverse field in terms of the propagation direction field
\begin{equation}\label{eqn:transverseField:260b}
F_t = j \frac{ \frac{\omega}{c} + k \Be_3 }{ \omega^2 \mu \epsilon – k^2 } \spacegrad_t F_z.
\end{equation}

Transverse field in terms of propagation

Let’s expand \ref{eqn:transverseField:260b} in terms of component electric and magnetic fields. First note that
\begin{equation}\label{eqn:transverseField:280}
\begin{aligned}
\spacegrad_t F_z
&= \spacegrad_t \Be_3 \lr{ E_z + I \eta H_z } \\
&= -\Be_3 \spacegrad_t \lr{ E_z + I \eta H_z }.
\end{aligned}
\end{equation}
so
\begin{equation}\label{eqn:transverseField:300}
F_t = -j \frac{ \frac{\omega}{c} \Be_3 + k }{ \omega^2 \mu \epsilon – k^2 } \spacegrad_t \lr{ E_z + I \eta H_z }.
\end{equation}
This may now be split into electric and magnetic fields, but first note that the multivector operator
\begin{equation}\label{eqn:transverseField:320}
\begin{aligned}
\Be_3 \spacegrad_t
&=
\Be_3 \cdot \spacegrad_t + \Be_3 \wedge \spacegrad_t \\
&=
\Be_3 \wedge \spacegrad_t,
\end{aligned}
\end{equation}
has only a bivector component.

For the transverse electric field component, we have
\begin{equation}\label{eqn:transverseField:340}
\begin{aligned}
\gpgradeone{ \lr{ \frac{\omega}{c} \Be_3 + k } \spacegrad_t \lr{ E_z + I \eta H_z } }
&=
k \spacegrad_t E_z + \frac{\omega}{c} \Be_3 \wedge \spacegrad_t \lr{ I \eta H_z } \\
&=
k \spacegrad_t E_z – \frac{\eta \omega}{c} \Be_3 \cross \spacegrad_t H_z.
\end{aligned}
\end{equation}
and for the magnetic field component
\begin{equation}\label{eqn:transverseField:360}
\begin{aligned}
\gpgradetwo{ \lr{ \frac{\omega}{c} \Be_3 + k } \spacegrad_t \lr{ E_z + I \eta H_z } }
=
\frac{\omega}{c} \Be_3 \wedge \spacegrad_t E_z + I \eta k \spacegrad_t H_z
\end{aligned}
\end{equation}

This means that
\begin{equation}\label{eqn:transverseField:380}
\begin{aligned}
\BE_t &= \frac{j}{\omega^2 \mu \epsilon – k^2 } \lr{ -k \spacegrad_t E_z + \frac{\eta \omega}{c} \Be_3 \cross \spacegrad_t H_z } \\
\eta I \BH_t &= -\frac{j}{\omega^2 \mu \epsilon – k^2 } \lr{ \frac{\omega}{c} \Be_3 \wedge \spacegrad_t E_z + I \eta k \spacegrad_t H_z }
\end{aligned}
\end{equation}

Cancelling out the \( \eta I \) factors in the magnetic field component, and substituting \( \eta/c = \mu, 1/(c\eta) = \epsilon \), leaves us with
\begin{equation}\label{eqn:transverseField:400}
\begin{aligned}
\BE_t &= \frac{j}{\omega^2 \mu \epsilon – k^2 } \lr{ -k \spacegrad_t E_z + \mu \omega \Be_3 \cross \spacegrad_t H_z } \\
\BH_t &= -\frac{j}{\omega^2 \mu \epsilon – k^2 } \lr{ \epsilon \omega \Be_3 \cross \spacegrad_t E_z + k \spacegrad_t H_z }.
\end{aligned}
\end{equation}

Propagation field in terms of transverse.

Now let’s invert \ref{eqn:transverseField:260a}. We seek the grade selections
\begin{equation}\label{eqn:transverseField:420}
\gpgrade{ \lr{ \frac{\omega}{c} + k \Be_3 } \spacegrad_t F_t }{1,2}
\end{equation}

Performing each of these four grade selections in turn, for the \( \spacegrad_t F_t \) products we have
\begin{equation}\label{eqn:transverseField:440}
\begin{aligned}
\gpgradeone{ \spacegrad_t F_t }
&=
\gpgradeone{ \spacegrad_t \lr{ \BE_t + I \eta \BH_t } } \\
&=
\eta \gpgradeone{ I \spacegrad_t \BH_t } \\
&=
\eta I \lr{ \spacegrad_t \wedge \BH_t } \\
&=
-\eta \lr{ \spacegrad_t \cross \BH_t }.
\end{aligned}
\end{equation}
Because \( \spacegrad_t \BE_t \) has only 0,2 grades, so the grade-one selection was zero, leaving us with only \( \BH_t \) dependence.

For the grade two selection of the same, we have
\begin{equation}\label{eqn:transverseField:460}
\begin{aligned}
\gpgradetwo{ \spacegrad_t F_t }
&=
\gpgradetwo{ \spacegrad_t \lr{ \BE_t + I \eta \BH_t } } \\
&=
\spacegrad_t \wedge \BE_t \\
&=
I \lr{ \spacegrad_t \cross \BE_t }.
\end{aligned}
\end{equation}
This time we note that the vector-bivector product \( \spacegrad_t (I \BH_t) \) has only 1,3 grades, and is killed by the grade-2 selection.

For the \( \Be_3 \spacegrad_t F_t \) products, we have
\begin{equation}\label{eqn:transverseField:480}
\begin{aligned}
\gpgradeone{ \Be_3 \spacegrad_t F_t }
&=
\gpgradeone{ \Be_3 \spacegrad_t \lr{ \BE_t + I \eta \BH_t } } \\
&=
\gpgradeone{ \lr{ \Be_3 \cdot \spacegrad_t + \Be_3 \wedge \spacegrad_t } \BE_t }
+
\eta \gpgradeone{ I \Be_3 \lr{ \spacegrad_t \cdot \BH_t + \spacegrad_t \wedge \BH_t } } \\
&=
\gpgradeone{ I \lr{ \Be_3 \cross \spacegrad_t } \BE_t } \\
&=
-\lr{ \Be_3 \cross \spacegrad_t } \cross \BE_t.
\end{aligned}
\end{equation}
Observe that we’ve made use of \( \Be_3 \cdot \spacegrad_t = 0 \), regardless of what it operates on. For the \( \BH_t \) dependence, we had a bivector-scalar product \( (I \Be_3) (\spacegrad_t \cdot \BH_t) \), and a bivector-bivector product \( (I \Be_3) (\spacegrad_t \wedge \BH_t) \), neither of which have any vector grades.

Finally
\begin{equation}\label{eqn:transverseField:500}
\begin{aligned}
\gpgradetwo{ \Be_3 \spacegrad_t F_t }
&=
\eta \gpgradetwo{ I \Be_3 \spacegrad_t \BH_t } \\
&=
-\eta \gpgradetwo{ \lr{\Be_3 \cross \spacegrad_t} \BH_t } \\
&=
-\eta I \lr{\Be_3 \cross \spacegrad_t} \cross \BH_t.
\end{aligned}
\end{equation}
Here we’ve discarded the \( \BE_t \) dependent terms, since the bivector-vector product \( \lr{ \Be_3 \wedge \spacegrad_t } \BE_t \) has only grades 1,3, and we seek grade 2 only.

Putting all the pieces together, noting that \( \eta/c = \mu \) and \( 1/(c \eta) = \epsilon \), we have
we have
\begin{equation}\label{eqn:transverseField:520}
\BE_z = -\frac{j}{\omega^2 \mu \epsilon – k^2 } \lr{ \omega \mu \lr{ \spacegrad_t \cross \BH_t } + k \lr{ \Be_3 \cross \spacegrad_t } \cross \BE_t },
\end{equation}
and
\begin{equation}\label{eqn:transverseField:540}
\BH_z = \frac{j}{\omega^2 \mu \epsilon – k^2 } \lr{ \omega \epsilon \lr{ \spacegrad_t \cross \BE_t } – k \lr{\Be_3 \cross \spacegrad_t} \cross \BH_t }.
\end{equation}

Average power for circuit elements

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

[Click here for a PDF of this post with nicer formatting] or [Click here for my notes compilation for this class]

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:

\begin{equation}\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}
\end{equation}

Here’s a recap of where these come from

Energy lost to resistance

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

the average power lost to a resistor is

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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}
\end{equation}

Energy stored in an inductor

The inductor energy is found the same way, with

\begin{equation}\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}
\end{equation}

which leads to

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

Energy lost due to conductance

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

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

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

\begin{equation}\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}
\end{equation}

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

February 3, 2016 ece1236 , , , , , , , , , , , , , , , , , , , , , , , ,

[Click here for a PDF of this post with nicer formatting and figures] or [Click here for my notes compilation for this class]

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

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

or in the \( \Delta z \rightarrow 0 \) limit

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

The KCL equation at the interior node is

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

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

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

\begin{equation}\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}
}
\end{equation}

The second derivatives are

\begin{equation}\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}
\end{equation}

which allow the \( V, I \) to be decoupled
\begin{equation}\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}
}
\end{equation}

With a complex propagation constant

\begin{equation}\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}
\end{equation}

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

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

We write the solutions to these equations as

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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}
\end{equation}

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

Introducing the characteristic impedance \( Z_0 \) of the line

\begin{equation}\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}
\end{equation}

we have

\begin{equation}\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}
\end{equation}

where

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

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

\begin{equation}\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}
\end{equation}

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

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

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

\begin{equation}\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}
\end{equation}

Therefore the capacitance per unit length is

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

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

\begin{equation}\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}
\end{equation}

The flux is

\begin{equation}\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}
\end{equation}

The inductance per unit length is

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

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

\begin{equation}\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}
\end{equation}

This gives

\begin{equation}\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}
\end{equation}

So

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

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

\begin{equation}\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}
\end{equation}

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

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

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

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

the characteristic impedance is

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

and the signals are
\begin{equation}\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}
\end{equation}

In the time domain for an infinite line, we have

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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}
\end{equation}

so
\begin{equation}\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}
\end{equation}

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

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

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

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

so
\begin{equation}\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}
\end{equation}

The phase factor can be written

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

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

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

When that is done we have
\begin{equation}\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}
\end{equation}

We have

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

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

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

Terminated lossless line.

Consider the load configuration sketched in fig. 9.

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

Recall that

\begin{equation}\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}
\end{equation}

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

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

So

\begin{equation}\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}
\end{equation}

where

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

is the reflection coefficient at the load.

The phasors for the signals take the form

\begin{equation}\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}
\end{equation}

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

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

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

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

Power

The average (time) power on the line is

\begin{equation}\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}
\end{equation}

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

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

where

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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}
\end{equation}

Insertion loss (dB) is defined as

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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}
\end{equation}

The absolute voltage at this point is
\begin{equation}\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}
\end{equation}

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

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

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

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

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

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

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

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

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 \( d_1 \)

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

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

    For \( k = 1 \), let the max occur at \( d_2 \)

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

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

    The difference is

    \begin{equation}\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}
    \end{equation}

    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

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

so at the load and input we have

\begin{equation}\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}
\end{equation}

where

\begin{equation}\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}
\end{equation}

Similarly

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

Define an input impedance as
\begin{equation}\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}
\end{equation}

This is analogous to

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

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

\begin{equation}\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}
\end{equation}

or
\begin{equation}\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 ) }.
}
\end{equation}

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

May 6, 2015 math and physics play , , , ,

[Click here for a PDF of this post with nicer formatting]

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.

Answer

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

\begin{equation}\label{eqn:ExPlaneWave:40}
\spacegrad \cross \BE = -j \omega \BB
\end{equation}
\begin{equation}\label{eqn:ExPlaneWave:60}
\spacegrad \cross \BH = j \omega \BD
\end{equation}
\begin{equation}\label{eqn:ExPlaneWave:80}
\spacegrad \cdot \BD = 0
\end{equation}
\begin{equation}\label{eqn:ExPlaneWave:100}
\spacegrad \cdot \BB = 0.
\end{equation}

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

\begin{equation}\label{eqn:ExPlaneWave:120}
-j \omega \BB
= \spacegrad \cross \BE
=
\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),
\end{equation}

or

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

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

\begin{equation}\label{eqn:ExPlaneWave:160}
j \omega \epsilon \xcap E
=
\inv{\mu} \spacegrad \cross \BB
=
-\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
}
\end{equation}

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

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

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

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

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

\begin{equation}\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} },
\end{equation}

or

\begin{equation}\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} }.
\end{equation}

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.