Class 3

Phasors, γ and the characteristic impedance

Class 2 left a wave equation with two time derivatives in it. Class 3 removes time from the problem, and what falls out is the propagation constant γ, the characteristic impedance Z₀, and a loss figure in decibels.

2026-09-03Slides (PDF)phasorspropagation constantcharacteristic impedancedecibels

Class 2 ended with a wave equation and a solution shaped like $f_1(t - z/v)$ , which allows any waveform at all. That freedom costs something. Every manipulation from here carries two variables, $z$ and $t$ , and second derivatives in both.

Class 3 buys the freedom back by giving up generality. Restrict the source to a single frequency, and the time dependence of every voltage and current on the line becomes the same $\cos(\omega t + \text{something})$ . Nothing is left to solve for in time. What survives is a function of $z$ alone, and the wave equation collapses from a partial differential equation into an ordinary one.

That restriction is the phasor method. This page assumes you have never used it.

The slide deck for this class runs to two slides, and both are algebra: slide 1 phasors the telegrapher equations and names $Z$ and $Y$ , slide 2 sets the two wave solutions against each other and reaches the characteristic impedance. The board went further, through the conversion examples, power and decibels, so this page says which source each part came from.

01Complex numbers, just enough

Three facts carry the whole method. If you already have them, skip to the next section.

A complex number is a pair of real numbers written as one object, $x + jy$ , where $j$ satisfies $j^2 = -1$ . Engineering writes $j$ where mathematics writes $i$ , because $i$ was taken by current. Read $x$ as the real part and $y$ as the imaginary part, and picture the number as a point in a plane with $x$ across and $y$ up.

The same point has a length and an angle. The length is $|x + jy| = \sqrt{x^2 + y^2}$ , called the magnitude. The angle measured anticlockwise from the positive real axis is $\theta = \arctan(y/x)$ , called the phase. Euler's formula connects the two descriptions:

e^{j\theta} = \cos\theta + j\sin\theta

so any complex number can be written $Ae^{j\theta}$ with $A$ its magnitude and $\theta$ its phase. That form matters because multiplying two of them multiplies the magnitudes and adds the phases, which turns trigonometry into arithmetic.

The third fact is the operation $\mathrm{Re}[\cdot]$ , which throws away the imaginary part and keeps the real one. From Euler,

\mathrm{Re}\!\left[Ae^{j\theta}\right] = A\cos\theta

02What a phasor is

Take a voltage that varies as a cosine at one frequency, at a fixed point on the line:

V(z,t) = |V_0|\cos(\omega t + \beta z)

The board writes this out and then strips it. Using the key result above with $\phi = \beta z$ :

V(z,t) = \mathrm{Re}\!\left[|V_0| e^{j(\omega t + \beta z)}\right] = \mathrm{Re}\!\left[\underbrace{|V_0| e^{j\beta z}}_{\text{depends on } z}\ \cdot\ \underbrace{e^{j\omega t}}_{\text{depends on } t}\right]

The factor $e^{j\omega t}$ carries every bit of the time dependence, and it is identical for the voltage, the current, and every other quantity in the problem, because they all oscillate at the same frequency. Agree once to carry it silently, cross it off, and what remains is the phasor:

\boxed{\ V_s(z) = |V_0|e^{j\beta z}\ }

The board where the time factor gets crossed off, with the conversion examples that follow it.

03Why removing time is worth anything

A time derivative of the real signal turns into a multiplication in phasor land. Differentiate $V_s e^{j\omega t}$ with respect to $t$ , treating $V_s$ as the constant in $t$ that it is:

\frac{\partial}{\partial t}\left[V_s\,e^{j\omega t}\right] = j\omega\, V_s\, e^{j\omega t}

The exponential survives untouched and picks up a factor $j\omega$ . Do it twice and the factor is $(j\omega)^2 = -\omega^2$ .

04The telegrapher equations in phasor form

Slide 1 does not start from Class 2's second-order wave equation. It goes back a step, to the two first-order telegrapher equations, and phasors those directly. The route is shorter, and it is the one to reproduce under exam conditions.

\frac{\partial V(z,t)}{\partial z} = -\left(R\,I(z,t) + L\,\frac{\partial I(z,t)}{\partial t}\right), \qquad \frac{\partial I(z,t)}{\partial z} = -\left(G\,V(z,t) + C\,\frac{\partial V(z,t)}{\partial t}\right)

These are Class 2's equations (1) and (2), with the minus sign pulled outside the bracket. Write $V(z,t) = V_s(z)e^{j\omega t}$ and $I(z,t) = I_s(z)e^{j\omega t}$ and substitute. Slide 1 keeps every $e^{j\omega t}$ on the page, which is worth copying while the method is new:

\frac{\partial}{\partial z}\left(V_s e^{j\omega t}\right) = -\left(R\,I_s e^{j\omega t} + L\,j\omega I_s e^{j\omega t}\right) \quad\Longrightarrow\quad \frac{dV_s}{dz} = -\left(R + j\omega L\right)I_s

\frac{\partial}{\partial z}\left(I_s e^{j\omega t}\right) = -\left(G\,V_s e^{j\omega t} + C\,j\omega V_s e^{j\omega t}\right) \quad\Longrightarrow\quad \frac{dI_s}{dz} = -\left(G + j\omega C\right)V_s

Two things happen on each line. The time derivative hits $I_s e^{j\omega t}$ and returns $j\omega I_s e^{j\omega t}$ , which is the replacement rule doing its one job. Then every surviving term carries the same $e^{j\omega t}$ , so it divides out and the derivative in $z$ turns from $\partial$ into $d$ .

Slide 1 names the two brackets:

Z \equiv R + j\omega L \quad [\Omega/\text{m}], \qquad Y \equiv G + j\omega C \quad [\text{S}/\text{m}]

$Z$ is the series impedance of one metre of line: the conductor resistance and the inductive reactance $\omega L$ , added as complex numbers because they act a quarter cycle apart. $Y$ is the shunt admittance of one metre: the leakage conductance and the capacitive susceptance $\omega C$ . Both stay per metre, the same bookkeeping Class 2 used for $R$ , $L$ , $G$ and $C$ .

05Two first-order equations become one

The pair is still coupled, so eliminate one unknown exactly as Class 2 did. Differentiate the voltage equation with respect to $z$ :

\frac{d^2 V_s}{dz^2} = -Z\,\frac{dI_s}{dz}

The second equation says what $dI_s/dz$ is. Substitute it:

\frac{d^2 V_s}{dz^2} = -Z\left(-Y\,V_s\right) = ZY\,V_s

Three lines, and the elimination that took half of Class 2 is finished.

\boxed{\ \frac{d^2 V_s}{dz^2} = \gamma^2 V_s\ } \qquad\text{with}\qquad \gamma^2 = ZY

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

The board reached $\gamma$ by a different road, starting from Class 2's equation (3) rather than from the first-order pair. It arrives at the same place, and following it is worthwhile once, because it shows where the factorisation comes from.

The equation now says that differentiating $V_s$ twice gives back $V_s$ times a constant, and the functions that do that are exponentials.

06The solution, and what α and β each do

An equation reading $d^2V_s/dz^2 = \gamma^2 V_s$ has two independent exponential solutions, $e^{-\gamma z}$ and $e^{+\gamma z}$ , and the general solution is any combination of them:

V_s(z) = V_0^+ e^{-\gamma z} + V_0^{-} e^{+\gamma z}

$V_0^+$ and $V_0^-$ are complex constants fixed by whatever sits at the two ends of the line. Class 2 called the same pair $f_1$ and $f_2$ . The exponentials are what those arbitrary shapes become once the source is restricted to one frequency.

Figure 1 The two waves on one length of line. The forward wave shrinks as

z

grows; the backward wave was launched at the far end, so it is largest there and shrinks as it travels the other way. Hover either arrow to light its term in the equations.

Now split $\gamma$ into its real and imaginary parts, which is the step the whole class was heading for:

\gamma = \alpha + j\beta

Then $e^{-\gamma z} = e^{-(\alpha + j\beta)z} = e^{-\alpha z}\,e^{-j\beta z}$ , and the general solution becomes

\boxed{\ V_s(z) = \class{hl hl-fw}{V_0^{+} e^{-\alpha z} e^{-j\beta z}} + \class{hl hl-bw}{V_0^{-} e^{+\alpha z} e^{+j\beta z}}\ }

The two factors do different jobs, and separating them is the point of writing $\gamma$ this way.

$e^{-\alpha z}$ is a real number between 0 and 1. It shrinks the amplitude as the wave moves along, and it carries no phase. That is loss.

$e^{-j\beta z}$ is a complex number of magnitude exactly 1. It changes the phase and leaves the size alone. That is propagation.

Put the time factor back to see the wave move. Multiply by $e^{j\omega t}$ and take the real part, using $\mathrm{Re}[Ae^{j\phi}] = A\cos\phi$ on each term:

V(z,t) = |V_0^{+}|\,e^{-\alpha z}\cos(\omega t - \beta z) + |V_0^{-}|\,e^{\alpha z}\cos(\omega t + \beta z)

The first term is a wave heading toward $+z$ , fading as it goes. The second heads toward $-z$ . Sign of the $\beta z$ term names the direction, and Class 2's argument still applies: a feature at fixed $\omega t - \beta z$ has to move to larger $z$ as $t$ grows.

07Turning a phasor back into a wave

The board works four conversions. Each one follows the same recipe: multiply by $e^{j\omega t}$ , collect every real exponent into an amplitude, collect every imaginary exponent into the cosine's argument, and read off the answer.

Worked example 1 Pure phase, no loss

\mathrm{Re}\!\left[5e^{-j7z}e^{j\omega t}\right] = 5\cos(\omega t - 7z)

Nothing multiplies the amplitude, so

\alpha = 0

and the wave keeps its size forever. Here

\beta = 7

rad/m, so the wavelength is

2\pi/\beta = 0.898

m. The minus sign in front of

7z

says the wave runs toward

+z

Worked example 2 Loss and phase together

\mathrm{Re}\!\left[5e^{-2z}e^{-j5z}e^{j\omega t}\right] = 5e^{-2z}\cos(\omega t - 5z)

The real exponent

-2z

carries no

j

, so it never enters the cosine. It stays outside as an amplitude, giving

\alpha = 2

Np/m. The wave loses half its amplitude every

\ln 2/2 = 0.347

m. With

\beta = 5

rad/m the wavelength is

1.26

Worked example 3 A constant phase term

5e^{(-3z\;+\;j7z\;+\;j10)} \ \longrightarrow\ 5e^{-3z}\cos(\omega t + 7z + 10)

Three exponents, sorted by whether they carry a

j

. The

-3z

has none, so it becomes the envelope with

\alpha = 3

Np/m. The

+j7z

gives

\beta = 7

rad/m and a plus sign, so this wave runs toward

-z

. The

+j10

has no

z

in it at all, so it is a fixed phase offset of 10 radians, which is

573.0

° and lands at

147.0

° once wrapped into one turn. It shifts the whole wave sideways and changes nothing else.

Worked example 4 No β at all

5e^{-5z} \ \longrightarrow\ 5e^{-5z}\cos(\omega t)

Every exponent is real, so

\beta = 0

. Nothing on the line is ever out of phase with anything else, and the whole line rises and falls together while its amplitude decays with distance. No wave travels. This is what a lumped circuit looks like in this notation, and Class 1's criterion is the statement that

\beta \ell

stays small.

08The characteristic impedance

Solving the second-order equation gave the voltage. It says nothing yet about the current, and the two are not independent: they still have to satisfy the first-order pair. Slide 2 is that reckoning.

Start from both wave solutions, each with a forward and a backward term:

V_s(z) = V_0^{+}e^{-\gamma z} + V_0^{-}e^{+\gamma z}, \qquad I_s(z) = I_0^{+}e^{-\gamma z} + I_0^{-}e^{+\gamma z}

Differentiate $V_s$ , then set it against $dV_s/dz = -Z I_s$ with the current solution substituted:

-\gamma V_0^{+}e^{-\gamma z} + \gamma V_0^{-}e^{+\gamma z} = -Z\left(I_0^{+}e^{-\gamma z} + I_0^{-}e^{+\gamma z}\right)

Slide 2 stops on that line. Finish it by comparing coefficients. The functions $e^{-\gamma z}$ and $e^{+\gamma z}$ are independent: no multiple of one equals the other at every $z$ . So an equation that holds for all $z$ needs the $e^{-\gamma z}$ parts to balance by themselves, and the $e^{+\gamma z}$ parts likewise.

Forward terms, cancelling $e^{-\gamma z}$ :

-\gamma V_0^{+} = -Z\,I_0^{+} \qquad\Longrightarrow\qquad \frac{V_0^{+}}{I_0^{+}} = \frac{Z}{\gamma}

Backward terms, cancelling $e^{+\gamma z}$ :

+\gamma V_0^{-} = -Z\,I_0^{-} \qquad\Longrightarrow\qquad \frac{V_0^{-}}{I_0^{-}} = -\frac{Z}{\gamma}

One quantity appears twice, once with each sign. Name it:

Z_0 = \frac{Z}{\gamma} = \frac{Z}{\sqrt{ZY}} = \sqrt{\frac{Z}{Y}}

The board reached the same Z₀ from the forward wave alone.

The two coefficient equations now read $I_0^{+} = V_0^{+}/Z_0$ and $I_0^{-} = -V_0^{-}/Z_0$ . That negative is the whole content of the backward wave's sign, and it was earned rather than assumed. Put both back into the current solution and write it in terms of the voltage amplitudes:

I_s(z) = \frac{1}{Z_0}\left[V_0^{+}e^{-\gamma z} - V_0^{-}e^{+\gamma z}\right]

This is Class 2's minus sign again, in its permanent form. Voltage adds its two waves and current subtracts them, because a backward wave pushes charge the other way along the line.

09The lossless line, and the 377 Ω that is not yours

Set $R = 0$ and $G = 0$ , the perfect conductor and the perfect insulator from Class 2. The general expressions simplify hard:

\gamma = \sqrt{(j\omega L)(j\omega C)} = j\omega\sqrt{LC}, \qquad \alpha = 0, \qquad \beta = \omega\sqrt{LC}

Z_0 = \sqrt{\frac{j\omega L}{j\omega C}} = \sqrt{\frac{L}{C}}

The $j\omega$ cancels top and bottom, so $Z_0$ comes out real, frequency independent, and equal to the quantity Class 2 met as the factor $1/(Lv)$ in its equation (8). For the worked coax that is $50$ Ω.

Worked example 5 A lossy coax at 1 MHz

Take the Class 2 cable, $L = 250$ nH/m and $C = 100$ pF/m, and add realistic loss: $R = 0.50$ Ω/m and $G = 1\times 10^{-6}$ S/m. At $f = 1$ MHz, $\omega = 2\pi f$ , so

Z = R + j\omega L, \qquad Y = G + j\omega C

Working through $\gamma = \sqrt{ZY}$ and $Z_0 = \sqrt{Z/Y}$ gives

quantity	lossless	with loss
$\alpha$ (Np/m)	0	0.00496
loss (dB/m)	0	0.0431
$\beta$ (rad/m)	0.0314	0.0318
$\lambda$ (m)	200	198
$v$ (m/s)	2.00×10⁸	1.98×10⁸
$\lvert Z_0\rvert$ (Ω)	50.0	51.2
$\theta$ (°)	0	-8.78

The losses are small, so the answers sit close to the lossless ones. $Z_0$ moves from 50 Ω to $51.2\,\angle\,-8.78°$ Ω, and $\alpha$ climbs off zero to 0.0431 dB per metre. Treating this cable as lossless costs little over a metre and a great deal over a kilometre.

10Power, and why it decays twice as fast

The board starts from DC, where everything is real:

V = IR, \qquad P = VI = I^2R = \frac{V^2}{R}

At one frequency the same product needs care, because $V$ and $I$ are cosines that peak at different moments. Averaging $v(t)i(t)$ over a cycle gives the result the board writes down:

P_{\text{av}} = \tfrac{1}{2}\,\mathrm{Re}\!\left[V_s\,I_s^{*}\right]

Average power, its exponential decay, and the conversion to decibels.

Two features of that expression matter. The factor $\tfrac12$ converts a peak amplitude into an average, the same $\tfrac12$ that makes an RMS value $A/\sqrt2$ . The conjugate star on $I_s$ subtracts the current's phase instead of adding it, which is what leaves the average and drops the part of the product that oscillates at $2\omega$ and averages to nothing.

Apply it to a forward wave. Write $V_s = V_0^{+}e^{-\alpha z}e^{-j\beta z}$ and use $I_s = V_s/Z_0$ , so $I_s = (V_0^{+}/Z_0)e^{-\alpha z}e^{-j\beta z}$ . Conjugating flips the sign of every imaginary exponent:

V_s I_s^{*} = \frac{\lvert V_0^{+}\rvert^2}{Z_0^{*}}\,\underbrace{e^{-\alpha z}e^{-\alpha z}}_{e^{-2\alpha z}}\,\underbrace{e^{-j\beta z}e^{+j\beta z}}_{1}

Writing $Z_0 = \lvert Z_0\rvert e^{j\theta}$ and taking the real part gives the board's line in its current form, using $\lvert V_0^{+}\rvert = \lvert I_0^{+}\rvert \lvert Z_0\rvert$ :

\langle P\rangle = \tfrac{1}{2}\,\lvert I_0^{+}\rvert^{2}\,\lvert Z_0\rvert\, e^{-2\alpha z}\cos\theta

$\cos\theta$ is the power factor. On a lossless line $\theta = 0$ , so $\cos\theta = 1$ and the expression reduces to $\tfrac12 I^2 R$ with $Z_0$ playing the part of $R$ .

Figure 2 Power launched at

z = 0

and what is left at

z

. Nothing reflects here, so the only thing happening is decay.

Since only $e^{-2\alpha z}$ depends on position, the power at any two points is related by that factor alone, which is Figure 2:

\boxed{\ P(z) = P(0)\,e^{-2\alpha z}\ }

11Decibels

Engineers report loss as a ratio in decibels rather than as a neper count. The definition for power is

\text{loss in dB} = 10\log_{10}\!\frac{P(z)}{P(0)}

Substitute the decay and use $\log_{10}(e^{x}) = x\log_{10}e$ :

10\log_{10}\!\left(e^{-2\alpha z}\right) = -20\,\alpha z\,\log_{10}e = -8.686\,\alpha z

Worked example 6 The board's cable

A line carries $P(0) = 1$ mW and has $\alpha = 0.0100$ Np/m. How much arrives 100 m along?

P(z) = P(0)e^{-2\alpha z} = (0.001)\,e^{-2(0.0100)(100)} = 1.35\times 10^{-4}\unit{W}

which is 0.135 mW, about 14% of what went in. In decibels,

10\log_{10}\frac{P(z)}{P(0)} = -8.69(0.0100)(100) = -8.69\unit{dB}

Check the two against each other: $10^{-8.69/10} = 0.135$ , matching the ratio $e^{-2\alpha z} = 0.135$ . The amplitude, which falls as $e^{-\alpha z}$ rather than its square, is down to 0.368 of what it was.

12Traps

13Check yourself

1Start from the two first-order phasor equations and reach

\gamma^2 = ZY

in three lines.

Differentiate the first: $d^2V_s/dz^2 = -Z\,dI_s/dz$ . The second says $dI_s/dz = -Y V_s$ . Substitute: $d^2V_s/dz^2 = -Z(-Y V_s) = ZY V_s$ . Comparing with $d^2V_s/dz^2 = \gamma^2 V_s$ gives $\gamma^2 = ZY$ , so $\gamma = \sqrt{ZY}$ . This is slide 1 into slide 2, and it is the version to write in an exam.

2Slide 2 writes the current solution with a plus between its terms. What does that force

I_0^{-}

to be, and why?

Matching the $e^{+\gamma z}$ coefficients in $dV_s/dz = -ZI_s$ gives $+\gamma V_0^{-} = -Z I_0^{-}$ , so $I_0^{-} = -V_0^{-}/Z_0$ . Writing the sum with a plus means the backward wave's direction reversal has to be carried inside the constant. Write the sum with a minus instead and $I_0^{-} = +V_0^{-}/Z_0$ ; the physical current is the same either way.

3Why can the

e^{-\gamma z}

and

e^{+\gamma z}

coefficients be matched separately?

Because the two exponentials are independent functions of $z$ . No constant multiple of one equals the other at every point, so the only way a combination of them can vanish for all $z$ is for each coefficient to vanish on its own. It is the same argument that lets you match powers of $x$ in a polynomial identity.

4Why does restricting the source to one frequency remove time from the problem?

Every voltage and current then varies as $\cos(\omega t + \phi)$ with the same $\omega$ . Writing each as $\mathrm{Re}[V_se^{j\omega t}]$ puts the identical factor $e^{j\omega t}$ on every term of every equation, so it divides out. What is left is the amplitude and the phase, held together in one complex number.

5Convert

V_s(z) = 12e^{-0.4z}e^{-j3z}

to the time domain, and give α, β and the wavelength.

$V(z,t) = 12e^{-0.4z}\cos(\omega t - 3z)$ . The real exponent gives $\alpha = 0.4$ Np/m and the imaginary one gives $\beta = 3$ rad/m, so $\lambda = 2\pi/3 = 2.09$ m. The minus sign on $3z$ sends it toward $+z$ .

6A wave has α = 0.05 Np/m. How far before its amplitude halves, and what is that in dB?

Amplitude follows $e^{-\alpha z}$ , so halving needs $z = \ln 2/\alpha = 13.9$ m. Halving the amplitude quarters the power, which is $10\log_{10}(0.25) = -6.02$ dB.

7Why does the power expression have no

\beta

in it?

The conjugate on $I_s$ flips the sign of its phase, so $e^{-j\beta z}$ meets $e^{+j\beta z}$ and the pair multiplies to 1. Power depends on the size of the wave and never on where it sits in its cycle.

8A lossless line has L = 400 nH/m and C = 160 pF/m. Give Z₀, β at 100 MHz, and λ.

$Z_0 = \sqrt{L/C} = 50$ Ω. With $v = 1/\sqrt{LC} = 1.25\times 10^{8}$ m/s, $\beta = \omega/v = 5.03$ rad/m and $\lambda = v/f = 1.25$ m.

9A cable is quoted at 0.26 dB/m. What is α in Np/m?

Divide by 8.686: $\alpha = 0.26/8.686 = 0.0299$ Np/m. Over 20 m the power ratio is $e^{-2\alpha z} = 0.302$ , a drop of -5.20 dB.

10On the lossy coax in the worked example, Z₀ came out at an angle of about -8.8°. What does a negative angle mean physically?

$Z_0 = \lvert Z_0\rvert e^{j\theta}$ with $\theta < 0$ puts the current ahead of the voltage, so the line looks slightly capacitive to a travelling wave. It also drags the power factor $\cos\theta$ below 1, to $0.98827$ here, which costs a little of the power the same voltage and current would carry on a lossless line.

11Why can a phasor never describe a square pulse?

A phasor holds one amplitude and one phase at one frequency. A square pulse contains a whole spectrum of frequencies, each needing its own phasor. Class 2's $f_1(t - z/v)$ handles any shape; the phasor form trades that away for algebra.

Answer out loud before opening one.

14Cheat sheet

Cheat sheet

quantity	expression	notes
phasor of a cosine	$A\cos(\omega t+\phi) = \mathrm{Re}\!\left[Ae^{j\phi}e^{j\omega t}\right]$	phasor is $Ae^{j\phi}$
time derivative	$\partial/\partial t \to j\omega$	twice gives $-\omega^2$
series impedance	$Z \equiv R + j\omega L$	Ω/m, slide 1
shunt admittance	$Y \equiv G + j\omega C$	S/m, slide 1
phasor telegrapher pair	$\dfrac{dV_s}{dz} = -Z I_s$ , $\dfrac{dI_s}{dz} = -Y V_s$	slide 1, start here
phasor wave equation	$\dfrac{d^2V_s}{dz^2} = \gamma^2 V_s$	differentiate one, substitute the other
propagation constant	$\gamma = \sqrt{ZY} = \sqrt{(R+j\omega L)(G+j\omega C)} = \alpha + j\beta$	$\alpha$ in Np/m, $\beta$ in rad/m
voltage on the line	$V_s = V_0^{+}e^{-\alpha z}e^{-j\beta z} + V_0^{-}e^{\alpha z}e^{j\beta z}$	forward plus backward
time domain	$V = \lvert V_0^{+}\rvert e^{-\alpha z}\cos(\omega t-\beta z) + \lvert V_0^{-}\rvert e^{\alpha z}\cos(\omega t+\beta z)$	sign of $\beta z$ names the direction
current, slide form	$I_s = I_0^{+}e^{-\gamma z} + I_0^{-}e^{+\gamma z}$	plus sign; $I_0^{-} = -V_0^{-}/Z_0$
current, voltage form	$I_s = \dfrac{1}{Z_0}\left[V_0^{+}e^{-\gamma z} - V_0^{-}e^{\gamma z}\right]$	note the minus
characteristic impedance	$Z_0 = \dfrac{Z}{\gamma} = \sqrt{\dfrac{Z}{Y}} = \lvert Z_0\rvert e^{j\theta}$	Ω, complex
lossless line	$\alpha = 0$ , $\beta = \omega\sqrt{LC}$ , $Z_0 = \sqrt{L/C}$	set $R = G = 0$
wavelength and speed	$\lambda = 2\pi/\beta$ , $v = \omega/\beta$
average power	$\langle P\rangle = \tfrac12\mathrm{Re}\!\left[V_sI_s^{*}\right] = \tfrac12\lvert I_0^{+}\rvert^2\lvert Z_0\rvert e^{-2\alpha z}\cos\theta$	$\cos\theta$ is the power factor
power decay	$P(z) = P(0)e^{-2\alpha z}$	the 2 comes from squaring
decibels	$10\log_{10}\left(P(z)/P(0)\right) = -8.69\,\alpha z$	1 Np = 8.686 dB
free space	$\eta_0 = \sqrt{\mu_0/\varepsilon_0} = 376.7$ Ω	fields, not cables

Constants: $\eta_0 = 376.7$ Ω, $8.686$ dB per neper. Class 2 coax: 250 nH/m and 100 pF/m give 2.00×10⁸ m/s and 50 Ω. Reading: Hayt chapter 10, Ulaby chapters 1 and 2.

Symbols and notation

press G to toggle

symbol	say it	what it means	units
$j$	"jay"	the imaginary unit, $j^2 = -1$	none
$\equiv$	"is defined as"	a name being introduced, not a result derived
$\omega$	"omega"	angular frequency, $\omega = 2\pi f$	rad/s
$V(z,t)$	"vee of zee tee"	the real voltage an oscilloscope reads	V
$V_s(z)$	"vee sub ess"	the phasor: a complex number holding amplitude and phase	V
$\mathrm{Re}[\ ]$	"real part of"	keep the real part, discard the imaginary one
$\gamma$	"gamma"	propagation constant, $\gamma = \alpha + j\beta$	1/m
$\alpha$	"alpha"	attenuation constant, how fast the amplitude dies	Np/m
$\beta$	"beta"	phase constant, radians of phase gained per metre	rad/m
$Z$	"zee"	series impedance of one metre, $R + j\omega L$	Ω/m
$Y$	"why"	shunt admittance of one metre, $G + j\omega C$	S/m
$Z_0$	"zee naught"	characteristic impedance, $V_0^+/I_0^+$	Ω
$\theta$	"theta"	the phase angle of $Z_0$	rad or °
$V_0^+$	"vee naught plus"	complex amplitude of the forward wave at $z = 0$	V
$V_0^-$	"vee naught minus"	complex amplitude of the backward wave at $z = 0$	V
$( \ )^*$	"star" or "conjugate"	flip the sign of the imaginary part
Np	"neper"	unit of $\alpha$ : one neper of decay divides amplitude by $e$
dB	"decibel"	$10\log_{10}$ of a power ratio