# The Expression for Gradient in Spherical Coordinates

The gradient of a scalar field $V$, written $\nabla V$, has the property that $dV = \nabla V \cdot\,d\bar{P}$ where $\bar{P}$ is a point in the domain of $V$ and $dV$ is the differential change produced by the differential change in the argument, $d\bar{P}$.

If we take as the definition of $\nabla V$ to be the vector field such that the above dot formula holds, then the expressions for $\nabla V$ in cylindrical & spherical coordinates become obvious.

Let’s derive the expression in spherical coordinates:
$$d\bar{P} = (dr)\hat{r} + (r\,d\phi)\hat{\phi} + (r\sin\theta\,d\theta)\hat{\theta}$$
You can see this by drawing a differential block in spherical coordinates and marking up the edges with differential lengths.  $d\bar{P}$ will go diagonally through the differential block.

Using the definition $dV = \frac{\partial V}{\partial r}\,dr + \frac{\partial V}{\partial \theta}\,d\theta + \frac{\partial V}{\partial \phi}\,d\phi$, the obvious choice for the gradient of $V$ is
$$\nabla V = (\frac{\partial V}{\partial r}) \hat{r} + (\frac{1}{r}\frac{\partial V}{\partial \theta})\hat{\theta} + (\frac{1}{r\sin\theta}\frac{\partial V}{\partial \phi})\hat{\phi}$$
(in order to cancel out the coefficients in $d\bar{P}$ when dotting).

# Tip on Integration-by-parts

See the Wikipedia article on integration-by-parts.  Apparently it can be derived by integrating both sides of the product rule.  Now, you can do things the hard way and by trial-and-error choose different options for $u$ and $v$ or you could just apply the product rule to the integrand, integrate both sides and solve for your original antiderivative.  Example:
$$\int re^{ar}\,dr$$
Doing it the hard way, first try guessing a factorization of the integrand (choosing a $u$ and a $v$ function such that the integrand is $u\,dv/dx$).  You may get lucky or can easily guess from experience.  But if you have my luck, you’ll fail.  The easier way is to differentiate:
$$\frac{d}{dx}(re^{ar}) = are^{ar} + e^{ar}$$
and then integrate (apply the antiderivative operator to both sides):
$$re^{ar} = a\int re^{ar}\,dr + \int e^{ar}\,dr$$
and solve algebraically for our original antiderivative:
$$\int re^{ar}\,dr = \frac{1}{a}[re^{ar} - \int e^{ar}\,dr]$$
the integral on the RHS being easier to compute (u,du substitution or from memory).
The moral of the story: don’t use the formal description of integration-by-parts and trial-and-error search, just use the method described in the derivation of integration-by-parts, and $u$ and $v$ are chosen for you automatically.

(Note: in the above, “antiderivative” = “integral”)

# The Signs of Gamma in TG’s Equations

The transmission line parameters: $R, G, C, L$ are all positive or zero, clearly.  The propagation constant $\gamma$ is defined as the positive square root of $ZY$ (since there are two $z\in\mathbb{C}-\{0\}$ such that $z^2 = a$).  This can be done because the context in which $\gamma$ is used does not ask for both solutions and if it does, it simply uses $-\gamma$ which is the other solution.
$$\gamma := \sqrt{ZY}$$
where
$$Z := R + jL \\ Y := G + jC$$
are the impedance and admittance per unit length, resp.  Let’s take a look at the components of $\gamma$.
$$ZY = (RG – \omega^2 LC) + \omega(RC + GL)j = \gamma^2 = (\alpha^2 – \beta^2) + (2\alpha\beta)j$$

The following table summarizes the different cases of line parameters $R, G, L, C$ and what result they have on $\gamma$’s components $\alpha, \beta$.  In the table, “$R = 0$” will mean all other parameters are nonzero.

 $R,G,C,L > 0$ $\alpha\beta > 0$ $\alpha, \beta>0$ only one of $R,G,C,L$ is $0$ $\alpha\beta > 0$ $\alpha, \beta > 0$ $R = G = 0$ $\gamma = j \omega \sqrt{LC}$ $\alpha = 0, \beta > 0$ $R = L = 0$ xor $G=C= 0$ $\gamma^2 = 0$ $\alpha = \beta = 0$ $R = C = 0$ $\gamma = \sqrt{GL\omega}\frac{1}{\sqrt{2}}(1 + j)$ $\alpha = 0, \beta > 0$ $G=L=0$ $\gamma = \sqrt{RC\omega}\frac{1}{\sqrt{2}}(1 + j)$ $\alpha = 0, \beta> 0$ $C = L=0$ $\gamma = \sqrt{RC}$ $\alpha > 0$, $\beta= 0$ exactly 3 of $R,G,C,L$ are zero $\gamma^2 = 0$ $\alpha = \beta = 0$

The table makes it clear that $\gamma$ can be chosen such that its components are $\geq 0$.  A more succinct way of showing this would probably be showing the above is the case for $\gamma^2 = x + jy$ with $y \geq 0$.

# Transmission-Line Equivalent Circuit Example

An equivalent circuit is a simpler model of a circuit such that the measurable (or algebraically derivable) quantities of interest (a voltage between two points say) is the same.  Below is an example of an equivalent circuit.  According to Schaum’s Outline of Electromagnetics, Third Edition, this can be used to find the phasor coefficient $V_0^+$ from which you can then construct the voltage and current functions of the transmission line.

# A “Calc One” Intro to RMS Voltage

Some book problems specify a voltage as root-mean-square (rms), eg “most household power in the US is 120 V rms”.  This means that the actual voltage measured from your wall outlet can be modeled by
$$v(t) = 120\sqrt{2} \sin(\omega t)$$
where $\omega = 2\pi f$.

An rms voltage is the effective voltage in the context of measuring power, meaning that if we replace our alternating voltage source with a dc source with voltage equal to the rms value, the same power will be delivered.  Say we have the above sine voltage as a source and a load R.  Then the instantaneous power dissipated in R is $v(t)^2 / R$.  The average power over a period $T = 1/f$ will be $\frac{1}{RT}\int^T_0 v(t)^2 dt$.  Let’s calculate this definite integral.

Let $V_m$ be our sinusoid peak value, $p(t)$ instantaneous power, $\omega = 2\pi f$ our angular frequency.  Realize that measuring the average power of a periodic function over any interval of length $T$, $[t_0, t_0 + T]$, will be the same as doing it over $[0, T]$.  So we have average power dissipated into a resistive load R, using a sinusoidal power source with peak $V_m$ as
$$\frac{1}{T}\int^T_0 p(t) dt = \frac{1}{T}\int^T_0 (\frac{V_m^2}{R}\sin^2{\omega t}) dt \\$$
How do we deal with the $\sin^2$ integrand?  One way is with a trig identity.  We can use the trig identity $sin^2{u} = \frac{1 – \cos{2u}}{2}$ to get
$$integral = C \int^T_0{\frac{1 – \cos{2\omega t}}{2}} dt$$
Make the substitution $u = \omega t, du = \omega dt$ to get
$$integral = C \int^{u(T)}_{u(0)}{\frac{1 – \cos{2u}}{2}}\frac{du}{\omega}$$
Finish expanding this and the result is that the average power dissipated into R over a time-period of length $1/f$ is $V_m^2 / (2 R)$.  Setting this equal to a $V_{dc}^2 / R$ for the purpose of finding the effective voltage we get $V_{dc} = V_m / \sqrt{2}$.

Some authors use the convention that phasor voltages are given as their rms ($V_{dc}$ or $V_{rms})$ and not their peak ($V_m$) voltage.

# Deriving Trig Identities Using Euler’s Identity

Euler’s identity is true for all complex numbers $\theta$:
$$e^{j\theta} = \cos(\theta) + j\sin(\theta)$$
where $j^2 = -1$.

One neat thing you can do is derive trig identities way more easily and succinctly than without it.

First know that
$$\sin(u) = Im(e^{ju}) = \frac{e^{ju} – e^{-ju}}{2j} \\ \cos(u) = Re(e^{ju}) = \frac{e^{ju} + e^{-ju}}{2}$$
To derive these you could use a formula for the real and imaginary parts of $z\in \mathbb{C}$ that uses the conjugate $z^*$ and then find out what the conjugating an exponential does to it.

Example: $\sin^2(u) = \frac{1 – \cos(2u)}{2}$.
$$\sin^2(u) = (\frac{e^{ju} – e^{-ju}}{2j})^2 = \frac{e^{j2u} + e^{-j2u} – 2}{4(-1)} = \frac{1 – \cos(2u)}{2}$$
Exercise: derive two other trig identities using the exponential.  In the future, when in need of a trig identity, instead of referring to a table, maybe try a quick derivation using the exponential.

(This was directly copied from the Math Stack Exchange site.)

This renders LaTeX on your page:

$$\sum_{i=1}^{\infty}{a_i} = \dots$$

# Electromagnetic Field Theory Note-Taker’s Abbreviations

###### (Each abbreviation in this table could depend on context.)
 straight = strgt becomes = becms time = t uniformly = unifmly, unif mathematical = matmtcl time-varying = t-vryg operator = op current = crnt derivative = deriv value = val electromagnetics = EM product = prod free space = fs determine = det field = fld scalar = sclr, scal Laplacian = Lapl, $\nabla^2$ single = sgl force = frc electromagnetism = EM rectangular = rect are = r similarly = simly, simlrly, sim point = pt vector = vect, vec, vctr research = rsrch test = tst quantities = quants, qtys displacement = dsplcmt placed = plcd coplanar = copl specified = specfd, spec’d with = w where = wh their = thr magnitude = mag, magtd density = dnsty, dens, dsty distance = dist direction = dir of = o, $\circ$ induced = indcd previous = prev identity = idty, id induces = indcs section = sect algebra = alg the = th chapter = ch within = w/in without = w/o through = thrgh peak = pk thrghout = thr/o, thrgh/o and = & which = wch important = imptnt, impt respectively = resp from = frm Cartesian = Cart measured = meas’d measure = measr coordinate = coord Cartesian Coord System = CCS flux linkage = $\phi$, fl system = sys increased = inc’d, $\uparrow$’d subject = subj increasing = $\uparrow$’g, inc’g however = h/e factor = fctr need = nd become = becm such as = sa, : identities = idents electric force = efrc Coulomb’s Law = CL electric field = efld superposition property = SPP, SP position = pos positive = + flux density = fd Gauss’s Law = GL electric potential = ep called = cld potential field = pfld magnetic field strength = mfs line = ln flux = flx magnetic flux density = mfd particle = ptcl electro-motive force = emf supplemented = spplmtd, suplmtd would = wd special = spcl Faraday’s Law = FL Maxwell’s Equations = ME equation = eqn solution = soln Ampere’s Law = AL AL spplmtd by ME = AML Gauss’s Law = GL Gauss’s Law f th efld = GLE Gauss’s Law f th mfld = GLM charge = chrg comprised = cmprsd static = stat dynamic = dyn generally = gen harmonic = harm representation = rep property = ppty partial differential eqns = PDEs source = src propagation = ppgtn, prop intensity = intnsty, intens following = folwg substitute = subst propagate = ppgt, prop essentially = ess polar coordinate system = PCS constant = const parallel = || conductor = cdctr, cond for = f negligible = negl additional = addl, add’l basic = bas symbol = sym infinity = infty, $\infty$ decimal = dec multiple = mult diameter = diam calculate = calc determinant = det each = ea perform = prfm Biot-Savart Law = BSL motion = motn more = m media = med dielectric = dielec moving = movg, mvg root-mean-square = rms inductor = idctr, ind inductance = indnc, ind capacitance = capnc, cap resistance = resnc, res will = wl, ‘ll proof = pf theorem = thm predict = prdct, pdct that = tht above = abv below = blw order = ord small = sm particular = part, ptclr then = th, thn written = wrtn could = cld can = cn be = b permittivity = pmtvty permeability = pmblty condition = cdtn, cond wave = wv voltage = vtg also = als as well as = awa in other words = iow with resp to = wrt trans(mission / mitter) = tx receiver = rx together with = tog w lossless = losls, lossls connects = cncts, conns $\hspace{3cm}$

## Telegrapher’s Equations Intro

### Gallery

An easy way to remember the telegrapher’s equations is $\partial_x v = R i + L \partial_t i$ together with its dual (know the duality $i\leftrightarrow v, R \leftrightarrow G, L \leftrightarrow C$). To derive them using the … Continue reading