Class 2

Slicing the line into circuits

Class 1 showed that voltage changes along a wire. Class 2 builds the circuit model that predicts how, turns it into a wave equation, and reads the speed straight off the cable.

2026-09-02Slides (PDF)telegrapher equationslumped element modelwave equation

Class 1 ended with a problem. Once $\ell/\lambda$ passes $0.01$ , the voltage at one end of a connection differs from the voltage at the other, so a single node in a circuit diagram cannot represent the wire. Class 2 fixes that by cutting the line into slices short enough for circuit laws to hold, writing KVL and KCL on one slice, and shrinking the slice to zero.

01Why the line gets sliced

A circuit law like KVL assumes every point in a loop sees the same signal at the same instant. Class 1 showed when that assumption dies: the loop has to be small compared with a wavelength.

So take a length $\Delta z$ of the line, small enough that $\Delta z/\lambda \ll 1$ holds no matter how high the frequency goes. Inside that slice the circuit laws still work. Model the slice with four components, apply KVL and KCL, then let $\Delta z \to 0$ and collect what survives.

Figure 1 The line as a chain of slices. Each slice is short enough for circuit laws, and the line is what you get by joining them. Only

L

and

C

are drawn here, so the picture is the lossless case.

02The four parameters

The slice on slide 1 carries four elements, drawn in Figure 2. Each is quoted per metre of line, so a slice of length $\Delta z$ holds $R\,\Delta z$ ohms, $L\,\Delta z$ henries and so on.

Figure 2 One slice of the line, length

\Delta z

. Hover any component to light up the term it contributes to the equations below, and hover a term to light up its component.

Notation

symbol	say it	what it is physically	units
$\class{hl hl-R}{R}$	"are"	resistance of the two conductors, one metre of each	Ω/m
$\class{hl hl-L}{L}$	"ell"	inductance from the magnetic field around the pair	H/m
$\class{hl hl-G}{G}$	"gee"	leakage conductance through the insulation between them	S/m
$\class{hl hl-C}{C}$	"see"	capacitance between the two conductors	F/m
$\Delta z$	"delta zee"	length of the slice being modelled	m
$V(z,t)$	"vee of zee tee"	voltage across the pair, at position $z$ and time $t$	V
$I(z,t)$	"eye of zee tee"	current along the line at that position and time	A
$\partial$	"partial"	derivative holding the other variable fixed

Two variables now govern everything, so the derivatives turn partial. Read $\partial V/\partial t$ as "how fast the voltage at this fixed position changes with time" and $\partial V/\partial z$ as "how much the voltage differs between neighbouring positions at this frozen instant".

03KVL across the slice

Walk the top conductor from $z$ to $z + \Delta z$ , following Figure 2. The current $I(z,t)$ passes through the resistance and the inductance, and each drops some voltage. The slide writes it as

V(z,t) = \class{hl hl-R}{R\,\Delta z\, I(z,t)} + \class{hl hl-L}{L\,\frac{\partial I(z,t)}{\partial t}\,\Delta z} + V(z + \Delta z, t)

The resistor drop is Ohm's law, $R\Delta z$ times current. The inductor drop is $L\Delta z$ times the rate of change of current, since an inductor resists change rather than current itself. Whatever survives both drops arrives at the far side as $V(z+\Delta z, t)$ .

Move that last term over and divide by $\Delta z$ :

\frac{\Delta V}{\Delta z} = -\left[R\,I(z,t) + L\,\frac{\partial I(z,t)}{\partial t}\right]

where $\Delta V = V(z+\Delta z,t) - V(z,t)$ . The minus sign says voltage falls as you move along the line in the direction the current flows.

Now shrink the slice. As $\Delta z \to 0$ the ratio $\Delta V/\Delta z$ becomes a derivative:

\frac{\partial V(z,t)}{\partial z} = -\left[R\,I(z,t) + L\,\frac{\partial I(z,t)}{\partial t}\right] \tag{1}

04KCL at the node

Some of the current that arrives never reaches the far end. It leaks through $G$ and it charges $C$ . Slide 1 writes

I(z,t) = \class{hl hl-G}{G\,V(z+\Delta z,t)\,\Delta z} + \class{hl hl-C}{C\,\frac{\partial V(z + \Delta z,t)}{\partial t}\,\Delta z} + I(z + \Delta z, t)

The leakage current is conductance times voltage. The capacitor current is $C\Delta z$ times the rate of change of voltage, since a capacitor passes current only while the voltage across it moves.

Same two steps. Divide by $\Delta z$ , then take the limit:

\frac{\partial I(z,t)}{\partial z} = -\left[G\,V(z,t) + C\,\frac{\partial V(z,t)}{\partial t}\right] \tag{2}

Equations (1) and (2) are the telegrapher's equations. They are coupled: the slope of $V$ depends on $I$ , and the slope of $I$ depends on $V$ .

05Getting one equation instead of two

Coupled equations are awkward, so eliminate one unknown. Differentiate (1) with respect to $z$ :

\frac{\partial^2 V}{\partial z^2} = -\left[R\,\frac{\partial I}{\partial z} + L\,\frac{\partial^2 I}{\partial t\,\partial z}\right]

Both terms on the right contain $\partial I/\partial z$ , and (2) says what that is. Substitute it, and substitute its time derivative for the second term:

\frac{\partial^2 V}{\partial z^2} = R\left[G V + C\frac{\partial V}{\partial t}\right] + L\left[G\frac{\partial V}{\partial t} + C\frac{\partial^2 V}{\partial t^2}\right]

Collect terms by which derivative of $V$ they multiply, and slide 1's equation (3) appears:

\frac{\partial^2 V(z,t)}{\partial z^2} = LC\,\frac{\partial^2 V(z,t)}{\partial t^2} + (LG + RC)\,\frac{\partial V(z,t)}{\partial t} + RG\,V(z,t) \tag{3}

Doing the same elimination the other way round gives the identical equation for current:

\frac{\partial^2 I(z,t)}{\partial z^2} = LC\,\frac{\partial^2 I(z,t)}{\partial t^2} + (LG + RC)\,\frac{\partial I(z,t)}{\partial t} + RG\,I(z,t) \tag{4}

06Dropping the losses

Slide 2 sets $R = 0$ and $G = 0$ . A perfect conductor carries current without dissipating, and perfect insulation leaks nothing. Every term carrying $R$ or $G$ disappears from (3):

\frac{\partial^2 V}{\partial z^2} = LC\,\frac{\partial^2 V}{\partial t^2}

That is the wave equation, the same equation that governs a plucked string and a sound pulse in air. Its arrival is the point of the whole derivation.

07What the solution looks like

Slide 2 substitutes equation (6) into the wave equation. Equation (6) is the proposed solution

V(z,t) = f_1\!\left(t - \frac{z}{v}\right) + f_2\!\left(t + \frac{z}{v}\right) \tag{6}

Read $f_1$ as any shape at all, with the single restriction that it depends on $t$ and $z$ only through the combination $t - z/v$ . Consider what that forces. Sit at $z = 0$ and watch some feature of the shape at time $t_0$ . To see the same feature at position $z$ , you need $t - z/v = t_0$ , which happens at $t = t_0 + z/v$ . The feature arrives later by exactly $z/v$ , so the shape moves in the $+z$ direction at speed $v$ without changing form.

$f_2$ carries $t + z/v$ , so the same argument runs backwards and that shape travels in the $-z$ direction.

08Watch both waves at once

09Verifying the solution

Slide 2 checks (6) against the wave equation by differentiating it. Each derivative uses the chain rule, since $f_1$ and $f_2$ take a combination of $t$ and $z$ rather than either alone. Write $u = t - z/v$ , so $\partial u/\partial z = -1/v$ and $\partial u/\partial t = 1$ .

In space:

\frac{\partial V}{\partial z} = -\frac{1}{v}f_1' + \frac{1}{v}f_2', \qquad \frac{\partial^2 V}{\partial z^2} = \frac{1}{v^2}\left(f_1'' + f_2''\right)

The first derivative keeps the sign difference, because differentiating $f_2$ brings down $+1/v$ . Squaring removes it, so the second derivatives add.

In time:

\frac{\partial V}{\partial t} = f_1' + f_2', \qquad \frac{\partial^2 V}{\partial t^2} = f_1'' + f_2''

Put both into $\partial^2 V/\partial z^2 = LC\,\partial^2 V/\partial t^2$ :

\frac{1}{v^2}\left(f_1'' + f_2''\right) = LC\left(f_1'' + f_2''\right)

The same bracket sits on both sides. Cancel it, and the shapes drop out of the problem entirely:

\frac{1}{v^2} = LC \qquad\Longrightarrow\qquad v = \frac{1}{\sqrt{LC}} \tag{7}

Worked example 1 The velocity of a 50 Ω coaxial cable

A common coax runs at $L = 250$ nH/m and $C = 100$ pF/m. In base units, $L = 2.50\times 10^{-7}$ H/m and $C = 1.00\times 10^{-10}$ F/m.

v = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{(3\times 10^{-7})(1\times 10^{-10})}}

which gives 2.00×10⁸ m/s, a velocity factor of 0.667 against the speed of light. Class 1's cable ran slower still, at $1\times10^{8}$ m/s. Holding $L$ fixed, that would need $C = 4.00\times 10^{-10}$ F/m, four times the capacitance per metre of this one.

10The current that goes with it

Slide 2 then asks what current accompanies that voltage. Take the lossless form of equation (1), which loses its $R$ term:

\frac{\partial V}{\partial z} = -L\,\frac{\partial I}{\partial t}

The left side is already known from the derivative table:

-\frac{1}{v}f_1' + \frac{1}{v}f_2' = -L\,\frac{\partial I}{\partial t}

Divide by $-L$ and the current's time derivative appears alone:

\frac{\partial I}{\partial t} = \frac{1}{Lv}\left(f_1' - f_2'\right)

Integrate with respect to $t$ . Since $f_1'$ is the $t$ derivative of $f_1$ , integrating returns $f_1$ , and the same for $f_2$ :

I(z,t) = \frac{1}{Lv}\left[f_1\!\left(t - \frac{z}{v}\right) - f_2\!\left(t + \frac{z}{v}\right)\right] + C \tag{8}

$C$ here is a constant of integration, a steady current that never changes with time. Physical boundary conditions set it to zero, which the slide states. Watch the symbol: this $C$ is not the capacitance per metre.

11Read a cable's speed off its datasheet

12Run the line and watch

The widget below solves the two telegrapher equations on a chain of 160 slices, the same chain as Figure 1, stepping voltage and current forward in time. Nothing about the answer is assumed. The shape you see is what the equations do.

13Traps

14Check yourself

1Why does the derivation slice the line rather than apply KVL to the whole thing at once?

KVL assumes every point in the loop sees the same signal at the same instant. Class 1 showed that fails once $\ell/\lambda$ passes $0.01$ . A slice of length $\Delta z$ can always be made short enough for the assumption to hold, and calculus reassembles the slices.

2A line has

L = 400

nH/m and

C = 160

pF/m. What is

v

, and what fraction of

c

is it?

$v = 1/\sqrt{LC} = 1.25\times 10^{8}$ m/s, which is 0.417 of the speed of light.

3In equation (8), why does

f_2

carry a minus sign when

V

adds both waves?

Current has a direction along the line. The backward wave carries its current toward $-z$ , so at a fixed point it subtracts from the forward wave's current while its voltage still adds.

4What happens to equations (3) and (4) if you keep

R

but set

G = 0

The $RG$ term vanishes and the $(LG + RC)$ coefficient collapses to $RC$ , leaving $\partial^2 V/\partial z^2 = LC\,\partial^2 V/\partial t^2 + RC\,\partial V/\partial t$ . The first-order time derivative is the loss term, and it damps the wave as it travels.

5A pulse shape

f_1

is 2 μs wide in time. How long is it in space on the coax from the worked example?

Length equals speed times duration: $2.00\times 10^{8} \times 2\times10^{-6} = 400$ m. Wide pulses occupy hundreds of metres of cable, which is why short cables look lumped to slow signals.

6Two cables have the same

L

, but one has twice the capacitance per metre. Compare their velocities and impedances.

$v$ scales as $1/\sqrt{C}$ , so the second is slower by $\sqrt{2}$ , a factor of 0.707. $Z_0$ scales the same way, so its impedance is lower by the same factor.

Answer out loud before opening one.

15Cheat sheet

Cheat sheet

result	equation	note
telegrapher, voltage	$\dfrac{\partial V}{\partial z} = -\left[RI + L\dfrac{\partial I}{\partial t}\right]$	KVL on one slice
telegrapher, current	$\dfrac{\partial I}{\partial z} = -\left[GV + C\dfrac{\partial V}{\partial t}\right]$	KCL on one slice
general wave equation	$\dfrac{\partial^2 V}{\partial z^2} = LC\dfrac{\partial^2 V}{\partial t^2} + (LG+RC)\dfrac{\partial V}{\partial t} + RGV$	equations (3) and (4)
lossless	$\dfrac{\partial^2 V}{\partial z^2} = LC\dfrac{\partial^2 V}{\partial t^2}$	set $R = G = 0$
solution	$V = f_1(t - z/v) + f_2(t + z/v)$	any shape, forward plus backward
velocity	$v = 1/\sqrt{LC}$	equation (7)
current	$I = \dfrac{1}{Lv}\left[f_1 - f_2\right] + C$	equation (8), constant set to zero
impedance, coming later	$Z_0 = \sqrt{L/C} = Lv$	makes (8) read $I = (f_1 - f_2)/Z_0$

Units: $R$ in Ω/m, $L$ in H/m, $G$ in S/m, $C$ in F/m. Worked cable: 250 nH/m and 100 pF/m give $2\times10^{8}$ m/s and 50 Ω.