Class 1

When a wire stops being a wire

Slide 1 sets up a circuit and asks what the voltage is at each end. Slide 2 answers it at nine frequencies. Both slides turn on one ratio.

2026-09-02Slides (PDF)transmission lineswavelengthphase

Two slides, one idea. A connection you would draw as a plain line in a circuit diagram stops behaving like one as soon as the signal on it gets fast enough. Slide 1 builds the argument. Slide 2 puts nine numbers on it. This page takes both apart and rebuilds every number from scratch.

01The picture on slide 1

A voltage source sits on the left. A load sits on the right. Two horizontal wires of length $\ell$ join them.

Figure 1 The circuit on slide 1. The source drives a pair of wires of length

\ell

into a load

Z

. The red curve is the signal along the top wire at one frozen instant, drawn here for a line that spans about one wavelength.

The slide marks two terminal pairs. $XX'$ is the pair at the source end: the point $X$ on the top wire and the point $X'$ directly below it on the bottom wire. $YY'$ is the pair at the load end. A prime mark is a naming convention and carries no arithmetic. Read $X'$ as "ex prime".

Voltage always needs two points. When the slide writes $V_{XX'}$ it means the voltage of $X$ measured with respect to $X'$ , in volts. Same for $V_{YY'}$ at the far end.

The red curve in Figure 1 is the signal itself, caught at one instant. Slide 1 draws about one full ripple across the length $\ell$ . That picture is the answer to the whole class, and the rest of the slides work out when you are allowed to ignore it.

Notation

symbol	say it	what it means	units
$V_0$	"vee nought"	the peak voltage the source reaches	V
$\omega$	"omega"	angular frequency, how fast the cosine turns	rad/s
$f$	"eff"	ordinary frequency, cycles per second, $\omega = 2\pi f$	Hz
$t$	"tee"	time	s
$\ell$	"ell"	length of the connection between the two ends	m
$c$	"see"	speed of light in vacuum, $3.00\times 10^{8}$ m/s	m/s
$v$	"vee"	speed the signal actually travels on this line	m/s
$\lambda$	"lambda"	distance the signal travels in one full cycle	m
$XX'$	"ex ex prime"	the terminal pair at the source end
$\ll$	"much less than"	smaller by a factor of roughly 100 or more

02Reading the three expressions

Slide 1 gives three lines. Take them one at a time.

V_s(t) = V_0\cos(\omega t)

The source. At $t = 0$ the cosine equals 1, so the source sits at its peak $V_0$ . As $t$ grows, $\omega t$ grows, and the cosine walks through its cycle. One full cycle takes the time that makes $\omega t = 2\pi$ .

V_{XX'} = V_0\cos(\omega t)

The voltage right at the source terminals. Identical to the source, because those terminals are the source.

V_{YY'} = V_0\cos\!\left[\omega\left(t - \frac{\ell}{c}\right)\right]

The voltage at the far end. Look at what changed: $t$ became $t - \ell/c$ . The quantity $\ell/c$ has units of metres divided by metres per second, which leaves seconds. It is a travel time.

03Freezing the clock

Slide 1 then picks a single instant, $t = 0$ , and asks what each end reads.

At the source end the cosine argument is zero:

V_{XX'} = V_0\cos(0) = V_0

At the load end, substitute $t = 0$ into the delayed expression:

V_{YY'} = V_0\cos\!\left[\omega\left(0 - \frac{\ell}{v}\right)\right] = V_0\cos\!\left(-\frac{\omega\ell}{v}\right) = V_0\cos\!\left(\frac{\omega\ell}{v}\right)

The minus sign vanishes because cosine is even, meaning $\cos(-x) = \cos(x)$ . The slide writes the result without the sign for that reason.

Now swap $\omega$ for $2\pi f$ :

V_{YY'} = V_0\cos\!\left(\frac{2\pi f\ell}{v}\right)

04The step that makes it about wavelength

Slide 2 rewrites the cosine argument. Follow the chain:

\frac{\omega\ell}{v} = \frac{2\pi f\ell}{v} = 2\pi\frac{\ell}{v/f} = 2\pi\frac{\ell}{\lambda}

The middle step divides top and bottom by $f$ . The last step names $v/f$ as the wavelength $\lambda$ , the distance the signal covers during one cycle.

So the load-end voltage at $t = 0$ is

V_{YY'} = V_0\cos\!\left(2\pi\frac{\ell}{\lambda}\right)

Nothing survives except $\ell/\lambda$ . Frequency alone tells you nothing. Length alone tells you nothing. The class is about their ratio, and $2\pi$ turns that ratio into an angle: $\ell/\lambda = 1$ means a full turn of $360^\circ$ across the line.

05The velocity the table actually uses

Slide 1 writes the delay as $\ell/c$ , which says the signal travels at the speed of light. Slide 2 lists a wavelength column. Test the two against each other.

At 10 kHz the slide gives $\lambda = 10\,000$ m. Then

v = f\lambda = (10^{4})(10^{4}) = 1.00\times 10^{8} \ \text{m/s}

Every other row agrees: 1 MHz with 100 m, 25 MHz with 4 m, 100 MHz with 1 m, all give 1.00×10⁸ m/s. That is $c/3$ , a velocity factor of 0.334, which is ordinary for a cable with a plastic dielectric.

06Slide 2, recomputed

Every value below comes from $V_{YY'} = V_0\cos(2\pi f\ell/v)$ with $\ell = 1$ m and $v = 1.00\times 10^{8}$ m/s. The build recomputes each one and checks it against the printed slide value before this page will publish.

$f$	$\lambda$ (m)	$\ell/\lambda$	phase across the line (°)	$V_{YY'}$ (V)	slide says
10 kHz	10 000	1×10^-4	0.0360	0.999999803	0.999999802
1 MHz	100	0.010	3.60	0.998026728	0.998026728
5 MHz	20	0.050	18	0.951057	0.951056
10 MHz	10	0.10	36	0.809	0.809
20 MHz	5	0.20	72	0.309	0.309
25 MHz	4	0.25	90	0.000	0
30 MHz	3.33	0.30	108	-0.309	−0.309
50 MHz	2	0.50	180	-1.000	−1
100 MHz	1	1	360	1.000	+1

The 10 kHz and 5 MHz rows differ from the slide in the last digit, because the slide truncates where this page rounds. Every other row matches exactly.

Read the phase column and the table stops being a list. At 25 MHz the line spans a quarter of a wavelength, the phase is $90^\circ$ , and the cosine of $90^\circ$ is zero. At 50 MHz it spans half a wavelength, the phase is $180^\circ$ , and the far end sits at $-V_0$ while the near end sits at $+V_0$ . At 100 MHz the wave fits exactly once along the line, the phase is a full $360^\circ$ , and the two ends agree again.

Worked example 1 The first row, worked line by line

Take $f = 10$ kHz.

\lambda = \frac{v}{f} = \frac{1\times 10^{8}}{10^{4}} = 10 000 \ \text{m}

The line is 1 m of that, so

\frac{\ell}{\lambda} = 1\times 10^{-4}

The phase across the line is $360^\circ \times 1\times 10^{-4} = 0.0360^\circ$ , a rounding error away from zero. The cosine of a tiny angle is a hair under 1:

V_{YY'} = \cos(0.0360^\circ) = 0.999999803 \ \text{V}

The two ends differ by 197 parts per billion. No instrument in your lab will see that, which is what "no transmission line effects" means in practice.

07Turn the frequency up yourself

08The same thing, moving

09Every frequency at once

10The two cases

Slide 2 closes with a split.

At $\ell = 1$ m and $v = 1\times 10^{8}$ m/s, the boundary sits at $f = 0.01 v/\ell = 1.00\times 10^{6}$ Hz, which is the 1 MHz row. A 1 m connection is a circuit element below 1 MHz and a transmission line above it.

11Traps

12Check yourself

1At 25 MHz the table says

V_{YY'} = 0

. Where has the energy gone?

Nowhere. The load end is passing through zero at that instant on its way up or down, the same as any cosine crossing. A quarter wavelength fits along the line, so the far end runs $90^\circ$ behind the near end. Wait a quarter period and the far end sits at its own peak.

2A 5 cm PCB trace carries a 200 MHz clock at

v = 1.5\times10^{8}

m/s. Which case?

$\lambda = 1.5\times10^{8}/2\times10^{8} = 0.75$ m, so $\ell/\lambda = 0.05/0.75 = 0.067$ . That sits above $0.01$ , so it is Case 2 and the trace needs transmission-line treatment.

3Same slide setup, but the frequency is mains, 50 Hz. What is

\ell/\lambda

$\lambda = 1\times 10^{8}/50 = 2.00\times 10^{6}$ m and $\ell/\lambda = 5.0\times 10^{-7}$ . Deeply lumped, which is why nobody worries about transmission lines in house wiring at mains frequency.

4Why does

V_{YY'}

return to

+1

V at 100 MHz after sitting at

-1

V at 50 MHz?

At 50 MHz the line holds half a wavelength, a $180^\circ$ phase difference. At 100 MHz it holds a full wavelength, $360^\circ$ , and $360^\circ$ of phase brings the cosine back to where it started. The two ends agree again by coincidence of geometry, not because the line got easier.

5Two engineers argue about whether a connection needs transmission-line analysis. What single number settles it?

$\ell/\lambda$ , which needs the length, the frequency and the velocity on that particular line. Compare it against $0.01$ .

6At what frequency does a 1 m line with

v = 1\times10^{8}

m/s hit the quarter-wave point?

A quarter wave means $\ell/\lambda = 0.25$ , so $f = 0.25v/\ell = 2.50\times 10^{7}$ Hz. That is the 25 MHz row, the one where the far end reads zero.

Answer out loud before opening one.

13Cheat sheet

Cheat sheet

quantity	relation	why you reach for it
wavelength	$\lambda = v/f$	first line of nearly every problem
electrical length	$\ell/\lambda = f\ell/v$	the one number that decides the case
phase across the line	$360^\circ\,\ell/\lambda$	turns the ratio into degrees
far-end voltage at $t=0$	$V_{YY'} = V_0\cos(2\pi\ell/\lambda)$	slide 2 in one line
travel time	$t_d = \ell/v$	how late the far end hears about a change

Threshold on these slides: $\ell/\lambda \ll 1$ is Case 1, lumped. $\ell/\lambda \ge 0.01$ is Case 2, transmission line.

Slide values: $\ell = 1$ m, $V_0 = 1$ V, $v = 1\times10^{8}$ m/s, so the boundary falls at 1 MHz and the far end reads zero at 25 MHz.