Impedance: what decides how much comes back

The puzzle: why the gel¶

Before an ultrasound scan, the technician squirts cold gel on your skin and presses the probe into it. Without the gel the image is not grainy or dim — it is black. The instrument works perfectly; sound simply refuses to enter you. A layer of air a fraction of a millimetre thick between the probe and your skin sends essentially all of it straight back.

Two more everyday facts, from the other end of the apparatus scale. Tie a rope to a wall, send a pulse down it, and the pulse comes back upside down. Tie the same rope to a light thread instead, send the same pulse, and it comes back right side up. Nobody changed the rope.

Predict before you calculate¶

Commit to an answer for each before running anything.

A pulse travels along a rope tied to a wall. Upright or inverted on the way back?
A heavy rope is knotted to a light string. A pulse arrives from the heavy side. Is the reflected pulse upright or inverted — and is the transmitted pulse taller or shorter than the one that made it?
Can a transmitted pulse be taller than the incident pulse? If it can, where did the extra energy come from?
Is there a junction between two different strings that produces no echo at all?

Explore the model¶

Three animations, all computed from the same wavelab.waves functions the laboratory calls.

The same pulse at a wall and at a free end — one string each, differing in a single word passed to the solver. Against the wall the returning pulse is inverted and the end never moves. At the free end, a ring sliding on a frictionless rod, the pulse returns upright and the end itself swings to twice the incident amplitude while the incoming and outgoing pulses sit on top of each other. The dotted guides mark $\pm A$ and $2A$ . Measured off these runs, the two returns are $-1.0000\,A$ and $+1.0000\,A$ , and neither end takes any energy.

One junction, approached from each side. The densities differ fourfold, so the impedances differ twofold and the coefficients are exactly thirds. From the heavy side (top, shaded): the echo is upright at $+\tfrac13$ and the transmitted pulse is taller than the incident one, $\tfrac43$ , and twice as wide because it travels twice as fast. From the light side (bottom): the echo is inverted at $-\tfrac13$ and the transmitted pulse is shorter and narrower. Both panels are in step, so the two outcomes can be compared rather than remembered.

A six-cycle packet at a 3:1 impedance step, bare (top) and with a quarter-wave matching section (bottom, the green sliver). The bare step sends a quarter of the energy back, $R = 0.2504$ against the formula’s 0.2500. The sliver — one piece of string of impedance $\sqrt{Z_1Z_3}$ , 57.7 mm long — drops that to $R = 0.0077$ , an echo thirty-two times weaker in energy. Nothing absorbed it. This is the advanced section, and the direct ancestor of the coating on a camera lens.

Model specification

System: two semi-infinite ideal strings, densities $\mu_1$ and $\mu_2$ , under one common tension $T$ , joined at $x = 0$ — numerically a single grid whose density jumps at a point.
Dynamics: the wave equation $\mu\,y_{tt} = T\,y_{xx}$ on each side, and at the junction the two matching conditions: $y$ continuous (the string is unbroken) and $-T\,y_x$ continuous (the joint is massless, so no net transverse force may act on it).
Boundary: the junction itself. The outer ends — fixed or free — are kept far enough away that nothing measured here has heard from them.
Ensemble: deterministic; detector noise, where added, is Gaussian, independent and seeded.
Ignored: junction mass, stiffness and damping; loss of any kind; oblique incidence, which has no meaning in one dimension.
Valid when: slopes are small, $|y_x| \ll 1$ , on both sides — the transmitted wave has its own slope — and the junction is a point. A density that changes gradually over a length comparable with the wavelength is a different problem, and the problem set makes it one.
Failure modes: reading $t > 1$ as a gain of energy; applying the amplitude coefficients to power; forgetting the flux factor $Z_2/Z_1$ ; a junction placed on too coarse a grid, where the measured $r$ is wrong in the third decimal for reasons that have nothing to do with impedance.

Run the laboratory now, then come back. Four things are worth doing before you read on:

Plot the transverse force against the transverse velocity on a travelling pulse, and read the impedance off the slope.
Measure $r$ and $t$ across a 625-fold range of density and find where the sign changes.
Weigh the energy on the two sides of a junction and watch $R + T$ come out at 1.
Measure a reflection coefficient through 5% noise, twice, with two estimators that are both honest arithmetic. One of them is off by a fifth of the answer.

Derive the result¶

What the medium feels like to push¶

Take hold of the end of a semi-infinite string and wave it. Everything you launch travels away and never comes back, so the string at your hand is a pure right-mover and obeys module 08’s $y_t = -v\,y_x$ . The transverse force you must supply is the transverse component of the tension, $-T\,y_x$ , so

F = -T\,\frac{\partial y}{\partial x} = \frac{T}{v}\,\frac{\partial y}{\partial t} = Z\,\frac{\partial y}{\partial t}, \qquad Z \equiv \frac{T}{v} = \sqrt{T\mu} = \mu v .

(1)

Impedance of a string

\boxed{\;Z = \sqrt{T\mu} = \mu v = \frac{T}{v}\;}

(2)

in kilograms per second — force per unit transverse velocity, N/(m/s).

Force proportional to velocity and in phase with it: the string resists like a dashpot, not like a spring and not like a mass. That is the whole content of the word. You do work at every instant and none of it is stored where you can get it back, because what you paid for is travelling away. An ideal infinite string is a perfect absorber, and an oscillator with one tied to it is a damped oscillator — module 02’s $\gamma$ , supplied by geometry rather than by friction.

$Z$ and $v$ are separate properties: $T$ and $\mu$ are two knobs, and $v = \sqrt{T/\mu}$ and $Z = \sqrt{T\mu}$ can be set independently. On a string they are not independent in practice, because a junction between two densities carries one tension; there $Z_2/Z_1 = \sqrt{\mu_2/\mu_1} = v_1/v_2$ , and the heavier side is both the slower and the harder one.

The laboratory reads $Z$ straight off a simulated pulse: plot $-T\,y_x$ against $y_t$ at every point and the cloud is a straight line of slope +0.199991 kg/s against $\sqrt{T\mu} = 0.2$ kg/s. A left-moving pulse gives -0.199991: the sign is the direction the energy goes, not a different medium.

The junction: two conditions and nothing else¶

Now join two strings at $x = 0$ and send a wave in from the left. Write the three waves in the course convention,

y_{\text{inc}} = \Real\!\left[A\,e^{\ii(k_1x - \omega t)}\right], \qquad y_{\text{ref}} = \Real\!\left[rA\,e^{\ii(-k_1x - \omega t)}\right], \qquad y_{\text{tr}} = \Real\!\left[tA\,e^{\ii(k_2x - \omega t)}\right],

(3)

with the same $\omega$ on both sides — the junction is driven at the frequency it is fed, and nothing about it changes in time — and $k_{1,2} = \omega/v_{1,2}$ adjusting instead. The wavelength changes at the junction; the frequency cannot.

Two statements about the joint finish the problem.

The string is unbroken. $y$ is continuous at $x = 0$ , so $1 + r = t$ .
The joint is massless. A point with no mass cannot survive a net force, so the transverse force $-T\,y_x$ is continuous too. With $T$ common to both sides that gives $k_1(1 - r) = k_2 t$ , and since $k = \omega/v$ and $Z = T/v$ , it is $Z_1(1 - r) = Z_2 t$ .

Reflection and transmission of displacement

Solving those two lines together:

\boxed{\;r = \frac{Z_1 - Z_2}{Z_1 + Z_2}, \qquad t = \frac{2Z_1}{Z_1 + Z_2}\;}

(4)

Both are ratios of displacement amplitudes, and both are real: at a lossless junction a wave can only come back in step or exactly out of step, never anywhere in between. The identity $1 + r = t$ was the first condition and survives as the check to run on any numbers claiming to be $r$ and $t$ .

The sign of $r$ is the sign of $Z_1 - Z_2$ , and it is the whole of prediction 2:

$Z_2 > Z_1$ — a harder far side — gives $r < 0$ : the echo is inverted.
$Z_2 < Z_1$ — a softer far side — gives $r > 0$ : the echo is upright.
$Z_2 = Z_1$ gives $r = 0$ and $t = 1$ : no echo at all.

And $t$ is never negative and never zero: whatever the junction, the far side moves the same way the incident wave pushed it. Only the echo can flip.

The wall and the free end were never special¶

Let $Z_2$ grow without bound — a far side too heavy to move:

r \to -1, \qquad t \to 0 .

(5)

That is a fixed end. The displacement there is $A(1 + r) = 0$ at all times, which is what “fixed” means, and the reflected pulse is the incident one turned upside down.

Let $Z_2 \to 0$ instead — a far side too light to resist:

r \to +1, \qquad t \to 2 .

(6)

That is a free end. The displacement there is $2A$ : incident and reflected pulses arrive on top of each other and add, which is why the ring on the rod overshoots. The slope there is zero, which is what “free” means.

The energy, and the factor that settles the paradox¶

A transmitted amplitude larger than the incident one looks like something for nothing. It is not, and the bookkeeping says exactly why.

Module 09 found that a sinusoidal travelling wave carries mean power $\langle P\rangle = \tfrac12 Z\,\omega^2|A|^2$ . All three waves here share $\omega$ , so dividing each power by the incident one leaves the squared amplitude ratio and, for the transmitted wave, the impedance it travels on:

R = r^2, \qquad T = \frac{Z_2}{Z_1}\,t^2 .

(7)

Now the paradox. Onto a string a hundred times lighter, $Z_2/Z_1 = 1/10$ , so

r = +0.818, \qquad t = 1.818, \qquad R = 0.669, \qquad T = 0.331 .

(10)

The far side swings almost twice as far as the incident wave did and carries only a third of its power. There is no contradiction: the light string is cheap to move. Power is $\tfrac12 Z(\omega|A|)^2$ , and the amplitude went up by 1.8 while the $Z$ in front of it went down by 10. The flux factor $Z_2/Z_1$ is the entire difference between how far something swings and what it cost to make it swing, and confusing the two is this module’s standing trap.

Matching: the case where nothing comes back¶

$r = 0$ requires $Z_1 = Z_2$ , and that is the whole criterion. It says something slightly surprising: two media may differ in every other respect and still be invisible to each other, as long as $\sqrt{T\mu}$ — or its analogue — agrees.

On a string this is a blunt instrument, because tension is one number from end to end: equal impedance means equal density, and the “matched junction” is a string with no junction in it. The idea earns its keep where the two constants are free of each other. Ultrasound gel has roughly the acoustic impedance of tissue and fills the place where air would otherwise sit; a 50 Ω cable terminated in a 50 Ω resistor absorbs its signal instead of echoing it back.

Matching two given media is a harder problem: you cannot choose $Z_2$ when both media are given. You insert a third. A section of impedance $\sqrt{Z_1Z_3}$ and a quarter-wavelength long sends back two reflections that cancel, and the advanced section builds one. That is the anti-reflection coating on every camera lens, and module 24 does it properly.

Verify computationally¶

Each check below is also a test in the project’s suite, so these claims cannot silently rot.

1. The split is the one the impedances predict. A pulse is fired at a jump in density and the two pieces it becomes are measured off the final frame. The solver integrates displacements; it has no notion of an amplitude ratio.

$\mu_2/\mu_1$	$Z_2/Z_1$	$r$ formula	$r$ measured	$t$ formula	$t$ measured
0.04	0.2	+0.6667	+0.6669	1.6667	1.6660
0.25	0.5	+0.3333	+0.3336	1.3333	1.3330
1.00	1.0	0.0000	+0.0002	1.0000	1.0002
4.00	2.0	-0.3333	-0.3336	0.6667	0.6666
25.00	5.0	-0.6667	-0.6668	0.3333	0.3332

Worst disagreement over the laboratory’s full eleven-point sweep: $6.9\times10^{-4}$ . The sign column is the falsifier for “reflection always inverts” — it changes at $\mu_2 = \mu_1$ and nowhere else.

2. A wall and a free end land on the limits. The same pulse under boundary="fixed" returns at $-1.0000\,A$ ; under boundary="free" it returns at $+1.0000\,A$ and the end itself reaches $+1.999\,A$ . Both runs keep their energy to one part in 10⁹. The solver was told about ghost points and clamped displacements, never about impedance.

3. The books balance, and they balance the same way from either side.

$\mu_2/\mu_1$	$R$ formula	$R$ weighed	$T$ formula	$T$ weighed	$R + T$
0.04	0.444444	0.444778	0.555556	0.555222	0.9999999
0.25	0.111111	0.111320	0.888889	0.888679	0.9999990
4.00	0.111111	0.111320	0.888889	0.888679	0.9999991
25.00	0.444444	0.444778	0.555556	0.555222	0.9999999

Two things are visible at once. Nothing is lost — no mechanism for loss was put into the model, so a leak would have been the scheme’s. And the rows for $1/4$ and 4 are identical: the same fraction of the energy comes back whichever side the pulse arrives from, though the amplitudes in the two directions are completely different (+0.33 against -0.33, 1.33 against 0.67). Energy cannot tell which way the wave was going.

4. The measured $r$ converges on the formula at second order. A junction is a step the grid can only place to within $\Delta x$ , so agreement is a question about resolution. With 10, 20, 40 and 80 points across the incident pulse, the measured reflected amplitude misses $-1/3$ by $8.9\times10^{-4}$ , $2.2\times10^{-4}$ , $5.6\times10^{-5}$ and $1.4\times10^{-5}$ — a factor of four per halving of the spacing, the same second order the solver obeys everywhere else. This is the gate the module’s animations sit behind: a junction rendered on too coarse a grid returns an amplitude that is wrong for reasons that have nothing to do with physics.

5. $t > 1$ , measured, with the energy accounted for. At $\mu_2/\mu_1 = 0.01$ the transmitted pulse stands 1.818 times as tall as the incident one, and the weighed energy fractions are $R = 0.669$ and $T = 0.331$ . Taller and weaker, in the same run.

6. A reflection coefficient with an error bar. The laboratory corrupts the final frame with Gaussian noise of 5% of the incident amplitude and estimates $r$ twice from the same data: by projection onto the expected pulse shape, and by picking the largest excursion in the window. At $\mu_2/\mu_1 = 4$ , where $r = -0.3333$ , the projection gives $-0.3355 \pm 0.0009$ over 64 frames and the peak-pick gives $-0.4311 \pm 0.0029$ — off by 29%, and no number of frames repairs it.

Why one honest estimator is wrong

The projection is linear in the data: noise enters it with a plus sign as often as a minus and averages away, leaving a scatter of about 0.011 per frame, close to the $\sigma/\sqrt{\sum g_i^2}$ the noise level predicts.

Picking the largest excursion is not linear. The maximum of many noisy samples is not an estimate of the signal at any of them; it is a search for the noise’s best excursion, and it finds one. The bias is away from zero, it does not shrink with more frames, and it grows as the pulse gets smaller relative to the noise — so the flattest, most interesting junctions are where it lies worst. Module 09 met the same trap in a wattmeter that squared a noisy signal. It will be waiting again at every photodetector in the optics half of this course.

7. And what a real rope can be asked. Knot a length of thick rope to a thin one, stretch the pair, and film a pulse crossing the knot from each side with a phone. The signs need no calibration at all — upright from the thin side, inverted from the thick — and that alone falsifies the misconception. Amplitudes need a ruler in frame and a fixed camera position; frame-by-frame, a heavy-to-light knot with a 4:1 mass ratio should give you $r \approx +1/3$ and $t \approx 4/3$ to perhaps 10%, with the pulse width changing visibly as it crosses. The problem set sets this up as a measurement with an uncertainty rather than a demonstration.

Transfer the idea¶

One calculation, many costumes

The two lines that produced $r$ and $t$ — the field is continuous, and so is the thing that carries force across the boundary — are the template for every boundary in this course. The algebra below is the same algebra, and it is worth recognising it now rather than deriving it four more times.

Light at a surface (module 18). For normal incidence the Fresnel coefficients are $r = (n_1 - n_2)/(n_1 + n_2)$ and $t = 2n_1/(n_1 + n_2)$ — this module’s formulas letter for letter, with the refractive index sitting in the slot where $Z$ sits here. Light reflecting off glass comes back inverted for the same reason a pulse off a wall does: in the slot that decides the sign, glass is the harder side. (The index is not light’s impedance, which is $Z = Z_0/n$ and falls as $n$ rises. The advanced section says why the formula still reads this way.)
Anti-reflection coatings (module 24). A quarter-wave layer of index $\sqrt{n_1n_3}$ is the advanced section of this page with $Z \to n$ .
Cavities and interferometers (module 26). Two junctions facing each other, with the wave bouncing between them, is a Fabry–Pérot etalon; everything about it is these coefficients applied repeatedly.
Cables, ultrasound, loudspeakers. A transmission line has $Z = \sqrt{L'/C'}$ , a fluid has $\rho c$ , and an unterminated cable, a missing gel layer and a badly matched horn are one engineering failure with three names.

One calculation, many costumes. This is among the largest unifications the course offers, and it is available here, on a rope, before any of the fields arrive.

Standing waves. 11-standing-waves puts a junction at each end. With $r = \mp 1$ there, a wave cannot escape and can only interfere with itself — which is what a mode is. The quantisation of allowed frequencies comes from these two reflection coefficients and nothing else.
Light in matter. 16-light-in-matter keeps the frequency fixed across a boundary and lets $k$ adjust, exactly as here. That is why the colour of light does not change when it enters glass, though its wavelength does.
The tsunami. A long ocean wave entering shallow water meets a rising impedance continuously rather than at a point. Almost nothing reflects — which is the problem set’s smeared junction — and the amplitude climbs as the depth falls. The wave that arrives at the coast is the one that was never turned back.
Impedance matching as engineering. Between an amplifier and a speaker, an antenna and a feedline, an optical fibre and a detector (module 47), the same transformer idea recurs: you cannot change either end, so you put something in between.

Check your understanding¶

A pulse travels along a heavy rope and reaches a knot where the rope becomes a light thread; the tension is the same throughout. A second, identical pulse travels along the same heavy rope toward a wall. Which pair of reflections do you see?
A. Upright off the light thread, inverted off the wall. B. Inverted in both cases, since reflection always flips a pulse. C. Upright in both cases, since the string is continuous through the knot and cannot flip. D. Inverted off the light thread, upright off the wall.
Check your answer
Correct answer: A — Upright off the light thread, inverted off the wall.
- A. Right, and one formula covers both. The reflected amplitude is r = (Z1 - Z2)/(Z1 + Z2) with Z = sqrt(T mu), so the sign of r is the sign of Z1 - Z2: the echo inverts only when the far side is harder than the near one. A light thread has a smaller Z, so r > 0 and the pulse returns upright; a wall is the limit Z2 -> infinity, so r = -1 and the pulse returns upside down. Simulated, the same junction gives +1/3 from the heavy side and -1/3 from the light side of a 4:1 density pair, with the change of sign exactly at equal densities.
- B. This is the module’s misconception, and it comes from meeting reflection first at a wall, which is one end of the range rather than the rule. Run the formula: r = (Z1 - Z2)/(Z1 + Z2) is negative only for Z2 > Z1. Onto a lighter medium r is positive and the echo is upright — measured at +0.6669 for a 25:1 density drop, against +2/3 predicted. The wall is the special case, not the general one, and the general case turns over at the match rather than at some threshold of hardness.
- C. Continuity of the string is one of the two junction conditions, and it gives 1 + r = t — it does not fix the sign of r on its own. The other condition, that the massless joint carries no net force, is what produces r = (Z1 - Z2)/(Z1 + Z2). At a wall the displacement must stay zero at all times, which forces the returning pulse to cancel the incoming one point by point: that is exactly an inverted copy.
- D. This has the two cases exactly swapped. The useful mental image is what the far side does back to the string. A wall cannot move, so it pulls back with whatever force is needed, and that reaction is the inverted pulse leaving. A thread offers almost no resistance, so the end flies up — nearly twice as far as the incident pulse alone would take it — and the echo it launches is upright.
A string with mu1 = 4.0 g/m is joined to one with mu2 = 36 g/m under a common tension. A wave arrives from the light side. What is the transmitted displacement amplitude coefficient t?
Check your answer
Correct answer: 0.5 (± 0.02)
Impedances first: Z = sqrt(T mu), so Z2/Z1 = sqrt(36/4.0) = 3 whatever the tension is — the tension cancels out of every coefficient at a junction, because it is one number for both sides. Then t = 2 Z1/(Z1 + Z2) = 2/(1 + 3) = 0.5 and r = (1 - 3)/(1 + 3) = -0.5. Check with 1 + r = t: 1 - 0.5 = 0.5. The transmitted wave is half the height of the incident one and the echo is inverted at half its height, which does not mean half the energy goes each way — see the power coefficients, where R = 0.25 and T = 0.75.
A pulse crosses from a string onto one a hundred times lighter. The transmitted pulse is 1.82 times taller than the incident pulse. What happened to the energy?
A. Most of it reflected. The light string is cheap to move, so a large motion of it carries little power — the transmitted fraction is 0.33 and the reflected 0.67. B. Energy was gained at the junction, because the far string adds its own motion to the wave. C. The energy split evenly, since amplitude 1.82 and amplitude 0.82 are close to equal shares. D. It is stored at the junction and released slowly afterwards.
Check your answer
Correct answer: A — Most of it reflected. The light string is cheap to move, so a large motion of it carries little power — the transmitted fraction is 0.33 and the reflected 0.67.
- A. Right. Power is (1/2) Z (omega A)^2, so the transmitted power carries the far side’s impedance, not the near side’s: T = (Z2/Z1) t^2. Here Z2/Z1 = 1/10 and t = 1.818, so T = 0.1 x 3.31 = 0.331, and R = r^2 = 0.818^2 = 0.669. They sum to 1, as they must. A large amplitude on a medium that is cheap to move is not much energy — the same reason a loudspeaker cone moves far in air and hardly at all in water.
- B. Nothing at the junction can add energy: the joint is massless, lossless and carries no store of its own. R + T = 1 follows from the two matching conditions by algebra — substitute r and t and the numerators (Z1 - Z2)^2 + 4 Z1 Z2 collapse onto (Z1 + Z2)^2 — so the bookkeeping was fixed before any energy question was asked. What changed at the junction is the price of a given amplitude, not the total.
- C. Amplitudes are not energies, and the two do not even go in the same order here: the taller pulse is the weaker one. Squaring the amplitudes is not enough either, because the two pulses live on different media. The transmitted fraction needs the flux factor Z2/Z1 in front of t^2, and it is that factor — one tenth here — that turns a squared amplitude ratio of 3.31 into a power fraction of 0.33.
- D. A massless, lossless joint has nowhere to store anything and no time constant with which to release it. Simulating the run and weighing both sides after the pulses separate gives R + T = 1 to seven decimal places, with no residue sitting at the junction. A junction that did store energy would be a different model — a bead of finite mass at the joint, which is a good exam question and not this one.
Two strings under common tension have densities in the ratio mu2/mu1 = 9. What fraction of an incident wave’s energy is transmitted across the junction? Give T as a decimal.
Check your answer
Correct answer: 0.75 (± 0.02)
Z2/Z1 = sqrt(9) = 3, so r = (1 - 3)/(1 + 3) = -0.5 and t = 2/(1 + 3) = 0.5. Then R = r^2 = 0.25 and T = (Z2/Z1) t^2 = 3 x 0.25 = 0.75, which sum to 1. Two traps to avoid. T is not t^2 = 0.25: the transmitted wave travels on the other medium, and the flux factor Z2/Z1 has to come with it. And T is not 1 - t = 0.5: a fraction of energy is never one minus a fraction of amplitude. Note that the answer is the same if the wave arrives from the heavy side instead, where r = +0.5 and t = 1.5 — different amplitudes, identical energy split.
Which junction between two media produces no reflection at all?
A. One where the two impedances are equal, whatever the individual material constants are. B. One where the two wave speeds are equal. C. One where the two densities are equal, since equal mass per metre means nothing changes. D. Any junction, provided the wave arrives slowly enough that the second medium has time to respond.
Check your answer
Correct answer: A — One where the two impedances are equal, whatever the individual material constants are.
- A. Right: r = 0 requires exactly Z1 = Z2 and nothing else. The individual constants may differ — a fluid’s density and sound speed, or a cable’s inductance and capacitance per metre, can both change as long as the product that makes Z comes out the same. On a string this is a blunt instrument, because tension is one number from end to end, so equal impedance means equal density and the matched junction is a string with no junction in it. Matching two media that are both given takes an intermediate layer — the quarter-wave section of the advanced section.
- B. Equal speeds are neither necessary nor sufficient. Z = sqrt(T mu) and v = sqrt(T/mu) are independent combinations of the same two constants, so two media can share a speed and differ in impedance — and then a wave crosses without changing wavelength and still reflects. On a string with one tension the two happen to travel together, which is why this answer feels right and stops being right the moment the tension is allowed to differ, as it does for sound in two different fluids.
- C. True on a string under uniform tension, and only there — which makes it a fact about this apparatus rather than the criterion. The criterion is on Z. Ultrasound gel is matched to tissue while being a completely different material; a 50 ohm resistor terminates a 50 ohm cable while being neither a cable nor a wave-carrying medium at all.
- D. There is no such threshold: r and t contain no frequency at all, so a mismatched junction reflects the same fraction at every frequency. What can depend on frequency is a junction that is not a point — a density that changes gradually over a distance comparable with the wavelength reflects less and less as the ramp widens, which is the problem set’s smeared junction and the reason a tsunami grows instead of bouncing.
Explain what the impedance of a string is, in words, to someone who knows F = ma and has never met the word. Say what it is a property of, what it is not, and why it — rather than the density or the wave speed alone — is what decides reflection.
Check your answer
Impedance is what the medium feels like to push sideways. Take the end of a very long string and wave it: everything you launch travels away and never comes back, so at your hand the string is a pure right-mover with y_t = -v y_x, and the transverse force you must supply is F = -T y_x = (T/v) y_t = Z y_t with Z = sqrt(T mu) = mu v. Force proportional to velocity, in phase with it — the string resists like a dashpot, not like a spring (force proportional to displacement) and not like a mass (force proportional to acceleration). Its unit is kg/s, which is N per m/s: force per unit transverse velocity.

It is a property of the medium alone, of the same standing as the wave speed, and it is a separate property: T and mu are two knobs, so v = sqrt(T/mu) and Z = sqrt(T mu) can be set independently. Two media can share a wave speed and differ in impedance. It is not a property of the wave — no amplitude, width, shape or frequency appears in it — and it is not a resistance in the dissipative sense, even though it behaves like one: the work you do is not converted to heat, it leaves as a travelling wave and does not come back. Reflection is decided by Z because the junction conditions are about force and displacement, not about mass or speed. Continuity of displacement and continuity of the transverse force -T y_x are the two statements available at a massless joint, and the second one is where T/v = Z enters. Solving them gives r = (Z1 - Z2)/(Z1 + Z2), in which the medium constants appear only through that one combination. Density alone cannot decide it, because doubling both tension and density leaves the reflection unchanged while doubling Z on both sides; speed alone cannot decide it either, for the same kind of reason. This is also why the result travels: for sound Z = rho c, for a cable Z = sqrt(L’/C’), for light Z = Z0/n, and the same formula for r applies with the local name substituted.


7. Ultrasound gel has an acoustic impedance close to that of soft tissue (about 1.6e6 against 1.5e6 kg m^-2 s^-1); air's is about 430. What does the gel accomplish?

A. It replaces the air gap, whose enormous impedance mismatch with tissue would reflect essentially all of the sound.
B. It lubricates the probe so that the transducer can be moved smoothly across the skin.
C. It slows the sound down so that more of it can be transmitted into the body.
D. It absorbs the reflected sound so that it does not re-enter the probe and confuse the image.

:::{dropdown} Check your answer
**Correct answer: A** — It replaces the air gap, whose enormous impedance mismatch with tissue would reflect essentially all of the sound.

- **A.** Right, and the numbers are dramatic. For a tissue-air interface, r = (1.5e6 - 430)/(1.5e6 + 430) = 0.9994, so R = 0.9989: about 99.9% of the sound comes straight back and the image is black, not merely dim. With gel in the gap, Z2/Z1 = 1.6/1.5 gives r = -0.032 and R = 0.001 — one part in a thousand reflected. The gel is not a lubricant and conducts sound no better than tissue does; it is there to be nearly the same impedance as what is on both sides of it.
- **B.** It does do that, and it is not why the image fails without it. A dry probe pressed firmly still leaves a microscopic air layer, and the tissue-air mismatch reflects about 99.9% of the sound at that layer. Reflectance depends on the impedance ratio and not at all on friction: R = ((Z1 - Z2)/(Z1 + Z2))^2 = 0.9989 for tissue against air.
- **C.** Nothing in r or t depends on the speed by itself. The coefficients contain the impedances only, Z = rho c, and a medium can be made slower without changing Z at all — raise rho and lower c in proportion. The gel works by matching the product, not by changing either factor in a particular direction.
- **D.** The gel is very nearly lossless at these frequencies over a fraction of a millimetre — it absorbs almost nothing. It also does not need to: matching means the reflection is never created, rather than created and then mopped up. That distinction matters, since absorbing the echo would cost the same energy the image needs, while matching keeps it and sends it into the patient.

Exam-style problems for this module — the ones worth doing with a pen — are collected in the problem set.

Explain it in your own words¶

Answer in a few sentences each. No number will save you.

Explain to someone who is certain that reflection always inverts a pulse why a rope tied to a thread returns it upright, without writing down a formula. Then say what a wall and a thread have in common.
A transmitted pulse is twice as tall as the incident one. Say where the energy is, and why the string on the far side being “cheap to move” is the whole answer.
Ultrasound gel is not a lubricant and not a conductor of sound in any special sense. Say in one sentence what it is for.
Two strings of different densities are joined, and someone claims they can make the junction echo-free by pulling harder on one side. Explain what is wrong with that, using the fact that a string at rest has one tension.

Advanced: the flux derivation in full, the analogue table, and a section that erases an echo¶

$R$ and $T$ from the flux directly. The derivation above divided mean powers, which needed module 09’s sinusoidal result. The same answer comes straight from the instantaneous flux $P = -T\,y_x\,y_t$ , and it is worth seeing once because it never assumes a sinusoid.

On the left of the junction the field is incident plus reflected, so the flux there has three terms: one from each wave and one cross term. For pulses that do not overlap in time the cross term integrates to nothing, and the energies are $\int P\,\mathrm{d}t$ over each pulse:

E_{\text{inc}} = Z_1\!\int (y_t^{\text{inc}})^2\,\mathrm{d}t, \qquad E_{\text{ref}} = Z_1 r^2\!\int (y_t^{\text{inc}})^2\,\mathrm{d}t, \qquad E_{\text{tr}} = Z_2 t^2\!\int (y_t^{\text{inc}})^2\,\mathrm{d}t,

(11)

using $P = Z\,y_t^2$ for a travelling wave in each medium — module 09’s $P = vu$ written with the impedance. Divide by $E_{\text{inc}}$ and $R = r^2$ , $T = (Z_2/Z_1)t^2$ drop out with no mention of frequency. That is why the laboratory can weigh pulses and still land on formulas derived for sinusoids.

The analogue table. Impedance is not a string concept that others borrow. It is what any medium answers when asked for force per unit velocity, and only the names change:

System	Displacement-like	Force-like	Impedance
string	transverse displacement $y$	transverse force $-T y_x$	$Z = \sqrt{T\mu} = \mu v$
sound in a fluid	particle displacement	pressure $p$	$Z = \rho c$
transmission line	charge	voltage $V$	$Z = \sqrt{L'/C'}$
light (normal incidence)	electric field $E$	magnetic field $H$	$Z = \sqrt{\mu_0/\varepsilon} = Z_0/n$

The last row is module 14’s, and it carries a subtlety worth meeting once. Light’s impedance falls as the index rises, $Z = Z_0/n$ , and yet the Fresnel coefficient at normal incidence reads $r = (n_1 - n_2)/(n_1 + n_2)$ — the index in the slot where this module puts $Z$ . Both statements are true. The amplitude quoted for light is the electric field, which is the force-like member of its pair, where a string’s $y$ is the motion-like member of ours; swapping which member you track inverts the mismatch ratio, $Z o 1/Z$ , and the two inversions cancel into one identical formula. Module 18 boxes the two side by side.

A junction with no echo, on a string. Insert a section of impedance $Z_2 = \sqrt{Z_1Z_3}$ and length $\lambda_2/4$ between media 1 and 3, where $\lambda_2$ is the wavelength in the section. Its front face reflects, its back face reflects, and the second reflection makes a round trip of $2 \times \lambda_2/4 = \lambda_2/2$ — half a wavelength, so it returns exactly out of step with the first. Equal sizes, opposite signs, nothing comes back.

The animation above does this for a 9:1 density step: $R$ falls from 0.2504 to 0.0077 with a 57.7 mm sliver of string. The cancellation is exact only at the design frequency, and the residual there is the packet’s own bandwidth, not a numerical error — which is exactly why a lens coating is optimised for the middle of the visible spectrum and leaves the purple fringe that tells you it is there. The problem set asks for $R(\omega)$ .

And the kick the wall takes. A pulse reflecting off a wall reverses the transverse momentum it carries, $\int \mu\,y_t\,\mathrm{d}x$ , so the wall receives twice that in impulse. This is honest and unremarkable. The longitudinal question — whether the wall is also pushed along the string’s direction — is the one module 09 flagged as open, and this module does not reopen it: the answer is second order in the slope, and everything computed here is first order in $y_x^2$ and agrees across every version of the model.