4  Ultrasound

The instrument used to perform an ultrasound scan of the heart is called an echocardiogram. It is noninvasive, operates in real time, and resolves the chambers contracting and the blood streaming through the valves, all from a probe pressed against the chest (Grogan and Mount 2023). The device that transmits and receives is a transducer: a probe, applied to the skin, that converts an electrical signal into a mechanical vibration — launching the pressure wave — and converts the returning pressure wave back into an electrical signal. It is both source and detector. This section discusses how a pressure wave is transmitted into the body, how its echoes are converted into a measurement of depth, what governs the sharpness of the resulting image, and how a Doppler measurement of blood velocity is obtained from the frequency shift in a returning echo.

The frequency is chosen for a specific engineering purpose. Human hearing spans approximately 20 Hz to 20,000 Hz (20 kHz); ultrasound is defined as any sound above this upper limit, inaudible to the human ear (Grogan and Mount 2023). Medical imaging operates at frequencies in the megahertz range, typically 2 to 10 MHz — hundreds of times higher than the highest audible tone (Grogan and Mount 2023; Aldrich 2007). In soft tissue the speed of sound is fixed at \(c \approx 1540~\text{m/s}\), so \(v = f\lambda\) rearranges to \(\lambda = c/f\), and a megahertz frequency yields a wavelength below one millimeter. Because a wave resolves detail no finer than its own wavelength, the elevated frequency produces a wavelength small enough to resolve important features of the heart.

The code below computes the wavelength across this range, from audible sound to the pulses emitted by an ultrasound probe.

import numpy as np
import matplotlib.pyplot as plt
plt.style.use("../../theme/bioe10.mplstyle")

c = 1540.0                      # speed of sound in soft tissue, m/s

# --- wavelength lambda = c / f, from audible frequencies up to ultrasound ---
points = {"20 Hz": 20, "20 kHz": 20e3, "2 MHz": 2e6, "10 MHz": 10e6}
print("wavelength in soft tissue (c = 1540 m/s):")
for label, f in points.items():
    lam = c / f
    shown = f"{lam*1e3:.3g} mm" if lam < 1 else f"{lam:.3g} m"
    print(f"  {label:>7}  ->  lambda = {shown}")

# --- at a fixed speed, doubling the frequency halves the wavelength ---
f1, f2 = 2e6, 4e6                        # 2 MHz and 4 MHz
lam1, lam2 = c / f1, c / f2
print(f"\n2 MHz -> {lam1*1e3:.3f} mm ;  4 MHz -> {lam2*1e3:.3f} mm "
      f"(doubling f halved lambda)")

fig, (axA, axB) = plt.subplots(1, 2, figsize=(10, 4.2))

Running it prints

wavelength in soft tissue (c = 1540 m/s):
    20 Hz  ->  lambda = 77 m
   20 kHz  ->  lambda = 77 mm
    2 MHz  ->  lambda = 0.77 mm
   10 MHz  ->  lambda = 0.154 mm

2 MHz -> 0.770 mm ;  4 MHz -> 0.385 mm (doubling f halved lambda)

At megahertz frequencies the wavelength falls below a millimeter: a 2 MHz pulse has a wavelength of \(0.77~\text{mm}\), and a 10 MHz pulse just \(0.15~\text{mm}\).

Echoes and acoustic impedance

The sound wave reflects off medium that has different properties from the one it is traveling through. The greater the difference, the stronger the reflection. A mechanical analogy may be helpful in understanding this. Imagine tying a light rope to a heavy rope and flicking a pulse into the light one: at the knot where the ropes are connected, part of the pulse continues into the heavy rope (transmitted), and part bounces back along the light rope (reflected), because the heavy rope resists motion almost like a wall (Figure 4.1).

Figure 4.1: A light rope tied to a heavy rope, a mechanical stand-in for an acoustic impedance mismatch. A pulse flicked into the light rope splits at the knot: part transmits into the heavy rope (upright, and slower, since the heavy rope carries waves more sluggishly), and part reflects back along the light rope (inverted). Here the heavy rope has four times the mass per unit length, giving an impedance ratio \(Z_2 = 2Z_1\) and reflecting 11% of the pulse’s energy. Generated by scripts/gen_ultrasound_rope.py.

A wave reflects at a boundary where the properties of the medium change, and the property that governs the reflection is the acoustic impedance \(z\). The impedance is the medium’s resistance to being set in motion by the incoming pressure wave. It is the product of the medium’s density \(\rho\) and its speed of sound \(c\) (Urone and Hinrichs 2012):

\[z = \rho\,c .\]

Its units are \(\text{kg}\,\text{m}^{-2}\,\text{s}^{-1}\), also termed the rayl. Impedance combines two properties already assigned to the medium in the waves section. An echo is produced wherever the impedance changes, and the magnitude of the echo increases with the size of the mismatch. When a wave travels from a medium of impedance \(z_1\) into one of impedance \(z_2\), the fraction of the incident intensity reflected at the boundary is (Urone and Hinrichs 2012)

\[R = \left(\frac{z_2 - z_1}{z_2 + z_1}\right)^{2}.\]

When the media are closely matched, \(z_2 \approx z_1\), the numerator approaches zero and \(R \approx 0\): almost no intensity reflects and the wave is transmitted. When the media are strongly mismatched, \(R\) approaches 1 and nearly all the intensity reflects. Table 4.1 lists representative impedances for the body’s tissues, adapted from OpenStax College Physics:

Table 4.1: Representative acoustic impedances, \(z=\rho c\) (Urone and Hinrichs 2012).
Medium Acoustic impedance \(z\) (\(10^{6}~\text{kg}\,\text{m}^{-2}\text{s}^{-1}\))
Air 0.000429
Fat 1.34
Blood 1.66
Soft tissue (average) ~1.63
Muscle 1.70
Bone 5.7 – 7.8

Let’s apply the reflection formula to Table 4.1 for a typical ultrasound scan. The soft tissues (fat, blood, muscle) have impedances that differ only slightly, so the boundaries between them reflect only about one percent of the wave. This weak reflection is the essential feature rather than a limitation: because most of the wave is transmitted at each soft-tissue boundary, the pulse penetrates further and retains sufficient intensity to echo off successive structures. A small fraction reflects at each interface, enough to register it, while the remainder continues onward. This is why a blood–tissue boundary — the inner surface of a heart chamber, the edge of a valve leaflet — appears on an echocardiogram: blood (\(1.66\)) against muscle (\(1.70\)) is a small but genuine mismatch that can be rendered as a bright line.

Air has an impedance thousands of times smaller than tissue, so a tissue–air boundary reflects about 99.9% of the wave, an almost total reflection. A thin layer of air trapped between the probe and the skin would therefore transmit essentially none of the wave into the body. To avoid this, coupling gel is applied to exclude the air and replace the tissue–air interface with a tissue–gel match through which the wave can pass. At the other extreme, bone has an impedance several times that of soft tissue, so a tissue–bone boundary reflects a large fraction of the wave; bones and calcifications appear bright and cast an acoustic shadow behind them. This is one reason the probe for a cardiac scan is angled to image between the ribs rather than through them.

The device that transmits and receives is a transducer: a probe, applied to the skin, that converts an electrical signal into a mechanical vibration that launches a pressure wave. It also converts the returning pressure wave back into an electrical signal. It is both source and detector. Figure 4.2 simulates a transducer firing a pulse across exactly such a boundary, with an impedance ratio \(Z_2 = 4Z_1\) comparable to tissue against bone, and records what the transducer sees.

Figure 4.2: A transducer fires a pulse into a soft, low-impedance medium (\(Z_1\)) that meets a stiffer, denser medium (\(Z_2 = 4Z_1\)), comparable to a tissue–bone boundary. Top: the pulse splits at the interface, part reflecting and part transmitting onward, faster, into medium 2. Bottom: the transducer’s own record — the outgoing pulse, then the returning echo — with the round-trip time giving the boundary’s depth via \(d=\tfrac{1}{2}ct\). This impedance ratio reflects 36% of the pulse’s energy, matching the tissue–bone case worked out in the exercises below. Generated by scripts/gen_ultrasound_echo.py.

Depth from timing

The transducer sends a pulse and records the echo, measuring the time \(t\) it takes to return. Because the pulse travels to the reflecting structure and back at the known tissue speed \(c \approx 1540~\text{m/s}\), the round-trip time can be converted directly into a distance. The pulse traverses a total path \(c\,t\), and the structure lies at half that distance:

\[d = \tfrac{1}{2}\,c\,t .\]

The factor of one-half reflects the round trip: the echo traverses twice the depth, so the depth is half the total path. A structure twice as deep returns its echo twice as late (Figure 4.3). The transducer transmits a pulse and then receives; each interface the pulse crosses returns a faint echo, and the arrival time of each echo encodes the depth of the interface that produced it.

import numpy as np

c = 1540.0                 # speed of sound in soft tissue, m/s
d1, d2 = 0.05, 0.12         # interface depths, m (5 cm and 12 cm)
t1, t2 = 2 * d1 / c, 2 * d2 / c    # round-trip time to each interface, t = 2d/c
# ... simulate a short windowed-sinusoid pulse at t=0, t1, and t2, and plot the
# summed trace against time
Figure 4.3: A single transmit pulse (at \(t=0\)) followed by two echoes returning from interfaces at 5 cm and 12 cm in tissue. The deeper interface returns its echo proportionally later; reading the round-trip time and applying \(d=\tfrac{1}{2}ct\) recovers each depth. Generated by scripts/gen_ultrasound_pulse_echo.py.

A pulse and its sequence of timed echoes, constitutes one line of an image: a profile of the structures directly beneath the probe at every depth along the beam. An echocardiogram is assembled by sweeping the beam across the heart and stacking many such lines into a two-dimensional image updated tens of times per second. The mechanism by which the scanner steers and combines these lines (beamforming and image reconstruction) is beyond the present scope of this text.

NoteKey result: pulse–echo depth

An ultrasound scanner locates a structure by timing its echo. With the tissue speed of sound \(c \approx 1540~\text{m/s}\) fixed and known, a round-trip time \(t\) places the reflector at a depth

\[\boxed{\; d = \tfrac{1}{2}\,c\,t \;}\]

the factor of one-half arising because the pulse travels out and back. Many such timed lines, placed side by side, compose an image.

ImportantKnowledge check

An ultrasound pulse is sent into soft tissue and its echo returns \(80~\mu\text{s}\) later (\(1~\mu\text{s} = 10^{-6}~\text{s}\)). Using \(c = 1540~\text{m/s}\), how deep is the structure that reflected it?

The round-trip distance is \(c\,t = 1540 \times 80\times10^{-6} = 0.123~\text{m}\), so the depth is half of that: \(d = \tfrac{1}{2}(0.123) = 0.0616~\text{m} \approx \mathbf{6.2~cm}\). The structure sits about 6 cm beneath the probe.

The imaging trade-off: resolution versus penetration

Resolution and penetration depend on frequency in opposing directions. Raising the frequency sharpens the image but reduces the depth to which the wave penetrates. The choice of frequency is therefore a compromise between the desired resolution and the required penetration depth.

Consider resolution first. A wave resolves detail approximately as fine as its wavelength, so the finest structure a beam can distinguish along its direction of travel — its axial resolution — improves as the wavelength decreases. Since \(\lambda = c/f\) at fixed tissue speed, higher frequency yields shorter wavelength and finer detail. A 3 MHz pulse has \(\lambda \approx 0.5~\text{mm}\); a 7.5 MHz pulse has \(\lambda \approx 0.2~\text{mm}\), resolving features two and a half times smaller.

Consider penetration second. As a pressure wave travels through tissue it is steadily attenuated (its energy absorbed and scattered), and the attenuation increases with frequency, at a rate of roughly \(0.5~\text{dB}\) per centimeter per megahertz (Grogan and Mount 2023; Aldrich 2007). A high-frequency pulse is sharp but shallow, decaying within a few centimeters; a low-frequency pulse is coarser but reaches further.

import numpy as np

c = 1540.0                      # speed of sound in soft tissue, m/s

for f in (3e6, 7.5e6):          # a deep cardiac probe vs a superficial vascular probe
    lam = c / f                 # wavelength = axial-resolution scale
    penetration = 500 * lam     # effective imaging depth, rule of thumb ~500 wavelengths
    print(f"f = {f/1e6:4.1f} MHz  ->  lambda = {lam*1e3:5.3f} mm ,"
          f"  penetration ~ {penetration*100:4.1f} cm")

which prints

f =  3.0 MHz  ->  lambda = 0.513 mm ,  penetration ~ 25.7 cm
f =  7.5 MHz  ->  lambda = 0.205 mm ,  penetration ~ 10.3 cm
import numpy as np
import matplotlib.pyplot as plt
plt.style.use("../../theme/bioe10.mplstyle")

c = 1540.0                        # speed of sound in soft tissue, m/s
f = np.linspace(1e6, 15e6, 400)   # diagnostic band, 1-15 MHz
lam = c / f                       # wavelength = axial-resolution scale
penetration = 500 * lam           # effective imaging depth, rule of thumb
# ... plot lambda vs. f (left) and penetration vs. f (right), marking the
# 3 MHz and 7.5 MHz probes from the printed values above
Figure 4.4: The ultrasound imaging trade-off. Left: axial resolution, set by the wavelength \(\lambda = c/f\), sharpens as frequency rises (smaller \(\lambda\), finer detail). Right: penetration depth (rule of thumb \(\approx 500\lambda\)) falls with frequency, because attenuation worsens at higher frequency. The increase in frequency that sharpens the image also shortens the reach. Markers show a 3 MHz cardiac probe and a 7.5 MHz vascular probe. Generated by scripts/gen_ultrasound_resolution.py.

The choice of frequency is therefore effectively a choice of depth. To image the heart, several centimeters within the chest, an echocardiogram uses a low-frequency probe near 2–3 MHz and accepts coarser detail as the cost of that depth. To image a blood vessel or tendon immediately beneath the skin, a probe of 7–15 MHz provides fine detail without requiring substantial reach (Figure 4.4).

Doppler: measuring blood velocity from a frequency shift

The methods so far image structure — the location walls and valves. However, echocardiography can also measure the speed of the blood itself, by the Doppler effect.

The Doppler effect is the shift in the observed frequency of a wave when its source or observer is in motion: motion toward the observer compresses the waves to a higher frequency, and motion away stretches them to a lower one. The pitch of a passing ambulance siren, rising on approach and falling on recession, is the familiar illustration. The same effect applies to an ultrasound pulse reflected from a moving target, and blood presents many such targets in the red cells streaming through a valve. When the beam is directed at blood moving toward the probe, the echo returns at a slightly higher frequency than transmitted, and the magnitude of this Doppler shift \(f_D\) is a direct measure of the blood’s speed.

The moving blood produces the shift twice, once as a receiver and once as a source.

NoteDerivation — the Doppler shift happens twice

Take a red blood cell moving straight along the beam, toward the probe, at speed \(v\), with the probe transmitting frequency \(f_0\) and tissue sound speed \(c\).

Stage 1 — the cell is a moving receiver. Figure 4.5 shows the general case: a moving observer closes in on wavefronts that a stationary source emits at the ordinary wavelength \(\lambda = c/f_0\). Because the observer meets each successive front sooner than a stationary one would, it receives wavefronts at an elevated rate,

\[f_1 = \frac{c+v}{\lambda} = f_0\,\frac{c+v}{c}.\]

Figure 4.5: The Doppler shift from a moving receiver alone: a stationary probe emits wavefronts at a fixed period \(T\), and a fixed reference point meets each one every \(T\). The blood cell, moving toward the probe, closes part of the gap to each new wavefront before it arrives, so it meets successive wavefronts spaced by a shorter period \(T' < T\) — a higher frequency than the probe transmitted. Generated by scripts/gen_doppler_period.py.

Stage 2 — the cell is a moving source. Having absorbed the beam at frequency \(f_1\), the cell immediately re-emits it as an echo, and it does so while still advancing toward the probe. Figure 4.6 shows what a moving source does to the waves it emits: each successive wavefront is launched from a point closer to the probe than the last, compressing them into a shorter wavelength ahead of the source,

\[\lambda_2 = (c-v)\,T_1 = \frac{c-v}{f_1}, \qquad\text{so the probe receives}\qquad f_2 = \frac{c}{\lambda_2} = f_1\cdot\frac{c}{c-v}.\]

Figure 4.6: The Doppler shift from a moving source alone: a source advancing toward a fixed point emits wavefronts that bunch together ahead of it, so a receiver in front measures a higher frequency than the source’s own — exactly the situation of the blood cell re-emitting the beam as it continues toward the probe. Generated by scripts/gen_doppler_source.py.

Combining the two stages. Substituting \(f_1\) from Stage 1 into \(f_2\) gives the frequency the probe finally records,

\[f_2 = f_0\,\frac{c+v}{c-v},\]

so the total shift is

\[\Delta f = f_2 - f_0 = f_0\left(\frac{c+v}{c-v}-1\right) = f_0\,\frac{2v}{c-v}.\]

Blood speeds are a few meters per second against a tissue sound speed of \(c\approx1540~\text{m/s}\), so \(v\ll c\) and the small \(v\) in the denominator may be dropped, leaving \(\Delta f \approx 2f_0 v/c\): each stage contributes a shift of \(f_0v/c\), and the two add.

For a beam that meets the blood’s motion at an angle \(\theta\), only the velocity component along the beam, \(v\cos\theta\), contributes to either stage, giving the general result (Urone and Hinrichs 2012)

\[f_D = \frac{2 f_0\, v \cos\theta}{c},\]

where:

  • \(f_0\) — the transmitted frequency
  • \(c\) — the tissue sound speed
  • \(\theta\) — the angle between the beam and the blood’s direction of travel

The factor \(\cos\theta\) enters because only the velocity component along the beam compresses the waves; the shift is largest when the beam is aligned with the jet, \(\theta \approx 0\) and \(\cos\theta \approx 1\). Alignment therefore matters in practice: a beam offset by an unmeasured angle \(\theta\) is treated as aligned (\(\cos\theta \approx 1\)), so the machine reports \(v\cos\theta\), an under-estimate of the true speed. At \(\theta = 60^\circ\), \(\cos\theta = 0.5\) and the reported speed is halved, which is why a sonographer aligns the beam with the jet. Rearranging the relation gives the velocity, the quantity otherwise inaccessible:

\[v = \frac{c\, f_D}{2 f_0 \cos\theta}.\]

The frequency shift of the returning echo thus yields the blood’s speed. For a cardiac probe at \(f_0 = 2~\text{MHz}\) and a fast valve jet, \(f_D\) evaluates to a few kilohertz, within the audible range. The machine plays the Doppler shift through a loudspeaker, and the clinician hears the blood velocity as a tone whose pitch tracks the flow. The heart section began by listening to the heart with a stethoscope; the Doppler method renders a different quantity audible — the measured velocity itself.

A velocity alone is not yet a diagnosis: the quantity that indicates whether a valve is dangerously narrowed is the pressure difference the heart must generate to drive blood through it, and reaching pressure from this velocity requires two further conservation laws, taken up in the sections that follow.

NoteKey result: sound, echoes, and the echocardiogram
  • Ultrasound is a high frequency pressure wave (≈2–10 MHz) designed so that its sub-millimeter wavelength can resolve anatomy; a transducer both transmits the pulse and receives its echoes.
  • Echoes come from impedance mismatches. With \(z = \rho c\), the reflected fraction is \(R = \big((z_2-z_1)/(z_2+z_1)\big)^2\): soft-tissue boundaries reflect weakly (so the pulse sees deep), while air and bone reflect strongly (hence coupling gel and acoustic shadows).
  • Depth comes from timing, \(d = \tfrac{1}{2}ct\); stacking timed lines makes an image.
  • Resolution and penetration trade off. Higher frequency \(\to\) shorter \(\lambda\) \(\to\) finer detail, but \(\to\) more attenuation \(\to\) shallower reach.
  • Doppler measures velocity. The echo’s frequency shift gives the blood-jet velocity, \(v = c f_D / (2 f_0 \cos\theta)\) — the quantity the sections that follow convert into the valve pressure gradient.
NoteFurther reading

Exercises

TipProblem 1

Two ultrasound probes are used on soft tissue, where the speed of sound is \(c = 1540~\text{m/s}\): probe A transmits at \(f = 4~\text{MHz}\), probe B at \(f = 15~\text{MHz}\).

  1. Find the wavelength of each in tissue.
  2. Axial resolution is limited to about one wavelength. Which probe resolves finer detail, and by what factor?
  1. \(\lambda_A = c/f = 1540/(4\times10^{6}) = 3.85\times10^{-4}~\text{m} = \mathbf{0.385~mm}\); \(\lambda_B = 1540/(15\times10^{6}) = 1.03\times10^{-4}~\text{m} = \mathbf{0.103~mm}\).

  2. The finer resolution belongs to the shorter wavelength, so probe B (15 MHz). The ratio is \(\lambda_A/\lambda_B = 15/4 = \mathbf{3.75}\) — probe B resolves detail about 3.75 times finer, because at fixed \(c\) the wavelength scales as \(1/f\).

TipProblem 2

A radiologist can choose a \(2.5~\text{MHz}\) probe or a \(10~\text{MHz}\) probe (\(c = 1540~\text{m/s}\)). Take the effective imaging depth to be roughly \(500\) wavelengths.

  1. Find the wavelength and the effective penetration depth for each probe.
  2. One patient needs an abdominal scan reaching about 20 cm deep; another needs a superficial scan of a wrist tendon about 2 cm deep at the finest possible detail. Which probe suits each, and why?
  1. At 2.5 MHz, \(\lambda = 1540/(2.5\times10^{6}) = 0.616~\text{mm}\), so penetration \(\approx 500\times0.616~\text{mm} = 308~\text{mm} = \mathbf{30.8~cm}\). At 10 MHz, \(\lambda = 1540/(10\times10^{6}) = 0.154~\text{mm}\), so penetration \(\approx 500\times0.154~\text{mm} = 77~\text{mm} = \mathbf{7.7~cm}\).

  2. The abdominal scan needs reach, so use the 2.5 MHz probe (penetration \(\approx 31\) cm easily covers 20 cm; the coarser \(0.6\) mm detail is the price). The wrist tendon is shallow and wants detail, so use the 10 MHz probe (its \(7.7\) cm reach is more than enough for 2 cm, and its \(0.15\) mm wavelength resolves four times finer). This is the resolution–penetration trade-off in practice: match the frequency to the depth you must reach.

TipProblem 3

During an echocardiogram a pulse is sent into the chest (\(c = 1540~\text{m/s}\)). The front wall of the heart returns an echo at \(t_1 = 130~\mu\text{s}\), and the back wall returns one at \(t_2 = 156~\mu\text{s}\) (\(1~\mu\text{s} = 10^{-6}~\text{s}\)).

  1. How deep is the front wall?
  2. How deep is the back wall?
  3. What is the front-to-back dimension of the heart at this position?
  1. \(d_1 = \tfrac{1}{2}c\,t_1 = \tfrac{1}{2}(1540)(130\times10^{-6}) = 0.1001~\text{m} \approx \mathbf{10.0~cm}\).

  2. \(d_2 = \tfrac{1}{2}c\,t_2 = \tfrac{1}{2}(1540)(156\times10^{-6}) = 0.1201~\text{m} \approx \mathbf{12.0~cm}\).

  3. The chamber spans the difference, \(d_2 - d_1 = 12.0 - 10.0 = \mathbf{2.0~cm}\). Equivalently, work from the extra round-trip time \(\Delta t = 26~\mu\text{s}\): \(\tfrac{1}{2}c\,\Delta t = \tfrac{1}{2}(1540)(26\times10^{-6}) = 0.020~\text{m}\) — the same 2 cm.

TipProblem 4

Acoustic impedance is \(z = \rho c\). Fat has \(\rho = 950~\text{kg/m}^3\) and \(c = 1450~\text{m/s}\); muscle has \(\rho = 1050~\text{kg/m}^3\) and \(c = 1580~\text{m/s}\).

  1. Compute the acoustic impedance of each.
  2. Compute the fraction of intensity reflected at a fat–muscle boundary, \(R = \big((z_2 - z_1)/(z_2 + z_1)\big)^2\).
  3. In one sentence, explain why such a small reflection is exactly what lets ultrasound image structures deep inside soft tissue.
  1. \(z_{\text{fat}} = 950 \times 1450 = 1.38\times10^{6}~\text{kg}\,\text{m}^{-2}\text{s}^{-1}\); \(z_{\text{muscle}} = 1050 \times 1580 = 1.66\times10^{6}~\text{kg}\,\text{m}^{-2}\text{s}^{-1}\).

  2. \(R = \left(\dfrac{1.66\times10^{6} - 1.38\times10^{6}}{1.66\times10^{6} + 1.38\times10^{6}}\right)^2 = \left(\dfrac{0.281}{3.037}\right)^2 = (0.0925)^2 = 0.0086 = \mathbf{0.86\%}\).

  3. Because only about 0.9% of the wave reflects at each soft-tissue boundary, the other 99% is transmitted and keeps traveling, so the pulse still has energy to reach — and echo off — structures much deeper in the body.

TipProblem 5

Use \(z_{\text{soft tissue}} \approx 1.63\times10^{6}\), \(z_{\text{air}} \approx 429\), and \(z_{\text{bone}} \approx 6.5\times10^{6}\) (all in \(\text{kg}\,\text{m}^{-2}\text{s}^{-1}\)), with \(R = \big((z_2-z_1)/(z_2+z_1)\big)^2\).

  1. Find the fraction reflected at a soft-tissue–air boundary. What clinical practice does this near-total reflection force?
  2. Find the fraction reflected at a soft-tissue–bone boundary. What two visual consequences does this large reflection have on the image?
  1. \(R = \left(\dfrac{1.63\times10^{6} - 429}{1.63\times10^{6} + 429}\right)^2 \approx (0.99947)^2 = 0.9989 = \mathbf{99.9\%}\). Almost the entire wave reflects at a tissue–air gap, so a trapped film of air would block the scan entirely — which is why the operator applies coupling gel to exclude all air between the probe and the skin.

  2. \(R = \left(\dfrac{6.5\times10^{6} - 1.63\times10^{6}}{6.5\times10^{6} + 1.63\times10^{6}}\right)^2 = (0.599)^2 = 0.359 = \mathbf{36\%}\). About a third of the wave reflects, so bone appears bright on the image, and because so little passes through it, structures behind the bone fall into an acoustic shadow.

TipProblem 6

A Doppler study uses \(f_0 = 2.0~\text{MHz}\) with the beam aligned along the flow (\(\cos\theta = 1\)), tissue speed \(c = 1540~\text{m/s}\). The returning echo is shifted by \(f_D = 5.2~\text{kHz}\).

  1. Find the blood velocity from \(v = c f_D / (2 f_0 \cos\theta)\).
  2. The formula carries a factor of 2. In one sentence, what physical fact puts it there?
  1. \(v = \dfrac{c\,f_D}{2 f_0 \cos\theta} = \dfrac{(1540)(5.2\times10^{3})} {2(2.0\times10^{6})(1)} = \dfrac{8.008\times10^{6}}{4.0\times10^{6}} = \mathbf{2.0~m/s}\).

  2. The moving blood is Doppler-shifted twice — once as a moving observer receiving the incoming wave, and again as a moving source re-emitting the echo — so the two shifts add and the total is double a single shift.

TipProblem 7

A diagnostic ultrasound probe emits at \(f = 5~\text{MHz}\) into soft tissue, where the speed of sound is \(c = 1540~\text{m/s}\).

  1. What is the wavelength of this pulse in tissue?
  2. The probe is switched to \(f = 10~\text{MHz}\). Without a full recalculation, state what happens to the wavelength, then confirm it by computing the new value.
  1. \(\lambda = c/f = 1540 / (5\times10^{6}) = 3.08\times10^{-4}~\text{m} = \mathbf{0.308~mm}\).

  2. The medium is unchanged, so \(c\) is fixed and \(\lambda \propto 1/f\): doubling the frequency halves the wavelength. Check: \(\lambda = 1540/(10\times10^{6}) = 1.54\times10^{-4}~\text{m} = \mathbf{0.154~mm}\), exactly half of the answer in (a).

TipProblem 8

Imaging resolution is limited to roughly one wavelength: a wave can only resolve detail about as small as its own \(\lambda\). Work in soft tissue, \(c = 1540~\text{m/s}\).

  1. To resolve detail as fine as \(0.5~\text{mm}\), roughly what frequency must the probe use?
  2. Explain, using \(v = f\lambda\), why you cannot image at this resolution with an audible \(20~\text{kHz}\) sound.
  1. Set \(\lambda = 0.5~\text{mm} = 5\times10^{-4}~\text{m}\) and solve \(f = c/\lambda = 1540/(5\times10^{-4}) = 3.08\times10^{6}~\text{Hz} \approx \mathbf{3~MHz}\), squarely in the diagnostic ultrasound range.

  2. At \(20~\text{kHz}\) the wavelength in tissue is \(\lambda = c/f = 1540/20000 = 0.077~\text{m} = \mathbf{7.7~cm}\). Detail finer than about 8 cm is simply invisible to such a wave, so an audible sound could never resolve a half-millimeter feature; you must raise \(f\) into the megahertz to shrink \(\lambda\) enough, which is exactly what ultrasound does.

TipProblem 9

An ultrasound probe sends a short pulse into tissue (\(c = 1540~\text{m/s}\)), and \(130~\mu\text{s}\) later (\(1~\mu\text{s} = 10^{-6}~\text{s}\)) it detects the echo bouncing back from a structure directly beneath it. Because the pulse travels out to the structure and then back, the echo’s total path is twice the depth.

  1. How far did the pulse travel in total during those \(130~\mu\text{s}\)?
  2. How deep is the reflecting structure?
  3. This “time it, then convert to distance” is the core of ultrasound imaging. Which single wave property made the conversion possible?
  1. Distance is speed times time: \(d_{\text{total}} = c\,\Delta t = 1540 \times 130\times10^{-6} = 0.200~\text{m} = \mathbf{20.0~cm}\).

  2. The round trip is twice the depth, so the depth is \(d = d_{\text{total}}/2 = 0.200/2 = 0.100~\text{m} = \mathbf{10~cm}\).

  3. The speed of sound in the medium, \(c \approx 1540~\text{m/s}\) (fixed by the tissue, and the same for every pulse), is what lets a measured time be turned into a distance — exactly the pulse–echo principle developed earlier in this section.

References

Aldrich, John E. 2007. “Basic Physics of Ultrasound Imaging.” Critical Care Medicine 35 (5): S131–37. https://doi.org/10.1097/01.CCM.0000260624.99430.22.
Grogan, Scott P., and Cristin A. Mount. 2023. “Ultrasound Physics and Instrumentation.” In StatPearls. StatPearls Publishing, Treasure Island (FL). https://www.ncbi.nlm.nih.gov/books/NBK570593/.
Heys, Jeffrey J., Nicole Holyoak, Anna M. Calleja, Marek Belohlavek, and Hari P. Chaliki. 2010. “Revisiting the Simplified Bernoulli Equation.” The Open Biomedical Engineering Journal 4: 123–28. https://doi.org/10.2174/1874120701004010123.
Maier, Andreas, Stefan Steidl, Vincent Christlein, and Joachim Hornegger, eds. 2018. Medical Imaging Systems: An Introductory Guide. Springer. https://www.ncbi.nlm.nih.gov/books/NBK546147/.
Thapaliya, Arbin, Alec Sithole, Michael Welsh, and Gaston Dana. 2022. Ultrasound Physics and Its Application in Medicine. https://pressbooks.palni.org/ultrasoundphysicsanditsapplicationinmedicine/.
Urone, Paul Peter, and Roger Hinrichs. 2012. College Physics 1e. OpenStax, Rice University. https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/17%3A_Physics_of_Hearing.