Ultrasound & Acoustic Impedance
Specification: OCR A H556 | Section: 6.5.3 Ultrasound (including Doppler, 6.5.3(g)) | Teaching time: ~3 hours
- Explain what ultrasound is and why it is used in medical imaging
- Describe the piezoelectric effect and how transducers generate and detect ultrasound
- Calculate acoustic impedance Z = ρc for different materials
- Calculate the intensity reflection coefficient at a boundary between two media
- Explain why coupling gel is needed and perform impedance matching calculations
- Compare A-scan and B-scan techniques and evaluate ultrasound against other imaging methods
- Use reflected Doppler frequency shifts to calculate blood speed, including beam/flow angle (6.5.3(g))
What is Ultrasound?
Ultrasound refers to sound waves with frequencies above the upper limit of human hearing (above 20 kHz). In medical applications, frequencies in the range 1–15 MHz are used — much higher than in industrial or cleaning applications.
Ultrasound is a longitudinal mechanical wave: the particles of the medium oscillate parallel to the direction of energy transfer, forming compressions and rarefactions. Unlike X-rays, ultrasound is non-ionising, so it avoids the ionising dose of X-rays. Diagnostic use controls exposure when imaging foetuses and soft tissues.
Medical ultrasound uses pulsed waves — a short pulse is sent, then the transducer waits to receive echoes before sending the next pulse. The time delay between sending and receiving tells you the depth of the reflecting surface.
Ultrasound scanning relies on two key facts:
- Ultrasound is reflected at boundaries between materials of different densities (e.g. tissue–organ, tissue–bone)
- Ultrasound is pulsed — the transducer sends a pulse, then listens for returning echoes
Because ultrasound is strongly reflected at an air–tissue boundary (although sound can propagate through air), a coupling gel must be used between the transducer and the skin. More on this later.
The Piezoelectric Effect
The heart of any ultrasound machine is the piezoelectric transducer. It contains a piezoelectric crystal (often made from lead zirconate titanate, PZT) that can convert electrical signals into mechanical vibrations and vice versa.
How it works
Transmitting (generating ultrasound): When a high-frequency alternating potential difference is applied across the piezoelectric crystal, the crystal expands and contracts at the same frequency as the AC signal. These rapid mechanical deformations produce longitudinal ultrasound waves.
Receiving (detecting echoes): When an ultrasound echo reaches the crystal, the pressure of the compression causes the crystal to deform. This deformation generates a p.d. across the crystal (the reverse piezoelectric effect). This small voltage is amplified and processed to build the image.
The same crystal acts as both transmitter and receiver. The transducer cannot transmit and receive at the same time — it pulses, then listens.
Acoustic Impedance
Illustrative approximate tissue data are used throughout this page. Each displayed impedance is derived from the listed density and speed, Z = ρc; real values vary with tissue and conditions.
Acoustic impedance is a property of a material that describes how much it resists the passage of sound waves. It depends on two things: how dense the material is, and how fast sound travels through it.
Where:
- Z = acoustic impedance, measured in kg m⁻² s⁻¹ (also called the Rayl)
- ρ = density of the medium (kg m⁻³)
- c = speed of sound in the medium (m s⁻¹)
A material with high density and a high speed of sound will have a high acoustic impedance. Bone, for example, has a very high Z because it is dense and sound travels quickly through it. Air has an extremely low Z because it has very low density and sound travels relatively slowly.
Typical Values of Acoustic Impedance
| Material | Density ρ (kg m⁻³) | Speed of sound c (m s⁻¹) | Acoustic impedance Z (kg m⁻² s⁻¹) |
|---|---|---|---|
| Air | 1.2 | 343 | 411.6 |
| Water | 1000 | 1480 | 1.48 × 10⁶ |
| Fat | 920 | 1450 | 1.334 × 10⁶ |
| Blood | 1060 | 1570 | 1.6642 × 10⁶ |
| Muscle | 1050 | 1580 | 1.659 × 10⁶ |
| Soft tissue (average) | 1060 | 1540 | 1.6324 × 10⁶ |
| Lung | 400 | 650 | 0.26 × 10⁶ |
| Bone | 1900 | 4000 | 7.6 × 10⁶ |
Notice the enormous difference between air (411.6) and soft tissue (~1.6 × 10⁶) — they differ by a factor of nearly 4000. This is why virtually no ultrasound would pass from air into the body without coupling gel.
Worked Examples: Calculating Acoustic Impedance
Calculate the acoustic impedance of muscle given its density is 1050 kg m⁻³ and the speed of sound in muscle is 1580 m s⁻¹.
Z = ρc = 1050 × 1580 = 1,659,000
Z = 1.659 × 10⁶ kg m⁻² s⁻¹ = 1.66 × 10⁶ kg m⁻² s⁻¹ (to 3 s.f.)
Calculate the acoustic impedance of bone given its density is 1900 kg m⁻³ and the speed of sound in bone is 4000 m s⁻¹.
Z = ρc = 1900 × 4000 = 7,600,000
Z = 7.6 × 10⁶ kg m⁻² s⁻¹ (to 2 s.f.)
Calculate the acoustic impedance of air given its density is 1.2 kg m⁻³ and the speed of sound in air is 343 m s⁻¹.
Z = ρc = 1.2 × 343 = 411.6
Z = 411.6 kg m⁻² s⁻¹ ≈ 410 kg m⁻² s⁻¹ (to 2 s.f.)
Blood has an acoustic impedance of 1.6642 × 10⁶ kg m⁻² s⁻¹. If the density of blood is 1060 kg m⁻³, calculate the speed of sound in blood.
c = Z / ρ = (1.6642 × 10⁶) / 1060 = 1570
c = 1570 m s⁻¹ (to 3 s.f.)
The speed of sound in fat is 1450 m s⁻¹. If the acoustic impedance of fat is 1.334 × 10⁶ kg m⁻² s⁻¹, calculate the density of fat.
ρ = Z / c = (1.334 × 10⁶) / 1450 = 920
ρ = 920 kg m⁻³ (to 3 s.f.)
Intensity Reflection Coefficient
When ultrasound reaches a boundary between two materials with different acoustic impedances, some energy is reflected and some is transmitted. The fraction reflected depends on the difference between the two impedances.
Where:
- Iᵣ = reflected intensity (W m⁻²)
- I₀ = incident intensity (W m⁻²)
- Z₁ = acoustic impedance of the first medium
- Z₂ = acoustic impedance of the second medium
The result is always between 0 (no reflection, perfect transmission) and 1 (total reflection, nothing transmitted). The larger the difference between Z₁ and Z₂, the greater the fraction reflected.
The order of Z₁ and Z₂ doesn't matter for the ratio — the result is the same whether ultrasound goes from medium 1 → 2 or 2 → 1. The fraction (Z₂ − Z₁)/(Z₂ + Z₁) is squared, so the sign cancels out.
Worked Examples: Reflection Coefficient
Calculate the fraction of ultrasound intensity reflected at a boundary between soft tissue (Z = 1.6324 × 10⁶ kg m⁻² s⁻¹) and bone (Z = 7.6 × 10⁶ kg m⁻² s⁻¹).
Iᵣ/I₀ = ((Z₂ − Z₁)/(Z₂ + Z₁))²
= ((7.6 × 10⁶ − 1.6324 × 10⁶)/(7.6 × 10⁶ + 1.6324 × 10⁶))²
= (5.9676/9.2324)²
= (0.6463756)²
Iᵣ/I₀ = 0.4178 — so about 42% of the intensity is reflected
Use Zmuscle = 1.659 × 10⁶ and Zfat = 1.334 × 10⁶ kg m⁻² s⁻¹.
Iᵣ/I₀ = ((1.334 − 1.659)/(1.334 + 1.659))² = (−0.325/2.993)² = 0.011791 ≈ 1.18%.
Calculate the fraction of ultrasound intensity reflected at a boundary between blood (Z = 1.6642 × 10⁶ kg m⁻² s⁻¹) and muscle (Z = 1.659 × 10⁶ kg m⁻² s⁻¹).
Iᵣ/I₀ = ((1.659 − 1.6642)/(1.659 + 1.6642))² (all in units of 10⁶)
= (−0.0052/3.3232)²
= (−0.001565)²
Iᵣ/I₀ = 2.45 × 10⁻⁶ — so only 0.000245% is reflected (nearly all transmitted)
When two materials have similar acoustic impedances (e.g. blood and muscle, or muscle and fat), very little ultrasound is reflected. This means most of the energy continues deeper into the body, allowing imaging of structures beyond the first boundary.
Coupling Gel & Impedance Matching
Now let's answer the critical question: why is gel essential?
Calculate the fraction of ultrasound reflected at an air–skin boundary. Air has Z = 411.6 kg m⁻² s⁻¹ and soft tissue (skin) has Z = 1.6324 × 10⁶ kg m⁻² s⁻¹.
Iᵣ/I₀ = ((Z₂ − Z₁)/(Z₂ + Z₁))²
= ((1.6324 × 10⁶ − 411.6)/(1.6324 × 10⁶ + 411.6))²
Substitute the actual values rather than replacing a nearly complete reflection by exactly 1:
((1 632 400 − 411.6)/(1 632 400 + 411.6))² = 0.998992 ≈ 99.9%
Iᵣ/I₀ ≈ 0.998992 — virtually 100% of the ultrasound is reflected!
Ultrasound gel has Z ≈ 1.5 × 10⁶ kg m⁻² s⁻¹ (similar to soft tissue). Calculate the fraction reflected at a gel–skin boundary.
Iᵣ/I₀ = ((1.6324 × 10⁶ − 1.5 × 10⁶)/(1.6324 × 10⁶ + 1.5 × 10⁶))²
= (0.1324/3.1324)²
= (0.0422679)²
Iᵣ/I₀ = 0.00179 — only 0.179% reflected (99.8% transmitted into the body)
Without gel (air gap): ~99.9% reflected — almost nothing enters the body.
With gel: ~0.179% reflected — nearly all the ultrasound enters the body.
This is called impedance matching: the gel provides an acoustic bridge between the transducer and the skin, eliminating the air gap that would otherwise cause nearly complete reflection.
The coupling gel also serves other practical purposes:
- Eliminates tiny air pockets between the transducer and skin
- Allows the transducer to glide smoothly across the skin
- Has similar acoustic impedance to soft tissue (Z ≈ 1.5 × 10⁶ kg m⁻² s⁻¹)
Select two materials to see how much ultrasound reflects at their boundary. Try air → soft tissue vs gel → soft tissue to see why gel matters.
Ultrasound Scanning Techniques
A-Scan (Amplitude Scan)
In an A-scan, a single ultrasound pulse is sent along one beam direction. The transducer receives echoes from boundaries at different depths and displays them as vertical spikes on a cathode ray oscilloscope (CRO). The horizontal position of each spike gives the time (and therefore depth), while the height gives the amplitude of the echo.
The depth of a reflecting surface is calculated from the time delay using:
Where d is depth, c is the speed of sound in the medium, and t is the time for the pulse to travel to the boundary and back (hence the factor of ½).
An ultrasound pulse is sent into the body and an echo is received 4.0 μs later. If the speed of sound in the tissue is 1540 m s⁻¹, calculate the depth of the reflecting surface.
d = ½ct = ½ × 1540 × 4.0 × 10⁻⁶
d = 3.08 × 10⁻³ m = 3.1 mm
B-Scan (Brightness Scan)
A B-scan is much more common in clinical practice. Instead of a single beam, the transducer is swept across the patient's body (or an array of transducers is used). Each beam direction produces a line of brightness values, and these are combined to form a 2D cross-sectional image.
In a B-scan, the brightness of each pixel corresponds to the amplitude of the reflected echo at that point. Strong reflections (large impedance mismatch, e.g. at bone boundaries) appear as bright spots, while weak reflections (similar impedances) appear dim.
A-Scan
- Single beam direction
- Displays amplitude spikes on a CRO
- Gives depth measurements
- No image produced
- Used for measuring distances (e.g. eye length before cataract surgery)
B-Scan
- Multiple beam directions (swept or array)
- Produces a 2D cross-sectional image
- Brightness = echo amplitude
- Real-time imaging possible
- Used for foetal scanning, abdominal imaging, echocardiography
Doppler Ultrasound: Measuring Blood Speed
Blood cells moving along a vessel reflect ultrasound. Their motion changes the frequency of the returning echo. Define the frequency shift as Δf = fecho − f, where f is the emitted frequency. A component of motion towards the probe produces a higher-frequency echo (Δf > 0); motion away produces a lower-frequency echo (Δf < 0).
Here v is blood speed, c is sound speed in the tissue, and θ is the angle between the ultrasound beam axis and blood flow. Only the velocity component along the beam contributes. The factor 2 accounts for two Doppler changes: the moving cell first receives the incident wave, then acts as a moving source of the reflected wave. This is a first-order approximation for v ≪ c.
At θ = 90°, the velocity component along the beam is zero, so the ideal Doppler shift is zero even though the blood is moving. Blood speed therefore cannot be inferred from a zero shift without knowing the beam/flow angle.
Use the emitted frequency, convert MHz to Hz, include the reflection factor 2 and retain the beam/flow angle. A pulse-echo depth measurement uses time delay; a blood-flow measurement uses a frequency shift.
Ultrasound vs Other Imaging Techniques
Advantages of ultrasound
- Non-ionising — avoids X-ray ionising radiation dose, including in foetal imaging
- Real-time imaging — can observe moving structures (heartbeat, blood flow)
- Relatively inexpensive compared to CT and MRI
- Portable — machines can be brought to the patient
- Controlled exposure — diagnostic use controls power and exposure duration
- Good contrast between different soft tissues
Disadvantages of ultrasound
- Cannot penetrate bone — high impedance mismatch causes strong reflection
- Cannot image through air — lungs and gas-filled organs are problematic
- Lower resolution than CT or MRI for fine structural detail
- Operator-dependent — image quality depends heavily on the skill of the sonographer
- Limited depth of penetration — higher frequencies give better resolution but less penetration
The choice of ultrasound frequency involves a trade-off: higher frequency gives better resolution (shorter wavelength) but less penetration depth (more attenuation). Lower frequencies penetrate deeper but produce less detailed images.
Knowledge Check
- Z = ρc where ρ is the density of the medium and c is the speed of sound in the medium (1 mark)
- Units: kg m⁻² s⁻¹ (or Rayl) (1 mark)
- Air has a very low acoustic impedance (~430) compared to skin (~1.6 × 10⁶) (1 mark)
- This large mismatch means nearly all ultrasound would be reflected at the air–skin boundary (1 mark)
- Gel has a similar impedance to soft tissue, providing impedance matching so most ultrasound enters the body (1 mark)
- Transmit: An alternating p.d. applied across the crystal causes it to vibrate at the same frequency, emitting ultrasound (1 mark)
- Receive: When an ultrasound echo reaches the crystal, the pressure causes it to deform, generating a p.d. across it (1 mark)
- The same crystal can perform both functions because the piezoelectric effect is reversible (1 mark)
- A-scan uses a single beam; B-scan uses multiple beams (or a swept transducer) (1 mark)
- A-scan produces a voltage-time trace (no image); B-scan produces a 2D image (1 mark)
- Accept also: A-scan measures depth; B-scan shows structure. Or: A-scan uses amplitude; B-scan uses brightness.
- Z = ρc = 1050 × 1600 = 1,680,000 (1 mark)
- Z = 1.68 × 10⁶ kg m⁻² s⁻¹ (1 mark for correct standard form and units)
Exam-Style Questions
Four original exam-practice questions: 30 marks. With the 12-mark knowledge check, this lesson provides 42 practice marks.
(b) Ultrasound is directed at a boundary between muscle (Z = 1.659 × 10⁶ kg m⁻² s⁻¹) and bone (Z = 7.6 × 10⁶ kg m⁻² s⁻¹). Calculate the fraction of the incident intensity that is reflected at this boundary.
(c) State and explain one reason why ultrasound is preferred to X-rays for imaging a foetus.
(a) [2 marks]
- The product of the density of a medium and the speed of sound in that medium (1 mark)
- Z = ρc with units kg m⁻² s⁻¹ (1 mark)
(b) [3 marks]
- Iᵣ/I₀ = ((Z₂ − Z₁)/(Z₂ + Z₁))² (1 mark for formula)
- = ((7.6 − 1.659)/(7.6 + 1.659))² (substitution)
- = (5.941/9.259)² = (0.641646)² (1 mark for substitution and method)
- = 0.4117 or 41.2% (1 mark for answer)
(c) [3 marks]
- Ultrasound is non-ionising / X-rays are ionising (1 mark)
- Ionising radiation can damage DNA / cause mutations / harm developing foetal cells (1 mark)
- Diagnostic ultrasound uses controlled exposure to obtain images without the ionising radiation dose of X-rays (1 mark)
(b) Explain why a coupling gel is applied between the transducer and the skin. Your answer should include a calculation to support your explanation.
(c) In a B-scan, the brightness of a pixel on the image depends on a particular property of the reflected ultrasound. State what this property is.
(a) [3 marks]
- Distance travelled = 2 × 0.050 = 0.100 m (round trip) (1 mark)
- t = d/c = 0.100 / 1540 = 6.49 × 10⁻⁵ s (1 mark)
- t = 65 μs (1 mark for correct answer with appropriate unit)
(b) [4 marks]
- Without gel, there is an air gap between transducer and skin (1 mark)
- Air (Z = 411.6) and tissue (Z ≈ 1.6 × 10⁶) have very different impedances (1 mark)
- Calculation: Iᵣ/I₀ ≈ ((1.6 × 10⁶ − 411.6)/(1.6 × 10⁶ + 411.6))² ≈ 1.0, so nearly all ultrasound is reflected (1 mark)
- Gel has impedance similar to tissue, so most ultrasound enters the body (impedance matching) (1 mark)
(c) [2 marks]
- The amplitude of the reflected echo (2 marks)
- Accept "intensity" — amplitude and intensity are related (amplitude² ∝ intensity)
(i) fat → muscle
(ii) soft tissue → bone
(iii) soft tissue → lung
Data: fat Z = 1.334 × 10⁶, muscle Z = 1.659 × 10⁶, soft tissue Z = 1.6324 × 10⁶, bone Z = 7.6 × 10⁶, lung Z = 0.26 × 10⁶ (all kg m⁻² s⁻¹)
(b) At which boundary is the most ultrasound reflected? Explain the significance of this for medical imaging.
(a)(i) fat → muscle [2 marks]
- Iᵣ/I₀ = ((1.659 − 1.334)/(1.659 + 1.334))² = (0.325/2.993)²
- = 0.011791 (1.18%) (2 marks: 1 for method, 1 for answer)
(a)(ii) soft tissue → bone [2 marks]
- Iᵣ/I₀ = ((7.6 − 1.6324)/(7.6 + 1.6324))² = (5.9676/9.2324)²
- = 0.4178 (41.8%) (2 marks: 1 for method, 1 for answer)
(a)(iii) soft tissue → lung [2 marks]
- Iᵣ/I₀ = ((0.26 − 1.6324)/(0.26 + 1.6324))² = (−1.3724/1.8924)²
- = 0.5259 (52.6%) (2 marks: 1 for method, 1 for answer)
(b) [3 marks]
- The most ultrasound is reflected at the soft tissue → lung boundary (52.6%) (1 mark)
- This means ultrasound cannot effectively image structures behind air-filled lungs (1 mark)
- Similarly, the large reflection at the tissue → bone boundary (42%) means ultrasound struggles to image structures behind bone (1 mark)
(a) Calculate the blood speed. [3 marks]
(b) For the same speed, explain what happens to the ideal frequency shift when the beam becomes perpendicular to the flow. [1 mark]
(a) [3 marks]
- Rearrange: v = |Δf|c/(2f|cos θ|) [1].
- Substitute v = 780 × 1540 / (2 × 4.0 × 10⁶ × cos 45°) [1].
- v = 0.212344… m s⁻¹ ≈ 0.21 m s⁻¹ [1].
(b) [1 mark]
- At 90°, cos θ = 0, so the velocity component along the beam and hence the ideal shift are zero [1].
Topic Summary
Acoustic Impedance
Z = ρc — a measure of how much a medium resists the passage of sound. Depends on density and speed of sound in the medium.
Reflection Coefficient
Iᵣ/I₀ = ((Z₂ − Z₁)/(Z₂ + Z₁))² — the fraction of ultrasound reflected depends on the impedance mismatch at a boundary.
Coupling Gel
Eliminates the air gap (Z ≈ 430) between transducer and skin (Z ≈ 1.6 × 10⁶). Without it, ~100% of ultrasound reflects.
A-Scan vs B-Scan
A-scan: single beam, amplitude spikes, depth measurements. B-scan: swept beam, 2D image, brightness = echo amplitude.
Doppler · 6.5.3(g)
Reflection gives two frequency changes. |Δf|/f = 2v|cos θ|/c for v ≪ c. Towards the probe raises echo frequency; away lowers it. Perpendicular flow has zero ideal shift.