Question 1

What is acoustic impedance and why does it matter?

Accepted Answer

Acoustic impedance Z = ρ·c determines how much opposition a material offers to a propagating sound wave. When two materials have different impedances, sound cannot cross the boundary without partial reflection. The mismatch determines the fraction of energy transmitted: for small mismatches (like water to soft tissue, Z₁ = 1.48 MRayl, Z₂ = 1.63 MRayl) nearly all energy crosses; for large mismatches (air to tissue) almost none does, which is why ultrasound gel is applied to skin before medical scans.

Question 2

What does the reflection coefficient R represent?

Accepted Answer

R = (Z₂ − Z₁)/(Z₂ + Z₁) is the pressure reflection coefficient, ranging from −1 to +1. Its square R² is the fraction of incident power that is reflected. When Z₂ > Z₁ (harder medium), R is positive (no phase inversion); when Z₂ < Z₁ (softer medium), R is negative (180° phase shift on reflection). For a perfect impedance match (Z₁ = Z₂), R = 0 and 100% of sound is transmitted.

Question 3

What is the standing wave ratio (SWR)?

Accepted Answer

The SWR = (1 + |R|)/(1 − |R|) measures the ratio between the maximum and minimum acoustic pressure amplitudes on the incident side, where the incident and reflected waves interfere. An SWR of 1 means no reflection (perfect match). An SWR approaching infinity means total reflection (no transmission). In the simulation, air–steel gives SWR ≈ ∞ while water–soft tissue gives SWR ≈ 1.06, showing near-perfect matching.

Question 4

Why is the air-to-tissue impedance mismatch such a problem for ultrasound?

Accepted Answer

Air has Z ≈ 415 Pa·s/m while soft tissue has Z ≈ 1.63 MRayl — a mismatch of roughly 3,500:1. Inserting this into the power reflection formula gives R² ≈ 0.9989, meaning only about 0.1% of sound energy enters the body. Ultrasound gel (with Z close to tissue) eliminates the air gap at the skin surface, raising transmission to nearly 100% and making medical imaging possible.

Question 5

How does impedance matching work in engineering?

Accepted Answer

In sonar and non-destructive testing, transducers are matched to the load medium using quarter-wavelength matching layers with impedance Z_match = √(Z₁·Z₂). A layer exactly one quarter-wavelength thick creates destructive interference for reflected waves, minimising the overall reflection. Multiple matching layers are stacked in modern piezoelectric transducers to achieve broadband transmission across large impedance mismatches.

Question 6

What is the difference between pressure transmission and energy transmission?

Accepted Answer

The pressure transmission coefficient T = 2Z₂/(Z₁+Z₂) gives the amplitude ratio of transmitted to incident pressure. However, energy (intensity) transmission is T_energy = (Z₁/Z₂)·T² = 1 − R². When Z₂ > Z₁, the transmitted pressure can exceed the incident pressure (T > 1) even though energy is conserved, because the higher-impedance medium carries the same power at lower particle velocity but higher pressure.

Question 7

What does transmission loss in dB measure?

Accepted Answer

Transmission loss (TL) = −20·log₁₀(|T|) in decibels measures how much weaker the transmitted signal is compared to the incident signal, in logarithmic scale. A TL of 0 dB means no loss; 6 dB means the transmitted amplitude is halved; 20 dB means a 10-fold amplitude reduction. Air-to-steel gives TL ≈ 33 dB, while water-to-soft tissue gives TL ≈ 0.4 dB.

Question 8

How does frequency affect acoustic transmission?

Accepted Answer

In the simple planar wave model used here, the reflection and transmission coefficients at a single boundary are frequency-independent. However, frequency matters greatly in layered systems (where resonance and anti-resonance between layers produce frequency-dependent transmission), in lossy materials (where higher frequencies are absorbed more strongly), and in biological tissue where frequency determines both resolution and penetration depth in ultrasound imaging.

Question 9

What are typical acoustic impedances of common materials?

Accepted Answer

Air: 415 Pa·s/m; water: 1.48 MRayl; soft tissue: 1.63 MRayl; fat: 1.38 MRayl; bone: 6–8 MRayl; steel: 47 MRayl; lead: 24 MRayl; rubber: 0.9–1.8 MRayl. The large mismatch between bone and soft tissue (R² ≈ 0.68) is why ultrasound cannot image through bone — most energy is reflected at the cortical bone surface.

Question 10

What does the standing wave envelope show?

Accepted Answer

On the incident side, the total acoustic field is the superposition of the incident wave and the reflected wave travelling in opposite directions. When |R| > 0, they interfere to form a standing wave pattern with spatial maxima (antinodes) at λ/2 intervals. The envelope plot shows this amplitude distribution — wider envelope means stronger partial standing wave. When R = 0 (perfect match), the envelope is flat, confirming no standing wave forms.

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