Early in any acoustics or RF course, you learn a clean rule: when a wave hits a change in impedance, part of it reflects. Easy to memorise. But why must it reflect? The answer is more interesting than most textbooks let on — reflection is not a flaw in the system; it is the only outcome consistent with the boundary conditions.

In a previous post we looked at reflection and refraction phenomenologically — what happens at boundaries, with animations and simulations. This post goes deeper: it asks why reflection must occur, and proves it from first principles.

This post works through the answer in two steps:

1. A physical analogy — mapping every acoustic quantity onto an electrical circuit element, so the problem becomes visual.
2. A mathematical proof — three equations at the boundary that, when impedances differ, leave no room for anything but a reflected wave.

The Acoustic–Electrical Analogy

Before any equations, we need a shared vocabulary. The electro-acoustic analogy rests on one key fact: the equations governing sound in a duct and signals in a transmission line are mathematically identical. Every acoustic quantity has an electrical counterpart:

This is not just a loose metaphor — the correspondence is quantitatively exact. Three everyday objects show how it works:

A Short Tube → Inductor

A short, narrow tube open at both ends — like a bottle neck — contains a plug of air. Sound pushing on one end accelerates the whole plug as a rigid mass. It resists sudden changes in flow, just like an inductor resists sudden changes in current. Impedance rises with frequency: $Z_L = j\omega M_a$.

A Sealed Cavity → Capacitor

A closed volume of air (the body of a bottle, or the sealed cabinet of a loudspeaker) acts as a spring. Compress air into it and the pressure rises; release and it springs back. This is a capacitor: it stores potential energy. Its impedance falls with frequency: $Z_C = \dfrac{1}{j\omega C_a}$.

Absorptive Material → Resistor

Dense foam, felt, or a micro-perforated sheet forces air through tiny passages, converting kinetic energy to heat via viscous friction. This is a pure resistor — frequency-independent energy dissipation.

Putting It Together: Acoustic Impedance

These three elements — mass, compliance, and resistance — combine into a single complex number: the acoustic impedance $Z_a$. It mirrors how $R$, $L$, and $C$ combine into electrical impedance $Z$:

$$ Z = R + j\left(\omega L - \frac{1}{\omega C}\right) \qquad \longleftrightarrow \qquad Z_a = R_a + j\left(\omega M_a - \frac{1}{\omega C_a}\right) $$

The real part ($R$ or $R_a$) is energy dissipation — power lost as heat. The imaginary part (reactance) is energy storage: $\omega L$ (or $\omega M_a$) stores kinetic energy (inertia), $\dfrac{1}{\omega C}$ (or $\dfrac{1}{\omega C_a}$) stores potential energy (springiness). Neither term consumes net power; they just shuttle energy back and forth.

When we discuss impedance mismatch later, $Z_1$ and $Z_2$ at the boundary are exactly these complex impedances. A mismatch in either the real or imaginary part causes reflection.

From Analogy to Transmission Lines

The lumped-element picture works when the structure is much smaller than the wavelength. A long duct — exhaust pipe, ventilation shaft, waveguide — is different: it is a distributed system. Each infinitesimal slice has its own mass (inductance per unit length) and compliance (capacitance per unit length). Chain infinitely many slices together and you have a transmission line.

A transmission line has a characteristic impedance:

$$ Z_0 = \frac{p}{U} = \frac{\rho c}{S} $$

where $\rho$ is the air density, $c$ is the speed of sound, and $S$ is the cross-sectional area. In the electrical world, a coaxial cable has $Z_0 = \sqrt{L’/C’}$, where $L’$ and $C’$ are inductance and capacitance per unit length.

The key point: $Z_0$ is a property of the medium, not of the signal. A 50 Ω cable always enforces a 50 Ω ratio between voltage and current, regardless of frequency or amplitude. A duct of cross-section $S$ filled with air always enforces $p = Z_0 U$.

So what happens when two transmission lines — each with its own $Z_0$ — are joined end-to-end?

The Three-Equation Proof

Here is the central argument. When $Z_1 \neq Z_2$, a forward-travelling wave alone cannot satisfy the physics at the junction — a reflected wave must appear.

Setting the Scene

Consider two semi-infinite transmission lines (i.e., long enough that we need not worry about further reflections) joined at a single junction:

• Medium 1 (left): characteristic impedance $Z_1$. An incident wave of amplitude $V_i$ travels to the right.
• Medium 2 (right): characteristic impedance $Z_2$. Whatever gets through appears as a transmitted wave $V_t$.

All voltages and currents (equivalently, pressures and volume velocities) are evaluated at the junction.

The Three Inescapable Equations

Equation 1 — Impedance relation in Medium 1. For any forward-travelling wave in Medium 1, voltage and current are locked in the ratio $Z_1$:

$$ V_i = Z_1 I_i \tag{1} $$

Equation 2 — Impedance relation in Medium 2. For any forward-travelling wave in Medium 2, the ratio must be $Z_2$:

$$ V_t = Z_2 I_t \tag{2} $$

Equation 3 — Continuity at the junction. At the physical junction, voltage (pressure) cannot have a discontinuity, and current (flow) cannot appear from nowhere. If there is no reflected wave, then at the boundary:

$$ V_i = V_t, \qquad I_i = I_t \tag{3} $$

The Contradiction

Substitute Equation (3) into Equations (1) and (2):

$$ Z_1 I_i = V_i = V_t = Z_2 I_t = Z_2 I_i $$

This simplifies to:

$$ Z_1 = Z_2 $$

But we assumed the two media have different impedances! The three equations are mutually inconsistent — no solution satisfies all three simultaneously when $Z_1 \neq Z_2$.

Resolution: The Reflected Wave

The way out is to allow a reflected wave ($V_r$, $I_r$) travelling backwards in Medium 1. The boundary conditions become:

$$ V_i + V_r = V_t \tag{3a} $$

$$ I_i - I_r = I_t \tag{3b} $$

Note the minus sign in (3b): the reflected current flows in the opposite direction.

Each wave still obeys its own medium’s impedance law:

$$ V_i = Z_1 I_i, \qquad V_r = Z_1 I_r, \qquad V_t = Z_2 I_t $$

Substituting into (3a) and (3b):

$$ Z_1 I_i + Z_1 I_r = Z_2(I_i - I_r) $$

Solving for the ratio $V_r / V_i$:

$$ \boxed{\Gamma = \frac{V_r}{V_i} = \frac{Z_2 - Z_1}{Z_2 + Z_1}} $$

This is the reflection coefficient — not an assumption, but the only value consistent with all three constraints.

A closely related quantity is the transmission coefficient:

$$ T = \frac{V_t}{V_i} = \frac{2 Z_2}{Z_2 + Z_1} = 1 + \Gamma $$

$\Gamma$ and $T$ together describe how incident power splits at any boundary.

Limiting Cases

The formula $\Gamma = \dfrac{Z_2 - Z_1}{Z_2 + Z_1}$ covers every case. Three extremes are worth examining:

Impedance Match ($Z_1 = Z_2$) → $\Gamma = 0$

The numerator vanishes. No reflection at all — every watt of incident power crosses the boundary. The junction is acoustically (or electrically) invisible.

This is the design goal in every impedance-matching application: antenna matching networks, audio amplifier output stages, and the horn on a loudspeaker.

Hard Wall / Open Circuit ($Z_2 \to \infty$) → $\Gamma = +1$

When Medium 2 is infinitely stiff (a rigid wall in acoustics, or an open-circuited cable in electronics), $\Gamma = +1$. Total reflection, in phase. The reflected pressure adds to the incident pressure, doubling it at the wall — which is why sound is loudest right next to a hard boundary.

• Acoustics: sound hitting a concrete wall.
• Electronics: a signal reaching an unterminated (open) cable end.

Soft Boundary / Short Circuit ($Z_2 = 0$) → $\Gamma = -1$

When Medium 2 offers no resistance at all (an open pipe end radiating into free space, or a short-circuited cable), $\Gamma = -1$. Total reflection, phase-inverted. Peaks become troughs. The pressure at the boundary is forced to zero.

• Acoustics: a flute’s open end — the sound reflects back with inverted phase, yet the pipe still resonates.
• Electronics: a short circuit at the end of a coaxial cable.

Real-World Implications

Impedance mismatch is not just a textbook topic. Engineers deal with it constantly — sometimes fighting reflections, sometimes using them on purpose.

The Horn: An Acoustic Transformer

A horn (megaphone, trumpet bell, loudspeaker horn) is a duct whose cross-section expands gradually. Inside the narrow throat, the impedance $Z = \frac{\rho c}{S}$ is high (high pressure, low flow). At the wide mouth, it is low (low pressure, high flow). The smooth taper transforms the impedance continuously, avoiding any abrupt mismatch. The result: far more sound power radiates into the room than a bare speaker cone could ever produce alone.

This is exactly the same principle as an electrical transformer matching a high-impedance source to a low-impedance load.

The Expansion-Chamber Muffler

Conversely, sometimes we want reflection. An expansion-chamber muffler deliberately creates a sudden area change in the exhaust pipe. The impedance mismatch reflects engine noise back toward the combustion chamber, preventing it from reaching the tailpipe. By cascading several such chambers, each tuned to a different frequency band, broadband noise attenuation is achieved. (A wave-level simulation of this effect can be found in the expansion muffler section of Acoustics-03 .)

Cable Termination in Signal Engineering

In high-speed digital circuits and RF systems, unterminated cables produce reflections that corrupt data. The standard fix is a termination resistor equal to the cable’s characteristic impedance ($Z_0 = 50,\Omega$ or $75,\Omega$), placed at the far end. This enforces $Z_2 = Z_1$ → $\Gamma = 0$, absorbing the signal cleanly with no reflection.

Conclusion

Reflection at an impedance boundary is not a side-effect — it is a mathematical necessity. At any junction, three equations must hold:

1. The impedance relation in Medium 1.
2. The impedance relation in Medium 2.
3. Continuity of voltage (pressure) and current (flow).

When $Z_1 \neq Z_2$, these three constraints are mutually contradictory unless we allow a reflected wave. The reflection coefficient $\Gamma = \dfrac{Z_2 - Z_1}{Z_2 + Z_1}$ is the unique value that restores consistency.

Whether you are designing a concert hall, a muffler, or a gigabit Ethernet link, the rule is the same: match the impedance and the wave passes through; mismatch it and part of it comes back.

References

• He Lin et al., Theoretical Acoustics and Engineering Applications, Science Press, 2006.
• Ma Dayou, Modern Acoustics Theory Basis, Science Press, 2004.
• David M. Pozar, Microwave Engineering, 4th Edition, Wiley, 2011.