Reflection, Transmission and Impedance
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You should be comfortable with the wave equation for a taut string, the definition of wave impedance \(Z = \sqrt{T\mu}\), and the expression for average power carried by a harmonic wave.
Why this matters
Understanding how waves behave at boundaries is essential for designing everything from musical instruments to optical coatings.
When a wave encounters a change in medium, it does not simply stop or pass through unchanged. Part of the wave is reflected back into the original medium, and part is transmitted into the new one. The relative amounts of reflection and transmission are determined by the mismatch in wave impedance between the two media. This concept is fundamental to acoustics, optics, and electrical engineering, where impedance matching is used to maximize energy transfer and minimize unwanted reflections.
The Physical Picture
Imagine a heavy rope tied to a light rope. If you send a pulse down the heavy rope, what happens when it reaches the junction? The heavy rope has a high impedance because it is difficult to move (high mass per unit length). The light rope has low impedance. When the pulse hits the junction, the heavy rope "pushes" back against the light rope more strongly than the light rope can push back. This imbalance causes part of the pulse to bounce back (reflection) and part to continue forward (transmission). If the two ropes had identical impedance, the pulse would pass through without any reflection, as if the junction did not exist. The key insight is that the boundary conditions—continuity of displacement and force—dictate how the wave splits.
Impedance and Boundary Conditions
The wave impedance \(Z\) of a medium is defined as the ratio of the transverse force to the transverse velocity of the wave. For a string under tension \(T\) with linear mass density \(\mu\), the impedance is given by \(Z = \sqrt{T\mu}\). At the junction between two media with impedances \(Z_1\) and \(Z_2\), two physical conditions must be met: the transverse displacement must be continuous (the string cannot break), and the transverse force must be continuous (Newton's third law). These conditions lead to the reflection and transmission coefficients for amplitude.
Here, \(r\) is the amplitude reflection coefficient and \(\tau\) is the amplitude transmission coefficient. Note that \(\tau = 1 + r\): this is simply the continuity of displacement at the junction. If \(Z_1 = Z_2\), then \(r=0\) and \(\tau=1\), meaning perfect transmission. If \(Z_2 \gg Z_1\) (close to a fixed end), then \(r \approx -1\) and \(\tau \approx 0\): almost total reflection with a phase inversion. If \(Z_2 \ll Z_1\) (close to a free end), then \(r \approx +1\) and \(\tau \approx 2\): the transmitted amplitude exceeds the incident amplitude, yet the transmitted power is small because it is carried by a low-impedance medium.
Power Coefficients and Energy Conservation
While amplitude coefficients describe the height of the wave, power coefficients describe the energy flow. The average power carried by a wave is proportional to the square of the amplitude and the impedance of the medium. Therefore, the power reflection coefficient \(R\) and power transmission coefficient \(\mathcal{T}\) are given by:
A crucial check is that energy is conserved. The sum of the reflected and transmitted power must equal the incident power. Indeed, adding the two expressions yields:
This result confirms that no energy is lost at the boundary; it is simply partitioned between the reflected and transmitted waves.
Impedance Matching
In many applications, we want to maximize the transmission of energy from one medium to another. This is achieved by impedance matching, where we insert an intermediate medium with an impedance \(Z_m\) a quarter of a wavelength thick, whose impedance is the geometric mean of the two original impedances: \(Z_m = \sqrt{Z_1 Z_2}\). This removes the reflection completely at that one frequency. By using multiple layers with gradually changing impedance, we can create an "impedance transformer" that minimizes reflection. This principle is widely used in optics (anti-reflection coatings), acoustics (microphone diaphragms), and electronics (transmission lines).
The derivation
Reflection, Transmission and Impedance Matching
For a transverse wave on a taut string that crosses an abrupt junction between two media of differing wave impedance \(Z_1=\sqrt{T\mu_1}\) and \(Z_2=\sqrt{T\mu_2}\), continuity of transverse displacement and of transverse force at the junction fixes the amplitude reflection and transmission coefficients as \(r=\dfrac{Z_1-Z_2}{Z_1+Z_2}\) and \(\tau=\dfrac{2Z_1}{Z_1+Z_2}\), and the corresponding power coefficients \(R=r^2\), \(\mathcal{T}=\dfrac{4Z_1Z_2}{(Z_1+Z_2)^2}\) satisfy \(R+\mathcal{T}=1\).
A common misconception is that the transmission coefficient \(\tau\) must be less than or equal to 1. However, \(\tau\) is an amplitude coefficient, not a power coefficient. If \(Z_2 < Z_1\), the transmitted amplitude can be larger than the incident amplitude (\(\tau > 1\)), but the transmitted power is still less than the incident power because the impedance is lower. Always distinguish between amplitude and power coefficients to avoid confusion. C
Worked example
Problem. A wave on a string with impedance \(Z_1 = 10\ \mathrm{kg/s}\) encounters a junction with a second string of impedance \(Z_2 = 40\ \mathrm{kg/s}\). Calculate the amplitude reflection coefficient \(r\), the amplitude transmission coefficient \(\tau\), and the power transmission coefficient \(\mathcal{T}\).
- Calculate the amplitude reflection coefficient: \(r = \frac{Z_1 - Z_2}{Z_1 + Z_2} = \frac{10 - 40}{10 + 40} = \frac{-30}{50} = -0.6\).
- Calculate the amplitude transmission coefficient: \(\tau = \frac{2Z_1}{Z_1 + Z_2} = \frac{2(10)}{50} = \frac{20}{50} = 0.4\).
- Calculate the power transmission coefficient: \(\mathcal{T} = \frac{4Z_1 Z_2}{(Z_1 + Z_2)^2} = \frac{4(10)(40)}{50^2} = \frac{1600}{2500} = 0.64\).
- Verify energy conservation: \(R = r^2 = (-0.6)^2 = 0.36\). Then \(R + \mathcal{T} = 0.36 + 0.64 = 1.0\).
Units check. The mechanical impedance of a string, \(Z = \sqrt{T\mu}\), is in \(\mathrm{kg/s}\), which is \(\mathrm{N \cdot s / m}\). The coefficients \(r\), \(\tau\), and \(\mathcal{T}\) are dimensionless, as expected.
Three quick tests
- What happens to the reflection coefficient if \(Z_1 = Z_2\)?
Answer
It becomes zero, meaning no reflection occurs. - Why is the sum of the power coefficients \(R + \mathcal{T}\) always equal to 1?
Answer
Because energy is conserved at the boundary; no energy is lost, only redistributed. - Can the amplitude transmission coefficient \(\tau\) be greater than 1?
Answer
Yes, if \(Z_2 < Z_1\), the transmitted amplitude can be larger than the incident amplitude, but the power is still conserved.
Problem set for PU-103 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.