Reflection, Transmission and Impedance Matching

Derivation

Reflection, Transmission and Impedance Matching

D-069 Home PU-103 Threads waves · energy Depends on The Wave Equation as a Continuum Limit, Energy Density and Power Transport

Statement

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 \(R+\mathcal{T}=1\).

Why it matters

Whenever a wave meets a change in the medium it partly reflects and partly transmits, and the split is governed by a single material property, the impedance. The string junction is the cleanest possible arena for this: one tension, two mass densities, two boundary conditions, and an answer that is exact rather than approximate.

The same algebraic form recurs across physics: transmission lines, acoustic tubes, optical thin films, and quantum step potentials all reduce to matching a field and its "flux" across an interface. Mastering the string case gives you the template for all of them, and it explains why impedance matching, minimising reflection by eliminating the mismatch, is the central design principle from loudspeakers to antenna feeds.

Assumptions

The junction is idealised as a massless, dimensionless point.If the joint carries mass or stiffness, an extra inertial/elastic boundary term appears, the force condition acquires a \(\partial^2 y/\partial t^2\) contribution, and \(r,\tau\) become frequency-dependent.

The tension \(T\) is common to both segments and constant in time.If tension differs across the junction the transverse-force balance changes; a static longitudinal force imbalance would also accelerate the joint sideways, breaking the steady wave picture.

Small-amplitude (linear) motion, so slopes \(|\partial y/\partial x|\ll 1\).Drop it and the transverse force is \(T\sin\theta\) not \(T\,\partial y/\partial x\); harmonics are generated and superposition of incident, reflected and transmitted waves fails.

Both segments are non-dispersive and lossless, with a single sharp interface.With damping or dispersion \(k\) becomes complex or frequency-dependent, \(Z\) is no longer real, and part of the incident power is absorbed so that \(R+\mathcal{T}<1\).

Steady-state, monochromatic excitation at angular frequency \(\omega\).For a pulse or a switch-on transient the same coefficients hold frequency-by-frequency, but only after Fourier decomposition; a single \(r\) does not describe the full transient at the joint.

Derivation

\[ y_{<}=A_i\cos(\omega t-k_1x)+A_r\cos(\omega t+k_1x),\qquad y_{>}=A_t\cos(\omega t-k_2x) \]

Place the junction at \(x=0\); the left segment (\(x<0\)) carries an incident \(+x\) wave and a reflected \(-x\) wave, the right segment (\(x>0\)) carries only a transmitted \(+x\) wave. The junction cannot change frequency, so all three share \(\omega\); the wavenumbers differ as \(k_1=\omega/v_1\), \(k_2=\omega/v_2\). A

\[ \lim_{x\to0^-}y_{<}=\lim_{x\to0^+}y_{>}\;\Longrightarrow\; A_i+A_r=A_t \]

The string is unbroken, so its displacement is single-valued at the joint for all \(t\). Evaluating both expressions at \(x=0\) removes the spatial phase and the common \(\cos\omega t\) cancels. A

\[ \left.T\frac{\partial y_{<}}{\partial x}\right|_{0^-}=\left.T\frac{\partial y_{>}}{\partial x}\right|_{0^+} \]

Newton's law for the massless junction: the net transverse force on a segment of vanishing mass must vanish, else its acceleration diverges. The transverse force in a taut string is \(T\,\partial y/\partial x\) (small slope), so its value must be continuous across the joint. B

\[ k_1A_i-k_1A_r=k_2A_t \]

Differentiate each wave in \(x\) and evaluate at \(x=0\): \(\partial_x y_i\to A_ik_1\sin\omega t\), \(\partial_x y_r\to -A_rk_1\sin\omega t\), \(\partial_x y_t\to A_tk_2\sin\omega t\). Cancel the common \(T\sin\omega t\). B

\[ k=\frac{\omega}{v}=\frac{\omega}{T}\,\sqrt{T\mu}=\frac{\omega}{T}\,Z\quad\Longrightarrow\quad \frac{k_1}{k_2}=\frac{Z_1}{Z_2} \]

Use \(v=\sqrt{T/\mu}\) and \(Z=\sqrt{T\mu}=T/v\) from the loaded-string result. At fixed \(T,\omega\) the wavenumber is proportional to the impedance, so every \(k\) may be replaced by the corresponding \(Z\). B

\[ Z_1(A_i-A_r)=Z_2(A_i+A_r)\;\Longrightarrow\; r\equiv\frac{A_r}{A_i}=\frac{Z_1-Z_2}{Z_1+Z_2} \]

Substitute \(A_t=A_i+A_r\) (Step 2) into the force condition (Step 4, now in \(Z\)), then collect \(A_r\) on one side. Purely algebraic. A

\[ \tau\equiv\frac{A_t}{A_i}=1+r=\frac{2Z_1}{Z_1+Z_2} \]

Displacement continuity (Step 2) gives \(A_t/A_i=1+A_r/A_i=1+r\) directly. A

\[ \langle P\rangle=\tfrac12 Z\omega^2A^2\;\Rightarrow\; R=\frac{\langle P_r\rangle}{\langle P_i\rangle}=r^2,\qquad \mathcal{T}=\frac{\langle P_t\rangle}{\langle P_i\rangle}=\frac{Z_2}{Z_1}\tau^2=\frac{4Z_1Z_2}{(Z_1+Z_2)^2} \]

Time-averaged power carried by a wave is \(\tfrac12 Z\omega^2A^2\) (energy-and-power result). The reflected and incident waves share \(Z_1\), so \(R=r^2\); the transmitted wave carries \(Z_2\), so the ratio picks up \(Z_2/Z_1\). Adding, \(\dfrac{(Z_1-Z_2)^2+4Z_1Z_2}{(Z_1+Z_2)^2}=1\), confirming energy conservation. C

Result

\[ r=\frac{Z_1-Z_2}{Z_1+Z_2},\qquad \tau=\frac{2Z_1}{Z_1+Z_2},\qquad R=r^2,\qquad \mathcal{T}=\frac{4Z_1Z_2}{(Z_1+Z_2)^2},\qquad R+\mathcal{T}=1 \]

Reading. The amplitude split depends only on the impedance mismatch \(Z_1-Z_2\) measured against the total \(Z_1+Z_2\). A wave entering a heavier (higher-\(Z\)) medium reflects with a sign flip (\(r<0\), a half-cycle phase shift); entering a lighter medium it reflects in phase (\(r>0\)). The transmitted amplitude can be as large as twice the incident (\(\tau\to2\) as \(Z_2\to0\)) yet no energy paradox arises, because the power that flows is weighted by \(Z\), and \(\mathcal{T}\) never exceeds one.

Units check. \(Z=\sqrt{T\mu}\) has units \(\sqrt{(\mathrm{N})(\mathrm{kg\,m^{-1}})}=\sqrt{\mathrm{kg\,m\,s^{-2}}\cdot\mathrm{kg\,m^{-1}}}=\sqrt{\mathrm{kg^2\,s^{-2}}}=\mathrm{kg\,s^{-1}}\). Both \(r\) and \(\tau\) are ratios of like quantities, hence dimensionless; \(R\) and \(\mathcal{T}\), ratios of powers, are likewise dimensionless. In \(\langle P\rangle=\tfrac12 Z\omega^2A^2\): \(\mathrm{kg\,s^{-1}}\cdot\mathrm{s^{-2}}\cdot\mathrm{m^2}=\mathrm{kg\,m^2\,s^{-3}}=\mathrm{W}\). Consistent.

Limiting cases

\(Z_2=Z_1\) (matched): \(r=0\), \(\tau=1\), \(\mathcal{T}=1\) — no reflection, full power transfer.
\(Z_2\to\infty\) (rigid/fixed end): \(r\to-1\), \(\tau\to0\), total reflection with a half-cycle phase flip.
\(Z_2\to0\) (free end): \(r\to+1\), \(\tau\to2\), total reflection in phase; large transmitted amplitude but zero transmitted power.
\(Z_2\gg Z_1\) but finite: \(r\approx-1+2Z_1/Z_2\), most power reflected, \(\mathcal{T}\approx4Z_1/Z_2\).
Reversing the direction (\(Z_1\leftrightarrow Z_2\)) flips the sign of \(r\) but leaves \(R\) and \(\mathcal{T}\) unchanged — reflectance is symmetric in the mismatch.

Breaks when

Lossy or dispersive media. With internal damping the wavenumber is complex, \(Z\) becomes complex and frequency-dependent, and absorbed power makes \(R+\mathcal{T}<1\); the real-\(Z\) formulas no longer apply.
Junction with mass or stiffness. A bead, clamp, or finite-length graded region adds an inertial term \(m\,\partial^2 y/\partial t^2\) (or a spring term) to the force condition, so \(r\) and \(\tau\) acquire explicit \(\omega\)-dependence and pick up a phase even between real media.
Large amplitude. Once slopes are not small the transverse force is \(T\sin\theta\), the wave equation is nonlinear, harmonics appear, and a single amplitude ratio ceases to describe the interface.
Graded (smooth) transition. If \(\mu(x)\) varies over a length comparable to or greater than a wavelength, the abrupt-junction analysis fails and one solves the WKB/adiabatic problem instead, where reflection is exponentially suppressed.

Failure modes

Writing \(R+\mathcal{T}=1\) as \(r+\tau=1\). Amplitudes are not conserved; only power is. The correct amplitude identity is \(\tau=1+r\).
Setting \(\mathcal{T}=\tau^2\) and forgetting the \(Z_2/Z_1\) weighting — this makes power appear non-conserved and lets \(\mathcal{T}\) exceed 1.
Using mass density \(\mu\) directly in the coefficients, e.g. \(r=(\mu_1-\mu_2)/(\mu_1+\mu_2)\). The impedance goes as \(\sqrt{\mu}\), not \(\mu\).
Assuming the transmitted wave has the same wavelength as the incident. Frequency is preserved, wavelength is not: \(\lambda_2=\lambda_1 v_2/v_1\).
Getting the reflection sign backwards — expecting a phase flip going into a lighter string. The flip occurs entering the heavier (higher-\(Z\)) medium.
Believing \(\tau>1\) violates energy conservation; it does not, because power depends on \(Z\) as well as amplitude.

Discussion

The result says something deceptively simple: reflection is caused not by a change in mass density as such, nor by a change in wave speed as such, but by the change in the single combination \(Z=\sqrt{T\mu}=T/v\). Two very different-looking strings with equal impedance are invisible to each other — a wave crosses without a hint of reflection. This is why impedance, not any one material parameter, is the physically meaningful "matching" quantity.

The two boundary conditions are the whole story: displacement continuity (the string does not tear) and force continuity (the joint does not accelerate infinitely). These are precisely a "continuity of the field" and a "continuity of its conjugate flux," and every wave-interface problem shares this pair. In transmission lines the field is voltage and its partner is current; in acoustics, pressure and volume velocity; in optics, the tangential \(\mathbf{E}\) and \(\mathbf{H}\); in the quantum step, the wavefunction \(\psi\) and its derivative \(\psi'\). In each the reflection amplitude takes the identical Möbius form \((Z_1-Z_2)/(Z_1+Z_2)\) once the correct impedance is identified. The string is the Rosetta stone.

The practical payoff is impedance matching. To connect two mismatched media \(Z_1,Z_3\) with zero reflection at a chosen frequency, insert a quarter-wavelength section of the geometric-mean impedance \(Z_2=\sqrt{Z_1Z_3}\). The wave reflected at the first interface and the wave reflected at the second return exactly out of phase — the extra half-wavelength round trip supplies a \(\pi\) shift — and cancel. This "quarter-wave transformer" is how anti-reflection optical coatings, ultrasound coupling gel, radio antenna matching networks, and audio horns all work; the string version can be built with a short length of intermediate rope. Note its narrowband nature: cancellation is exact only near the design frequency, because the phase condition is frequency-dependent.

Common misconceptions. The transmitted amplitude exceeding the incident (up to a factor of two at a free end) routinely alarms students, who suspect energy creation. It is a genuine amplitude increase, but the transmitting medium is "softer" (lower \(Z\)), so the same displacement carries less power; the power transmittance \(\mathcal{T}=4Z_1Z_2/(Z_1+Z_2)^2\le1\) always. A second trap is thinking reflection needs a "hard wall": any mismatch reflects, and the reflectance depends only on the ratio \(Z_2/Z_1\), symmetric under swapping the two media.

Worked examples

Two strings under common tension \(T=100\ \mathrm{N}\), with \(\mu_1=1.0\ \mathrm{g\,m^{-1}}\) and \(\mu_2=4.0\ \mathrm{g\,m^{-1}}\). Find \(r,\ \tau,\ R,\ \mathcal{T}\).

Symbolic first: since \(Z=\sqrt{T\mu}\propto\sqrt{\mu}\), the ratio \(Z_2/Z_1=\sqrt{\mu_2/\mu_1}\). A

\[ \frac{Z_2}{Z_1}=\sqrt{\frac{\mu_2}{\mu_1}}=\sqrt{\frac{4.0}{1.0}}=2.0,\qquad Z_1=\sqrt{(100)(1.0\times10^{-3})}=0.316\ \mathrm{kg\,s^{-1}} \]

Insert the numbers; \(Z_2=0.632\ \mathrm{kg\,s^{-1}}\). A

\[ r=\frac{Z_1-Z_2}{Z_1+Z_2}=\frac{1-2}{1+2}=-\frac{1}{3},\qquad \tau=1+r=\frac{2}{3} \]

The mismatch ratio alone fixes \(r\); the sign is negative because segment 2 is heavier. A

\[ R=r^2=\frac{1}{9}=0.111,\qquad \mathcal{T}=1-R=\frac{8}{9}=0.889 \]

Cross-check \(\mathcal{T}=4Z_1Z_2/(Z_1+Z_2)^2=4(1)(2)/3^2=8/9\). Consistent. B

\[ r=-\tfrac13,\quad \tau=\tfrac23,\quad R=11.1\%,\quad \mathcal{T}=88.9\% \]

Reading. Going onto a string four times as heavy, one third of the amplitude bounces back with a phase flip, but only about \(11\%\) of the power is reflected — most of the energy still gets through.

A wave of impedance \(Z_1=0.30\ \mathrm{kg\,s^{-1}}\) must feed a load segment \(Z_3=1.20\ \mathrm{kg\,s^{-1}}\). Quantify the direct reflection, then design a quarter-wave matching section held at tension \(T=120\ \mathrm{N}\), matched at \(f=150\ \mathrm{Hz}\).

First the un-matched reflectance from the master result. B

\[ R_{\text{direct}}=\left(\frac{Z_1-Z_3}{Z_1+Z_3}\right)^2=\left(\frac{0.30-1.20}{0.30+1.20}\right)^2=\left(\frac{-0.90}{1.50}\right)^2=0.36 \]

Butt-joining the media reflects \(36\%\) of the incident power — unacceptable. B

\[ Z_2=\sqrt{Z_1Z_3}=\sqrt{(0.30)(1.20)}=\sqrt{0.36}=0.60\ \mathrm{kg\,s^{-1}} \]

Zero net reflection requires the intermediate impedance to be the geometric mean of the two it bridges. C

\[ v_2=\frac{T}{Z_2}=\frac{120}{0.60}=200\ \mathrm{m\,s^{-1}},\quad \lambda_2=\frac{v_2}{f}=\frac{200}{150}=1.33\ \mathrm{m},\quad \ell=\frac{\lambda_2}{4}=0.33\ \mathrm{m} \]

Use \(Z=T/v\) to get the speed in the matching rope, then its wavelength at the design frequency; the section is a quarter of that. C

\[ Z_2=0.60\ \mathrm{kg\,s^{-1}},\quad \ell=\lambda_2/4=0.33\ \mathrm{m}\ \Rightarrow\ R\to0\ \text{at }150\ \mathrm{Hz} \]

Reading. A \(33\ \mathrm{cm}\) intermediate segment of the geometric-mean impedance turns a \(36\%\) reflection into (ideally) none at the design frequency — the string analogue of an anti-reflection coating. Away from \(150\ \mathrm{Hz}\) some reflection returns.

Problems

A wave travels from a string of density \(\mu_1\) onto one of density \(\mu_2=9\mu_1\) at the same tension. Find \(r,\tau,R,\mathcal{T}\).
Solution
\(Z_2/Z_1=\sqrt{9}=3\). \(r=(1-3)/(1+3)=-\tfrac12\); \(\tau=1+r=\tfrac12\); \(R=r^2=\tfrac14=25\%\); \(\mathcal{T}=1-R=\tfrac34=75\%\) (check \(4(1)(3)/4^2=\tfrac{12}{16}=\tfrac34\)). A half-cycle phase flip on reflection since medium 2 is heavier.
Show from the general result that a rigidly fixed end (\(Z_2\to\infty\)) gives total reflection with a phase inversion and no transmitted power, and interpret the sign physically.
Solution
As \(Z_2\to\infty\), \(r=(Z_1-Z_2)/(Z_1+Z_2)\to -Z_2/Z_2=-1\), so \(|r|=1\) (total reflection) with a minus sign (\(\pi\), half-cycle, phase flip). \(\tau=2Z_1/(Z_1+Z_2)\to0\) and \(\mathcal{T}=4Z_1Z_2/(Z_1+Z_2)^2\to4Z_1/Z_2\to0\): no power crosses. Physically the wall cannot move, so the incident and reflected displacements must sum to zero there, forcing the reflected wave to be inverted. The reaction force of the wall does no work, and \(R=1\), consistent with an ideal fixed end being lossless.
String tension \(T=80\ \mathrm{N}\); \(\mu_1=0.50\ \mathrm{g\,m^{-1}}\), \(\mu_2=2.0\ \mathrm{g\,m^{-1}}\). An incident wave carries \(\langle P_i\rangle=2.0\ \mathrm{W}\). Find the reflected and transmitted powers.
Solution
\(Z_1=\sqrt{(80)(5.0\times10^{-4})}=\sqrt{0.040}=0.20\ \mathrm{kg\,s^{-1}}\); \(Z_2=\sqrt{(80)(2.0\times10^{-3})}=\sqrt{0.16}=0.40\ \mathrm{kg\,s^{-1}}\). \(r=(0.20-0.40)/0.60=-\tfrac13\), \(R=\tfrac19=0.111\), \(\mathcal{T}=0.889\). Reflected \(=0.111\times2.0=0.22\ \mathrm{W}\); transmitted \(=0.889\times2.0=1.78\ \mathrm{W}\). Sum \(=2.0\ \mathrm{W}\), conserved.
A wave enters a heavier string and you require exactly \(4\%\) of the power to be reflected. What ratio \(\mu_2/\mu_1\) (at fixed tension) achieves this?
Solution
\(R=r^2=0.04\Rightarrow r=-0.20\) (negative, since heavier). With \(x\equiv Z_2/Z_1\): \((1-x)/(1+x)=-0.20\Rightarrow1-x=-0.20-0.20x\Rightarrow1.20=0.80x\Rightarrow x=1.5\). Thus \(Z_2/Z_1=1.5\) and \(\mu_2/\mu_1=x^2=2.25\). (The lighter-medium alternative \(r=+0.20\) gives \(x=2/3\), \(\mu_2/\mu_1=0.444\), but that is a lighter string.)
Design a quarter-wave matching segment to join \(Z_1=0.25\ \mathrm{kg\,s^{-1}}\) to \(Z_3=1.00\ \mathrm{kg\,s^{-1}}\) at \(f=200\ \mathrm{Hz}\), with the matching rope held at \(T=100\ \mathrm{N}\). Give its linear density, wave speed, and length.
Solution
Required impedance \(Z_2=\sqrt{Z_1Z_3}=\sqrt{(0.25)(1.00)}=0.50\ \mathrm{kg\,s^{-1}}\). Linear density \(\mu_2=Z_2^2/T=(0.50)^2/100=2.5\times10^{-3}\ \mathrm{kg\,m^{-1}}=2.5\ \mathrm{g\,m^{-1}}\). Wave speed \(v_2=T/Z_2=100/0.50=200\ \mathrm{m\,s^{-1}}\). Wavelength \(\lambda_2=v_2/f=200/200=1.0\ \mathrm{m}\); length \(\ell=\lambda_2/4=0.25\ \mathrm{m}\). The un-matched reflectance would have been \(((0.25-1.00)/1.25)^2=0.36\), i.e. \(36\%\), reduced to zero at \(200\ \mathrm{Hz}\).