RF-06.04 · Distributed resonators and microwave filters

01Traveling waves and finite-line resonance

A finite uniform transmission line sustains two traveling phasor waves, one forward and one reflected. Their superposition establishes standing-wave voltage and current profiles when boundary conditions select a discrete frequency. At an ideal open end current vanishes; at a shorted end voltage vanishes. Those are physical electrical conditions, not arbitrary resonance rules.

Adopt exp(+jωt), a real characteristic impedance and distributed L′ and C′. In the lossless TEM approximation β=ω√(L′C′), and the guided wavelength depends on phase velocity. All dimensions must use the relevant guided wave, not the free-space wavelength by default.

V(z)=V^+e^{-j\beta z}+V^-e^{j\beta z}

I(z)=\frac{V^+}{Z_0}e^{-j\beta z}-\frac{V^-}{Z_0}e^{j\beta z}

\beta=\omega\sqrt{L'C'},\qquad Z_0=\sqrt{\frac{L'}{C'}},\quad \lambda_g=\frac{2\pi}{\beta}

02Open–open and short–short half-wave modes

Both ends shorted require voltage nodes at the ends; a sine standing wave meets these boundaries when the electrical length is an integer number of half wavelengths. Both ends open require current nodes at the ends and voltage antinodes there, yielding the same nonzero modal frequency series but complementary field patterns.

The RF fundamental is a half-wave guided path in either case. For nondispersive propagation, higher longitudinal modes scale approximately with the integer mode index. The uniform static solution of an open-open line is not a nonzero-frequency microwave resonance, and practical fields at open ends are not exactly zero outside the conductor.

Boundary conditions: open–open half-wave line and short–open quarter-wave line. The bottom conductor is the RF return; open far ends are intentionally not grounded.

V(0)=V(l)=0\;\Longrightarrow\;\sin(\beta l)=0

I(0)=I(l)=0\;\Longrightarrow\;\sin(\beta l)=0

f_n^{\mathrm{SS,OO}}=\frac{nv_p}{2l},\qquad n=1,2,\ldots

03Mixed-boundary quarter-wave modes

A line shorted at one end and open at the other must have zero voltage at the short and zero current at the open. Referencing a sinusoidal voltage distribution to the short, the open-end current condition requires its spatial derivative to vanish. The allowed electrical lengths are odd multiples of one quarter wavelength.

The quarter-wave fundamental is half the frequency of the corresponding half-wave mode for equal physical length and phase velocity; its next longitudinal mode is the third harmonic. A real ground via is inductive, and an open microstrip end carries fringing capacitance, so electromagnetic corrections shift resonance from this ideal model.

Left short: voltage node. Right open: current node. Magnitudes shown for an ideal lossless line.

V(0)=0,\quad I(l)=0\;\Longrightarrow\;\cos(\beta l)=0

f_n^{\mathrm{SO}}=\frac{(2n+1)v_p}{4l},\qquad n=0,1,\ldots

Z_{\mathrm{in,SC}}=jZ_0\tan(\beta l),\quad Z_{\mathrm{in,OC}}=-jZ_0\cot(\beta l)

04Stored energy and unloaded quality factor

The unloaded quality factor is defined as angular frequency times stored electromagnetic energy divided by internal mean dissipated power. Using peak phasor amplitudes, electric and magnetic line energies each carry a quarter factor after time averaging. Their spatial integrals must both be included to obtain the energy of the standing-wave mode.

For weak independent conductor, dielectric and radiation losses, inverse partial Q contributions add at fixed mode energy. This intrinsic Q characterizes dissipative mechanisms before strong external excitation. A measured resonance connected to feed lines instead exhibits a loaded Q that must also account for escaping energy.

U=\frac14\int_0^l\!\left(C'|V(z)|^2+L'|I(z)|^2\right)\,dz

Q_0=\frac{\omega_0 U}{P_{\mathrm{int}}},\qquad Q_{e,i}=\frac{\omega_0 U}{P_{e,i}}

\frac1{Q_L}=\frac1{Q_0}+\sum_i\frac1{Q_{e,i}}

05External coupling and loaded resonance

Each coupled measurement port removes energy from the resonant mode and is assigned its own external Q, defined by emitted power relative to the same stored energy. The reciprocals of unloaded and all external quality factors sum to the reciprocal loaded Q. Under an isolated Lorentzian approximation, frequency divided by loaded Q estimates half-power bandwidth.

Temporal coupled-mode theory gives a simple complex two-port transmission when one mode dominates and no direct bypass path exists. Symmetrical external couplers still suffer finite insertion loss whenever intrinsic dissipation is nonzero. Direct feedthrough, non-Lorentzian Fano features or strongly varying couplers require a more general response model.

\kappa=\kappa_0+\kappa_{e1}+\kappa_{e2},\quad\kappa_i=\frac{\omega_0}{Q_i}

S_{21}(\omega)=\frac{\sqrt{\kappa_{e1}\kappa_{e2}}}{j(\omega-\omega_0)+\kappa/2}

|S_{21}(\omega_0)|=\frac{2\sqrt{Q_{e1}^{-1}Q_{e2}^{-1}}}{Q_0^{-1}+Q_{e1}^{-1}+Q_{e2}^{-1}}

06Numerical guided-line dimensions at 2.45 GHz

With guided phase velocity 2.00×10^8 m/s at 2.45 GHz, the guided wavelength is 81.63265 mm. A half-wave open/open or short/short resonator then has ideal electrical length 40.81633 mm, while a mixed short/open quarter-wave resonator has length 20.40816 mm. These are line-model values at the stated frequency.

Physical masks require corrections for microstrip effective permittivity, bends, fringing fields, coupling to adjacent traces, via inductance and discontinuity capacitance. Synthesize electrical length first and then refine using field simulation or calibrated reference structures rather than treating the ideal numbers as fabrication-ready geometries.

f_0=2.45\,\mathrm{GHz},\quad v_p=2.00\times10^8\,\mathrm{m\,s^{-1}}

l_{\lambda/2}=\frac{v_p}{2f_0}=40.81633\,\mathrm{mm}

l_{\lambda/4}=\frac{v_p}{4f_0}=20.40816\,\mathrm{mm}

07Two-port resonator: linewidth, return and insertion loss

Let Q0=200 and two external Q values each equal 100 for one isolated resonance at 2.45 GHz. The reciprocal sum gives QL=40 and an approximate half-power width 61.25 MHz. This explicitly separates intrinsic energy dissipation from energy emitted into the input and output transmission lines.

For a single-pole two-port device without a nonresonant path, equal couplers yield on-resonance transmission magnitude 0.8 and reflected amplitude 0.2. Therefore insertion loss is about 1.94 dB and return loss about 13.98 dB. These are model-specific values; a multi-pole filter has a different transfer function.

Q_0=200,\quad Q_{e1}=Q_{e2}=100

Q_L=\left(\frac1{200}+\frac1{100}+\frac1{100}\right)^{-1}=40

\Delta f_{3\mathrm{dB}}\approx\frac{f_0}{Q_L}=61.25\,\mathrm{MHz},\quad |S_{21}(f_0)|=0.8

08Mode splitting and pairwise coupling

Two weakly interacting identical resonators exchange stored electromagnetic energy and their formerly coincident modes split into two distinct eigenfrequencies. The squared-frequency splitting estimates the magnitude of coupling in the standard symmetric two-mode approximation. Electrical and magnetic coupling can produce different physical signs under explicit modal conventions.

Gap distance alone does not determine a universal coupling coefficient. Extract two eigenfrequencies with weak probes or electromagnetic eigenmode simulation while avoiding heavy external loading. The observed splitting is meaningful only when the two uncoupled resonators are sufficiently similar and the mode identification is reliable.

k=\frac{f_h^2-f_l^2}{f_h^2+f_l^2}\quad\mathrm{(magnitude;\ ideal\ identical\ pair)}

\mathrm{FBW}=\frac{f_2-f_1}{f_0},\quad f_0=\sqrt{f_1f_2}

Q_{e1}=\frac{g_0g_1}{\mathrm{FBW}},\quad Q_{eN}=\frac{g_Ng_{N+1}}{\mathrm{FBW}},\quad |k_{i,i+1}|=\frac{\mathrm{FBW}}{\sqrt{g_i g_{i+1}}}

09Two-pole Butterworth bandpass design

For a narrowband synchronous second-order bandpass filter, begin with a normalized lowpass prototype and a specified fractional bandwidth. The Butterworth two-pole prototype has outer g values equal to one and both inner values equal to √2. Standard weak-coupling synthesis maps these values to target external quality factors and inter-resonator coupling magnitude.

With 10% fractional bandwidth the external target on each port is about 14.142 and the target coupling magnitude is 0.07071. Those numbers are not physical gaps or line widths: feed position, gap and resonator spacing must be tuned using electromagnetic simulations or extraction measurements. Unloaded loss shifts actual insertion loss.

Two open-ended microstrip resonators: capacitive feed gaps to source/load and field-mediated inter-resonator coupling; there is NO galvanic connection between the two strips.

g_0=g_3=1,\qquad g_1=g_2=\sqrt2\quad\mathrm{(2\ pole\ Butterworth)}

\mathrm{FBW}=0.10\quad\Longrightarrow\quad Q_{e1}=Q_{e2}=14.1421

|k_{12}|=\frac{0.10}{\sqrt2}=0.0707107

10Microstrip fabrication and calibrated VNA validation

Open microstrip resonators exhibit strong electric field near their ends; quarter-wave shorts exhibit significant current near the grounded connection. Parallel-coupled-line and hairpin filters exploit these field distributions to obtain electric and magnetic coupling. Dielectric tangent, metal roughness, bends, open-end fringes and packaging perturb the ideal resonance.

Characterize a weakly coupled standalone resonator before a coupled pair, then set input/output feed coupling to the external-Q targets. Calibrate a two-port VNA at documented reference planes and measure complex S11 and S21, bandwidth, insertion loss and group delay. Compare both lossy field simulation and measured tolerance rather than one center-frequency number.

\alpha\ll\beta,\quad Q_{\mathrm{prop}}\simeq\frac{\omega_0}{2\alpha v_g}

\frac1{Q_0}\simeq\frac1{Q_c}+\frac1{Q_d}+\frac1{Q_r},\quad Q_d\simeq\frac1{p_d\tan\delta}

\mathrm{IL}=-20\log_{10}|S_{21}|,\quad\mathrm{RL}=-20\log_{10}|S_{11}|

Ten mastery exercises and worked derivations

Derive the nonzero resonant frequencies of a uniform short–short line and compare its voltage with an open–open line.
For a short at z=0 and an open at z=l, derive quarter-wave eigenfrequencies and explain the missing even harmonics.
At 2.45 GHz and phase velocity 2.00×10^8 m/s, compute the ideal half-wave and quarter-wave lengths in millimeters.
Find the loaded Q and three-decibel width for an isolated 2.45 GHz resonance with Q0=200 and two external Q values of 100.
Calculate on-resonance two-port transmission magnitude and insertion loss for the same single-pole resonator.
Calculate reflection magnitude and return loss on the input port, stating the assumed lack of direct feedthrough.
Derive external Q and nearest-neighbor coupling magnitude for a two-pole Butterworth filter with ten percent fractional bandwidth.
Given a mode-splitting coupling magnitude of 0.07071 and geometric mean frequency 2.45 GHz, predict the two eigenfrequencies.
With phase and group velocity both 2.00×10^8 m/s and distributed attenuation 0.1 Np/m, estimate propagation-limited Q.
At 1 GHz, design a 10 nH series RLC resonator with 2 Ω series resistance and find capacitance, Q and ideal half-power bandwidth.

Show complete solutions

1. Derive the nonzero resonant frequencies of a uniform short–short line and compare its voltage with an open–open line.

At a short circuit, the line voltage vanishes. Choose the voltage phasor as a sine whose origin is one shorted end, and then require it also to vanish at the other. A nontrivial oscillatory solution exists when the accumulated electrical phase is nπ with positive integer n, giving n times the velocity divided by twice the length.

An open-ended line instead requires current zero at both ends, producing voltage extrema at the terminations and the same nonzero half-wave frequency series. Their current and voltage patterns are complementary; a zero-frequency static uniform potential is not counted as an RF resonant mode.

V(z)=A\sin(\beta z),\quad V(l)=0\Rightarrow\beta l=n\pi

f_1=\frac{v_p}{2l}

2. For a short at z=0 and an open at z=l, derive quarter-wave eigenfrequencies and explain the missing even harmonics.

At the shorted boundary set voltage zero, so the standing-wave voltage starts with a sine. At the open boundary the current vanishes, which means the spatial voltage derivative is zero for an ideal TEM line. The cosine of the total electrical phase must therefore vanish.

Those solutions lie at odd multiples of π/2, yielding successive frequencies proportional to one, three and five. The even harmonics of the quarter-wave fundamental are absent in the ideal non-dispersive longitudinal mode series. A real via and an open-end fringing field cause resonance-frequency shifts.

V(z)=A\sin(\beta z),\quad I(l)\propto\cos(\beta l)=0

\beta l=\frac{\pi}{2}+n\pi,\quad f_0=\frac{v_p}{4l}

3. At 2.45 GHz and phase velocity 2.00×10^8 m/s, compute the ideal half-wave and quarter-wave lengths in millimeters.

First divide phase velocity by the operating frequency to obtain a guided wavelength of approximately 81.63265 millimeters. The calculation assumes a single quasi-TEM phase velocity that does not vary significantly across the narrow design region.

Divide by two for the fundamental open–open or short–short line and by four for a mixed-boundary resonator, obtaining about 40.81633 and 20.40816 millimeters. These are electrical lengths before microstrip end and bend corrections.

\lambda_g=\frac{2.00\times10^8}{2.45\times10^9}=81.63265\,\mathrm{mm}

l_{\lambda/2}=40.81633\,\mathrm{mm},\quad l_{\lambda/4}=20.40816\,\mathrm{mm}

4. Find the loaded Q and three-decibel width for an isolated 2.45 GHz resonance with Q0=200 and two external Q values of 100.

Power escaping into each external port contributes separately to the decay of stored energy. Add the three inverse quality factors under a common modal stored-energy normalization: the reciprocal contributions are 0.005, 0.01 and 0.01, hence their sum is 0.025.

The resulting loaded quality factor is forty. In an isolated weakly dispersive Lorentzian, center frequency divided by loaded Q gives a full half-power width of 61.25 megahertz. One must not use the unloaded Q of two hundred as the width denominator after attaching two ports.

\frac1{Q_L}=\frac1{200}+\frac1{100}+\frac1{100}=0.025

Q_L=40,\qquad\Delta f=\frac{2.45\,\mathrm{GHz}}{40}=61.25\,\mathrm{MHz}

5. Calculate on-resonance two-port transmission magnitude and insertion loss for the same single-pole resonator.

Use energy decay rates proportional to inverse quality factors. For a single isolated two-port mode without a direct bypass, the resonant transmission amplitude equals twice the geometric mean of external inverse-Q values divided by the total inverse-Q sum.

With both external inverse Q values 0.01 and intrinsic inverse Q 0.005, the expression gives an amplitude of 0.8. The corresponding insertion loss is minus twenty times the base-ten logarithm of 0.8, or approximately 1.9382 decibels, even for symmetric couplers.

|S_{21}(\omega_0)|=\frac{2\sqrt{(1/100)(1/100)}}{1/200+2/100}=0.8

\mathrm{IL}=-20\log_{10}(0.8)=1.9382\,\mathrm{dB}

6. Calculate reflection magnitude and return loss on the input port, stating the assumed lack of direct feedthrough.

For the same one-mode two-port model, the complex input reflection at resonance equals unity minus twice the input coupling fraction of total decay. That coupling fraction is 0.01 divided by 0.025, giving a reflected amplitude of positive 0.2.

Return loss is minus twenty times the logarithm of the reflected amplitude, or about 13.9794 decibels. These definitions use power-normalized 50-ohm S parameters and presume negligible coherent direct transmission; a real Fano-type background changes this result.

S_{11}(\omega_0)=1-\frac{2/Q_{e1}}{1/Q_0+1/Q_{e1}+1/Q_{e2}}=0.2

\mathrm{RL}=-20\log_{10}(0.2)=13.9794\,\mathrm{dB}

7. Derive external Q and nearest-neighbor coupling magnitude for a two-pole Butterworth filter with ten percent fractional bandwidth.

Normalize the second-order Butterworth lowpass prototype. Its outer coefficients are one while the two inner coefficients both equal the square root of two. Under the ideal narrowband synchronous mapping, each external Q is an adjacent outer-inner product divided by fractional bandwidth.

For ten percent fractional bandwidth this gives 14.1421 at each feed. The nearest-neighbor coupling magnitude equals bandwidth divided by the square root of the product of both inner coefficients, or 0.0707107. Extract feed gap and resonator gap geometrically; the coupling is not itself a distance.

g_1=g_2=\sqrt2,\quad g_0=g_3=1,\quad\mathrm{FBW}=0.10

Q_{e1}=Q_{e2}=\frac{\sqrt2}{0.10}=14.1421,\quad |k_{12}|=\frac{0.10}{\sqrt2}

8. Given a mode-splitting coupling magnitude of 0.07071 and geometric mean frequency 2.45 GHz, predict the two eigenfrequencies.

The usual weakly coupled identical-mode extraction defines coupling magnitude as the positive squared-frequency difference normalized by its squared-frequency sum. Algebraically solve the resulting ratio for the ratio of upper and lower eigenfrequencies. The geometric-mean frequency fixes their product.

At the stated coupling and 2.45 gigahertz geometric mean, the lower mode lies near 2.36475 gigahertz and the upper near 2.53832 gigahertz. The spacing is about 173.57 megahertz. Physical coupling sign cannot be determined merely by sorting the two observed frequencies.

r=\frac{f_h}{f_l}=\sqrt{\frac{1+k}{1-k}}

f_l=f_0\left(\frac{1-k}{1+k}\right)^{1/4},\quad f_h=f_0\left(\frac{1+k}{1-k}\right)^{1/4}

9. With phase and group velocity both 2.00×10^8 m/s and distributed attenuation 0.1 Np/m, estimate propagation-limited Q.

A weakly attenuated traveling wave decays in amplitude with distance as the exponential of minus αz, so its transported energy diminishes at twice that spatial rate. Convert attenuation per distance to energy-decay rate using group velocity, and divide angular frequency by this decay rate.

For group and phase velocities both equal to two hundred million meters per second, the propagation-limited Q equals the propagation phase constant divided by twice α. At 2.45 gigahertz and attenuation 0.1 Np/m it is approximately 384.845. Strong dispersion invalidates the replacement of group velocity by phase velocity.

Q_{\mathrm{prop}}=\frac{\omega_0}{2\alpha v_g}\quad(\mathrm{weak\ loss})

v_g=v_p\Longrightarrow Q_{\mathrm{prop}}=\frac{\beta}{2\alpha}

10. At 1 GHz, design a 10 nH series RLC resonator with 2 Ω series resistance and find capacitance, Q and ideal half-power bandwidth.

At series resonance the inductive and capacitive reactances cancel, so angular frequency squared times inductance times capacitance equals one. Solve for capacitance using one gigahertz and ten nanohenries to obtain approximately 2.53303 picofarads.

With a series resistance of two ohms, the unloaded quality at resonance equals angular frequency times inductance divided by resistance, about 31.4159. Under the high-Q series-resonator approximation the full half-power frequency width is center frequency divided by Q, approximately 31.8310 megahertz.

f_0=\frac1{2\pi\sqrt{LC}},\quad C=\frac1{(2\pi f_0)^2 L}

Q_{\mathrm{series}}=\frac{\omega_0 L}{R_s},\quad\Delta f=\frac{f_0}{Q_{\mathrm{series}}}

Next: RF-06.05, losses, bandwidth and implementation constraints.