RF-06.03 · Single- and double-stub impedance matching

01Transmission lines, phasors and the matching reference plane

At microwave frequencies the voltage and current on a uniform transmission line are superpositions of forward and backward propagating waves. In the lossless TEM or quasi-TEM approximation, they are related by a real characteristic impedance and a real propagation constant. The relevant wavelength is the guided wavelength, set by phase velocity rather than by the vacuum value of the speed of light.

Set the phase convention to exp(+jωt), choose a real reference impedance and place the distance coordinate at the load, increasing toward the generator. The input admittance after a section of line is obtained from its two-port wave solution, not by simply adding impedances. Normalize all admittances by the same main-line characteristic admittance before employing the Smith chart.

Z_{\mathrm{in}}(d)=Z_0\frac{Z_L+jZ_0\tan(\beta d)}{Z_0+jZ_L\tan(\beta d)}

y(d)=\frac{y_L+j\tan(\beta d)}{1+jy_L\tan(\beta d)}

\lambda_g=\frac{v_p}{f},\qquad \beta=\frac{2\pi}{\lambda_g}

02From telegrapher equations to lossless and lossy transformation

For a uniform line, the distributed resistance, inductance, shunt conductance and capacitance per unit length give the telegrapher equations. Their harmonic solutions propagate with a generally complex propagation constant. In the lossless limit the hyperbolic input-impedance formula becomes the familiar tangent expression, reflecting a periodically repeating impedance transformation.

Attenuation causes the reflection magnitude to decrease as one moves toward the generator, while the phase changes by twice the propagation phase. Consequently an exact lossless stub length is not necessarily an exact match on a lossy microstrip. Do not use the real phase constant as a replacement for the full complex propagation constant in a loss-sensitive calculation.

Z_{\mathrm{in}}(l)=Z_0\frac{Z_L+Z_0\tanh(\gamma l)}{Z_0+Z_L\tanh(\gamma l)}

\gamma=\alpha+j\beta,\qquad \alpha=0\ \Longrightarrow\ \tanh(j\beta l)=j\tan(\beta l)

\Gamma(d)=\Gamma_L\,e^{-2\gamma d},\qquad \Gamma_L=\frac{Z_L-Z_0}{Z_L+Z_0}

03Open- and short-circuited shunt stub susceptance

A stub is a finite transmission-line section connected at a tee junction. Its far end is deliberately terminated in an open circuit or a short circuit. In the lossless case the input impedance of a short is purely inductive for sufficiently short positive electrical length, while its input admittance is negative imaginary; an open stub has positive imaginary input admittance for a sufficiently short electrical length.

A shunt stub adds admittance rather than impedance to the main line. If its characteristic impedance differs from the main line, normalize its admittance with the main-line characteristic admittance instead of inserting its dimensionless tangent directly. Both ideal stubs repeat their susceptance every half guided wavelength, although open-end fringing and short-via inductance break exact physical ideality.

z_{\mathrm{stub,sc}}=j\tan(\beta l_s),\quad y_{\mathrm{stub,sc}}=-j\cot(\beta l_s)

z_{\mathrm{stub,oc}}=-j\cot(\beta l_s),\quad y_{\mathrm{stub,oc}}=j\tan(\beta l_s)

b_{\mathrm{stub}}=\frac{Z_{0,\mathrm{main}}}{Z_{0,\mathrm{stub}}}\begin{cases}-\cot(\beta_s l_s)&\mathrm{short}\\\tan(\beta_s l_s)&\mathrm{open}\end{cases}

04Single shunt-stub synthesis: unity conductance first

Let the normalized load admittance have real part g and imaginary part b. Propagation through a distance from the load toward the source traces a constant-reflection circle on a lossless admittance Smith chart. The line transforms both conductance and susceptance, but adding an ideal shunt stub changes only susceptance; therefore the junction must first be placed where the transformed conductance equals unity.

Solve the real-part equation for the tangent of electrical distance. In general two solutions exist within one half-wavelength period; evaluate the imaginary admittance at each location. An open or short stub with the opposite normalized susceptance then cancels the imaginary part exactly at the design frequency. Changing the sign of a stub susceptance without moving its physical junction will generally not preserve the match.

y_L=g+jb,\quad t=\tan(\beta d),\quad y(d)=\frac{g+j(b+t)}{1-bt+jgt}

\operatorname{Re}y(d)=\frac{g(1+t^2)}{(1-bt)^2+(gt)^2}=1

b_{\mathrm{stub}}=-\operatorname{Im}y(d),\quad y_{\mathrm{total}}=y(d)+jb_{\mathrm{stub}}=1

05Worked single-stub example with a resistive load

Consider a one-hundred-ohm resistive termination on a fifty-ohm lossless line, both referred to the same phase convention. Its normalized load admittance is one half. Enforcing unity transformed conductance leads to a tangent magnitude equal to the square root of two. The forward solution, measured from the load toward the generator, corresponds to about 0.152 guided wavelengths.

At this junction the normalized admittance is one plus j divided by the square root of two. The required shunt stub susceptance is therefore negative with the same magnitude. A short-circuit stub can provide that susceptance at the same normalized length as the main-line spacing; an open-circuit stub instead requires a length near 0.402 guided wavelengths in the ideal model.

Single shorted shunt stub on a 50 Ω line: tee 0.152 guided wavelengths before a 100 Ω load; stub terminates electrically on the common RF return.

Admittance Smith chart · single-shunt matching. Γ=(1−y)/(1+y), d/λg=0.152043.

y_L=\frac12,\quad t=\pm\sqrt2,\quad y(d)=1\pm j\frac{1}{\sqrt2}

\frac{d_+}{\lambda_g}=\frac{\arctan\sqrt2}{2\pi}\approx0.152043

\frac{l_{\mathrm{sc}}}{\lambda_g}=0.152043,\quad\frac{l_{\mathrm{oc}}}{\lambda_g}=0.402043

06Convert electrical lengths into fabrication dimensions

An electrical distance does not become a length in millimetres until the phase velocity is specified. In a homogeneous TEM approximation the guided wavelength equals phase velocity divided by frequency; a microstrip uses the relevant quasi-TEM effective permittivity, and its discontinuities may require additional electromagnetic corrections. Never substitute the free-space wavelength without checking the substrate and field distribution.

For a numerical illustration take a phase velocity of two hundred million metres per second at 2.45 gigahertz. The guided wavelength is approximately 81.63 millimetres. The positive-distance short-stub solution uses a line-to-stub junction approximately 12.41 millimetres from the load and a shorted stub of equal ideal electrical length; the open-stub option is about 32.82 millimetres.

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

\lambda_g=\frac{v_p}{f_0}\approx81.63265\,\mathrm{mm}

d_+\approx12.41170\,\mathrm{mm},\quad l_{\mathrm{sc}}\approx12.41170\,\mathrm{mm},\quad l_{\mathrm{oc}}\approx32.81987\,\mathrm{mm}

07Double shunt stubs with fixed separation

If two stub tee junctions have a fixed spacing, the first shunt stub adjusts the imaginary part of the admittance at its load-side junction. Propagation along the line between junctions then converts that adjustment into a change in both real and imaginary admittance at the source-side junction. The second stub only cancels the remaining imaginary part if the conductance arriving at that junction is exactly unity.

Write the normalized load-side admittance plus the first stub susceptance. Apply the lossless line-admittance transform over the fixed physical spacing, then impose unit real part and solve for the first stub susceptance. Finally set the second stub susceptance equal to minus the imaginary part. The ordering is crucial: the second stub is nearer the generator, and the separation is not a variable when the hardware has already been fabricated.

y_1=g+j(b_L+b_1),\quad t_\Delta=\tan(\beta\Delta),\quad y_2=\frac{y_1+jt_\Delta}{1+jy_1t_\Delta}

g_2=\frac{g(1+t_\Delta^2)}{[1-(b_L+b_1)t_\Delta]^2+(gt_\Delta)^2}=1

b_2=-\operatorname{Im}y_2,\quad y_{\mathrm{in}}=y_2+jb_2=1

08Double-stub numerical branches and short/open realization

For a normalized half-siemens load conductance and fixed inter-stub separation equal to one eighth of guided wavelength, the line tangent equals one. The first-stub conductance condition becomes a quadratic equation with two valid real roots. These yield two different possible first-stub susceptances and, after propagation, two different second-stub values.

For the smaller first-stub solution its normalized susceptance is approximately positive 0.133975; the second stub requires approximately negative 0.732051. An open first stub and a short second stub can realize these signs using suitable electrical lengths. The larger first-stub root is also valid but tends to require different stub lengths; neither choice means that a matched physical load has been inserted at the intermediate junction.

Fixed-spacing double-shunt stubs: load-side stub 1 is OPEN, source-side stub 2 is SHORTED; open end is deliberately isolated from common RF return.

g=\frac12,\quad b_L=0,\quad \Delta=\frac{\lambda_g}{8},\quad t_\Delta=1

(1-b_1)^2=\frac34,\quad b_1=1\pm\frac{\sqrt3}{2}

b_1=1-\frac{\sqrt3}{2}\approx0.133975,\quad b_2=-\left(\frac34-b_1^2\right)\approx-0.732051

09Forbidden regions of a fixed-spacing double-stub network

A double-stub network does not always admit a solution when both junctions are fixed. Even with an arbitrary first stub susceptance, the real part after the inter-stub line has an upper bound determined by the load-side conductance and the stub spacing. Matching requires that the maximum possible conductance at the second tee be at least unity.

For a separation of one eighth of guided wavelength, the bound is a load-side normalized conductance no greater than two. A purely resistive load having normalized admittance three therefore cannot be matched by this precise fixed arrangement using only two ideal shunt stubs and no additional line section before the first stub. Moving the first tee or modifying the spacing changes the feasible region.

\min_{b_1}\left([1-(b_L+b_1)t_\Delta]^2+(gt_\Delta)^2\right)=g^2t_\Delta^2

g\le\frac{1+t_\Delta^2}{t_\Delta^2}=\csc^2(\beta\Delta)

\Delta=\frac{\lambda_g}{8}\Longrightarrow g\le2,\qquad g=3\ \mathrm{has\ no\ lossless\ solution}

10Smith chart checks, bandwidth, loss and VNA measurement

Use the admittance Smith chart rather than the impedance form when reading shunt-stub junctions. Start at the load, rotate toward the generator to the unity-conductance circle for single-stub matching, and read the susceptance to cancel. For double-stub synthesis, rotate between the two fixed tee positions after applying the first stub and verify the real part at the second junction before tuning its susceptance.

Physical microstrip stubs add open-end capacitance, via inductance, tee discontinuities, attenuation and dispersion. Check phase velocity or effective permittivity across frequency, design widths for the proper characteristic impedances, and include losses in the model. Calibrate a VNA at defined reference planes, measure complex S11, preserve Touchstone data, and distinguish return loss from insertion loss measured with a true two-port fixture.

\Gamma_{\mathrm{in}}(f)=\frac{1-y_{\mathrm{in}}(f)}{1+y_{\mathrm{in}}(f)}

\mathrm{RL}(f)=-20\log_{10}|\Gamma_{\mathrm{in}}(f)|,\qquad \gamma=\alpha+j\beta

\varepsilon_{\mathrm{eff}}=\left(\frac{c}{v_p}\right)^2\quad\mathrm{(quasi\!-\!TEM\ homogeneous\ equivalent)}

Mastery problems with detailed derivations

Derive the normalized line-admittance transformation from the input-impedance formula and state the coordinate convention.
Find the lossless input admittance of a shorted stub and an open stub, including the correct signs under the positive phasor convention.
For a 50-ohm line terminated by 100 ohms, find both single-shunt-stub junction positions in the first half guided wavelength.
At the nearer single-stub junction, compute the required negative susceptance and realize it with a shorted stub.
Realize the same single-stub match with an open stub, explaining the required branch and its different physical length.
At 2.45 GHz with phase velocity 2.00×10^8 m/s, convert the chosen junction and stub lengths into millimetres.
At a double-stub spacing of one eighth wavelength, solve for both allowed first-stub susceptances for a normalized load admittance 0.5.
For the smaller double-stub root, derive the second stub susceptance and explain where both stubs must connect.
Derive the forbidden-region bound and determine whether a load conductance of three can be matched with two fixed shunt stubs separated by one eighth wavelength.
Outline an S11-based VNA validation strategy for an actual microstrip stub network, including limitations of ideal open and short terminations.

Show ten worked solutions

1. Derive the normalized line-admittance transformation from the input-impedance formula and state the coordinate convention.

Start from the transmission-line input impedance written as a bilinear fraction of load impedance and the tangent of propagation phase. Divide every impedance by the same main-line characteristic impedance. Invert the entire normalized expression, not its numerator alone; the resulting input admittance is a ratio involving load admittance and the tangent.

Distance is measured positively from the load toward the source, with voltage and current phasors using the same positive exponential time convention. At zero distance the result must equal the load admittance; at one quarter guided wavelength it gives reciprocal normalized admittance. Those two limits are an independent algebraic consistency check.

Z_{\mathrm{in}}=Z_0\frac{Z_L+jZ_0t}{Z_0+jZ_Lt},\qquad t=\tan(\beta d)

y(d)=\frac{y_L+jt}{1+jy_Lt}

2. Find the lossless input admittance of a shorted stub and an open stub, including the correct signs under the positive phasor convention.

For a shorted lossless stub, substitute a zero load impedance in the line equation. The input impedance is positive imaginary for the first eighth or quarter wavelength, and its reciprocal admittance is negative imaginary. For an open stub take the infinite load-impedance limit first, then calculate the reciprocal of the resulting input impedance.

A positive small open-stub length gives a positive imaginary shunt admittance. When both stubs share the main-line characteristic impedance, their dimensionless susceptances are simply the opposite cotangent and the tangent. If the characteristic impedances differ, multiply by the main-to-stub impedance ratio before applying any normalized KCL.

y_{\mathrm{sc}}=-j\cot(\beta l_s)

y_{\mathrm{oc}}=j\tan(\beta l_s)

3. For a 50-ohm line terminated by 100 ohms, find both single-shunt-stub junction positions in the first half guided wavelength.

The real normalized load conductance is one half, and its imaginary part is zero. Set the real part of the propagated normalized admittance equal to unity and substitute the tangent of the unknown electrical distance. The equation reduces to tangent squared equal to two, giving two possible signs.

Measure distance from the load toward the generator. The positive-tangent solution occurs at about 0.152043 guided wavelengths, while the negative-tangent solution occurs at 0.347957 guided wavelengths within the first half wavelength. The two junction choices demand susceptances of opposite sign, so the physical stub realization must be recalculated for each.

y_L=0.5,\qquad t^2=2

d_+/\lambda_g=0.15204336,\quad d_-/\lambda_g=0.34795664

4. At the nearer single-stub junction, compute the required negative susceptance and realize it with a shorted stub.

At the positive-distance solution, evaluate the line-admittance fraction rather than reading only the real part. The imaginary contribution is positive one divided by the square root of two, so the shunt stub must supply its negative. KCL then yields a total normalized input admittance exactly equal to unity.

A shorted stub contributes negative cotangent of electrical length. Equating its cotangent to positive one over the square root of two gives tangent equal to the square root of two, hence a stub length of about 0.152043 guided wavelengths. This length is measured along the stub, independently of the main-line separation though numerically equal in this example.

y(d_+)=1+j/\sqrt2

b_s=-1/\sqrt2,\quad l_{\mathrm{sc}}/\lambda_g=0.15204336

5. Realize the same single-stub match with an open stub, explaining the required branch and its different physical length.

An open stub contributes positive tangent of its own electrical length. For the same selected junction it must be negative one over the square root of two; therefore the open-stub electrical angle must lie in the second quadrant, not in the small-angle capacitive region.

Selecting the shortest positive solution modulo half a guided wavelength gives approximately 0.402043 guided wavelengths. Keeping the junction 0.152043 guided wavelengths away from the load, either this open stub or the previously calculated short stub cancels the positive junction susceptance in an ideal uniform lossless network.

b_s=-1/\sqrt2,\quad \tan(\beta l_{\mathrm{oc}})=-1/\sqrt2

l_{\mathrm{oc}}/\lambda_g=0.40204336

6. At 2.45 GHz with phase velocity 2.00×10^8 m/s, convert the chosen junction and stub lengths into millimetres.

Compute the guided wavelength from the given propagation velocity divided by the frequency, without using the speed of light in vacuum. The result is approximately 81.63265 millimetres. Multiplying this length by the junction fraction 0.152043 gives approximately 12.41170 millimetres.

A shorted stub uses the same 12.41170 millimetres in the ideal equal-velocity example, while the open-stub solution uses about 32.81987 millimetres. In a real microstrip the stub may have a different effective phase velocity and an open-end fringing correction; do not interpret these ideal dimensions as final photolithography mask dimensions.

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

d_+=l_{\mathrm{sc}}=12.41170\,\mathrm{mm}

7. At a double-stub spacing of one eighth wavelength, solve for both allowed first-stub susceptances for a normalized load admittance 0.5.

The first load-side stub changes only the load susceptance, leaving its real normalized conductance at one half. The intervening lossless line of length one eighth wavelength has unit tangent. Require the transformed conductance at the second junction to equal unity and multiply through the positive denominator.

The resulting quadratic condition is that one minus the first-stub susceptance, squared, equals three quarters. Taking both square roots gives approximately 0.133975 or 1.866025. Each root corresponds to a different legal first-stub reactance; the second stub remains to be computed rather than guessed or assumed equal.

y_2=\frac{(0.5+jb_1)+j}{1+j(0.5+jb_1)}

b_1=1-\frac{\sqrt3}{2},\quad b_2\approx-0.73205081

8. For the smaller double-stub root, derive the second stub susceptance and explain where both stubs must connect.

Select the smaller positive susceptance for the first, load-side tee. Compute the line transformation over one eighth wavelength using complex inversion; the real part is exactly one and the imaginary part is approximately positive 0.732051. The source-side second stub must contribute the negative of this value.

The first stub may be implemented as a short positive-length open-ended branch; the second can be a suitable short-circuited branch. Both stubs join the main signal conductor at distinct tee nodes, separated by the fixed main-line distance; their returns use the same reference conductor, and the open stub far end must remain physically disconnected.

b_1=1+\frac{\sqrt3}{2}\approx1.8660254

b_2=-\left(0.75-b_1^2\right)\approx2.73205081

9. Derive the forbidden-region bound and determine whether a load conductance of three can be matched with two fixed shunt stubs separated by one eighth wavelength.

Minimize the denominator of the transformed conductance over all real first-stub susceptances. The square term can vanish, but the residual positive term involving squared load conductance cannot, so the maximum achievable conductance at the second tee is finite.

The unity-conductance equation has a real solution only if the first tee conductance is at most the inverse squared sine of the fixed inter-stub phase. For an eighth-wavelength separation this upper bound is two. A load-side conductance of three violates it and lies inside the forbidden region unless the fixed geometry or an additional line section is changed.

g(1+t_\Delta^2)\ge g^2t_\Delta^2

\Delta=\lambda_g/8,\quad g_{\max}=2<3

10. Outline an S11-based VNA validation strategy for an actual microstrip stub network, including limitations of ideal open and short terminations.

Calibrate the vector network analyzer at clearly stated reference planes and measure complex reflection across a useful band surrounding the operating frequency. Compare the measured match frequency and phase with the lossless single- or double-stub model and with a circuit model that includes propagation attenuation.

Record the substrate, conductor width, effective permittivity or extracted propagation constant, tee parasitics, open-end capacitance and shorting-via inductance. Save calibrated Touchstone data and repeat measurements when evaluating tolerance. A one-port return-loss dip is not an independent measurement of dissipation; use a true two-port calibration for insertion loss.

\Gamma(f)=\frac{1-y_{\mathrm{in}}(f)}{1+y_{\mathrm{in}}(f)}

\mathrm{RL}=-20\log_{10}|\Gamma|

Next: RF-06.04, distributed resonators and filters.