3.23: Single-Stub Matching
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In Section 3.22, we considered impedance matching schemes consisting of a transmission line combined with a reactance which is placed either in series or in parallel with the transmission line. In many problems, the required discrete reactance is not practical because it is not a standard value, or because of non-ideal behavior at the desired frequency (see Section 3.21 for more about this), or because one might simply wish to avoid the cost and logistical issues associated with an additional component. Whatever the reason, a possible solution is to replace the discrete reactance with a transmission line “stub” – that is, a transmission line which has been open- or short-circuited. Section 3.16 explains how a stub can replace a discrete reactance. Figure 3.23.1 shows a practical implementation of this idea implemented in microstrip. This section explains the theory, and we’ll return to this implementation at the end of the section.

Figure 3.23.2 shows the scheme. This scheme is usually implemented using the parallel reactance approach, as depicted in the figure. Although a series reactance scheme is also possible in principle, it is usually not as convenient. This is because most transmission lines use one of their two conductors as a local datum; e.g., the ground plane of a printed circuit board for microstrip line is tied to ground, and the outer conductor (“shield”) of a coaxial cable is usually tied to ground. This is contrast to a discrete reactance (such as a capacitor or inductor), which does not require that either of its terminals be tied to ground. This issue is avoided in the parallel-attached stub because the parallel-attached stub and the transmission line to which it is attached both have one terminal at ground.

The single-stub matching procedure is essentially the same as the single parallel reactance method, except the parallel reactance is implemented using a short- or open-circuited stub as opposed a discrete inductor or capacitor. Since parallel reactance matching is most easily done using admittances, [admittance] it is useful to express Zin(l)=+jZ0tanβl and Zin(l)=−jZ0cotβl (input impedance of an open- and short-circuited stub, respectively, from Section 3.16) in terms of susceptance: Bp=−Y02cot(β2l2) short-circuited stub Bp=+Y02tan(β2l2) open-circuited stub As in the main line, the characteristic impedance Z02=1/Y02 is an independent variable and is chosen for convenience.
A final question is when should you use a short-circuited stub, and when should you use an open-circuited stub? Given no other basis for selection, the termination that yields the shortest stub is chosen. An example of an “other basis for selection” that frequently comes up is whether DC might be present on the line. If DC is present with the signal of interest, then a short circuit termination without some kind of remediation to prevent a short circuit for DC would certainly be a bad idea.
In the following example we address the same problem raised in Section 3.22 (Examples 3.22.1 and 3.22.2), now using the single-stub approach:
Design a single-stub match that matches a source impedance of 50Ω to a load impedance of 33.9+j17.6 Ω. Use transmission lines having characteristic impedances of 50Ω throughout, and leave your answer in terms of wavelengths.
Solution
From the problem statement: Zin≜ and Z_L = 33.9+j17.6~\Omega are the source and load impedances respectively. Z_0=50~\Omega is the characteristic impedance of the transmission lines to be used. The reflection coefficient \Gamma (i.e., Z_L with respect to the characteristic impedance of the transmission line) is
\Gamma \triangleq \frac{Z_L - Z_0}{Z_L + Z_0} \cong -0.142 + j0.239 \nonumber The length l_1 of the primary line (that is, the one that connects the two ports of the matching structure) is the solution to the equation (from Section 3.22):
\mbox{Re}\left\{Y_1\right\} = \mbox{Re}\left\{Y_{01} \frac{ 1 - \Gamma e^{-j2\beta_1 l_1} }{ 1 + \Gamma e^{-j2\beta_1 l_1} }\right\} \nonumber where here \mbox{Re}\left\{Y_1\right\}=\mbox{Re}\left\{1/Z_S\right\}=0.02 mho and Y_{01} = 1/Z_0 = 0.02 mho. Also note
2\beta_1 l_1 = 2 \left(\frac{2\pi}{\lambda}\right) l_1 = 4\pi \frac{l_1}{\lambda} \nonumber
where \lambda is the wavelength in the transmission line. So the equation to be solved for l_1 is:
1 = \mbox{Re}\left\{ \frac{ 1-\Gamma e^{-j4\pi l_1/\lambda } } { 1+\Gamma e^{-j4\pi l_1/\lambda } } \right\} \nonumber
By trial and error (or using the Smith chart; see “Additional Reading” at the end of this section) we find for the primary line l_1 \cong 0.020 \lambda, yielding Y_1 \cong 0.0200-j0.0116 mho for the input admittance after attaching the primary line.
We now seek the shortest stub having an input admittance of \cong +j0.0116 mho to cancel the imaginary part of Y_1. For an open-circuited stub, we need
B_p = +Y_0 \tan{2\pi l_2/\lambda} \cong +j0.0116~\mbox{mho} \nonumber
The smallest value of l_2 for which this is true is \cong 0.084\lambda. For a short-circuited stub, we need B_p = -Y_0 \cot{2\pi l_2/\lambda} \cong +j0.0116~\mbox{mho} \nonumber The smallest positive value of l_2 for which this is true is \cong 0.334\lambda; i.e., much longer. Therefore, we choose the open-circuited stub with l_2 \cong 0.084 \lambda. Note the stub is attached in parallel at the source end of the primary line.
Single-stub matching is a very common method for impedance matching using microstrip lines at frequences in the UHF band (300-3000 MHz) and above. In Figure \PageIndex{1}, the top (visible) traces comprise one conductor, whereas the ground plane (underneath, so not visible) comprises the other conductor. The end of the stub is not connected to the ground plane, so the termination is an open circuit. A short circuit termination is accomplished by connecting the end of the stub to the ground plane using a via; that is, a plated-through that electrically connects the top and bottom layers.