Question 2 of 8: Distance to the Unit-Conductance Circle on a Mismatched Line

Question 2: Distance to the Unit-Conductance Circle on a Mismatched Line (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A lossless line of characteristic impedance $Z_0$ is terminated in $Z_L = Z_0(0.5 + j2.0)$, so the normalised load impedance is $z_L = 0.5 + j2.0$ and the answer is required as a fraction of a wavelength.

Find. The distance $d$ (in wavelengths) back from the termination at which the input admittance has real part equal to $Y_0$, i.e. at which the normalised input admittance is $y_{\text{in}} = 1 + jb$.

Approach. Map the load to its reflection coefficient, note that moving along a lossless line only rotates $\Gamma$ at constant magnitude, and intersect that circle with the locus $\operatorname{Re}\{y\} = 1$, which in the $\Gamma$ plane is the circle of centre $-\tfrac12$ and radius $\tfrac12$.

1. Compute the load reflection coefficient. $$\Gamma_L = \frac{z_L - 1}{z_L + 1} = \frac{-0.5 + j2.0}{1.5 + j2.0} = 0.520 + j0.640 = 0.8246\,\angle\,50.91^\circ .$$ The standing-wave ratio is $S = (1+|\Gamma|)/(1-|\Gamma|) = 10.4$, so this is a badly mismatched line.
2. State the target locus. With $y = (1-\Gamma)/(1+\Gamma)$, the condition $\operatorname{Re}\{y\} = 1$ maps to the circle $$\left|\Gamma + \tfrac{1}{2}\right| = \tfrac{1}{2},$$ the unit-conductance circle used in single-stub matching.
3. Intersect the two circles. Writing $\Gamma = \rho e^{j\phi}$ with $\rho = 0.8246$ fixed and expanding $|\rho e^{j\phi} + \tfrac12|^2 = \tfrac14$ gives $\rho^2 + \rho\cos\phi = 0$, hence $$\cos\phi = -\rho = -0.8246 \quad\Longrightarrow\quad \phi = \pm 145.57^\circ .$$ There are exactly two crossings per half wavelength, as the figure shows.
4. Convert the rotation into a distance. Moving a distance $d$ toward the generator changes the phase by $-2\beta d$, so $\phi = \theta_\Gamma - 2\beta d$ with $\theta_\Gamma = 50.91^\circ$ and $2\beta d = 720^\circ (d/\lambda)$. Solving for each root, $$\frac{d}{\lambda} = \frac{\theta_\Gamma - \phi}{720^\circ} \pmod{0.5}.$$
5. Evaluate both roots. $$\boxed{\ \frac{d_1}{\lambda} = 0.273, \qquad \frac{d_2}{\lambda} = 0.369\ }$$ and both repeat every half wavelength, at $0.273 + 0.5k$ and $0.369 + 0.5k$ for integer $k$.
6. Check the admittances at those points. Back-substituting gives $y_{\text{in}} = 1 + j2.92$ at $d_1$ and $y_{\text{in}} = 1 - j2.92$ at $d_2$: the conductance is exactly $Y_0$ in both cases, and the two susceptances are equal and opposite as the symmetry of the construction requires.

The nearest answer to the termination is therefore $d = 0.273\lambda$. The practical significance is immediate: these are the only two places per half wavelength where a single shunt stub can complete a match, because a shunt element can cancel susceptance but cannot change conductance. Choosing $d_1$ needs a stub supplying $-j2.92$ and choosing $d_2$ needs one supplying $+j2.92$.