Takashi Ohira
IMAGE LICENSED BY INGRAM PUBLISHING
Plane geometry, in academia, often helps students intuitively understand something difficult to do with algebra. Graphics provide rich visual information rather than equations. This idea applies to postulates and theorems in a variety of scientific and technical fields. Although they are specially useful in understanding wave engineering, students sometimes find it difficult to first learn wave-specific concepts, such as S-parameters and voltage standing wave ratios (VSWRs). If we introduce those concepts by performing a persuasive graphical demonstration on a blackboard, students become familiar with them more smoothly than if we use only algebra. What concepts in our field, then, can we effectively translate into graphics? As an attempt to address the question, this article presents eight instructive figures signifying basic physics in wave engineering.
When students learn ac circuit theory, voltage and current are assumed to be in a sinusoidal waveform. A sinusoidal wave is characterized with its magnitude and phase, and is usually represented by a complex number for mathematical convenience. A complex number can be projected onto a Gaussian plane. For example, a voltage is projected as a point on the plane, as shown in Figure 1. The arrow that links origin O and voltage point V is called the voltage phasor. The plane also finds a current phasor from O to I.
Figure 1. Complementary triangles on a phasor chart. Their areas signify the effective power ${P}_{a}$ and reactive power $\text{P}{}_{b}$ observed at a load.
In this example, V is located on the clockwise side from I. This physically implies a load of inductive impedance so that V advances ahead of I. If the load exhibits capacitive impedance, V is located on the counterclockwise side from I instead. The angle VOI never exceeds 90° in any direction. This is because the load does not generate power but only consumes. Then, when we wonder how much power is consumed, a question arises: can we read out the power from the phasor chart?
To answer this question, we first focus on triangle VOI. This triangle’s area, denoted as $\text{P}{}_{b},$ physically signifies the reactive power observed at the load’s input port. This is explained by defining a complex power: \[{P}_{a} + {jP}_{b} = \frac{1}{2}{VI}^{\ast} \tag{1} \] where j stands for the imaginary unit number, and the superscript $\ast$ designates a complex conjugate. The real part $\text{P}{}_{a}$ means effective power, and the imaginary part $\text{P}{}_{b}$ means reactive power. The factor 1/2 stems from V and I denoting their sinusoidal peak values, not in root mean square. Note that $\text{P}{}_{a}$ is always nonnegative for a passive load. In contrast, $\text{P}{}_{b}$ is positive for an inductive load, negative for a capacitive load, and zero for a purely resistive load. We should properly choose this polarity even though the geometrical area stays nonnegative for any load.
We next demonstrate the effective power $\text{P}{}_{a}$ on the plane. This can be done by revolving the voltage phasor OV by 90° counterclockwise, which creates a new phasor ${O}{V'}{.}$ In a complex notation, the 90° revolution is expressed as ${V'} = {jV}{.}$ Between phasors ${V'}$ and I, we find a new triangle ${V'}{OI}{.}$ The area of this triangle, denoted as $\text{P}{}_{a},$ signifies the effective power observed at the load. It would be a moderate exercise for students to prove this area from (1).
Since the phasors OV and $\text{O}{V'}$ are orthogonal to each other, the areas $\text{P}{}_{a}$ and $\text{P}{}_{b}$ always complement each other, like sine and cosine. When V is in phase with I, the reactive power $\text{P}{}_{b}$ vanishes. Consequently, ${V'}{OI}$ shapes up as a right triangle, and, thus, the effective power $\text{P}{}_{a}$ is maximized. This result is exactly consistent with what we practically experience with a purely resistive load.
To display the input or output impedance of an RF circuit, professional engineers regularly use a Smith chart. The Smith chart can conveniently draw a complex locus in both immittance and reflectance domains on the same plane. However, the Smith chart needs curved coordinates based on hyperbolic geometry, which sometimes leads students to quite a labyrinth. Especially for undergraduates, it is recommended to first use a chart having simple Cartesian coordinates.
A typical Cartesian plane is shown in Figure 2, where the abscissa designates resistance R, and the ordinate designates reactance X. This is called the R-X plane or impedance plane, whose coordinates have a common linear scale in the unit of ohms. A complex impedance \[{Z} = {R} + {jX} \tag{2} \]
Figure 2. A rectangle on an impedance plane. The aspect ratio signifies Q.
is projected as a point on the plane. An instructive figure presented here is a rectangle with base length a and height b. Their ratio b/a signifies the quality factor Q of impedance Z. This might be quickly recognized by professional engineers [1] but must be a persuasive demonstration for students to visually grasp the concept of Q for the first time. Since a and b signify geometrical lengths, they are always nonnegative, even for negative values of X. Note that R cannot be negative because we usually assume a passive load impedance. As a result, Q stays nonnegative for any Z.
Another instructive figure on the same impedance plane is a triangle. Imagine again that an impedance Z is given anywhere on the plane. We draw a triangle with its vertex at Z and base ended at $\pm{50}\,\Omega$ on the abscissa, as shown in Figure 3(a). Let a and b denote the length of two slopes. Now, what does the side-to-side ratio b/a physically signify? The answer is the reflectance magnitude of load Z observed in a standard ${50}{-}\Omega$ system. Its proof is rather straightforward if students visually find that ${a} = \mid{Z} + {50}\mid$ and ${b} = \mid{Z}{-}{50}\mid$ from both sides of the triangle. Their ratio is thus written as \[\frac{b}{a} = \left|{\frac{{Z}{-}{50}}{{Z} + {50}}}\right|{.} \tag{3} \]
Figure 3. A triangle on an impedance plane. The side-to-side ratio signifies the reflectance. (a) A standard 50-Ω system. (b) A general complex-impedance system.
This right-hand side reminds us of a reflectance formula in terms of the impedance. Namely, the triangle’s side-to-side ratio signifies the reflectance magnitude.
We can extend the reference impedance ${50}\,\Omega$ to any complex number except for zero and infinity. The triangle in Figure 3(a) is replaced by that in Figure 3(b), and (3) is generalized as \[\frac{b}{a} = \left|{\frac{{Z}_{2}{-}{Z}_{1}^{\ast}}{{Z}_{2} + {Z}_{1}}}\right|{.} \tag{4} \]
Physically, the right-hand side signifies the reflectance of load ${Z}_{2}$ observed from source ${Z}_{1}{.}$ Be careful that the asterisk superscript appears on ${Z}_{1}$ only in the numerator, not in the denominator. See [2] for its rationale. Especially when ${Z}_{2} = {Z}_{1}^{\ast},$ (4) tells us that the reflectance becomes zero. This condition is known as the conjugate impedance matching between a source and a load. Note that the triangle’s vertical angle ${\theta}$ in Figure 3 signifies the phase of complex reflectance. Let its proof be left as homework for students. A vital clue is found in the next section. Another question is on how to measure the distance between impedances, such as a and b in Figures 2 and 3. In addition to the usual linear scale, wave engineers sometimes employ a different scale called the Poincaré metric to measure distance [2]. With this metric, distance is always invariant, no matter on which plane it is projected: the impedance plane, admittance plane, or even Smith chart.
Once students master the impedance plane, the class can proceed to the Smith chart. The Smith chart plays the same role as the impedance plane but with different coordinates. The Smith chart has two coordinate systems: impedance Z and reflectance $\Gamma{.}$ Under a standard ${50}{-}\Omega$ reference, the two complex quantities are related by a Möbius transformation: \[\Gamma = \frac{{Z}{-}{50}}{{Z} + {50}} \tag{5} \] or, reversely, \[{Z} = {50}\frac{{1} + \Gamma}{{1}{-}\Gamma}{.} \tag{6} \]
Looking at (5) and (6) in parallel, students may presume a mathematical analogy between Z and $\Gamma{.}$ That is absolutely right. What we find in the Z domain also appears in the $\Gamma$ domain. For instance, we saw a triangle on the impedance plane in Figure 3, so we should be able to see a similar triangle on the Smith chart that must be instructive as well.
Consider an arbitrary impedance Z projected onto the Smith chart, as shown in Figure 4. Let $\hat{Z}$ be the symmetrical point of Z with regard to the ${50}{-}\Omega$ origin. Now, focusing on triangle $\text{ZO}\hat{Z},$ what does its side-to-side ratio a/b physically signify? The answer is the ${50}{-}\Omega$ normalized impedance magnitude, formulated as \[\frac{a}{b} = \frac{1}{50}\left|{Z}\right|{.} \tag{7} \]
Figure 4. A triangle on a Smith chart. The inscribed angle signifies the impedance phase.
This can be directly proved from (6) because we know that $\Gamma = {-}{1}$ exactly corresponds to the point of ${Z} = {0},$ and, thus, ${a} = \mid{1} + \Gamma\mid$ and ${b} = \mid{1}{-}\Gamma\mid{.}$
In addition to the impedance magnitude, its phase also matters in wave engineering, as Z is a complex number. Looking at Figure 4 again in the reflectance domain, we find ${\theta} = \angle{(}{1} + \Gamma{)}{-}\angle{(}{1}{-}\Gamma{)}{.}$ Then, considering the phase of (6), we can conclude that the triangle’s inscribed angle ${\theta}$ signifies the phase of impedance Z, simply written as \[{\theta} = \angle{Z}{.} \tag{8} \]
For example, when ${Z} = {73} + {j}{42}\,\Omega,$ we find ${\theta}\approx{30}$ from (8). Remembering (2), we also notice ${Q} = {X}{/}{R} = \tan{\theta}{.}$
One more supplement worth noting about symmetric impedances is that their geometric mean converges to the origin. For a ${50}{-}\Omega$ centered circle, that is given as \[\sqrt{\text{Z}\hat{Z}} = {50}{.} \tag{9} \]
This is always true no matter where on the chart Z is located.
Instead of reflectance, we often use the VSWR ${\rho}$ as an index to evaluate the impedance match between a source and a load. From textbooks on RF circuit theory, students find a VSWR formula: \[{\rho} = \frac{{1} + {\gamma}}{{1}{-}{\gamma}} \tag{10} \] where ${\gamma}$ designates the reflectance magnitude stemming from (3). For example, ${\rho} = {3}$ when ${\gamma} = {1}{/}{2}{.}$ We can translate (10) into the graphics shown in Figure 5.
Figure 5. A right triangle demonstrates the relationship between the VSWR and reflectance.
Create a right triangle with its unity-length base and hypotenuse length ${\rho}{.}$ Draw a broken line that precisely bisects the left angle. Then, a small triangle appears inside. Create a square with its side common to the small triangle in height. The square’s area finally signifies ${\gamma}{.}$
Looking at this triangle, students can intuitively learn how ${\rho}$ behaves with ${\gamma}{.}$ When ${\gamma}$ increases, while keeping the triangle’s base length at unity, we find that ${\rho}$ monotonically goes up to infinity. Conversely, when ${\gamma}$ decreases, ${\rho}$ monotonically goes down but limited to unity. This is fully consistent with our common sense: VSWR ranges from unity to infinity, while reflectance ranges from zero to unity.
If students are interested in this geometry, its proof must be practice for them. One gentle clue is to focus on the half angle. If we let ${\theta}$ denote the half angle, everyone will be aware that the square’s area is \[{\gamma} = {\tan}^{2}{\theta} \tag{11} \] remembering that the triangle is based on the unity length. Substituting (11) into (10), the half-angle trigonometric rule helps us finally reach \begin{align*}{\rho} & = \frac{{1} + {\tan}^{2}{\theta}}{{1}{-}{\tan}^{2}{\theta}} \\ & = \sec{2}{\theta} \tag{12} \end{align*} which explains the triangle’s hypotenuse length.
In the development of radio and wireless systems, impedance matching always matters. If hardware components match each other well, they can be connected with a low reflectance or low VSWR between them. If not, undesirable wave reflection will take place between them, which leads to system performance degradation. Now, can we effectively connect two components of different impedances? The answer is yes. Although it might not directly be done, it is possible by way of impedance conversion.
The most primitive impedance converter is a coil-wound transformer. This is because the transformer exhibits an impedance conversion ratio simply derived from the number of winding turns. An ideal transformer of winding ratio 1:n is shown in Figure 6(a). Its output voltage is enhanced n times from the input voltage, and its current is reduced to one nth. Therefore, the impedance conversion ratio results in ${1}{:}{n}^{2}{.}$ Note that this conversion ratio is invariant to what is loaded at its port, unlike other impedance matching circuits. That is why students should learn the transformer prior to sophisticated matching networks.
Figure 6. An ideal transformer of the winding ratio 1:n. (a) A circuit diagram. (b) A unit semicircle with points Q and ${Q'}$ symmetrically lying on it. The central angle is specified from the coil-winding ratio. The coordinates read out the transformer’s two-port S-parameters.
In wave engineering classes, the transformer should be regarded as a two-port network of typical RF components, such as linear amplifiers and filters. Students may raise a very basic question: what are the two-port S-parameters of an ideal transformer? The students expect a rigorous explicit answer (not using a simulator) to be derived from such a simple circuit diagram as shown in Figure 6(a). That is exactly what we envision as we advise them on how to apply a transformer to RF systems. The answer is visually demonstrated in Figure 6(b).
Create a semicircle of unity radius based on the abscissa. Stand a perpendicular pole from (1, 0). On this pole, let P be the point of height n. Draw a straight line to link P and origin O. Draw another straight line of double slope from O. On this line, let Q be the intersection to cross the semicircle. Finally, let ${Q'}$ be the symmetrical point of Q with regard to the ordinate.
From this graph, we can directly read out the transformer’s ${S}_{11},\,{S}_{21},$ and ${S}_{22}$ on the coordinates. Needless to say, ${S}_{12}$ is always equal to ${S}_{21}$ thanks to reciprocity. Note that the four parameters are not complex but just real numbers. This is a natural result from the fact that an ideal transformer never stores reactive energy.
The proof of this geometry is a good exercise for students again. As depicted in Figure 6(b), we specify the central angle ${\theta}$ so as to meet \[{n} = \tan{\theta}{.} \tag{13} \]
According to the S-parameter definition for a standard ${50}{-}\Omega$ referenced system, ${S}_{11}$ is the reflectance observed at port 1 when port 2 is loaded with a ${50}{-}\Omega$ resistor. Since the impedance conversion ratio is ${1}{:}{n}^{2}$, we find the input impedance ${Z}_{\text{in}} = {50}{/}{n}^{2}$ at port 1. Recalling (5) and (13), we obtain \begin{align*}{S}_{11} & = \frac{{Z}_{in}{-}{50}}{{Z}_{in} + {50}} \\ & = \frac{{1}{-}{n}^{2}}{{1} + {n}^{2}} \\ & = \cos{2}{\theta}{.} \tag{14} \end{align*}
By interchanging ports 1 and 2, we also obtain \[{S}_{22} = {-}\cos{2}{\theta}{.} \tag{15} \]
Because the ideal transformer exhibits no power dissipation, the energy conservation law tells us the scalar S-parameter relation ${S}_{11}^{2} + {S}_{21}^{2} = {1}{.}$ Applying it to (14), we finally obtain \[{S}_{21} = \sin{2}{\theta}{.} \tag{16} \]
Note that we should take the negative polarity as ${S}_{21} = {-}\sin{2}{\theta}$ if the two coils are wound in opposite directions. From (14), (15), and (16), the transformer’s S-parameters are read out as the coordinates of points Q and ${Q'}$ moving along the semicircle in response to the specified winding ratio, as shown in Figure 6(b).
Students may next learn another solution to the impedance matching problem: a transmission line of a quarter-wavelength or ${\lambda}{/}{4}$ line. This plays a similar role to the aforementioned transformer but is effective at higher frequencies thanks to its distributed-element structure. A typical scheme where a ${\lambda}{/}{4}$ line works is a feeder system to link a transmitter and an antenna, as shown in Figure 7(a). Let us now focus on the characteristic impedances ${Z}_{1},\,{Z}_{2},$ and ${Z}_{3}$ of three components, which are all assumed to be real and positive for simplicity. To achieve impedance matching simultaneously at both ports before and after the ${\lambda}{/}{4}$ line, the three impedances must satisfy the well-known condition \[{Z}_{2} = \sqrt{{Z}_{1}{Z}_{3}}{.} \tag{17} \]
Figure 7. A quarter-wavelength feeder line acts as an intermediary between the transmitter and antenna. (a) The system configuration. (b) A semicircle with three interior line pieces, of which Z1 and Z3 cast a diameter.
This condition can be projected onto graphics, as shown in Figure 7(b).
Create a semicircle of diameter ${Z}_{1} + {Z}_{3}{.}$ Stand a vertical pole up from the junction of ${Z}_{1}$ and ${Z}_{3}$ to reach the ceiling. Let ${Z}_{2}$ denote the pole’s height. The power of a point theorem tells us that ${Z}_{1},\,{Z}_{2},$ and ${Z}_{3}$ exactly measure the three impedances to satisfy (17).
Students may imagine that the junction point horizontally moves along the diameter while keeping (17). Especially when the junction point just comes to the semicircle’s center, we apparently find ${Z}_{1} = {Z}_{2} = {Z}_{3},$ which takes place most often in practical systems. In conclusion, this graphic gives direct inspiration on the ${\lambda}{/}{4}$ matching section, particularly to students who are better at geometry than at algebra.
The final feature graphics to appear in this article explain the essence of wireless power transfer (WPT): one of the most attractive topics in recent wave engineering. Students will be impressed by a spectacular WPT demonstration exemplified by the photograph presented in Figure 8(a). An incandescent bulb is brightly lit by way of a wireless link, where large twin solenoids compose an inductive coupler as the key device.
Figure 8. The WPT. (a) An impressive photo courtesy of Tommy Reach. (b) The inductive coupler circuit equivalent. (c) The same triangle as in Figure 5 explains the relationship between kQ and ${\eta}_{max}{.}$
In the development of high-efficiency wireless couplers, a common figure of merit is called kQ. It would take long time to explain what kQ generally signifies [3]. However, if we assume the simple equivalent circuit shown in Figure 8(b), kQ can just be regarded as the algebraic product of k and Q, where ${k} = {M}{/}{L}$ denotes the coupling coefficient between coils, and ${Q} = {\omega}{L}{/}{R}$ denotes the coil’s quality factor. To maximize this figure of merit, we should choose such a material and shape of the coils that exhibit as high a kQ as possible.
Once kQ is given, the wireless coupler’s power transfer efficiency can be maximized by simultaneously adjusting the power source’s output impedance and load’s input impedance. This is rigorously proven by a linear passive two-port network theory [3]. The resultant maximum power transfer efficiency, denoted as ${\eta}_{\max},$ is exclusively dominated by the coupler’s kQ. This rule is expressed as \[{\eta}_{\max} = \frac{\sqrt{{1} + {k}^{2}{Q}^{2}}{-}{1}}{\sqrt{{1} + {k}^{2}{Q}^{2}} + {1}}{.} \tag{18} \]
Equation (18) may look complicated for students to imagine how ${\eta}_{\max}$ behaves as a function of kQ. Plane geometry should now come onstage. To project (18) onto a graphic, we introduce such an angle ${\theta}$ as to meet \[{kQ} = \tan{2}{\theta}{.} \tag{19} \]
Employing this angle, (18) becomes amazingly elegant as \[{\eta}_{\max} = {\tan}^{2}{\theta}{.} \tag{20} \]
That is to say, kQ and ${\eta}_{\max}$ are correlated by way of ${\theta}{.}$ In this sense, $\tan{\theta}$ should be called the efficiency tangent by analogy with so-called loss tangent. The square on $\tan{\theta}$ implies a quantity measurable on a power scale rather than an amplitude scale.
Thanks to (19) and (20), we can visualize the relation (18) as shown in Figure 8(c). In this figure, a triangle is drawn exactly by the same procedure as that in Figure 5, but it represents a different physical meaning. When kQ is given as the prime triangle’s height, the smaller triangle becomes $\tan{\theta}$ tall, and, thus, the square’s area results in ${\eta}_{\max}{.}$ Looking at these graphics, students realize a basic law of WPT: ${\eta}_{\max}$ ranges from zero to unity when kQ ranges from zero to infinity.
We have explored plane geometry and found eight figures that explain wave engineering essentials. The leading figures are all in elementary shapes, such as a square, rectangle, and triangle, so they are genuinely friendly even to undergraduates. Employing these graphics in classes, students will be richly inspired and, thus, become interested in our wave engineering world. To seek further stimulating discovery about the impedance plane, Smith chart, and other related geometry, students are encouraged to proceed to the three articles in the “References” section.
The author would like to thank Tommy Reach for providing an impressive photo of his WPT demonstration. This work is supported in part by the Cross-Ministerial Strategic Innovation Promotion Program of the Japan Cabinet Office.
[1] T. Ohira, “What in the world is Q? [Distinguished Microwave Lecture] ,” IEEE Microw. Mag., vol. 17, no. 6, pp. 42–49, Jun. 2016, doi: 10.1109/MMM.2016.2538512.
[2] T. Ohira, “A radio engineer’s voyage to double-century-old plane geometry [Educator’s Corner] ,” IEEE Microw. Mag., vol. 21, no. 11, pp. 60–67, Nov. 2020, doi: 10.1109/MMM.2020.3015136.
[3] T. Ohira, “Power transfer theory on linear passive two-port systems,” IEICE Trans. Electron., vol. E101.C, no. 10, pp. 719–726, Oct. 2018, doi: 10.1587/transele.E101.C.719.
Digital Object Identifier 10.1109/MMM.2022.3226534