Case Study 2 — Reading an Antenna's Radiation Pattern as a Polar Curve

Field: Electrical engineering / telecommunications Calculus used: The polar curve gallery, especially rose curves and the figure-eight (Section 26.2); the slope/pole-tangent idea for finding lobe directions (Section 26.3); the polar area formula applied to radiated power (Section 26.5)

Every antenna engineer who has ever opened a datasheet has met a polar plot. It is the single most important picture in the field: the radiation pattern, a curve $r = G(\theta)$ drawn around the antenna at the pole, where the radius $r$ at each angle $\theta$ shows how strongly the antenna radiates (or receives) in that direction. The shape of that curve decides whether your Wi-Fi reaches the back bedroom, whether a radar can tell two aircraft apart, and whether a satellite dish hears its faint signal against the noise. This case study takes a single, concrete antenna design problem and walks it through the exact machinery of Chapter 26 — the curve gallery of Section 26.2, the pole-tangent trick of Section 26.3, and above all the sector-area formula of Section 26.5, which is what turns a shape into a number: the radiated power.

Why the pattern is polar

An antenna sits at a point and launches energy outward in all directions, but not equally. A central question — "how much power goes that way?" — is a question about angle, measured from the antenna. That is precisely the situation Section 26.1 described as the home territory of polar coordinates: a problem organized around a center, with rotational structure. Writing the pattern as $y = f(x)$ would be perverse; writing it as $r = G(\theta)$ is natural, and the resulting curves are the very gallery of Section 26.2. The simplest real antenna, a half-wave dipole, has the pattern

$$G(\theta) = |\cos\theta|,$$

which is the figure-eight — two lobes nose-to-nose, radiating strongly broadside (perpendicular to the wire) at $\theta = 0$ and $\theta = \pi$, and nulling out completely along the wire's axis at $\theta = \pm\pi/2$. (The genuine dipole pattern is closer to $\cos(\tfrac{\pi}{2}\cos\theta)/\sin\theta$, but $|\cos\theta|$ captures the essential two-lobe geometry and keeps the calculus clean.) This is a near relative of the lemniscate of Section 26.2, and it is no accident: both are stories about two opposed angular lobes.

Finding the lobes and the nulls

The first thing an engineer wants from a pattern is the location of its main lobes (directions of strongest radiation) and its nulls (directions of zero radiation). The nulls are where $G(\theta) = 0$, and here Section 26.3's pole-tangent result earns its keep: a polar curve passes through the pole exactly where $r = 0$, and the curve leaves the pole tangent to the ray $\theta = \theta_0$. For the dipole, $|\cos\theta| = 0$ at $\theta = \pi/2$ and $\theta = 3\pi/2$ — the pattern pinches to the pole along the vertical axis, which is exactly the deep null along the antenna's own length. The main lobes sit at the maxima, $\theta = 0$ and $\theta = \pi$, where $G = 1$.

Now make the antenna directional. A four-element design can produce a four-lobe pattern modeled by the rose curve

$$G(\theta) = |\cos(2\theta)|,$$

which by the petal-counting rule of Section 26.2 has $2n = 4$ lobes (the even case), spaced $90°$ apart and separated by four sharp nulls at $\theta = \pi/4, 3\pi/4, 5\pi/4, 7\pi/4$. The rose curve, which looked like a decorative flower in Section 26.2, is here a literal map of where a four-port antenna array throws its energy. The same petal rule that told us $r = \cos(4\theta)$ has eight petals tells the engineer how many lobes a given array geometry will produce — a direct, load-bearing use of a "pretty curve" fact.

Turning the shape into power: the sector formula at work

Here is the payoff, and it is pure Section 26.5. An engineer cannot stop at the shape of the pattern; they must quantify it. The headline quantity is the total radiated power, and in the planar idealization of this case study it is

$$P = \frac{1}{2}\int_0^{2\pi} G(\theta)^2\,d\theta.$$

That is the polar area formula of Section 26.5, with the pattern $G(\theta)$ playing the role of the radius $r$. The factor of $\tfrac12$ and the squared gain are not borrowed by analogy — the radiated power in a thin angular wedge $d\theta$ really is proportional to $\tfrac12 G(\theta)^2\,d\theta$, the energy in a sector. Hearing "$\tfrac12 G^2$," an antenna engineer hears the same thing a calculus student hears: sector.

For the dipole $G(\theta) = |\cos\theta|$:

$$P = \frac{1}{2}\int_0^{2\pi}\cos^2\theta\,d\theta = \frac{1}{2}\int_0^{2\pi}\frac{1 + \cos 2\theta}{2}\,d\theta = \frac{1}{2}\cdot\frac{1}{2}\Big[\theta + \tfrac{\sin 2\theta}{2}\Big]_0^{2\pi} = \frac{1}{4}\,(2\pi) = \frac{\pi}{2}.$$

For the four-lobe rose $G(\theta) = |\cos(2\theta)|$, the squaring removes the absolute value, and by symmetry the total is four times one petal's contribution. One petal lives on $-\pi/4 \le \theta \le \pi/4$:

$$P_\text{one lobe} = \frac{1}{2}\int_{-\pi/4}^{\pi/4}\cos^2(2\theta)\,d\theta = \frac{1}{2}\cdot\frac{1}{2}\Big[\theta + \tfrac{\sin 4\theta}{4}\Big]_{-\pi/4}^{\pi/4} = \frac{1}{4}\cdot\frac{\pi}{2} = \frac{\pi}{8},$$

(the $\sin 4\theta$ term vanishes at $\theta = \pm\pi/4$), so the four lobes give $P = 4 \cdot \pi/8 = \pi/2$ as well. The two antennas radiate the same total power but pack it into very different directions — and the sector integral is what makes that comparison precise.

Beamwidth: how sharp is the beam?

A second number every datasheet reports is the half-power beamwidth (HPBW): the angular width of the main lobe between the two directions where the gain has fallen to $1/\sqrt2$ of its peak (the "$-3\,$dB" points). This is a pure root-finding problem on the polar curve. For the dipole, the peak is $G = 1$ at $\theta = 0$, and we want $|\cos\theta| = 1/\sqrt2$:

$$\cos\theta = \frac{1}{\sqrt2} \implies \theta = \pm\frac{\pi}{4} = \pm 45°.$$

The main lobe therefore spans from $-45°$ to $+45°$ — a half-power beamwidth of $90°$. A dipole is a broad radiator. The four-lobe rose $|\cos 2\theta|$ has its peaks at $\theta = 0, \pi/2, \dots$ and reaches half power at $|\cos 2\theta| = 1/\sqrt2$, i.e. $2\theta = \pm\pi/4$, so $\theta = \pm\pi/8 = \pm 22.5°$ around each peak: a beamwidth of $45°$, half as wide. Sharper lobes mean better angular resolution — the ability to tell two nearby targets apart — which is exactly why radar and point-to-point links use narrow-beam designs while broadcast antennas stay wide. The trade-off (narrow beam = high resolution but slow coverage; wide beam = fast coverage but blurry) is read directly off the polar curve's beamwidth.

The engineering loop

In practice the engineer runs this loop: choose an array geometry, write its pattern $G(\theta)$, plot it as a polar curve to see the lobes and nulls (Section 26.2), find the beamwidth by solving $G(\theta) = G_\text{max}/\sqrt2$ (root-finding on the curve), and integrate $\tfrac12\int G^2\,d\theta$ to get the radiated power and, with a normalization, the directivity — how concentrated the beam is compared to an antenna that radiates equally everywhere. Every one of those steps is a Chapter 26 operation. The same $\tfrac12\int r^2\,d\theta$ that gave a rose petal its area gives an antenna its coverage; the same petal-counting rule that decorated a flower predicts the number of radar lobes. Calculus appears in every quantitative field — and here the polar calculus of a sophomore course is, almost without translation, the working mathematics of a $300-billion antenna industry.

Discussion Questions

  1. Pattern to power. Why is the total radiated power $\tfrac12\int_0^{2\pi}G(\theta)^2\,d\theta$ rather than $\int_0^{2\pi}G(\theta)\,d\theta$? Point to the place in Section 26.5 where the $\tfrac12$ and the square come from.

  2. Nulls as pole crossings. Explain how Section 26.3's result "a polar curve leaves the pole tangent to $\theta = \theta_0$ where $r = 0$" lets you read an antenna's null directions straight off the equation $G(\theta) = 0$.

  3. Petal rule, lobe count. A design produces the pattern $G(\theta) = |\cos(3\theta)|$. How many lobes does it have, and how are they spaced? Use the petal-counting rule of Section 26.2 — and note the subtlety that taking $|\cdot|$ of an odd-$n$ rose changes the count.

  4. Beamwidth trade-off. The four-lobe rose has half the beamwidth of the dipole. State the resolution-versus-coverage trade-off in your own words and give one application that wants a wide beam and one that wants a narrow beam.

  5. Why polar at all? Rewrite the dipole pattern in Cartesian coordinates and argue, in one or two sentences, why no engineer would ever choose to.

Your Turn

  1. Plot $G(\theta) = |\cos\theta|$, $|\cos 2\theta|$, and $|\cos 3\theta|$ as polar curves in matplotlib. Count the lobes and confirm the petal-counting rule predicts each.
  2. Numerically evaluate $\tfrac12\int_0^{2\pi}G(\theta)^2\,d\theta$ for each pattern and compare. (You should find the dipole and the four-lobe rose both give $\pi/2$; what does $|\cos 3\theta|$ give?)
  3. For a "narrow forward beam" model $G(\theta) = \cos^{10}(\theta/2)$, find the half-power beamwidth numerically and compare it to the dipole's $90°$.

Annotated Further Reading

  • Balanis, C. A. (2016). Antenna Theory: Analysis and Design (4th ed.), Ch. 2. The standard reference; its opening chapter develops radiation patterns, beamwidth, and directivity as exactly the polar integrals used here.
  • Stutzman, W. L., & Thiele, G. A. (2012). Antenna Theory and Design (3rd ed.). A more applied companion, strong on reading and shaping real polar patterns.
  • Kraus, J. D., & Marhefka, R. J. (2002). Antennas for All Applications (3rd ed.). Classic, intuition-first treatment with many worked polar-pattern examples.
  • Stewart, J. (2020). Calculus: Early Transcendentals, §10.4. The polar-area section this case study leans on; the antenna power integral is the same machinery as the textbook's polar-area problems.