# Reading — Spectrum & Propagation

By the end of this lesson you should be able to:

1. Place a radar in its **frequency band** and state the tradeoffs of moving up or down the spectrum.
2. Relate wavelength and frequency through $\lambda f = c$.
3. Compute **free-space path loss (FSPL)** in decibels and apply the **6 dB rules** for range and frequency.
4. Explain how **polarization** and **antenna gain** shape the energy that reaches a target.

## The spectrum is the battlefield

Every radar lives at a frequency, and that choice drives everything else: antenna size, resolution, atmospheric loss, and how easy it is to jam. Wavelength and frequency are tied by the speed of light,

$$
\lambda f = c, \qquad c \approx 3\times10^{8}\ \text{m/s}.
$$

A handy form for radar work: at $f$ in GHz, $\lambda \approx 30/f$ in centimeters. So **X-band at 10 GHz has a 3 cm wavelength**; L-band at 1.5 GHz is about 20 cm.

### Radar bands

The standard IEEE letter bands and their rough character:

| Band | Frequency | Wavelength | Typical use / character |
| --- | --- | --- | --- |
| **L** | 1–2 GHz | ~15–30 cm | Long-range search; low atmospheric loss, large antennas, coarse resolution. |
| **S** | 2–4 GHz | ~7.5–15 cm | Search & moderate-range; weather radars. |
| **C** | 4–8 GHz | ~3.7–7.5 cm | Compromise band; some weather and fire control. |
| **X** | 8–12 GHz | ~2.5–3.7 cm | Fire control, airborne radar; good resolution, manageable antenna size. |
| **Ku / K / Ka** | 12–40 GHz | ~0.75–2.5 cm | Seekers, high-resolution imaging; small antennas, higher atmospheric loss. |

The general trend: **lower frequency** gives longer range and less weather loss but needs a big antenna and gives poor resolution; **higher frequency** gives fine resolution and small apertures but suffers more atmospheric attenuation. Fire-control radars cluster around X-band because it balances these — which is why the B-21 threat picture is dominated by X-band fire control.

:::{admonition} Key Concept
:class: key-concept

Frequency choice is a chain of consequences. Pick $f$ and you have implicitly chosen wavelength, antenna size for a given gain, resolution, atmospheric loss, and your exposure to particular jamming techniques.
:::

## Free-space path loss

Energy radiated into free space spreads over an expanding sphere. By the time it reaches range $R$, the power density has fallen as $1/R^2$ (one-way). Free-space path loss captures how much a signal weakens between two isotropic antennas. In decibels, with range in kilometers and frequency in GHz, the working formula is

$$
\boxed{\,L_\text{fs,dB} = 20\log_{10}(R_\text{km}) + 20\log_{10}(f_\text{GHz}) + 92.45\,}
$$

Two things to notice. First, loss grows with **both** range and frequency. Second, both appear as $20\log_{10}(\cdot)$ — so each behaves identically in dB.

### The 6 dB rules

Because $20\log_{10}(2) \approx 6.02$ dB:

- **Double the range → add 6 dB of loss.**
- **Double the frequency → add 6 dB of loss.**

These are worth memorizing because they let you reason about link budgets without a calculator. Going from 50 km to 100 km costs 6 dB. Moving a sensor from X-band (10 GHz) to Ku (20 GHz) costs another 6 dB of one-way path loss before you account for anything else.

::::{admonition} Quick Exercise
:class: quick-exercise

An X-band radar ($f = 10$ GHz) sees a target at 50 km. Using the FSPL formula and the 6 dB rules:

1. What is the one-way FSPL at 50 km?
2. The target flies out to 200 km. How much additional loss?
3. A second radar at 20 GHz looks at the same 200 km target. How does its FSPL compare to the 10 GHz radar at 200 km?

:::{admonition} Solution
:class: dropdown

1. $L = 20\log_{10}(50) + 20\log_{10}(10) + 92.45 = 33.98 + 20 + 92.45 \approx 146.4$ dB.
2. 50 → 200 km is two doublings (×4), so $+12$ dB.
3. Doubling frequency adds one more 6 dB, so the 20 GHz radar sees about **6 dB more** one-way path loss than the 10 GHz radar at the same range.
:::

::::

## Polarization

A radar's electric field oscillates in a particular orientation — its **polarization** (horizontal, vertical, or circular). The receiving antenna is matched to a polarization too. When transmit and receive polarizations disagree, the captured power drops. A full cross-polarization mismatch (horizontal into a vertical antenna) can cost **20–30 dB**.

This cuts both ways in EW. A target's RCS depends on polarization, so an adversary may choose a polarization that maximizes return. And polarization is a lever for EP and EA: polarization-agile radars resist some jammers, while a jammer that does not match the victim's polarization wastes most of its power.

## Antennas: turning power into gain

An antenna concentrates radiated power into a beam. Its gain over isotropic is set by its effective aperture $A_e$ and the wavelength:

$$
G \approx \frac{4\pi A_e}{\lambda^2}.
$$

For a fixed physical aperture, **shorter wavelength → higher gain** (and a narrower beam). The half-power beamwidth of an aperture of dimension $D$ is roughly

$$
\theta_\text{BW} \approx \frac{\lambda}{D}.
$$

So the same dish gives a tighter beam at higher frequency. This is the other half of the frequency tradeoff: high frequency buys gain and angular resolution from a small aperture, which is exactly why missile seekers — where size is at a premium — push into Ku/Ka band.

:::{admonition} Key Concept
:class: key-concept

Antenna gain and beamwidth are two sides of the same coin: $G \approx 4\pi A_e/\lambda^2$ and $\theta_\text{BW} \approx \lambda/D$. Concentrating energy into a narrow beam is what makes a radar's effective transmitted power far exceed what an isotropic radiator would deliver — and it is the $G_t$ and $G_r$ terms we meet next.
:::

## Wrap-Up

Frequency sets a radar's whole personality through $\lambda f = c$ and the band tradeoffs. Free-space path loss, $L_\text{fs,dB} = 20\log_{10}(R_\text{km}) + 20\log_{10}(f_\text{GHz}) + 92.45$, thins the signal with both range and frequency, and the 6 dB rules let you reason about it in your head. Polarization mismatch can cost 20–30 dB, and antenna gain $G \approx 4\pi A_e/\lambda^2$ concentrates power into a beam whose width is $\theta_\text{BW} \approx \lambda/D$.

These are the pieces of the constant $K$ from Lesson 1. The next lesson assembles them — transmit power, two antenna gains, target RCS, and the two-way spread — into the **radar range equation**, and shows where the fourth-power law comes from.