Radars operating in the X-band have a narrow beamwidth. Radars operating in the S and C bands have a bit more beamwidth than X band radars.

I know that antenna size also plays a major role. My doubt is:

  1. Whether the beamwidth is determined only by the operating frequency?
  2. Whether it is theoretically possible to have a narrow beam width for S band radar, like X band radar?
  3. Similarly, can I have a narrow beamwidth for operating at HF frequencies?

1 Answer 1


It's theoretically possible to have a narrow beam width at any frequency, but it's the ratio of size of the antenna to wavelength that determines the narrowest possible beam width. So you need either very large antennas, or very small wavelengths. Or both.

There are many relationships like this in physics and engineering.


Lasers have very narrow beam widths, so let's talk about those. Light is just a much higher frequency electromagnetic radiation, so the same physics applies.

You might have heard lasers produce a perfect beam of light, that is one with parallel sides which does not diverge at all. However, this is not possible.

If the objective is to have the smallest beam width, the best kind of physically possible beam is a Gaussian beam. Somewhere along this beam is a waist, the point at which it is narrowest. Usually the waist is the aperture from which light is emitted from the laser.

Away from the aperture, the beam diverges at some angle $\theta$ (in radians). If the wavelength is $\lambda$ and the waist diameter is $w_0$, then that angle is given by:

$$ \theta \simeq \frac{\lambda}{\pi w_0} $$

From this equation you can see to decrease the beam width (decrease $\theta$), one can either decrease the wavelength, or make the aperture bigger. It's because the wavelength of light is so small that a common laser pointer, which might have an aperture of only 1mm, can create a beam which appears "perfect" by casual observation.

Diffraction Limit

The diffraction limit in optical imaging has a similar relationship between sharpness and wavelength. The smallest resolvable feature size $d$ again depends on wavelength, as well as the index of refraction $n$, and the half-angle subtended by the optical objective lens $\theta$.

$$ d \simeq \frac {\lambda }{2n\sin {\theta }} $$

Again, making the lens bigger or decreasing the wavelength will allow smaller features to be resolved.

Antenna Gain and Aperture

Some kinds of antennas have a physical aperture. For example, dish antennas. If we assumed 100% of the energy hitting that aperture was collected by the antenna, then the relationship between aperture area $A$ and gain $G$ is:

$$ A = {\lambda^2 \over 4 \pi} G $$

And of course, a higher gain means a higher directivity and a narrower beam width. Real antennas don't capture energy 100% efficiently due to resistive losses, diffraction, shadowing of the dish, and so on, so there's an "aperture efficiency" factor that's added, which is typically around 40% to 80%.

However, there's still the general relationship: increasing gain requires increasing aperture area or decreasing wavelength.


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