# Difference between Uniform Linear Array (ULA) 3 dB beamwidth and bearing resolution

Is there a difference between linear array 3dB beam width and linear array bearing resolution ?

• A little more context is suggested to help readers answer your question – Laurent Duval Apr 6 '17 at 12:22

## 1 Answer

The answer will be NO, but to be precise especially in case of conventional beam forming.

# Theoretical Derivation

The antenna electric field pattern of array antenna can be given by

$\begin{vmatrix}&space;E(\Theta&space;)&space;\end{vmatrix}&space;=&space;\begin{vmatrix}&space;\frac{&space;sin\begin{bmatrix}&space;N\left&space;(&space;\pi&space;d&space;/\lambda\right&space;)sin\Theta&space;)&space;\end{bmatrix}}{sin\left&space;[&space;(\pi&space;d/\lambda&space;)sin\Theta&space;\right&space;]}&space;\end{vmatrix}$

where N is the number of antenna elements

d is the spacing between antenna elements

In order to find the beamwidth (3 dB), the above equation should be equated to $\frac{1}{\sqrt{2}}$ and solve for $\Theta$

The solution will come to be as $0.89\frac{\lambda&space;}{D}$

where D is the total aperture distance and can be approximated as $Nd$

For spacing of $d&space;=&space;\lambda&space;/2$ the equation simplifies and can be approximated as $2/N$

Thus beam width of planar antenna can be represented as

$\Delta&space;\theta_{3dB}=&space;2/N$

# Angular Resolution Using FFT Over Antenna Elements

In order the prove the lemma that the angular resolution obtained by performing the FFT across antenna dimension equals to the beamwidth of the antenna array, we have to obtain the angular resolution using FFT.

The below diagram shows the frequency obtained due to path difference between the antenna elements which occurred due to the angle of arrival other than the broadside angle

The frequency resolution using FFT can be represented as $\frac{2\pi}{N}$ where N is the number of samples but in our case it is equal to number of antenna elements.

The angular resolution $\Delta&space;\theta$ can be found by equating the difference in frequency of different angles $w_1&space;=&space;\frac{2&space;\pi&space;d&space;sin\Theta&space;}{\lambda}$ and $w_2&space;=&space;\frac{2&space;\pi&space;d&space;sin\left(\Theta&space;+&space;\Delta&space;\Theta&space;\right&space;)&space;}{\lambda}$

The frequency resolution becomes: $w_2&space;-&space;w_1&space;=&space;\frac{2\pi}{N}&space;=&space;\frac{2&space;\pi&space;d&space;sin\left(\Theta&space;+&space;\Delta&space;\Theta&space;\right&space;)&space;}{\lambda}&space;-&space;\frac{2&space;\pi&space;d&space;sin\Theta&space;}{\lambda}$

Solving we get $\Delta&space;\Theta&space;=&space;\frac{\lambda&space;}{Ndcos\Theta&space;}$ and if we put $\theta&space;=&space;0$ and $d&space;=&space;\lambda&space;/2$ we get the same resolution as the beamwidth i.e
$\Delta&space;\Theta&space;=&space;2/N&space;=&space;\Delta&space;\Theta_{3dB}$

# Summary

Thus for summarizing the above discussion using FFT we can only achieve the max angular resolution equals to the beamwidth of the antenna array which is equal to angular resolution = antenna array beamwidth = 2/N

Kindly note that this equality is true for the case of conventional beam forming only, there exist also some advance beam former where your angular resolution is better than the 3db beam width

This is the same answer which I provided on ULA beamwidth and angular resolution, might this will be helpful.