# Uniform Linear Array (ULA) beamwidth and angular resolution using FFT

I have a four element uniform linear array receiver and I want to know what will be the angular resolution if I perform perform Fast Fourier Transform (FFT) over antenna elements. As I think that it will be same as the beam width of the array.

The antenna element spacing is lambda/2

Question1: What is the beamwidth of the array structure

Question2: If I perform FFT over the element array will I get the same angular resolution as beamwidth

Question3: How each bin in antenna FFT will represent my region of space (e.g Frequency bin 1 = theta 0 to 30 degrees)

# Theoretical Derivation

The antenna electric field pattern of array antenna consists of isotropic radiator 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

Edit: *The derivation holds only for isotropic radiating element, in case of non isotropic element "array factor" should compensate the specific antenna's field pattern.

*Also the beam width is approximated to 2/N i.e. N should be large enough to get the equality

Reference: Fundamentals of Radar Signal Processing by Mark A. Richards, 2005

• I like your answer very much. I hope the answer 2/N is in radians and we have to convert it into degrees. I would also like to know whether you can provide some reference (Journal paper or textbook) for the above derivation, so that I can cite it in my reference list. Thanks. – Ashish Aug 13 '17 at 16:56
• Thanks for the appreciation. Yes, answer is in radians. I will share the reference shortly. Feel free to up vote the answer, if it helps you in any case. – Zeeshan Aug 13 '17 at 21:04
• You can see "Fundamentals of Radar Signal Processing by Mark A. Richards, 2005". – Zeeshan Aug 16 '17 at 5:30
• This derivation is for an isotropic radiator, this "array factor" will be modified by the electric field pattern of the actual array antenna elements used. – Captainj2001 Sep 20 '17 at 13:33
• sure, array factor incorporated into the note – Zeeshan Sep 20 '17 at 13:41

I think estimating the HPBW as 2/N is not that accurate. The more accurate answer would be 0.89*2/N. I have checked this by actually plotting the array response in MATLAB. For four antenna elements the answer is off by about 2.5% but the answer is very accurate as the number of elements is increased to ten. I guess the answer would get progressively more accurate as the number of elements is increased.

• For sure, its not exactly 2/N its approximated to 2/N for N large enough to estimate. This is also being quoted in my answer too, "For spacing of [d = \lambda /2] the equation simplifies and can be ~approximated~ as [2/N]" I think that is what you are also observing using the MATLAB plots for array length of less than 10. – Zeeshan Jan 25 '18 at 13:49
• In my own view, it should be a comment under the answer not an answer itself. – Zeeshan Jan 25 '18 at 13:55
• Often the approximation $\lambda/D$ is used for beamwidth, where D = length of array = $(N-1)*d$. If $d = \lambda/2$, then this simplifies to $2/(N-1)$. – Gillespie Feb 12 at 15:43