# beam pattern are fourier transform of the beam weight, is it true ? if it's true how?

what i understood is the equation of AF looks similar to DFT equation as we are multiplying with exponential term

• I assume "AF" refers to array factor, but many people won't immediately know that, so it would be good if you could define the term to make your question more clear. Feb 16 at 20:46
• See this answer as well: electronics.stackexchange.com/a/290101/276634 Feb 18 at 1:49
• Baddioes gave a really good answer so if that suffices please accept it to help the site. Feb 18 at 20:02

## 1 Answer

I'll work out some math that should hopefully capture the gist of the derivation of why the DFT can be used.

The magnitude squared of the array factor is the beampattern for an array of isotropically radiating elements. The array factor for a uniformly weighted ULA with normalized isotropic radiating elements is defined as

$$$$AF(\theta) = \sum_{n=0}^{N-1}e^{-jkdn\cos(\theta)}$$$$

Here, $$k$$ is the wavenumber which is typically $$\frac{2\pi}{\lambda}$$, $$d$$ is the element spacing which is typically $$\frac{\lambda}{2}$$. Plugging in we get

$$$$AF(\theta) = \sum_{n=0}^{N-1}e^{-j\pi n\cos(\theta)}$$$$

The definition of the DFT is

$$$$X(k) = \sum_{n=0}^{N-1}x(n)e^{-j\frac{2\pi}{N}kn}\; \forall \: k \in [0,N-1]$$$$

If we want to use the DFT, we need to use the equation

$$$$\pi\cos(\theta)=\frac{2\pi k}{N}$$$$

Solving this for $$\theta$$ we get

$$$$\theta = \arccos\left(\frac{2k}{N}\right)$$$$

This will not work for most values of $$k$$ as the domain of $$\arccos$$ is restricted to $$-1\leq x\leq 1$$. However, because $$\cos(\pi) = \cos(-\pi)$$, we can say

$$$$\theta=\arccos\left(\frac{l}{N}\right)\;\forall \: l \in [-N,N-1]$$$$

which equates to evaluating $$\theta \in [-\pi,\pi)$$.

Plugging this back into the array factor we get

$$$$AF(l) = \sum_{n=0}^{N-1}e^{-j\frac{\pi}{N} nl}\;\forall \: l \in [-N,N-1]$$$$

This is almost the form of the DFT, but we first need to get it back to the range $$[0,N-1]$$. We can shift $$l$$ to instead be from $$[0,2N-1]$$. This will give us

$$$$AF(l) = e^{j\frac{\pi}{N}l}\sum_{n=0}^{N-1}e^{-j\frac{\pi}{N} nl}\;\forall \: l \in [0,2N-1]$$$$

which is a $$2N$$-point DFT in the range $$[0,2\pi)$$. We need an $$N$$ point DFT in the same range, which we can get by plugging back in for $$k$$

$$$$AF(k) = e^{j\frac{2\pi}{N}k}\sum_{n=0}^{N-1}e^{-j\frac{2\pi}{N}nk}\;\forall \: k \in [0,N-1]$$$$

Now we have an equation that is in the form of a DFT. Since the beampattern is the magnitude squared of the array factor, we get

$$$$BP(k) = \|AF(k)\|^{2} = \|\sum_{n=0}^{N-1}e^{-j\frac{2\pi}{N}nk}\|^{2}$$$$

Now we see that if we weight the ULA of normalized isotropic radiators by $$\underline{w}$$, the array factor becomes

$$$$AF(k) = e^{j\frac{2\pi}{N}k}\sum_{n=0}^{N-1}w_{k}e^{-j\frac{2\pi}{N}nk}$$$$

If you have non-isotropic radiating elements, the beampattern is related to the product of the array factor and the element factor, where the element factor is the beampattern of an individual element.