in a linear phased array it is possible to move the peak angle ("beam steering") by applying proper phase shift between the excitation currents:

enter image description here enter image description here

The second picture represents the array factor for different beams angle. Last picture taken from here. What I see in the second picture is that beam steering increases all the lobes width but does not change the lobe peak. So, the peak value of the array factor is (or at least seems to be) the same for each scan angle.

Always there it is written that if beam steering is used, there will be some "scan losses":

Another characteristic of all active antennas is the loss of aperture gain as the beam is steered away from the boresight direction — defined as Ɵ=0. This characteristic, called scan loss, follows 10*log(cosN(Ɵ)) power, where Ɵ is the scan angle off boresight and N is a numeric value, typically in the 1.3 range, which accounts for the non-ideal isotropic behavior of the embedded element gain. Fig. 6 plots scan loss in dB vs. scan angle, measured in degrees. Note, at the origin, where the boresight angle is zero, there is no scan loss. As the scan angle is increased to 45 degrees, there is 2 dB scan loss. If you increase scan angle to a practical limit of 60 degrees, there is 4 dB scan loss.

The following picture is then shown: enter image description here

So, which is the scan loss? From my initial picture it doesn't seem to be caused by the array factor. But, if it were caused by the single element pattern, how could we say that it is linked to the scan angle with a fixed equation not dependent on the single element pattern?

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    \$\begingroup\$ If the individual elements emit more energy directly forward then to the sides (anisotropic emission), then as you scan off to the sides there is less total energy emitted. By symmetry, the same effect happens in reverse when receiving. \$\endgroup\$ – user1850479 May 17 at 19:12
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    \$\begingroup\$ The scan loss does depend on the single element pattern. "N" is how anisotropic each element is. 1 means it is isotropic. At least that is my reading of the quote. \$\endgroup\$ – user1850479 May 17 at 19:31
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    \$\begingroup\$ If N=1 it'll reduce to the Cos(theta), which is the length of the array projected into the direction the beam is pointed. I didn't read your link or derivation, but isn't that the expected geometric loss? \$\endgroup\$ – user1850479 May 17 at 19:54
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    \$\begingroup\$ Not quite. I think the N is supposed to be the an exponent, that is cos^N(Ɵ). The radiation pattern of many real elements follow this equation. N=1.4 is a typical value. And that does refer to an element pattern, not the array pattern. The array gain will also roll off away from boresite because the effective size of the aperture gets smaller (another cos(Ɵ) term, and so the beamwidth increases. This effect is independent of what the element does. \$\endgroup\$ – SteveSh May 17 at 20:10
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    \$\begingroup\$ Yes, N is the exponent. \$\endgroup\$ – user1850479 May 17 at 20:12

Steering any real array off bore-sight induces scan loss, or a loss in directivity. In the image you posted with the multiple beams, that graph is likely normalizing the patterns, hence why they all peak at 0 dB. It does however show the main beam broadening, which is a consequence of steering an array.

For now, forget about the term "array factor" as it might cause confusion. When speaking of an array composed of isotropic elements, the array factor and the antenna pattern are mathematically the same thing. This is not the case with real antennas. But again, terminology not too important to understand the concept of scan loss. We have an antenna pattern and that's that.


Following is an update that corrects and clarifies what others might have read already. It was stated that uniform linear arrays do have scan loss, and it was a result of a mistake in a model. See the comments in Jason's answer for some background.

For both linear and planar uniform arrays, there is no scan loss if the steering occurs along the principal axes, assuming an element spacing of \$\lambda/2\$. This is trivial with a 1-D linear array, and with a planar array, scanning in only azimuth or elevation will preserve the antenna peak gain. It does however, still broaden the beam. Note that this element spacing is a very special case and is not practically achievable.

Let's do a quick example with a 32 x 32 uniform planar array, whose isotropic elements are spaced \$\lambda/2\$ meters apart. Below is a plot showing the nominal unsteered array along with the same array steered 20°, 40°, and 60° off of bore-sight in azimuth only. Each of the arrays are normalized to the un-steered array's peak value. Since we're steering along one of the principal axes, we don't induce a scan loss:

![enter image description here

If we now apply the 20°, 40°, and 60° steering angles to both azimuth and elevation axes, it results in a composite off-boresight angle \$\Theta\$. In this case we do see scan loss. For the plot, an azimuth cut is taken at each pattern's peak location:

![enter image description here

The scan loss expression given by \$10\ log_{10}(cos^N(\Theta ))\$ is an approximation that's pretty good for initial analysis. After building the antenna you can characterize it more accurately from measurements. Below is a plot of the scan loss for this particular array and you can see that it follows a similar trend as the approximation. Since this is a theoretical array, a smaller value of \$N = 0.9\$ shows a good match:

![enter image description here

ULA Scan Loss Special Case: \$d = \lambda/2\$

Here is the reference from Optimum Array Processing, Part IV of Detection, Estimation, and Modulation Theory by Harry L. Van Trees regarding the ULA with \$\lambda/2\$ spacing:

enter image description here

  • \$\begingroup\$ Thank you for your description. What I still haven't understood is if this phenomenon occurs also with isotropic elements. An isotropic element shows an effective area equal on each direction, since the directivity is the same for each direction. How can there be scan loss? \$\endgroup\$ – Kinka-Byo May 25 at 9:33
  • \$\begingroup\$ @Kinka-Byo You're confusing an antenna composed of one element (the element is the antenna) vs an array composed of multiple elements. In a single-element antenna, there is no scan loss since you cannot steer a single element's pattern. \$\endgroup\$ – Envidia May 25 at 13:54
  • \$\begingroup\$ @Envidia, is there a reference for the approximation of scan loss expression given by 10 log10(cos^N(Θ)) ? \$\endgroup\$ – user289928 Jun 28 at 13:56
  • \$\begingroup\$ But uniform arrays of isotropic elements, even 1-d linear arrays do have scan loss, which is a result of the reduction in the effective size of the aperture. You seem to show that it your first set of patterns that shows the main beam width broadening when you scan off boresite, but it is "hidden" by the fact that you seem to be normalizing the patterns to the peak of the main beam. If the main beam gets broader, it's peak has to be reduced, all other things being kept the same. \$\endgroup\$ – SteveSh Jun 29 at 0:02
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    \$\begingroup\$ @SteveSh we should be clear, this is a mathematical peculiarity and is not practical (so your intuition is correct). Different element spacings on an isotropic 1-D array have different scan loss (some actually increase gain with scan angle). The moment you add area either through a non-isotropic element or 2-D array, you'll see a more intuitive scan roll-off. While the beamwidth in phi=0 increases, the area in phi=90 decreases (there's no beamwidth in phi=90 since it's horizon to horizon). Van Trees has a rigorous explanation but if you don't believe it or us, try it yourself. \$\endgroup\$ – Jason Jun 29 at 18:27

Array factor does not show scan loss. Array factor is merely a multiplier. It's actually a common mistake for engineers to use the array factor as a gain value. Array factor gain does not change with scan angle or element spacing!

To see scan loss, you need to calculate the directivity. However, for a uniform linear array (ULA) of isotropic elements you will not see any scan loss when you calculate directivity. You need to either use a 2-D array or use a non-isotropic element pattern.

Physically speaking, a loss in directivity is caused due to less effective area pointing in the direction of interest. Scan loss occurs when the array has area (e.g. not an isotropic ULA).

  • \$\begingroup\$ So then the description of the second graphic in OP's uestion, the one labeled "Beam Steering Angle" is not correct? It's not the array factor for different beam angles, but rather shows the effect of changing the operating freuency while keeping the phase shifter settings constant? \$\endgroup\$ – SteveSh May 21 at 1:11
  • \$\begingroup\$ @SteveSh The label is correct but OP's interpretation is incorrect. It is array factor showing beam squint of a phased array using phase delay. It is not intended to show scan loss. \$\endgroup\$ – Jason May 22 at 15:54
  • \$\begingroup\$ @Jason Perhaps we're bumping on semantics but there absolutely is scan loss in any array that is steered off boresight, even if it's using isotropic elements. \$\endgroup\$ – Envidia May 24 at 20:24
  • \$\begingroup\$ @Envidia you know, I got into an argument with someone recently about this and I lost (I argued that all arrays have scan loss). A ULA of isotropic elements spaced half-wavelength does not have scan loss. It's a mathematical special case but not useful practically. Check out page 62 of: books.google.com/books?id=J5TZDwAAQBAJ (and I've confirmed using multiple tools). I notice your plots below seem to conflict but I'm unable to replicate your results. \$\endgroup\$ – Jason May 25 at 21:48
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    \$\begingroup\$ @Jason I found an error in my model and updated my answer. I haven't rigorously proven the lack of scan loss of a 2-D array under the same special case. However my corrected model shows that it maintains the peak gain like a ULA when steered only along one of the principal axes. \$\endgroup\$ – Envidia Jun 28 at 21:54

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