I've found an old antenna that once came with an HP laptop with a TV (DVB-T?) card. I decided to see if I can use that antenna for something else, so I opened it up and found the following antenna design. However, I have not seen this before, so I'm curious to know:

  1. What is this antenna design?
  2. Is it possible to determine an approximate usable frequency range?

The element measurements are as follows:

  • The ground "clip" length is 5 cm.
  • The ground clip width is 2 mm.
  • The driven thin spacers leads are 2 mm in length.
  • The wide elements are all 8.3 mm in length
  • The wide elements are of the following widths (mm): 1.6, 7.2, 8.3, 10.1, 11.1, 12.5, 13.1, 14.0, 15.2

enter image description here

The HP part numbers for this is:

HP P/N:     411352-003
HP Spare:   412176-003

And it look like this in its original state:

enter image description here

EDIT: This seem a bit similar to the antenna designs of Resonant Rectangular Series-Fed Patch Array and the Traveling Series-Fed Patch Array, also described a bit in this paper.

enter image description here enter image description here

  • 1
    $\begingroup$ gut feeling: interpret the "U" as "slightly folded ground plane", interpret the connected patches as "electrically enlarged half of a Dipole". Maybe they "just" act as impedance matching network, with the smallest one matching let's say 75Ω (cable) to 80Ω, the next one 80Ω to 90Ω and so on, but then the short transmission lines in between wouldn't make sense. Hm. If you feel like you're really interested in this, get OpenEMS, build it, run the patch antenna octave example that comes with it, and modify the octave file until you get your geometry in simulation :) $\endgroup$ Jul 22, 2017 at 8:02
  • $\begingroup$ A patch array like the paper you have found requires a ground plane on the back of the board. I don't think you'll find one, but please check. Also an FR4 substrate wouldn't be good enough, it would need to be a microwave substrate. What kind of board is it? $\endgroup$
    – tomnexus
    Jul 25, 2017 at 15:54

1 Answer 1


It is an antenna made of different microstrip patches at different frequencies.

They might use this antenna to receive signals from different bands. An other option is that since the fractional bandwidth of a patch antenna is narrow ($ Bw/f_o$ is low), to achieve a large bandwidth they "concatenate" several patches at different but close central frequencies.

The width of each patch is calculated as:

$$W = \frac{c}{2f_0 \sqrt{\frac{\epsilon_r+1}{2}}} $$


so you can easily isolate the frequency at which they are tunned: $$f_0 = \frac{c}{2W \sqrt{\frac{\epsilon_r+1}{2}}} $$

$\epsilon_r$ is the relative dielectric constant of the substrate where the antenna is located. In this case, it seems fiberglass (FR-4), which has a $\epsilon_r$ between 4.35 and 4.8 (source).

The 1.6 mm I think is the microstrip transmission line to feed the antenna.

According to the formulas, 7.2, 8.3, 10.1, 11.1, 12.5, 13.1, 14.0, 15.2 would correspond to: 12.7 GHz, 11.05 GHz, 9.08 GHz, 8.26 GHz, 7.33 GHz, 7 GHz, 6.55 GHz, 6 GHz.

Do these numbers make any sense to you?

  • 2
    $\begingroup$ no, these numbers don't make sense for terrestrial TV at all. $\endgroup$ Jul 21, 2017 at 22:44
  • $\begingroup$ Nor for low-band Wifi, but perhaps for upper-band. But wideband antennas are certainly not the first choice when the frequency is known. $\endgroup$
    – SDsolar
    Jul 22, 2017 at 5:30
  • 1
    $\begingroup$ Your analysis, while interesting, does not match up to the DVB-T frequencies for which the OP stated the antenna is used. $\endgroup$
    – Glenn W9IQ
    Jul 22, 2017 at 10:34
  • 1
    $\begingroup$ This probably does not fit with the description of a "patch" antenna as those usually have a ground plane behind the patches. In my case the back side is just empty FR-4. Also the cheap antenna cable could probably never support more than ~1 GHz. $\endgroup$
    – not2qubit
    Jul 24, 2017 at 20:19

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