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I live next to a 100 kW FM transmitter, 600 feet away transmitting on 103.7 MHz.

It is south east and most other station I want to listen to are in other directions. This station comes though most of the FM band interfering with other stations.

Is there a way to block or weaken this station on regular FM radios with internal antennas and pick up other stations?

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Theoretically, it is possible: you need to buy or build a “notch” filter tuned exactly to 103.7 MHz, and insert this in line with the receiver's antenna port.

But if the receiver has only an internal antenna then it is inconvenient to connect the filter. Furthermore, any such receiver (even many with an external antenna port) will not actually have enough shielding to prevent the strong station from entering the circuit through paths other than the nominal antenna.

If I were trying to solve this problem, I would start by looking only at receivers with a metal enclosure (which will also have an external antenna port) to ensure that the circuit is properly shielded. Then the remaining questions are:

  • Is the enclosure actually complete without gaps or slots of significant length? (Can be fixed cheaply with foil tape or tidily by installing "RF gasket" strips.)
  • Are the shields of all input/output sockets or cables actually connected to the enclosure? (Can be fixed by soldering in short wires for the signal shields — if this doesn't prevent reassembly or cause the circuit to malfunction. And if one of the cables is an AC power cord then you need a line filter module instead of a simple wire.)

The receiver should also have a high selectivity — this measures how good it is at tuning in one station without also receiving adjacent stations.


Some other possibilities:

  • A directional antenna pointed at a distant station will receive it better and the local station worse. But this will be a large antenna, and again, it doesn't do much good if the receiver isn't shielded.
  • Even a receiver with ordinarily good selectivity may have trouble with a strong station due to overload. If this is the problem, then a filter will help by lowering the strong signal, but also just an attenuator, which reduces the level of all signals, may help, making it easier for the receiver to handle them precisely. But in your case the level difference may be just too much for this strategy.
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  • $\begingroup$ Thank you for your answer. Most of my radios are of the inexpensive AM/FM alarm type. As you say,seems like it would easier just to listen to the stations with internet. The station in question was nice enough to supply us with a good roof type TV antenna. $\endgroup$ – crip659 Oct 26 at 17:04
  • $\begingroup$ I feel your pain. I have a 50kW ERP station 1.5 miles away and have the same problem. Short of putting the radio in a Faraday cage and putting a sharp (likely large and expensive) filter on what's left, you're likely out of luck. You will find some radios are better at rejecting the problem (better selectivity, less prone to overload), but until you plug it in and try it out, there's no good way to find out in advance. $\endgroup$ – Duston Nov 22 at 15:14
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One thing you could try is to add a "shorting stub" tuned to 103.7 MHz to the antenna of your receiver. The point of this shorting stub is to "short out" (attenuate) the 103.7 MHz signal, and leave the rest of the frequencies (somewhat) untouched.

This configuration works best when you use an external antenna with your radio. If your radio only uses an internal antenna (such as a collapsible metal antenna), then this procedure may not work.

For now, I'll assume you can use an external antenna, and that your radio has two screw lugs for attaching one to the radio. We will want to connect this "shorting stub" directly across these two terminals for best results.

A piece of 300 ohm twin-lead, which you can probably still find at your local hardware store, should work best to make the shorting stub.

One end of this twin-lead will need to be attached to the radio (antenna and ground lugs). The other end of the twin-lead will be shorted together.

Start by shorting together the free end of the twin-lead. (Just strip and twist the two sides of the twin-lead together.)

Measure up from this now-shorted end exactly 3 foot 10 and 11/16 inches. Mark this point (don't cut it yet!).

Now figure out how much extra wire you will need to attach this stub to the radio's antenna and ground screw lugs. It is very important that you do your best to minimize the "extra" or "slack" here. Use the least "extra" wire as you can. Add the smallest practical amount of wire you need to connect to these lugs as you can to the stub you have made so far.

If you make a mistake, twin-lead is cheap, so start over.

Once you attach this stub to your radio, move your FM antenna around to see what position works best to null out the 103.7 MHz signal.

There is a strong possibility that your stub will reduce other signals on the FM band, but you might be surprised to find that you have at least some that will still work for you.

Good Luck!

[Edited to correct length. I used "low-pass" formula, needed "fundamental" formula: (492 / f(MHz)) * 0.82 - Where 0.82 = Velocity factor of 300 ohm twin-lead]

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  • $\begingroup$ Thank you, Don't know if it will work since the station is so close and powerful, it comes in on most of the FM range. Your suggestion is simple and cheap enough to try. $\endgroup$ – crip659 Nov 21 at 21:24
  • $\begingroup$ A λ/4 shorting stub is an excellent way to do this! However, I'm not sure that I follow your math. Can you tell us the stub material's velocity factor? IIRC, if you short the far end of a λ/4 stub, the end connected to the feedline will appear as an open circuit. Don't we want to leave it open? $\endgroup$ – Mike Waters Nov 21 at 21:27
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    $\begingroup$ 300 ohm twin-lead velocity factor is typically listed as .82. $\endgroup$ – Mike S. Nov 21 at 21:39
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    $\begingroup$ A shorted 1/4 wave stub placed ACROSS the antenna line is a shorting stub. You are right to question my math - I think I used the wrong formula from the table. I've calculated the HIGH IMPEDANCE length for the 103.7 MHz signal. I'll re-calculate and edit my original post. Thanks!! $\endgroup$ – Mike S. Nov 21 at 21:45
  • $\begingroup$ Starting with an open circuit, one quarter wavelength away you will "see" a short circuit. Starting from a short circuit, one quarter wave away you have an open circuit. From this webpage. Also check out the drawing here. $\endgroup$ – Mike Waters Nov 21 at 21:58

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