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Word has it these antennas are broadband, however my calculations show one may not need FM / AM BCB wave traps if not operating around those frequencies.

I have run the script here for:

f_lower = 88e6
f_upper = 108e6
step = 7500e3

# in meters
length = 0.01
diameter = 19.05e-3

0.01m is a dipole length for around 1 GHz operation. Got back a SWR on the order of 5777949500.79 which corresponds to a mismatch loss (attenuation?) of around 92 dB.

is this correct ?

And if it is, how do I calculate dBm level at a receiver location some 4 miles away from a 50 kW FM BC ? How about a 5kW AM BC ? The tower is in urban surroundings.

edit

Right @webmarc !

So, FM@ 1.4 GHz ~ 10 cm ~ SWR: 450 ~ ML: 20.5 dBm
And for 140 MHz ~ 1m ~ SWR: 1863499990 ~ ML: 86.7 dBm (AM frequencies)

I guess I should definitely use a FM trap, maybe AM too.

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    $\begingroup$ I think your math is off somewhere: 1 GHz is a wavelength of approx 0.3 meters and a half-wave dipole would be .15 meters aka 15 cm (quite a bit longer than 0.01 m / 1 cm) $\endgroup$
    – webmarc
    Commented May 16 at 11:16
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    $\begingroup$ “Word has it these antennas are broadband” compared to what? what is your interpretation of broadband in this context? $\endgroup$
    – webmarc
    Commented May 16 at 12:39
  • $\begingroup$ I think you may also be misunderstanding "trap" which I think doesn't make sense in this context. You may mean a high pass filter? $\endgroup$
    – webmarc
    Commented May 16 at 18:45

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Word has it these antennas are broadband,

As far as I can tell, that word would be wrong: Broadband antennas work "well" for a contiguous band larger than 1/5 of its center frequency. (This is not quite a universal definition. Some authors let that start at 1/2, others call antennas with a fractional bandwidth of 1/2 already "ultra-wideband"; anyways, the idea is that for an antenna to be broadband, we expect the ratio of bandwidth vs center frequency to be significant).

Rule of thumb is that a $0.48\lambda$ dipole has a bandwidth of about 5% (how much it really is depends on you writing down precisely what you mean with "working well" above). So, we're a factor of 4 too narrow to call a dipole broadband.

however my calculations show one may not need FM / AM BCB wave traps if not operating around those frequencies.

Exactly, the bandwidth is small enough; the antenna becomes inefficient outside its design frequency range quickly. As a corollary, FM transmitters can often legally operate with relatively nonlinear output stages (and hence, OOB emission) if their antenna is narrowband enough. (The RPitx community makes extensive use of that.)

Note, however, that commercial FM transmitters easily put out powers beyond 50 kW = 47 dBW = 77 dBm, and might be close to populated areas (where you are likely to be!), say 10 km, while you might want to receive signals that were emitted at 30 dBm from 100 km away, end hence are 67 dB weaker (and this is a benign example. Put your interferer into the switch mode supply 2m away from your receiver, and you see that transmit power only contributes linearly to your interference problem, and factor in distances quadratically).

You might hence still want to filter OOB reception. (I mean, that's why we rely on filters, and especially involved multi-stage architectures like the superhet receiver, instead of simply letting the antenna be the selector.) However, for powers that you'd still get via your antenna, it's not very likely you'd be deafening your LNA (which is usually the first thing you'd want to attach to your receive antenna, thanks to Friis' noise formula); so, you can add the filter later in the signal chain (e.g. on the IF after first mixing).

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  • $\begingroup$ Care to detail the "Put your interferer [...] and factor in distances quadratically)." part ? You have lost me entirely. Thanks $\endgroup$
    – kellogs
    Commented May 16 at 15:56
  • $\begingroup$ @kellogs sure! If your interferer (X) and your transmitter of interest (T) are the same distance from you, and transmit with the same power, then of course the same power reaches your antenna. And then the attenuation of the antenna against the out-of-band X might totally do. If I is half as far away as T and both transmit with the same power, then your antenna sees X as 4 times as powerful as T. The physics there is a bit straightforward: imagine a transmitter is a candle in the middle of a hollow sphere. Let's make that sphere the size that its surface is 1.(pick whatever area unit you want) $\endgroup$
    – Marcus Müller
    Commented May 16 at 16:59
  • $\begingroup$ you achieve some illumination strength on the inner surface of the sphere, right, all the light from the candle gets distributed (not necessarily evenly) across that surface area of 1. Now, double the diameter of the sphere, and you see that the surface of the sphere gets 4 times as large; the surface grows quadratically with distance. The same amount of light distributes across an area of 4, so the illumination per area goes down to ¼. Generally, the intensity of illumination hence goes down with the square of the distance from the source of illumination. Hence, the "square power law". Nice; $\endgroup$
    – Marcus Müller
    Commented May 16 at 17:02
  • $\begingroup$ Now, imagine you have an emitter X that's, say, putting out 10 µW (harmonics from your USB charger, a computer bus, your electric hamster, whatever; 10 µW = -20 dBm), 3 m away from your receiver. But you're interested in a transmitter T that's 30 km away, but emits 1 kW (=60 dBm). Let's do the math: the distance to T is 10,000 (=10⁴) times as large as to X, so the power of T gets distributed on a sphere with (10⁴)²=10⁸=80 dB times the surface, and that means it sees 80 dB more attenuation than X. So, we now compare the powers of X and T at the receiver, knowing that both T sees 80 dB more… $\endgroup$
    – Marcus Müller
    Commented May 16 at 17:08
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    $\begingroup$ …attenuation. Suddenly, your you need to compare what is left of -20 dBm to what is left to 60 dBm - 80 dB = -20 dBm. And that means both have the same incident power area density at your receive antenna! $\endgroup$
    – Marcus Müller
    Commented May 16 at 17:10

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