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Message-ID: <L$BgVDAJwKv7EwOR@pickmere.demon.co.uk>
Date: Thu, 4 Oct 2001 19:35:53 +0100
To: rsgb_lf_group@blacksheep.org
From: "M.J.Powell" <mike@pickmere.demon.co.uk>
Subject: Re: LF: Which effect is this?
References: <200110041316_MC3-E238-AB9B@compuserve.com>
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In message <200110041316_MC3-E238-AB9B@compuserve.com>, 'Geri' Kinzel,
DK8KW <DK8KW@compuserve.com> writes
>Hello LF group,
>
>I went to Holland recently, and listened to Deutschlandradio on 153 kHz. I
>have already noticed years ago that mediumwave and longwave signals are
>being attenuated or even completely faded out under bridges. Now, listening
>to that radio transmission not too far away from our 136 kHz band I asked
>myself some questions:
>
>Why is the signal faded out, even if the bridge over the highway is only a
>few meters wide? The wavelength is almost 1000 times larger than the
>obstacle. What does this tell us about the radiation pattern of the
>Deutschlandradio? Does it really mean that the waves are coming
>perpendicular from above (which seems to be a common explanation for this
>effect)? Why does this effect only occur on LF and MF, but not on HF (I
>have checked for example transmissions on 14 MHz, where I can see no effect
>at all under bridges, altough the wavelength is much smaller, and the
>obstace compared to the wavelength is much larger)? Could any similar
>vertical obstacle (e.g. a chinmey, etc) also lead to such an attenuation,
>for example if that chimney is in the direction of an LF station I want to
>hear (I have a power plant with a 200m high chinmney only 4 km = 2 x lamda
>away, 270 degress from here ...).

Is this not akin to microwaves in a waveguide, whereby waves longer than
a cutoff wavelength will not propagate? The cut-off frequency is:

  Fco = 2a  where a is the dimension in the direction of the electric
component.

This gives the longest wave that will propagate. If you try and send
longer waves through then only a small 'fringing' effect is observed.

 IIRC it is called 'evanescent propagation'.

The long waves will of course propagate round an obstacle by
diffraction, but not necessarily though a hole.

Mike
-- 
M.J.Powell