When it gets very narrow you get pictures like this - it doesn't say at what frequency other than

Thanks Kent, Can you do a Return Loss plot to determine if the feed impedance is 50 ohms? I can readily get the gap to between 1 to 2 mm witch surely is close enough for the 4GHz or so maximum

Hello All, I'm looking for a couple of small 10GHz ferrite isolators for a new WG input LNA I'm working-on to replace my 2011 design. Does anyone have any they'd be prepared to sell me at a sensible

I have perhaps a dozen Ridged horns of different sub-species. Most have a ridge several millimeters wide. But my 1-18 GHz one has the ridge ground to a fine edge along the top. I think the fine edge

Hi all, Thanks guys for the pictures of commercial wide band horns but... Can somebody answer the question please? The ridge thickness, can it be thin, one sheet of sheet metal, say 1mm ? (Just

Not sure I do either - in the traditional 2.D²/λ sense But when the transmitter is not a point source (not diffraction limited, I think, in optical terms) there must be some name for it. The

Hi Ben That's what I use with simple switch to reverse polarity 12vdc set your the centre to 0deg then You have equal throw either side , it was suggested to me by G8DMU and it works very well Cheers

Thinking on what someone said I was trying to figure out how to nod the dish up and down to get the best signal. In my case it needs to be very light as its manually raised. I was trying to think

Thinking on what someone said I was trying to figure out how to nod the dish up and down to get the best signal. In my case it needs to be very light as its manually raised. I was trying to think

<< ...in the sun's near field. >> Not sure I completely agree with that Andy. If the sun is a single source driving a sun diameter antenna, then yes indeed. If you consider the sun to be an enormous

Yes please, Chris Andy www.g4jnt.com

Andy, If it's the same as the one I have, and it looks suspiciously like it, I have the manufacturer's data sheet on that horn ... Let me know if you'd like it. Chris G4DGU

An instinctive feeling is nagging me that the fraction of the sun in the antenna beamwidth doesn't matter when working things out this way. But it's hard to quantify why, except that using a solar

Worked out the equation from first principles now, and it gives the same as Mike's example k is Boltzman's constant, 1.38E-23 and S in the equation is the actual solar flux in Watts per m^2/Hz, not

I'm also puzzled as to how that can work with arbitrary antenna beamwidth If the beamwidth is much larger than the sun's disc diameter then only a portion of the solar flux is contributimg, A very

Looks a bit like that, but it's smaller Mouth dimensions (internal) 104 x 69mm Andy www.g4jnt.com

To which I add I have a number of the 1.5 -15GHz ones in stock. Sam, G4DDK

Looks like a 1-18GHz horn similar to the ETS Lindgren / Emco 3115. Best Regards Tim Hague

I can't make that G/T equation work with your values. What is k ? And I assume it should read G/T = 10 * log [ 4 π k (10^(N/10) - 1) / ( λ^2 S) ] . dB/K Andy www.g4jnt.com

Forgot to add, if the transmitted noise power is roughly the same as the solar flux that gives a handle on the gain of the transponder once you do the link budget taking into account the distance, the