I don't know if there's an actual formula, but GE published some duplexer curves that go with some of their equipment, showing the amount of attenuation needed for certain configurations. They can be found here:
http://www.ka9fur.net/geduplex/duplex.html 50 watts TX power is +47dBm. 0.2uV sensitivity is -121dBm, for a difference of 168dB. This is, of course, on the same frequency, which is not how repeaters operate. The amount of attenuation necessary depends on the receiver's bandpass - sensitivity to a signal on the TX frequency, and how much noise the transmitter puts out on the RX frequency. The duplexer has to attenuate the unwanted signals enough so they don't affect the receiver's sensitivity while not losing too much TX signal going out to the antenna. Bob M. ====== --- Andy <[EMAIL PROTECTED]> wrote: > Hi all, > > I've been searching old posts and the RB page but > haven't found quite > what I'm looking for, so I'm resorting to spamming > the list :-) > > My question has to do with how much TX/RX isolation > is required for > different frequency spacings. I'm assuming that as > the TX/RX > frequency spacing increases, the amount of isolation > that the cavities > needs to provide decreases. My question is how do I > calculate this? > > For example, if a repeater puts out 50W, for that > 50W signal to be > attenuated to .2uV, something like 154 dB (I > think...someone check my > math) of attenuation would be required. Now, since > the TX frequency is > actually separated from the RX frequency by some > amount (let's say 600 > kHz), not all of the TX signal energy is going to be > seen by the > receiver. However, some of it will be seen...my > question is how much? > Since a lot of the traffic on this list mentions > numbers like 80 dB > of cavity isolation, I'm assuming that about 80 dB > is coming from the > cavities and the other 76 dB (from the 154 dB > mentioned above) is > coming from the fact that the TX and RX frequencies > are spaced apart. > So now if we expand the TX/RX spacing from 600 kHz > to 5 MHz, how much > less isolation is required from the cavities, since > the spacing is > contributing more? Obviously the transmitter and > receiver factor into > this overall equation, but are there some rules of > thumb that can be > used with "averagely clean" receivers and > transmitters? > > I'm sure this information is readily available...I'm > just having > trouble finding it. > > Thanks in advance! > > Andy KB9JOZ __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com
