Hi, I've got some questions about how the quarter wave cavity duplexer works in this configuration. From what I've read the cavities need to be connected together with 1/4 wavelength sections of cable. I'm not sure what the purpose of 1/2 cabling is. I'm told that its to transform the low impedance, but I'm not sure where. Also I thought the inputs to the receiver, transmitter and the cavities are all 50 ohms, so I'm not sure how the 1/4 wave cables would transform the impedance anyways. thanks
I am not sure that all of the cable lengths you have in your attachment are in fact 1/4 wave in the typical setup. It has been quite a while since I messed with a set of cans. Typically, a 1/4 wave line acts as a transmission line transformer. An open at one end of the 1/4 wave transmission line appears as a short circuit at the other end. And vice versa.
Yeah I understand about the transformation, but in this case I'm not sure where the impedance is that's supposed to be transformed. I thought that typically the receiver and transmitter ports are usually 50 ohms, and the input and outputs of the cans are 50 ohms.
For band pass or band reject filters, the cable lengths between cavities and between final cavities and the T- Connector to the single antenna feedline should ALL be electrical λ/2 long. Using λ/4 lengths of interconnect cables can cause very undesired impedance transformations This doesn't happen with λ/2 interconnects as the impedance seen at one end of a λ/2 will be seen at the other end Also the cable lengths between the TX or RX and their respective first cavities is NOT critical. They can be any convenient length. Cheers Dave
yes that is correct and that 's why in my last post I stated that its normal to use λ/2 lengths to avoid any impedance transformations if a cavity impedance changes due to detuning etc Don't forget that a λ/4 impedance transformer will double the impedance presented to one end of it Dave
Thanks Davenn. I keep reading though, especially on papers at a ham site, that the interconnections between the cavities are 1/4 λ to deepen the rejection notch of the "opposite frequency" in the band pass filter. So I kind of understand what what they are saying it's purpose it but I don't understand how it is accomplished in an EM or transmission line theory way, which is something I would really like to understand. There explanation are that the 1/4 λ transform a low impedance to a high impedance thereby blocking the unwanted signal energy. I'm not exactly sure how that works. Do they mean converting a short to an open? From my understanding the impedance into a 1/4 λ transmission line is Z_{in} = Z_{0}^{2}/Z_{L}, which normalized essentially means the input impedance is the reciprocal of the load impedance. And since, i think all impedance in the cavities are reactive, the transformer looks like it can only convert from a negative to a positive reluctance, or a short to and open, which from what I can tell both a zero and infinite impedance reflect 100% of the power at the input. Anyways, I'm confused about what's going on. By the way, here are some links that reference the 1/4 λ interconnections, I just don't understand they're explanations, at least from an EM, transmission line perspective. http://www.repeater-builder.com/antenna/quarter-wave-cables.html. http://www.repeater-builder.com/antenna/pdf/ve2azx-duplexerinfo.pdf http://www.seits.org/duplexer/duplexer.htm http://www.repeater-builder.com/tx-rx/tx-rx-duplexer-instructions-manual-number-7-9176-4-(4-inch).pdf
A duplexer contains a number of cavities X 2. Common is 4 and 6 cavity setups. A 6 cavity setup will have 3 cavities for the receive and 3 for transmit. Obviously each cavity is a bandpass filter for the appropriate frequency. Amateur 2 meter repeaters have a 600 KHz split so a hypothetical case would be a 146.76/146.16 Mhz machine. The cavities connected to the transmitter pass 146.76 and the receiver's cavities pass 146.16. But each cavity is a little more than just a bandpass filter. Each cavity also contains a notch filter. The receiver's cavities notch the transmitter frequency for obvious reasons as to not overload the receiver. And the transmitter's cavities notch the receive frequency. Here is where transmission line length comes into play. Whenever you place a filter of this type in-line, the input to the filter appears as an open circuit at the frequency we wish to attenuate. After all, wouldn't one wish to do this? At the Tee connector where the transmission line heads to the antenna we want the receive frequency to not even see the transmitter cavities. We want it to look like an open circuit. This is where line length comes into play.
OK, so I'm trying to understand how this set up works. Are the inputs to the cavities a short circuit, that the 1/4λ cable transforms to an open? Wouldn't an input that appears as a short circuit be just as effective for that matter since an open and short both have a reflection coefficient of unity? Thanks. I've had the standard EM, transmission lines, and microwave engineering courses, but I"ve never encountered the analysis for a repeater duplexer and I'm trying to reconcile how they work with what I've learned before.