1
\$\begingroup\$

Updated

I have a RF splitter/combiner. I am focusing on the black and red vectors. (Picture attached) enter image description here

  • When the black vector of 0.25V and 0 phase enters the box, it will split. 1/sqrt(2)*0.25V at 0 phase (half power split and no phase change) will go to port 1 and 1/sqrt(2)*0.25V 90 degree phase to port 2 (half power split and phase change)

  • When the red vector of 1/sqrt(2)*0.25V and 90 phase enters the box, it will split. 1/sqrt(2)*0.25V at 90 phase (half power split and no phase change) will go to port 2 and 1/sqrt(2)*0.25V with 180 phase at port 1 (half power split and phase change)

This means at port 1, we have perfect cancellation. A lot of the signal has gone. Also at port 2, we have perfect addition that leads to 0.35V. Lost 0.15V. In reality we get 0.5V output at port 2, what has gone on in this hybrid splitter/combiner that vector maths cannot show because half the signal is gone?

Please help

\$\endgroup\$
1
  • 1
    \$\begingroup\$ The mistake here is that you need to conserve power, not voltage. Power is V^2/R. So when all the power is directed to one port you expect a voltage of 0.25 * sqrt(2) or about .35 V which is the correct answer. \$\endgroup\$
    – Evan
    Nov 26 '18 at 18:20
0
\$\begingroup\$

It's not so simple at all. Cancellation and in-phase summing are useful tools to recall what was the output port, but actually the circuit works in more complex way. We can investigate it by simulating.

Here we have a simple and well known TEM transmission line 3dB hybrid + input signal sources + matched 50 Ohm loads.

enter image description here

Transmission lines T1...T4 have all 0,25us signal delay. They are lossless. 50 Ohm lines and 35 Ohm lines are used. Actually 35 Ohm should be 50/sqrt(2) ohms.

Voltage sources V1 and V2 have both 10V peak voltage and their frequency is 1MHz. Thus the lenght of all transmission lines is the wavelength/4. V2 lags 90 degrees behind V1.

I have inserted 10 ohm resistors to inputs of the hybrid for 2 reasons:

  1. The used free version of a commercial simulator program has many tools disabled. I must add a dummy component to measure currents. I cannot probe a wire nor insert an ammeter. With R5 I can get the output current of V1 as written function i(R5) to my signal graphs.

  2. At startup of the simulation there's a complex transient. The lines are not at all matched in the beginning. In simulation this is seen as ringing, when the waves travel around the transmission line loop. But resistors R5 and R6 eat gradually that part of the waves which are travelling back to signal sources. Without R5 and R6 they would reflect back from the sources completely and the ringing would continue forever.

At first we show that the inputs of the hybrid would be perfectly matched to 50 Ohm feeds altough the feeds have now zero lenght and only 10 Ohm impedance.

enter image description here

The currents in resistors R5 and R6 are (after the initial transient is dead) perfectly in phase with the signal source voltages. Thus both signal sources have resistive loads. The peak currents are 166,6 mA. That's 10 volts divided by 60 ohms. When the 10 Ohm resistors are subtracted, the remainder is 50 Ohm.

Next we prove that all signal (after the initial transient has died) is outputted only to R4. R3 gets nothing. We want to show that all power goes to R4.

Of course we could plot the voltages and calculate the powers, but as well we can plot the powers directly. It would in addition reveal mismatches, if there were them.

enter image description here

We have plots for the powers that go to loads, R3 get v(1)*i(R3) and R4 gets v(3)*i(R4). The green curve shows that R3 gets nothing after the initial transient has died. R4 gets as much as there's totally inputted to the ports.

The existence of mismatch would be seen as part time negative power. Full reflections would be 50% of time negative. That's only reactive power and as cumulated transported energy it would be zero.

So, nothing is lost, R4 gets all what's inputted, nothing reflects back to signal sources.

The phase cancellation and in-phase reasoning to determine the output port is handy, but it's physically nonsense, because the signals have different voltage levels due different line impedances if powers are same. In addition what comes out from T1 splits to T3 and R3. What comes out from T3 splits to R4 and T2. What comes to left from T2 splits to R6 and T4, which feeds again T1 and R5. The phase cancellations should be calculated iteratively several rounds along the transmission line loop.

A person with the needed math skills obviously could solve the wave equations in the circuit by handling the hybrid as whole and having the inputs and loads as constraints. I must skip it.

The simulator solves the circuit operation numerically, but the system is handled as circuit. T1...T4 are presented as matrices where the wave behaviour is ready to use component data, no wave equations need to be solved during the simulation.

What actually happens to mutually cancelling waves which come to same point from two joined lines?

Answer: Their voltages are summed at the joint but the waves continue their travel. If there happens to be a load just in that point where the lines meet, the load gets nothing, because the sum voltage is zero. It's just like a zero voltage point in a standing wave pattern in case of full reflection.

But how the wave can continue - for ex.those who come up through T3? How they can continue to T1 when the voltage is constantly zero at node 1?

Answer: The wave has 2 components. Electric and magnetic field in the space between the conductors. If the total electric field happens be zero in a point because the the incoming waves cancel each other, the magnetic field will not be cancelled if the waves travel to different directions. The magnetic vectors have different directions. That's enough to allow the wave to continue. Only such summed wave is nonexistent where both electric and magnetic fields are cancelled. You surely have read that a standing wave can have a zero voltage point, but there's a current amplitude maximum in the same place.

We electricians like simplifications which allow us to think currents and voltages and forget those 3D vector fields which are the actual waves. The voltage is the strength of the electric field squeezed to one number and the current in a transmission line is induced by the fields. It's much simpler than the magnetic field in the 3D space. One can follow the waves in a TEM mode transmission lines well with voltage and current (as we have done above).

\$\endgroup\$
2
  • \$\begingroup\$ Hi, are you able to elaborate please? @user287001 \$\endgroup\$ Nov 26 '18 at 12:00
  • \$\begingroup\$ @NatalieJohnson the story is rewritten and a hybrid is simulated. \$\endgroup\$
    – user287001
    Nov 27 '18 at 5:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.