Question about the Wilkinson Divider

Hi everyone:

When i'm studying the wilkinson divider part of the Microwave Engineering (by D.M. Pozar). There's one question that confuses me. In Chap7.3 (Wilkinson power divider), it says the Wilkinson divider can be lossless when outputs are matched, but in Chap7.1 (Basic properties of dividers and couplers) it says a three port network cannot be lossless, reciprocal and matched at all ports. These two statements seem to be inconsistant.

According to the Chap4.3(Reciprocal networks and lossless networks) , the S matrix of a lossless network should be unitary, but the ideal S matrix of Wilkinson Divider is not.

Could anyone answer this question for me ? Thanks.

Hi

It is true that a three port network cannot be lossless, reciprocal and matched at all ports, at the same time.

The Wilkinson is matched, reciprocal, but not lossless. It is between port 1and port 2 or 3. But the Wilkinson it is LOSSY between the two isolated ports. So it satisfy the condition.

Regards

Hi tyassin:

thanks for your reply, but what i still don't understand is that since the Sxx and the S32 are all zero , how come there will be reflected wave passing through the resistor and causing loss?

the S matrix is (assume port1 is input, port2 and port3 are outuputs)

0 -j/√2 -j/√2
-j/√2 0 0
-j/√2 0 0

Hi

If Sxx and S32 is zero then no reflected power will be dissipated in the resistor.
If S22 and S33 is not zero, then there will be reflected power and power will also be dissipated in the resistor.

Regards

Hi,

One can apply even-odd mode analysis to understand the operation of Willkinson divider.

Even mode: the middle point of resistor is open circuit and resistor is not contained in the circuit, so the even part is indeed losless.

Odd mode: the middle point of resistor is short circuited so the resitor is contained in the circuit and contributes to the losses.

For the generall case of the Willkinson divider operation, these two should be added so for the generall case of two circuits connected to the outputs there will be loss in the resistor that will contribute to the isolation between the ports.

For the particular case when the signal is split into two identical matched circuits (2x50Ohm termination, and accordingly nothing else!) there are no reflected signals, the circuit exhibits perfect symmetry and odd mode cannot exist at all, so all of the ports are isolated, so indeed it can be lossless for this particular case.

However, this is rearly ever the case of using the real Willkinson divider. When the terminations are mutually equal (symmetrical) but not matched, no reflected power will be disipated on the resistor (again, it is perfect symmetry) but it will appear on the input port.

If the terminations are not matched and are not mutually equal, part of the reflected power (imbalance, odd mode) will be spent in the resistor and part (balanced, even mode) reflected to the input.

When doing the acctual calculation about even/odd mode part quantities by hand, one must take into account that S parameters are complex quantities so the terminations must be considered as complex impedances too (phase of the reflected signals is of course very important). Any simulator does this automatically, so one just have to interpret the results)
flyhigh

Thanks ! flyhigh

Do you mean it is still possible for a three ports network to be reciprocal, lossless, and all-ports matched at the same time ?

TWang

My careless fault, It should be that "If you use the circulator, it is possible for lossless and matched".

Hello

It is NOT possible to have a: Lossless, Reciprocal and Matched three port.
A circulator is: Matched, lossless, but NOT reciprocal.

If you want to know the details check out the first few pages in chapter 2 in: Microwave engineering (2 edition) by David Pozar.

Hope this will help

Regards