Can i replace short-circuited stub L with open-circuited L+lambda/4?
I do calculations:
OpenStub=ATan(-i*50/Zs)OpenStub = -1.335176878e0 ShortStub=ATan(Zs/i*50)ShortStub = 1.569130208e0
For short stub i can understand it, it is inductive.
But how to interpret negative electrical length of OpenStub?
For a single frequency circuit, yes you can. The added lambda/4 (open end) behaves as a short circuit, so you can replace a short circuit for an open stub. You may experience some reduction in Q factor as the short circuit will not be zero Ohms (due to transmission line loss).
you should check your new solution over the full frequency range as the impedance changes somewhat faster compared to a real shorted lambda/4 stub. If your matching network provides some out of band filtering, you will get other behavior when using the open lambda/4 stub for creating a short circuit.
If transient energy may enter the system, a real shorted stub removes more low frequency energy then a stub with open end lamda/4 extension.
regarding negative electrical length, you may extend the negative length line with a half wave (so you make a full turn at the Smith Chart).
at ONE FREQUENCY they are equivalent. But the longer a line gets with respect to one wavelength, the farther its response departs from approximating a lumped element. The common result of this is smaller bandwidth. So one tries to stick with either lumped elements or short matching lines
In many designs a short-circuit stub is utilized to create a path for DC current, both bias (with capacitor to ground) and DC return to ground, as well as the obvious impedance matching effect it provides. So the general answer to this question is that you cannot always substitute different length open and short circuit stubs. It depends on the circuit requirements.