I did a thorough study of makeYbus() (both a code analysis and step-by-step executions) and was able to answer the majority of my questions. However, I still can't get the mathematical sense on how a diagonal element with multiple transformers is treated.
As an example consider case14.m bus 4, here is stripped down version of its portion of the branch input matrix: fbus | tbus | r | x | b | ratio | angle 2 | 4 | 0.05811 | 0.17632 | 0.034 | 0 | 0 3 | 4 | 0.06701 | 0.17103 | 0.0128 | 0 | 0 4 | 5 | 0.01335 | 0.04211 | 0 | 0 | 0 4 | 7 | 0 | 0.20912 | 0 | 0.978 | 0 4 | 9 | 0 | 0.55618 | 0 | 0.969 | 0 There are 5 lines that meet at bus 4, 2 have transformers (albeit with no phase shift, but that shouldn't be important) and of the remaining 3, 2 have line charging. What is the general formula for a diagonal element? Thanks a lot On Fri, Aug 26, 2016 at 7:58 PM, Abhyankar, Shrirang G. <abhy...@anl.gov> wrote: > Please take a look at the function makeYbus(). > > > On Aug 26, 2016, at 10:05 AM, davor sutic <davor.su...@gmail.com> wrote: > > > > I'm trying to understand the effects of having a transformer connected > to a branch on the implicated admittance matrix. In the MatPower manual, > section 3.2, equation 3.2 describes such a relation for a 2x2 branch > admittance matrix. > > > > I'm interested in the general form of the (diagonal and off-diagonal) > elements of an admittance matrix, when multiple branches have transformers > attached. Particularly the diagonal elements are confusing, how are the > elements summed up, if multiple have transformers? > > > > Further, in the test case files, the absence of a transformer is > indicated with its ratio and angle set to 0. In the mentioned equation > (3.2) that would cause problems when dividing with 0. So another question > is how are those elements treated in the light of formula 3.2? > > > > Thanks a lot. > > >
