Hi all,

I have seen the small-signal equation for output dc link of an inverter as can be seen in following figures in an IEEE paper.

I understand the equations until (14). But can't understand how the laplace transform of 14 has led to 15. Can you help me please?

 

Hello there,

Try first ignoring Delta i0 and then try to solve for the negative quantity in (15) alone. You should see that come out right, then you can include the part with Delta i0 in it and that will be simpler.
I believe you can also use the transform/substitution:
C*d(Dvdc)/dt=C*s*Dvdc

as the Laplace transform of the left hand side. The right hand side should be ok as is. "D" here is the delta symbol.
If you need the entire solution i'll post that later.

These kinds of equations often come up in perturbed, averaged AC analyses, but i have to admit i dont do these very much anymore. They do apply to converter circuits quite well. If you post the reference paper we can take a look at that also.

 

You can see my solution in the following link, which leads to an answer different from (15) in the first post.
http://s8.picofile.com/file/8291014400/0304201711753.jpg

I have the reference paper, but I don't know if it's ok to post it here with respect to copyright and such. Anyhow, here is the link of the paper:

http://ieeexplore.ieee.org/document/7469404/

 

Hello there,

Try first ignoring Delta i0 and then try to solve for the negative quantity in (15) alone. You should see that come out right, then you can include the part with Delta i0 in it and that will be simpler.
I believe you can also use the transform/substitution:
C*d(Dvdc)/dt=C*s*Dvdc

as the Laplace transform of the left hand side. The right hand side should be ok as is. "D" here is the delta symbol.
If you need the entire solution i'll post that later.

These kinds of equations often come up in perturbed, averaged AC analyses, but i have to admit i dont do these very much anymore. They do apply to converter circuits quite well. If you post the reference paper we can take a look at that also.

I would appreciate if you post your solution.

 

Hello again,

Yes you are right something is wrong here. I can only guess then that the transform on the left is different. The transform of dv/dt is really s*V(s)-v0 so that could mean we should have something more like C*(s*Dvdc-K) on the left. Maybe they are using K=Po/vdco^2. But then that would not explain why the second equation was required in the original set. So sorry but i am at a loss for the correct solution at the moment. I'll still continue to look around for some reference material i might have.

 

Hello again,

I took another quick look at this and realized there is something strange here right off the bat.

If i wrote an equation for this in continuous time with more circuit values like input resistance and load resistance i would write:
C*dv/dt=(Vin-Vout)/Rin-Vc/Rout
[which BTW makes for an interesting circuit for a comparative study]
which for the small schematic in this thread is just about the same thing as:
C*dv/dt=Idc-Io

and notice that in the original equations for this thread they have:
V*dDvdc/dt=Io-Idc

so Idc and Io are reversed sign wise. This is unusual in normal circuit analysis, and might even be outright incorrect. We'd have to see if it works with electron current flow instead of the more usual conventional current flow.

Side note for what it is worth, i found that
DP/vdco-Po/vdco^2*Dvdc=0

at some operating point for extremely small changes in time. That makes sense though because then also Dio=0 and so C*dv/dt=0 also, which then works out as 0=0 which is ok.