Hi again,
Yes that's right, and you're welcome
Swoosh:
Here's another way to do it using the ODE's and skipping the flow graph (which is cheating if you ask me ha ha):
Starting with the original set of ODE's again:
x1=-X2/C-X1/(Rs*C)-X1/(Rc*C)+Vs1/(Rs*C)
x2=X1/L-X2*RL/L
we note that xN is just the derivative of XN, so using the differential operator 's' we can write:
x1=s*X1, and
x2=s*X2
This cuts down on the work required but doesnt offer as much insight into what is actually happening.
When we substitute into the original set of ODE's we get:
s*X1=-X2/C-X1/(Rs*C)-X1/(Rc*C)+Vs1/(Rs*C)
s*X2=X1/L-X2*RL/L
and all we have to do now is solve this set of simultaneous equations in the variables X1 and X2, and all we want is the solution for X2, so we end up with:
X2=(Rc*Vs1)/(Rc*Rs*s*C*RL+Rs*RL+Rc*RL+Rc*Rs*s^2*C*L+Rs*s*L+Rc*s*L+Rc*Rs)
which is the same as when using the flow graph, and after rearranging a little we again get:
(RL*X2)/(Rs*C*L)+(RL*X2)/(Rc*C*L)+(s*RL*X2)/L+X2/(C*L)+(s*X2)/(Rs*C)+(s*X2)/(Rc*C)+s^2*X2=Vs1/(Rs*C*L)
which is again the same as before, so after regrouping and replacing all s^n*X2 with the n^th derivatives of iL, we again get the same solution as the paper.