Re: differential equation

[Thomas Shores]

On 05/16/2011 04:52 PM, Liam Groener wrote:
> On May 16, 2011, at 9:35 AM, CdeMills wrote:
>
> > Hello,
> >
> > I'm trying to simulate an electrical circuit whose time-domain differential
> > equation is
> > V" + k1 V' +k2 V = a1 I' + a2 I
> > Vs(t) - Rs I = V
> > where V and I are the device voltage and current, respectively, and the rest
> > are constants. Vs is the time-varying voltage source, and Rs its internal
> > resistance. Starting from equilibrium, all values and their derivatives are
> > set to zero.
> >
> > I don't see how to formulate the problem, in particular adding the source
> > equation into the picture.
> >
> > LSODE: let's say x1 = V, x2 = I, x3 = V'; so we have
> > x1' = x3
> > x2' = ??? =>  I' depends on Vs'(t) ???
> > x3' = a1 * ??? + a2 x2 - k1 x3 - k2 x1
> > How to fill the "???"
> >
> > DASSL: same variables
> > res1 = x1'-x3
> > res2 = ...
> > res3 = x3' - a1 x2' - a2 x2 - k1 x3 - k2 x1
> > In all cases, I need a relationship around I', and would like Vs(t) to be a
> > step source, in this case Vs'(t) would become infinite ???
> >
> > Any hints ?
>
> I would substitute I = (Vs-V)/Rs into your first equation and use lsode.
> You would then have 2 d.e.'s:
> x'(1) = x(2)
> x'(2) = aVs - k1 x(2) - b x(1)
> where:
> a = a1/Rs and b = k2+a1/Rs
>
> The initial conditions would be:
> x(1)=0, x(2)=0
> Don't worry about the infinite derivative, the imbalance in the initial voltage 
> should take care of that.
>
> _______________________________________________
> Help-octave mailing list
> address@hidden
> https://mailman.cae.wisc.edu/listinfo/help-octave
Interesting.  This works.  In any case, the three variable cannot 
possibly work, since lsode is based on methods (ABM, BDF) that assume 
continuous solutions, and this is all they will generate.  However, the 
formula Vs = V+Rs*I guarantees that one of V or I cannot possibly be 
continuous at any jump discontinuity of Vs.  Thus the 3-variable 
approach cannot possibly work in any formulation since it would mandate 
that V, V' and I be continuous initially and at every interior point.  
In that regard, the conditions that I and I' vanish initially make the 
system inconsistent if they are meant to be used as part of an initial 
value problem.   Your approach ignore those constraints and  banished 
the potential discontinuities due to the step source on the right hand 
side of the first order ode as simple jump discontinuities, so the 
problem remains well posed.  Here's a script that illustrates these 
points with an impulse function.

Thomas Shores

% DEtest.m

clear
global k = [1,2] # [k1,k2]
global a = [2,2] # [a1,a2]
global Rs = 0.5

function retval = heaviside(t)
retval = (t==0)*0.5 + (t>0);
end

function retval = Vs(t)
retval = heaviside(t-0.5)-heaviside(t-1);
end

function retval = Vsp(t)
retval = zeros(size(t)); # ignore any dirac delta function
end

function retval = fcn(x,t)
# here x = [V; Vp]
global k;
global a;
global Rs;

retval = [0;0];
retval(1) = x(2);
retval(2) = a(1)*(Vsp(t)-x(2))/Rs + a(2)*(Vs(t)-x(1))/Rs - k(1)*x(2) - 
k(2)*x(1);
end

% main: give it a spin
x0 = [0;0];
t = linspace(0,4,800);
x = lsode(@fcn,x0,t);
clf
plot(t',x(:,1),"-1;V(t);")
hold on, grid
plot(t',x(:,2),"-2;V'(t);")
plot(t',(Vs(t)'-x(:,1))/Rs,"-3;I(t);")
plot(t,Vs(t),"-4;Vs(t);")