Re: Help with mathematical function

[Stefan van der Walt]

You can vectorise your results by doing something like

d = angle(a) - angle(b);
d(some_condition) = d(some_condition) - 2*pi;

etc.

Also take a look at the function "unwrap".

Regards
Stefan

On Sun, Apr 10, 2005 at 12:22:48AM -0700, Robert A. Macy wrote:
> Stefan,
> 
> Thank you for your reply.  
> 
> The problem is the data only "looks" like a spiral.  There
> is actual information in the way the spiral decays,
> including "lumpiness".  
> 
> My problem was that the angle() function resets itself as
> the spiral decays.  It is very easy to get erroneously huge
> delta angle changes at those junctions.  
> 
> Luckily the angle steps are known to be small enough that
> one should NEVER get too large a change in angle.  
> 
> So my solution was to add some simple "if" tests.
> v(i,k) is an array of complex values...
> 
> for i=1:500
>   for k=2:100
> ...processing v(i,k)...
> delta_angle(i,k)=angle(v(i,k))-angle(v(i,k-1));
> if { delta_angle(i,k) > pi() )
>   delta_angle(i,k) = delta_angle(i,k) - 2*pi();
> endif
> if ( delta_angle(i,k) < -pi() )
>   delta_angle(i,k) = delta_angle(i,k) + 2*pi();
> endif
> ...more processing of results(i,k)...
>   endfor
> endfor
> 
> 
> The tests still fail when the spiral is really close to
> zero, but by then the contribution to error is quite small
> since the absolute vector length is so small.  
> 
> The only problem now is that the loops to do this, the
> for..for.. loops, take several minutes.  
> 
> Is there someway to use vector math?  And still include the
> checks on the angles?  Or, at least disable the constant
> testing of the sizes of the variables which goes on each
> time the loop is exercised?
> 
>         - Robert -
> 
> On Sat, 9 Apr 2005 18:34:49 +0200
>  Stefan van der Walt <address@hidden> wrote:
> > Hi Robert
> > 
> > I assumed that the radius of the spiral decreases
> > linearly towards
> > zero, as one moves through 360 degrees.
> > 
> > The following script draws the spiral and calculates the
> > area for
> > angles between 0 and 360 (0 and 2*pi radian).
> > 
> > Will this do, or do you need to take the measured data
> > into account?
> > In that case, your method is nearly correct.  You should,
> > however,
> > approximate the area as that of a rectangular triangle:
> > 
> > 1/2 * radius^2 * delta-angle
> > 
> > (radius * delta-angle is the length of one side of the
> > trangle, radius
> > the length of the other)
> > 
> > This is the same method used to solve the line integral
> > of the
> > theoretical function,
> > 
> > integral 1/2 (2 pi - theta)^2 R^2 dtheta
> > 
> > Regards
> > Stefan
> > 
> > ----
> > R = 1;
> > theta = linspace(0, 2*pi, 1000);
> > 
> > area = R^2 * theta / 6 .* (theta.^2 - 6*pi*theta +
> > 12*pi^2);
> > 
> > automatic_replot = 0;
> > subplot(121);
> > axis("square");
> > grid on;
> > polar(theta, (2*pi - theta)*R, ";Spiral;");
> > 
> > subplot(122);
> > plot(theta, area, ";Area under spiral;")
> > ----
> > 
> > On Fri, Apr 08, 2005 at 04:08:15PM -0700, Robert A. Macy
> > wrote:
> > > I am not a mathematician, so not sure I can ask this
> > > correctly, but...
> > > I need to calculate a "spiral" integral - calculate the
> > > area a vector, or "ray", sweeps out as a spiral
> > collapses.
> > >  
> > > 
> > > The tip of the vector monotonically decreases down to
> > near
> > > zero value.  with 500 of these spirals.  
> > > 
> > > The data array is composed of 500 by 100 complex
> > values.
> > >  The data along each row (500 rows) spirals down to
> > near
> > > zero values at the highest index value.  
> > > 
> > > I would like to calculate the area the ray sweeps out
> > as
> > > the spiral collapses.  
> > > 
> > > I tried a simple approach like using the area defined
> > by
> > > radius times the change in angle (delta angle)
> > > 
> > > average( abs(arrayvalues) ) * delta( angle(arrayvalues)
> > )
> > > 
> > > and then integrated these values along each row.  The
> > mesh
> > > plots looked promising, BUT...
> > > 
> > > The angle function has a limited range so their plus
> > minus
> > > transitions played havoc with the results.  The spirals
> > can
> > > be over 700 to 1000 degrees.  
> > > 
> > > Plus, there is noise in the data, so the spiral tends
> > to
> > > "lump" around 0,0 not necessarily converge to it, which
> > > again causes wild angle transitions.  However, along a
> > row
> > > between any two adjacent array values the data seems to
> > be
> > > monotonically decreasing (spiralling down) to zero
> > values.
> > >  
> > > 
> > > Does anyone know of a simple algorithm that will take
> > each
> > > vectors' end points, calculate the area, and sum that
> > up
> > > over the 100 values along each row?  
> > > 
> > >              - Robert -
> > > 



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