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center_finite_diff

Performs a centered finite difference operation on the rightmost dimension.

Prototype

	function center_finite_diff (
		q        : numeric,  
		r        : numeric,  
		rCyclic  : logical,  
		Option               
	)

	return_val [dimsizes(q)] :  numeric

Arguments

q

A multi-dimensional array.

r

A scalar, one-dimensional, or multi-dimensional array containing the coordinates along which q is to be differenced. Does need not be equally spaced from a computational point of view.

• scalar: r assumed to be the (constant) distance between adjacent points.
• one-dimensional (and the same size as the rightmost dimension of q): applied to all dimensions of q.
• multi-dimensional: then it must be the same size as q.

rCyclic

True: q treated as cyclic in r and the end values and all the returned values will be calculated via centered differences. False: q NOT treated as cyclic in r and the end values will use a one-sided difference scheme for the end points. q should not include a cyclic point.

   result(n) = (q(n+1)-q(n))/(r(n+1)-r(n))  for the initial value
   result(m) = (q(m)-q(m-1))/(r(m)-r(m-1))  for the last value

Option

Reserved for future use. Currently not used. Set to an integer.

Description

Performs a centered finite difference operation on the rightmost dimension. If missing values are present, the calculation will occur at all points possible, but coordinates which could not be used will set to missing.

result(n) = (q(n+1)-q(n-1))/(r(n+1)-r(n-1))

Examples

Example 1

   q = (/30,33,39,36,41,37/)     
   r = 2.                                 ; constant

   dqdr = center_finite_diff (q,r,False,0)

; result
   dqdr(0) = (33-30)/2 = 1.5
   dqdr(1) = (39-30)/4 = 2.25
   dqdr(2) = (36-33)/4 = 0.75
   dqdr(3) = (41-39)/4 = 0.5
   dqdr(4) = (37-36)/4 = 0.25  
   dqdr(5) = (37-41)/2 = -2.0

Example 2:

   theta = (/ 298,299,300,302,...,345,355,383/)   ; potential temp.
   p     = (/1000,950,900,850,...,200,150,100/)   ; pressure (hPa)

   dtdp  = center_finite_diff (theta,p,False,0)

Example 3

p is not equally spaced. Demonstrates defining the appropriate pressure levels when the r coordinate is not equally spaced. The end points are computed via a one-sided difference.

   theta = (/ 297, 298,300,302,...,345,350,355,383/)   ; potential temp.
   p     = (/1013,1000,900,850,...,200,175,150,100/)   ; pressure (hPa)

   dtdp  = center_finite_diff (theta,p,False,0)

   np    = dimsizes(p)
   pPlot = new ( np, "float" )
                                                       ; arithmetic mean
   pPlot(0)      = (p(0)+p(1))*0.5                     ; set bottom
   pPlot(np-1)   = (p(np-1)+p(np-2))*0.5               ; set top    
   pPlot(1:np-2) = (p(0:np-3) + p(2:np-1))*0.5         ; mid points 

; or 
                                                              ; log (mass) wgted
   pPlot(0)      = exp((log(p(0))+log(p(1)))*0.5)             ; set bottom
   pPlot(np-1)   = exp((log(p(np-1))+log(p(np-2)))*0.5)       ; set top    
   pPlot(1:np-2) = exp((log(p(0:np-3)) + log(p(2:np-1)))*0.5) ; mid points 

Example 4

Let T be four-dimensional with dimensions time,level,lat,lon. P is one-dimensional. Perform the finite differencing only in the vertical (level) dimension. This requires that T be reordered to put level in the rightmost dimension.

   dTdP = center_finite_diff(T(time|:,lat|:,lon|:,lev|:),P,False,0)
;  returns dTdP (time,lat,lon,lev). this variable can be reordered
;  again to place it back in the original order

Example 5

Now P is also four-dimensional and requires reordering:

   dTdP = center_finite_diff (T(time|:,lat|:,lon|:,lev|:)  \
                             ,P(time|:,lat|:,lon|:,lev|:), False,0)
; returns dTdP(time,lat,lon,lev)

Example 6 Assume that the longitude coordinate variable associated with T in the examples above is cyclic and is equally spaced in degrees but not in physical space.

   dlon = (lon(2)-lon(1))*0.0174533 ; convert to radians
                                    ; pre-allocate space
   dTdX = new ( (/ntim,klev,nlat,mlon/), typeof(T), T@_FillValue)

   do nl=0,nlat-1                      ; loop over each latitude
      dX = 6378388.*cos(0.0174533*lat(nl))*dlon  ; constant at this latitude
      dTdX(:,:,nl,:) = center_finite_diff (T(:,:,nl;,:), dX , True,0)
  end do
; result: dTdX(time,lev,lat,lon)