I need to find the orientation of a fingerprint.I have a code that uses gaussian filter that I got from mathworks.in but I'm not getting the required results.

function [orientim, reliability] = ... ridgeorient(im, gradientsigma, blocksigma, orientsmoothsigma)

    [rows,cols] = size(im);

    % Calculate image gradients.
    sze = fix(6*gradientsigma);   if ~mod(sze,2); sze = sze+1; end
    f = fspecial('gaussian', sze, gradientsigma); % Generate Gaussian filter.
    [fx,fy] = gradient(f);                        % Gradient of Gausian.

    Gx = filter2(fx, im); % Gradient of the image in x
    Gy = filter2(fy, im); % ... and y

    % Estimate the local ridge orientation at each point by finding the
    % principal axis of variation in the image gradients.

    Gxx = Gx.^2;       % Covariance data for the image gradients
    Gxy = Gx.*Gy;
    Gyy = Gy.^2;

    % Now smooth the covariance data to perform a weighted summation of the
    % data.
    sze = fix(6*blocksigma);   if ~mod(sze,2); sze = sze+1; end    
    f = fspecial('gaussian', sze, blocksigma);
    Gxx = filter2(f, Gxx); 
    Gxy = 2*filter2(f, Gxy);
    Gyy = filter2(f, Gyy);

    % Analytic solution of principal direction
    denom = sqrt(Gxy.^2 + (Gxx - Gyy).^2) + eps;
    sin2theta = Gxy./denom;            % Sine and cosine of doubled angles
    cos2theta = (Gxx-Gyy)./denom;

    sze = fix(6*orientsmoothsigma);   if ~mod(sze,2); sze = sze+1; end    
    f = fspecial('gaussian', sze, orientsmoothsigma);    
    cos2theta = filter2(f, cos2theta); % Smoothed sine and cosine of
    sin2theta = filter2(f, sin2theta); % doubled angles

    orientim = pi/2 + atan2(sin2theta,cos2theta)/2;

    % Calculate 'reliability' of orientation data.  Here we calculate the
    % area moment of inertia about the orientation axis found (this will
    % be the minimum inertia) and an axis  perpendicular (which will be
    % the maximum inertia).  The reliability measure is given by
    % 1.0-min_inertia/max_inertia.  The reasoning being that if the ratio
    % of the minimum to maximum inertia is close to one we have little
    % orientation information. 

    Imin = (Gyy+Gxx)/2 - (Gxx-Gyy).*cos2theta/2 - Gxy.*sin2theta/2;
    Imax = Gyy+Gxx - Imin;

    reliability = 1 - Imin./(Imax+.001);

    % Finally mask reliability to exclude regions where the denominator
    % in the orientation calculation above was small.  Here I have set
    % the value to 0.001, adjust this if you feel the need
    reliability = reliability.*(denom>.001);