实现对数 Gabor 滤波器组

Question

我正在阅读这篇论文 "Self-Invertible 2D Log-Gabor Wavelets" 它定义了 2D log gabor 过滤器：

论文还指出，滤波器只覆盖频率空间的一侧，并显示在这张图片中

在尝试实施过滤器时，我得到的结果与论文中所说的不匹配。让我从我的实现开始，然后我会陈述问题。

实施：

1. 我创建了一个包含过滤器的二维数组，并转换了每个索引，使频域的原点位于数组的中心，正 x 轴向右，正 y 轴向上。

   number_scales = 5         # scale resolution
   number_orientations = 9   # orientation resolution
   N = constantDim           # image dimensions
   
   def getLogGaborKernal(scale, angle, logfun=math.log2, norm = True):
       # setup up filter configuration
       center_scale = logfun(N) - scale          
       center_angle = ((np.pi/number_orientations) * angle) if (scale % 2) \
                   else ((np.pi/number_orientations) * (angle+0.5))
       scale_bandwidth =  0.996 * math.sqrt(2/3)
       angle_bandwidth =  0.996 * (1/math.sqrt(2)) * (np.pi/number_orientations)
   
       # 2d array that will hold the filter
       kernel = np.zeros((N, N))
       # get the center of the 2d array so we can shift origin
       middle = math.ceil((N/2)+0.1)-1
   
       # calculate the filter
       for x in range(0,constantDim):
           for y in range(0,constantDim):
               # get the transformed x and y where origin is at center
               # and positive x-axis goes right while positive y-axis goes up
               x_t, y_t = (x-middle),-(y-middle)
               # calculate the filter value at given index
               kernel[y,x] = logGaborValue(x_t,y_t,center_scale,center_angle,
           scale_bandwidth, angle_bandwidth,logfun)
   
       # normalize the filter energy
       if norm:
           Kernel = kernel / np.sum(kernel**2)
       return kernel
   

2. 为了计算每个索引处的过滤器值，我们在对数极坐标空间进行另一个变换

   def logGaborValue(x,y,center_scale,center_angle,scale_bandwidth,
                 angle_bandwidth, logfun):
       # transform to polar coordinates
       raw, theta = getPolar(x,y)
       # if we are at the center, return 0 as in the log space
       # zero is not defined
       if raw == 0:
           return 0
   
       # go to log polar coordinates
       raw = logfun(raw)
   
       # calculate (theta-center_theta), we calculate cos(theta-center_theta) 
       # and sin(theta-center_theta) then use atan to get the required value,
       # this way we can eliminate the angular distance wrap around problem
       costheta, sintheta = math.cos(theta), math.sin(theta)
       ds = sintheta * math.cos(center_angle) - costheta * math.sin(center_angle)    
       dc = costheta * math.cos(center_angle) + sintheta * math.sin(center_angle)  
       dtheta = math.atan2(ds,dc)
   
       # final value, multiply the radial component by the angular one
       return math.exp(-0.5 * ((raw-center_scale) / scale_bandwidth)**2) * \
               math.exp(-0.5 * (dtheta/angle_bandwidth)**2)
   

问题：

1. 角度：论文指出，从 1->8 索引角度会产生良好的方向覆盖，但在我的实现中，从 1->n 的角度除了一半之外不覆盖方向。即使是垂直方向也没有正确覆盖。这可以在此图中显示，其中包含比例为 3 和方向范围为 1->8 的过滤器集：

2. 覆盖范围：从上面的过滤器可以清楚地看出，过滤器覆盖了空间的两侧，这不是纸上说的。这可以通过使用从 -4 -> 4 的 9 个方向更明确。下图包含一张图像中的所有过滤器，以显示它如何覆盖光谱的两侧（此图像是通过在每个位置取最大值创建的来自所有过滤器）：

3. 中间列（方向 $\pi / 2$）： 在方向从 3 -> 8 的第一个图中可以看出，过滤器在方向 $ \pi / 2$ 处消失。这是正常的吗？当我将所有过滤器（所有 5 个比例和 9 个方向）组合在一张图像中时，也可以看到这一点：

更新： 在空间域中添加滤波器的脉冲响应，可以看到在 -4 和 4 方向上有明显的失真：

Answer 1

经过大量代码分析，我发现我的实现是正确的，但是getPolar 函数搞砸了，所以上面的代码应该可以正常工作。这是没有getPolar 函数的新代码，如果有人正在寻找它：

number_scales = 5          # scale resolution
number_orientations = 8    # orientation resolution
N = 128                    # image dimensions
def getFilter(f_0, theta_0):
    # filter configuration
    scale_bandwidth =  0.996 * math.sqrt(2/3)
    angle_bandwidth =  0.996 * (1/math.sqrt(2)) * (np.pi/number_orientations)

    # x,y grid
    extent = np.arange(-N/2, N/2 + N%2)
    x, y = np.meshgrid(extent,extent)

    mid = int(N/2)
    ## orientation component ##
    theta = np.arctan2(y,x)
    center_angle = ((np.pi/number_orientations) * theta_0) if (f_0 % 2) \
                else ((np.pi/number_orientations) * (theta_0+0.5))

    # calculate (theta-center_theta), we calculate cos(theta-center_theta) 
    # and sin(theta-center_theta) then use atan to get the required value,
    # this way we can eliminate the angular distance wrap around problem
    costheta = np.cos(theta)
    sintheta = np.sin(theta)
    ds = sintheta * math.cos(center_angle) - costheta * math.sin(center_angle)    
    dc = costheta * math.cos(center_angle) + sintheta * math.sin(center_angle)  
    dtheta = np.arctan2(ds,dc)

    orientation_component =  np.exp(-0.5 * (dtheta/angle_bandwidth)**2)

    ## frequency componenet ##
    # go to polar space
    raw = np.sqrt(x**2+y**2)
    # set origin to 1 as in the log space zero is not defined
    raw[mid,mid] = 1
    # go to log space
    raw = np.log2(raw)

    center_scale = math.log2(N) - f_0
    draw = raw-center_scale
    frequency_component = np.exp(-0.5 * (draw/ scale_bandwidth)**2)

    # reset origin to zero (not needed as it is already 0?)
    frequency_component[mid,mid] = 0

    return frequency_component * orientation_component