Computer Vision : CISC 4/689

Computer Vision : CISC 4/689
Linear Filters General process: Form new image whose pixels are a weighted sum of original pixel values, using the same set of weights at each point. Properties Output is a linear function of the input Output is a shift-invariant function of the input (i.e. shift the input image two pixels to the left, the output is shifted two pixels to the left) Example: smoothing by averaging form the average of pixels in a neighbourhood Example: smoothing with a Gaussian form a weighted average of pixels in a neighbourhood Example: finding a derivative Computer Vision : CISC 4/689

Convolution Represent these weights as an image, H H is usually called the kernel Operation is called convolution it’s associative Result is: Computer Vision : CISC 4/689

Example: Smoothing by Averaging
Here is the point to introduce some visual “notation”. I’ve given the kernel as an image on the top. Usually, these images are white for the largest value, and black for the smallest (which is zero in this case). The point of this pair of images is that if you convolve the one on the left with the kernel shown above you get the one on the right, which is nothing like a smoothed version of the one on the left (there’s some fairly aggressive ringing). Explain the ringing qualitatively in terms of the boundaries of the kernel, and point out that the effect would disappear if there weren’t sharp edges in the kernel. Computer Vision : CISC 4/689

Smoothing with a Gaussian
Smoothing with an average actually doesn’t compare at all well with a defocussed lens Most obvious difference is that a single point of light viewed in a defocussed lens looks like a fuzzy blob; but the averaging process would give a little square. I always walk through the argument on the left rather carefully; it gives some insight into the significance of impulse responses or point spread functions. A Gaussian gives a good model of a fuzzy blob Computer Vision : CISC 4/689

An Isotropic Gaussian The picture shows a smoothing kernel proportional to (which is a reasonable model of a circularly symmetric fuzzy blob) Computer Vision : CISC 4/689

Smoothing with a Gaussian
You want to point out the absence of ringing effects here. Computer Vision : CISC 4/689

Problem: Image Noise from Forsyth & Ponce Computer Vision : CISC 4/689

Solution: Smoothing (Low-Pass) Filters
If object reflectance changes slowly and noise at each pixel is independent, then we want to replace each pixel with something like the average of neighbors Disadvantage: Sharp (high-frequency) features lost Original image 7 x 7 averaging neighborhood Computer Vision : CISC 4/689

Smoothing Filters: Details
Filter types Mean filter (box) Median (nonlinear) Gaussian Can specify linear operation by shifting kernel over image and taking product 1 3 x 3 box filter kernel Computer Vision : CISC 4/689

Gaussian Kernel Idea: Weight contributions of neighboring pixels by nearness Smooth roll-off reduces “ringing” seen in box filter 5 x 5,  = 1 Computer Vision : CISC 4/689

Gaussian Smoothing In theory, the Gaussian distribution is non-zero everywhere, which would require an infinitely large convolution mask, but in practice it is effectively zero more than about three standard deviations from the mean, and so we can truncate the mask at this point Discrete approximation to Gaussian function with sd = 1.4 Source: Computer Vision : CISC 4/689

Gaussian Smoothing Once a suitable mask has been calculated, then the Gaussian smoothing can be performed using standard convolution methods. The convolution can in fact be performed fairly quickly since the equation for the 2-D isotropic Gaussian shown above is separable into x and y components. Thus the 2-D convolution can be performed by first convolving with a 1-D Gaussian in the x direction, and then convolving with another 1-D Gaussian in the y direction. (The Gaussian is in fact the only completely circularly symmetric operator which can be decomposed in such a way.) Below shows the 1-D x component mask that would be used to produce the full mask shown in previous slide. The y component is exactly the same but is oriented vertically. A further way to compute a Gaussian smoothing with a large standard deviation is to convolve an image several times with a smaller Gaussian. While this is computationally complex, it can have applicability if the processing is carried out using a hardware pipeline. Computer Vision : CISC 4/689

Gaussian Smoothing One of the principle justifications for using the Gaussian as a smoothing filter is due to its frequency response. Most convolution based smoothing filters act as lowpass frequency filters. This means that their effect is to remove low spatial frequency components from an image. The frequency response of a convolution filter, i.e. its effect on different spatial frequencies, can be seen by taking the Fourier transform of the filter. Frequency responses of Box (i.e. mean) filter (width 7 pixels) and Gaussian filter ( = 3 pixels). Computer Vision : CISC 4/689

Example: Gaussian Smoothing
Original image Box filter 7 x 7 kernel  = 1 Computer Vision : CISC 4/689  = 3

Convolution Formalization of idea of overlap (product) of two functions as one is shifted across the other The area under convolution is product of areas under the functions The convolution of two Gaussians is another Gaussian The animations above graphically illustrate the convolution of two rectangle functions (left) and two Gaussians (right). In the plots, the green curve shows the convolution of the blue and red curves as a function of t , the position indicated by the vertical green line. The gray region indicates the product as a function of t , so its area is precisely the convolution. courtesy of mathworld.wolfram.com Computer Vision : CISC 4/689

Convolution Notes Note the assumption that both f and g are continuous and defined everywhere Properties Commutative Associative Computer Vision : CISC 4/689

Discrete 2-D Convolution
Sum instead of integral: I’ = K ¤ I (book uses ¤¤ for 2-D convolution) is defined by: I’(u, v) = §x, y K(u ¡ x, v ¡ y) I(x, y) where the kernel K approximates the impulse function f by sampling from it where it is non-zero Computer Vision : CISC 4/689

Correlation Same as convolution, a dot product, thus it is largest when the pattern maches (vectors are parallel), this can be used to find texture patterns This yields a value that is +ve when the image region looks like the filter kernel, and small and –ve when it is opposite. Can be squared if pattern reversal doesn’t matter. Since value maybe large if image is bright, so divide by root sum of squares of image region and filter. (dot of unit vectors) Some ways to interpret what the kernel is doing As a template being matched by correlation As simply a set of weights on the corresponding image pixels Computer Vision : CISC 4/689

Normalized correlation
Think of filters of a dot product now measure the angle i.e normalised correlation output is filter output, divided by root sum of squares of values over which filter lies Tricks: ensure that filter has a zero response to a constant region (helps reduce response to irrelevant background) subtract image average when computing the normalizing constant (i.e. subtract the image mean in the neighbourhood) absolute value deals with contrast reversal Computer Vision : CISC 4/689

Positive responses Zero mean image, -1:1 scale Zero mean image, -max:max scale The filter is the little block in the top left hand corner. Notice this is a fair spot-detector. Computer Vision : CISC 4/689

Positive responses Zero mean image, -1:1 scale Zero mean image, -max:max scale The filter is the little block in the top left hand corner. Notice this is a fair bar-detector. Computer Vision : CISC 4/689

This is a figure from the book, figure Read the caption for the story! Figure from “Computer Vision for Interactive Computer Graphics,” W.Freeman et al, IEEE Computer Graphics and Applications, 1998 copyright 1998, IEEE Computer Vision : CISC 4/689

Dealing with Image Edges
Only convolve with interior Shrinks image Zero-padding Results in spurious gradients Border replication Symmetric: Reflect image at border b so that I(b + i) = I(b ¡ i) Results in spurious 2nd-derivatives Computer Vision : CISC 4/689

1 1 1 2 2 2 3 Step 1 -1 2 1 2 1 3 3 -1 -1 1 2 2 1 2 1 3 2 2 1 1 1 -1 3 2 1 4 2 5 -1 -2 1 Computer Vision : CISC 4/689

1 1 1 2 2 2 3 Step 2 -1 2 1 2 1 3 3 -1 -1 1 2 2 1 2 1 3 2 2 1 1 1 3 2 1 -2 4 2 5 4 -2 -1 3 Computer Vision : CISC 4/689

1 1 1 2 2 2 3 Step 3 -1 2 1 2 1 3 3 -1 -1 1 2 2 1 2 1 3 2 2 1 1 1 3 2 1 -2 4 3 4 5 -1 -3 3 Computer Vision : CISC 4/689

1 1 1 2 2 2 3 Step 4 -1 2 1 2 1 3 3 -1 -1 1 2 2 1 2 1 3 2 2 1 1 1 3 2 1 -2 6 1 4 -2 5 -3 -3 1 Computer Vision : CISC 4/689

1 1 1 2 2 2 3 Step 5 -1 2 1 2 1 3 3 -1 -1 1 2 2 1 2 1 3 2 2 1 3 2 1 2 2 4 9 -2 5 -1 4 1 -1 -2 2 Computer Vision : CISC 4/689

1 1 1 2 2 2 3 Step 6 -1 2 1 2 1 3 3 -1 -1 1 2 2 1 2 1 3 2 2 2 3 2 1 2 2 6 4 9 -2 5 -2 2 3 -2 -2 1 Computer Vision : CISC 4/689

and so on… Computer Vision : CISC 4/689

Final Result I I’ Why is I’ large in some places and small in others?
1 1 1 Final Result -1 2 1 -1 -1 1 2 3 1 12 7 6 4 8 14 5 9 11 -2 I I’ Why is I’ large in some places and small in others? Computer Vision : CISC 4/689

Linear Shift Invariance
Possible properties of f Superposition: f(I1 + I2) = f(I1) + f(I2) Scaling: f(®I) = ®f(I) Shift invariance: f(Shift(I, k)) = Shift (f(I), k) A system with these properties is performing convolution Computer Vision : CISC 4/689

Imaging Systems An imaging system describes a functional transformation f of an image due to… Physics: A real-world phenomenon such as blurring from defocus or fish-eye lens distortion Filtering: A transformation we apply in order to Undo or mitigate the bad effects of a physical system (e.g., deblur, undistort, etc.) Emphasize or highlight particular image properties (e.g., color similarity, edges, etc.) I I’ f Computer Vision : CISC 4/689

What’s Not a Convolution?
Nonlinear systems E.g., radial distortion of fish-eye lens is not LSI because geometric transformation depends on pixel location f courtesy of M. Fiala Computer Vision : CISC 4/689

Filtering in Matlab imfilter(I, K)filters image I with kernel K Default filtering is correlation (no kernel rotation) Can set options on border handling corr2, conv2 are the generic versions Kernel creation Custom (create a matrix) fspecial function Computer Vision : CISC 4/689

Differentiation and convolution
Recall Now this is linear and shift invariant, so must be the result of a convolution. We could approximate this as (which is obviously a convolution; it’s not a very good way to do things, as we shall see) I tend not to prove “Now this is linear and shift invariant, so must be the result of a convolution” but leave it for people to look up in the chapter. Computer Vision : CISC 4/689

Finite differences Partial derivative in x-direction: gives vertical stripes. mid-gray = 0, dark gray = -ve, light gray = +ve Kernel = (1,0,-1) Now the figure on the right is signed; darker is negative, lighter is positive, and mid grey is zero. I always ask 1) which derivative (y or x) is this? 2) Have I got the sign of the kernel right? (i.e. is it d/dx or -d/dx). Computer Vision : CISC 4/689

Noise Simplest noise model independent stationary additive Gaussian noise the noise value at each pixel is given by an independent draw from the same normal probability distribution Issues this model allows noise values that could be greater than maximum camera output or less than zero for small standard deviations, this isn’t too much of a problem - it’s a fairly good model independence may not be justified (e.g. damage to lens) may not be stationary (e.g. thermal gradients in the ccd) Computer Vision : CISC 4/689

sigma=1 Computer Vision : CISC 4/689

sigma=16 Computer Vision : CISC 4/689

Finite differences and noise
Finite difference filters respond strongly to noise obvious reason: image noise results in pixels that look very different from their neighbours Generally, the larger the noise the stronger the response What is to be done? intuitively, most pixels in images look quite a lot like their neighbours this is true even at an edge; along the edge they’re similar, across the edge they’re not suggests that smoothing the image should help, by forcing pixels different to their neighbours (=noise pixels?) to look more like neighbours Computer Vision : CISC 4/689

Finite differences responding to noise
Increasing noise -> (this is zero mean additive gaussian noise) Computer Vision : CISC 4/689

The response of a linear filter to noise
Do only stationary independent additive Gaussian noise with zero mean (non-zero mean is easily dealt with) Mean: output is a weighted sum of inputs so we want mean of a weighted sum of zero mean normal random variables must be zero Remember: Gaussian is symmetric, product of 2 Gaussians is Gaussian Variance: recall variance of a sum of random variables is sum of their variances variance of constant times random variable is constant^2 times variance then if s is noise variance and kernel is K, variance of response is Computer Vision : CISC 4/689

Filter responses (of noise) are correlated
over scales similar to the scale of the filter Filtered noise is sometimes useful looks like some natural textures, can be used to simulate fire, etc. Computer Vision : CISC 4/689

Smoothing reduces noise
Generally expect pixels to “be like” their neighbours surfaces turn slowly relatively few reflectance changes Generally expect noise processes to be independent from pixel to pixel Implies that smoothing suppresses noise, for appropriate noise models Scale the parameter in the symmetric Gaussian as this parameter goes up, more pixels are involved in the average and the image gets more blurred and noise is more effectively suppressed Computer Vision : CISC 4/689

The effects of smoothing Each row shows smoothing with gaussians of different width; each column shows different realisations of an image of gaussian noise. There are two notions of gaussian here, related only by a vague analogy. I did this slide like this to put them clearly in apposition, because this point always confused me. There’s no particular reason that gaussian noise should attract smoothing with a gaussian kernel, just an irritating coincidence. Computer Vision : CISC 4/689

Gradients and edges Points of sharp change in an image are interesting: change in reflectance change in object change in illumination noise Sometimes called edge points General strategy determine image gradient now mark points where gradient magnitude is particularly large wrt neighbours (ideally, curves of such points). Computer Vision : CISC 4/689

The Gradient and Edges Consider image intensities as a 2-D height function I(x, y). Then the image gradient is the vector field defined by: Definition of an edge Line segment separating regions of contrasting intensity Location: Where gradient magnitude is high Direction: Orthogonal to the gradient Computer Vision : CISC 4/689

Edge Causes Depth discontinuity Surface orientation discontinuity Reflectance discontinuity (i.e., change in surface material properties) Illumination discontinuity (e.g., shadow) Computer Vision : CISC 4/689

Edge Detection An edge point can be regarded as a point in an image where a discontinuity (in gradient) occurs across some line. A discontinuity may be classified as one of five types Searching for Edges: Filter: Smooth image Enhance: Apply numerical derivative approximation Detect: Threshold to find strong edges Localize/analyze: Reject spurious edges, include weak but justified edges Gradient Discontinuity -- where the gradient of the pixel values changes across a line. This type of discontinuity can be classed as roof edges, ramp edges convex edges concave edges, by noting the sign of the component of the gradient perpendicular to the edge on either side of the edge. Ramp edges have the same signs in the gradient components on either side of the discontinuity, while roof edges have opposite signs in the gradient components. A Jump or Step Discontinuity -- where pixel values themselves change suddenly across some line. A Bar Discontinuity -- where pixel values rapidly increase then decrease again (or vice versa) across some line. Source: LOCAL_COPIES/MARSHALL/node28.html Computer Vision : CISC 4/689

Step edge detection: First Derivative Operators
Method: Differentiate and find extrema Examples Sobel operator (Matlab: edge(I, ‘sobel’)) Prewitt, Roberts cross Derivative of Gaussian Book uses this format -1 1 -2 2 -1 -2 1 2 Sobel x Sobel y Computer Vision : CISC 4/689

Sobel Edge Filtering Example
1 -1 2 -2 2 Rotate 1 -1 2 -2 Computer Vision : CISC 4/689

-1 1 2 2 Step 1 -2 2 2 2 -1 1 2 2 2 2 -1 1 -2 1 2 -1 Computer Vision : CISC 4/689

-1 1 2 2 Step 2 -2 2 2 2 -1 1 2 2 2 2 -1 1 1 2 3 4 6 2 Computer Vision : CISC 4/689

-1 1 2 2 Step 3 -2 2 2 2 -1 1 2 2 2 2 -1 1 1 2 3 4 6 2 Computer Vision : CISC 4/689

-1 1 2 2 Step 4 -2 2 2 2 -1 1 2 2 2 2 -1 1 2 3 -4 2 6 -6 -2 1 edge effect from zero- padding Computer Vision : CISC 4/689

Sobel Edge Filtering Example: Result
6 8 -8 -6 (pad with zeroes again, the boundary) and then we threshold… Computer Vision : CISC 4/689

Sobel Edge Detection: Gradient Approximation
Note anisotropy of edge finding -1 1 -2 2 -1 -2 1 2 Horizontal Vertical Computer Vision : CISC 4/689

Sobel These can then be combined together to find the absolute magnitude of the gradient at each point and the orientation of that gradient. The gradient magnitude is given by: an approximate magnitude is computed using: which is much faster to compute. The angle of orientation of the edge (relative to the pixel grid) giving rise to the spatial gradient is given by: In this case, orientation 0 is taken to mean that the direction of maximum contrast from black to white runs from left to right on the image, and other angles are measured anti-clockwise from this. Computer Vision : CISC 4/689

Derivative of Gaussian
Computer Vision : CISC 4/689

Computer Vision : CISC 4/689

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