Lecture 1

Question 1

Digital Image

Answer 1

• N-dimensional matrix of intensity values
• Each cell of the matrix is a pixel or voxel
  o Voxel is 3D equivalent of a pixel
• Conversion from real world into digital images by:
  o Sampling: digitising the coordinate values
  ▪ How we divide the image into pixels
  ▪ Result of limited special resolution
  o Quantisation: digitising the amplitude values
  ▪ How we measure colour intensity (for example, blackness/whiteness of an
  image)

Question 2

Sampling

Answer 2

• Ideal sampling s(x,y) in a regular grid of MxN locations can be represented using a collection
  of Dirac functions
  o s(x,y) = ∑ ∑ 𝛿(𝑥 − 𝑗∆𝑥, 𝑦 − 𝑘∆𝑦)
  𝑁
  𝑘=1
  𝑀
  𝑗=1
  ▪ Δx := horizontal spacing
  ▪ Δy := vertical spacing
  o For each point in the grid, we can
  compute the Dirac function
  ▪ Translates to amplitude
  value
  ▪ Grid defines spacing
  between pixels
  ▪ Pixel coverage defines what
  area of the image has the
  same Dirac function value
• Larger pixel
  coverage -> larger
  pixel coverage -> lower resolution
  ▪ Parameters of grid are stored in image headers
• (e.g. DICOM format)

Question 3

Quantisation

Answer 3

• Limited intensity values -> require quantisation
  o Often map intensity value to grayscale
  values between 0 and 255
• Often measured in Hounsfield Units (HU)
  o Range of [-1000,1000]
  ▪ Increases with density (e.g.
  bone is 1000)

Question 4

Intensity Transformations

Answer 4

• Transformation T() of the original pixel value p from scale [p0,pk] into brightness from a new
  scale [q0,qk] given by:
  o 𝑞 = 𝑇(𝑝)
  o This depends on the pixel value, not on the position of the pixel
• Log transformation boosts low grey-level values
• Negative function = invert function

Question 5

Histogram matching

Answer 5

• Histograms provide frequency of frequency of
  the pixel value p in an image
  o Normalised: divide frequencies by the
  number of pixels/voxels in the image
• Histogram matching: transform intensity of input
  image to match the histogram of target image
  o Compute cumulative distribution
  function of histogram of both images

Question 6

Linear Contrast Stretching

Answer 6

• Add more contrast to image by stretching
  intensity frequency distribution
  o Rescale range [p0,pk] to [0,1]
  o Then rescale the result to the desired range [q0,qk]
  o q = q0 + (qk – q0) [(p-p0) / (pk – p0)]
• Used to reveal structures that are different but have very
  similar intensity values

Question 7

Spatial Filtering

Answer 7

• Operations that work with the values of the image pixels in a
  neighbourhood and corresponding values of a sub-image that has the same
  dimensions as the neighbourhood
  o Called: filter, mask, kernel, template, window
  o Values in a kernel are referred to as coefficients, no longer as pixels
• Filters should have an uneven width and height such that we have a centre pixel
• For example, calculating the correlation at each position
  o Pass input filter over input
  o Multiply input value by filter value, sum all multiplications
  o Move filter
• Convolutions are the same as first rotating the filter by 180 degrees and then using
  correlation
• Problems may occur at the border of an image: some parts
  of the filter may cover a part outside of the image
  o Valid convolution: only allow the filter to go over
  parts of the image where the entire filter covers
  part of the image
  ▪ Reduces size of the output
  o Same convolution: add padding around the image
  (typically 0), and perform filter operation
  ▪ Keeps output size the same if one
  row/column is added
  o Cycling padding: Copy rows/columns that are valid,
  and place them around the original image
  ▪ Helps to learn features without bias towards
  the centre of the image
• Padding does not add new
  information or artifacts to the image

Question 8

Types of filters

Answer 8

• Image smoothing -> blurring and noise reduction
  o Based on averaging brightness value in some
  neighbourhood -> i.e. multiply with filter that has
  coefficients of 1
  o Should be normalised again
• Edge detection
  o Kernel should have a sign change -> measures difference between
  adjacent pixels
  o If the kernel looks like an edge, it responds on edges
• Image derivative
  o Measure change of intensity in x or y direction
  o To find structures along a certain direction
  o Prewitt gradient kernel: derivative in one direction,
  smoothing in perpendicular direction
• Blob detection
  o Using the “Mexican hat” (Laplacian of Gaussian) filter
  o Region of a digital image in which all the points can be considered
  to be similar to each other, and different from surroundings (e.g.
  nodules)
  o Laplacian of Gaussian

Question 9

Gaussian Filters

Answer 9

• Gaussian operator in 1D:
  o Mu: mean value, controls position of the bell
  o Sigma: standard deviation, controls width of the bell
• Gaussian operator in 2D:
  o Circular filter
  o Now no longer has a Mu, only has sigma as a parameter
  o Typically a normal distribution in two dimensions
  o Convolving this function with an image blurs the function
  ▪ Changes in sigma influence how blurry the image becomes
• When you have to define the kernel size for the Gaussian function:
  o Rule of thumb: Gaussian function goes to zero after approximately
  3σ
• Derivative of Gaussian filters:
  o Linearity of convolution theorem states that convolutions are
  associative
  o Computing the derivative of an image is thus done by convolving with the derivate of
  a Gaussien -> introduces scale dependency in the computation of derivative
• Also used for edge detection via first order derivative (gradient direction)
  o Edge at maximum of the first-order derivative (local maxima of gradient magnitude)
  o Edge at zero-crossing of the second-order derivative

Question 10

Binary Morphology

Answer 10

• Structuring element (SE): point set that is used as image probe
  o Own local origin
• Morphological transformation: records locations where certain relations
  between image and SE are satisfied
• Dilation: combines two sets by vector addition
  o First two images on the right.
  ▪ Top one where origin is member of SE
  ▪ Bottom one where origin is not member of SE
  o Cell with X in it is the origin
  o Apply filter B, if the left cell contains a black square, add one to the
  right if it is not there yet
  o Used to fill in dotted lines
• Erosion: combines two sets by vector subtraction
  o Removes noise while keeping structures intact
• Opening: erosion followed by dilation
  o Smooth contours
  o Cut narrow bridges
  o Remove small islands and sharp corners
• Closing: dilation followed by erosion
  o Fill narrow channels
  o Fill small holes

Question 11

Segmentation

Answer 11

• Dividing image into segments
• Thresholding: grey value remapping operation that results in a binary
  image
  o Divides image into two segments
  o Usually to identify a single object (segment 1) and background (segment 0)
  o Threshold is tuneable parameter
  o Can use multiple thresholds to identify multiple objects that can be distinguished by
  their grey values
• Automatic threshold: objects have approximately the same grey level that differs from the
  grey level of the background

Question 12

Otsu algorithm

Answer 12

• Given a threshold, we can define two classes
  o Pixels belonging to the background
  o Pixels belonging to the foreground
• For each class, we can compute:
  o Mean value μ
  o Variance σ
  o Weight ω = #nr pixels per class / #total pixels
• Finds optimal threshold such that:
  o Intra-class variance is minimised
  o Inter-class variance is maximised