Lecture 4 - Filters Flashcards

Question

What are souces of Noise?

Answer 1

– Light Variations – Camera Electronics – Surface Reflectance – Lens

Answer 2

– Additive noise: 𝑓 ̂(i,j) = f(i,j) + n(i,j) – Multiplicative noise: 𝑓 ̂(i,j) = f(i,j) * n(i,j)

Answer 3

Salt and Pepper Gaussian Noise

Answer 4

Contains random occurrences of black and white pixels

Answer 5

Variations in intensity drawn from a Gaussian normal distribution

Answer 6

- Filtering: Use filters to remove noise - Filters (masks) operate on a neighborhood of pixels.

Answer 7

- Min: Set the pixel value to the minimum in the neighborhood - Max: Set the pixel value to the maximum in the neighborhood - Average: Set the pixel value to the average of the neighborhoods - Median: The median value of a set of numbers is the midpoint value in that set (e.g. from the set [1, 7, 15, 18, 24] 15 is the median). Sometimes the median works better than the average

Answer 8

- Remove noise - Sharpen contrast - Highlight contours - Detect edges Image filters - linear or nonlinear. - Linear filters are also know as convolution filters - Nonlinear operations such as median filter

Answer 9

- Linearity: filter(f1 + f2) = filter(f1) + filter(f2) - Shift invariance: same behavior regardless of pixel location. - The value of the output depends on the pattern in the image neighborhood, not the position of the neighborhood - filter(shift(f)) = shift(filter(f)) - Any linear, shift-invariant operator can be represented as a convolution - Commutative: a * b = b * a - Associative: a * (b * c) = (a * b) * c - Often apply several filters one after another: (((a * b1) * b2) * b3) - This is equivalent to applying one filter: a * (b1 * b2 * b3) - Distributes over addition: a * (b + c) = (a * b) + (a * c) - Scalars factor out: ka * b = a * kb = k (a * b) - Identity: unit impulse e = [0, 0, 1, 0, 0], a * e = a - Differentiation

Answer 10

One of the simplest spatial filtering operations we can perform is a smoothing operation (remove sharp features) - Simply average all of the pixels in a neighborhood around a central value - Especially useful in removing noise from images - Also useful for highlighting gross detail REFER TO SLIDES FOR EXAMPLE

Answer 11

More effective smoothing filters can be generated by assigning different pixels in the neighborhood different weights in the averaging function - Pixels closer to the central pixel are more important - Reduce smoothing effect - Often referred to as a weighted averaging REFER TO SLIDES FOR EXAMPLES

Answer 12

Weights center pixels more heavily, effective for smoothing while retaining edge detail REFER TO SLIDES WITH FORMULA

Answer 13

- A Median Filter operates over a window by selecting the median intensity in the window. - No new pixel values introduced - Removes spikes: good for impulse, salt & pepper noise - It is not as efficient at averaging away regular Gaussian noise REFER TO SLIDES FOR EXAMPLES

Answer 14

REFER TO SLIDES FOR EXAMPLES

Answer 15

Linear filters Remove “high-frequency” components from the image (low-pass filter) – Images become more smooth ... Convolution of a Gaussian with a Gaussian is another Gaussian – So can smooth with small-width kernel, repeat, and get same result as larger-width kernel would have – Convolving two times with Gaussian kernel of width 𝜎 is same as convolving once with kernel of width 𝜎sqrt(2) ... Separable kernels – Factors into product of two 1D Gaussians

Answer 16

REFER TO SLIDES Low-pass filtering -> convolve the image with a box / Gaussian filter High-pass filtering -> result equals 0, given a box filter/average filter

Answer 17

Here we take the original image and boost the high frequency components. Can think of HighPass = Original – LowPass. Thus HighBoost = b*Original – LowPass = (b-1)*Original + Original – LowPass = (b-1)*Original + HighPass b is the boosting factor. When b=1, HighBoost = HighPass

Answer 18

* Reduces image size by discarding rows/columns. * Compresses image and improves efficiency.

Answer 19

* Enables object detection at different scales. * Used in hierarchical object detection methods.

Answer 20

* Sampling frequency must be at least twice the highest frequency in the image to avoid aliasing. * Violating this results in loss of original information.

Answer 21

* Compares a template image to the input by sliding over it. * Detects locations with highest similarity.

Answer 22

* SSD (Sum of Squared Differences): ○ Lower values = better match * Normalized Correlation (NC): ○ Correlates intensity values, accounts for brightness variations * Normalized Cross-Correlation (NCC): ○ Normalized by mean and variance; robust to intensity scale

Lecture 4 - Filters Flashcards

(46 cards)