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2D Convolution Equation

For an image of size M x N and kernel of size k, a direct implementation would require O(MNk^2) time, which is infeasiblefor a real-time implementation. Instead, it is possible to exploit the separable property of certain convolution matricesh, e.g. that h can be represented as the product of two onedimensional matrices h1 and h2. This effectively reduces ourtwo-dimensional convolution of Equation (1) to two separate instances of one-dimensional convolution:

Separable Convolution

which is implemented as

Separable Convolution Equation

Execution times of direct and separable 2D Convolution

For the same M x N input and square kernel of size k, a separable implementation reduces the computational complexityto O(MNk). By comparison, an implementation using the FFT would cost O(MN logMN). In practice, considerthe separable convolution speed gain evident in the following results for a convolution of a 3 x 3 Gaussian filter kernel withimages of varying sizes (visualized in Figure 2).

Execution Time of Direct 2D Convolution

Execution Time of Separable 2D Convolution

As detailed in later sections, we take advantage of the fast computational complexity of separable convolution in ourimplementation, as all of the filters we apply are separable into one-dimensional matrices.

Noise elimination: two-dimensional gaussian filtering

The first task in our edge detection algorithm is denoising the input in order to reduce the amount of high-frequencycontent in the image by as much as possible without destroying critical information points in the image (e.g. the real edges).We filter out high-frequency noise so that random noise is not mistakenly interpreted as an edge, as edges correspond topoints in the image where the gradient has an above-threshold magnitude.

For example, consider the 5312 x 2988 pixel image (taken at a high ISO to deliberately introduce noise) of Figure 3and a plot of its grayscale intensity versus the image’s spatial dimensions in Figure 4. It is clear from the mesh that thereis much high-frequency noise, which can be removed with a low-pass filter.

Fig. 3. unfiltered noisy image

noisy image

Fig. 4 grayscale intensity of unfiltered noisy image at each pixel

noisy image intensity plot

Fig. 5. fft magnitude of unfiltered noisy image (log scale)

FFT magnitude of unfiltered noisy image (log scale)

To low-pass filter our image, we apply a discrete Gaussian filter. Generally, a Gaussian blur kernel of size 2n+1 x 2n+1(where n is a positive integer, and with parameter sigma) is given by

equation for Gaussian kernel

Our implementation of Gaussian filtering uses the constantsized (k = 3), constant-sigma kernel

Gaussian kernel

This kernel is implemented separably as

Separable Gaussian Kernel

Applying a Gaussian filter with parameters k = 5 and sigma = 5 to the above image significantly denoises the image withoutsacrificing edge precision, which can be seen from the spatial intensity plot in Figure 7.

Fig. 6. gaussian-filtered noisy image

Gaussian-filtered noisy image

Fig. 7. grayscale intensity of gaussian-filtered noisy image

Grayscale intensity of Gaussian-filtered noisy image

Fig. 8. fft magnitude of gaussian-filtered noisy image

FFT magnitude of Gaussian-filtered noisy image

Gradient computation: the sobel operator

Edges in the image correspond to pixel locations at which there is a rapid change in intensity with respect to the image’sspatial dimensions. Thus, edge pixels are defined as those whose gradient magnitude |G| is maximized along thegradient direction ThetaG.

Questions & Answers

given eccentricity and a point find the equiation
Moses Reply
12, 17, 22.... 25th term
Alexandra Reply
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Shirleen Reply
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
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salma
Commplementary angles
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
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Tamia
hii
Uday
hi
salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
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Source:  OpenStax, Elec 301 projects fall 2015. OpenStax CNX. Jan 04, 2016 Download for free at https://legacy.cnx.org/content/col11950/1.1
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