- 92 Views
- Uploaded on
- Presentation posted in: General

CMPB 345: IMAGE PROCESSING

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

CMPB 345: IMAGE PROCESSING

DISCRETE TRANSFORM 2

- We may want to modify the resulting spectrum after an image has been transformed into the frequency/sequency domain.
- High frequency information can be removed with a lowpass filter (effect of blurring an image).
- Low frequency information can be removed with a highpass filter (tend to sharpen an image).
- Bandpass filtering is to extract the frequency information in specific parts of the spectrum.
- Bandreject filtering is to eliminate specific parts of the spectrum (e.g. to remove unwanted noise).
- After we have transform the image into frequency or sequency domain, we may want to modify the resulting spectrum before apply any filtering to the image.
- 2 types of spectral symmetries result from taking the continuous functions & sampling them to develop the discrete transform:

- WE repeat the discrete NxN spectrum in all directions to infinity.
- The frequency in the spectral components only increasing up to N/2 and then decreases due to periodicity and conjugate symmetry properties of the Fourier transform.
- The origin of the spectrum is often shifted to the center for display & filtering purposes. This shifting is done by using a property of the Fourier transform called the frequency translation property by multiply the original image I(r,c) by (-1)r+c before we perform the transformation.
- Human Visual System give more responds to brightness. So we need to log-remapped the original image which has been shifted to the center by using this formula.:-
- log(u,v) = k log [ 1+|F(u,v)| ]

- The advantage of this formula is:
(a) It will compress the data into lower range

(b) It will remap the dynamic range of the Fourier spectrum which is greater than 256 gray levels (8 bits) into the 0-255 range, k will control the range.

- Example: (k=4)
300 -400= 4 log [ 1 + | -300 | ]

250 320= 4 log [ 1 + 300 ]

F(u,v)= 4 log 301

= 4(2.4786)

= 9.91

- The spectrum is folded about the origin, creating a 2Nx2N pattern.
- The frequency increasesfrom the origin outward & thus we do not have to shift it to the center (half of the frequency information will lost if we do so).
- The Walsh-Hadamard Transform also exhibit increasing sequency from the origin outward.

FILTERING METHODS

- Tend to visually blur images, considered an enhancement because it imparts a softer effect to the image.
- Pass low frequencies & attenuate/eliminate the high frequency information.
- Used for image compression or for hiding effects caused by noise.
- Using the below equation:
I(r,c) = T-1 [ T(u,v) H(u,v) ]

I (r,c) = The filtering image

T(u,v) = The transform

H(u,v) = The filter function

- It is called ideal because the transition from the passband to the stopband in the filter is perfect (it goes from 0 to 1 instantly). H(u,v) only have the value 1’s and 0’s.
- The frequency at which we start to eliminate the information is called the cutoff frequency.
- The frequencies in the spectrum that are not filtered out are in the passband while the spectral components that do get filtered out are in the stopband.
- However, ideal filter leaves undesirable artifacts in images.
- This artifact appears in the lowpass filtered image as ripples/waves, wherever there is a boundary in the image.
- This problem can be avoided by using a “nonideal filter” which does not have perfect transition.

- To overcome the problems in ideal filter.
- Makes the image become smoother.
- One of the nonideal types is called the Butterworth Filter.
- With this, the order of the filter can be specified which determines how steep the slope is in the transition of the filter function.
- Higher order to the filter create steeper slope & the closer we get to an ideal filter.

- Passes only high frequency information corresponding to places where gray levels are changing rapidly.
- Used for edge enhancement.
- The function for a special type of highpass filter, called a high-frequency emphasis filter, is to retain some of the low frequency information by adding an offset value to the function, so we do not lose the overall image info.
- The contrast will be added back to the image after using the high-frequency emphasis filter.
- Greater offset will give a sharper image.

- Bandpass & Bandreject are specified by two cutoff frequencies, low cutoff & high cut off frequencies.
- These filters can be modified into nonideal filters by making the transitions gradual (steeper slope) at the cutoff frequencies.
- A special form of these filters is called a notch filter because it only notches out or passes specific frequencies.
- These 3 types of filters are used in image restoration, enhancement & compression.
- The example of these 3 filters can be referred to page 124 in the text book.

WAVELET TRANSFORM

- The most common model for a wavelet transform uses the Fourier transform and highpass and lowpass filters.
- Wavelet Transform contain both spatial and frequency information.
- A transformation of an image which has its basis functions shifted and expanded
- The filters must be perfect resolution filters (distortion that exist would be canceled in the inverse transform) in order to satisfy the conditions of the wavelet transform.

- The Wavelet Transform is sufficient for most practical applications and for reconstruction of the signal.
- The Wavelet Transform provides enough information, and offers an enormous reduction in the computation time.
- The Wavelet Transform is considerably easier to implement when compared to the continuous wavelet transform.

- We can use Wavelet Transform for image compression when
we want to conserve time and space when handling large images.

- The Wavelet Transform breaks an image into four subsampled images.
- Subsampled by keeping every other pixel.
- Results in an image which has been highpass filtered both vertically and horizontally, one that has bees lowpass both vertically and horizontally, one which has bee highpass on the vertical side and lowpass on the horizontal side and one that has been lowpass on the vertical side and highpass on the horizontal side
- Transform implemented in the spatial domain using 1-D convolution filters.
- Circular convolution is a special type of convolution used in order to perform wavelet transformation

- Take the underlying image array and extend it in a periodic manner to match the symmetry implied by the discrete Fourier transform.
- Convolution starts with the origin of the image and the mask aligned so that the first value contains contributions from the “previous” copy of the extended image.
- Last value contains contributions from “next” copy of the extended, periodic image.

a. Extended, periodic image (x = origin)

One period

Center of filter

b. Extended, periodic 1-D convolution filter (x = origin)

Period

Of interest

Next

period

Previous

period

X

c. With circular convolution, the outer rows and columns include products of both previous and next periods.

Example:

Lowpass: 1/√2 [X1 X2]

Highpass: 1/√2 [X1 -X2]

- To use the Haar Basis vector to implement the wavelet transform, it must be zero padded to the same size as the subimage (in this case four).
- The origin of the basis vector is define in the center, corresponding to the value to the right of the center

- Form the example above if we want to implement the Haar basis vector we have to zero pads the vectors to have a length of four.
Lowpass: 1/√2 [X1 X2 0 0]

Highpass: 1/√2 [X1 -X2 0 0]

Origin

- Take note that after the vector has been zero-padded on the right the origin is no longer to the right of the center. The origin is determined before zero padding.

An example of Daubechies basis vectors :

LOWPASS : 1/4√2[ 1+√3, 3+√3, 3-√3, 1-√3]

HIGHPASS :1/4√2[ 1-√3, √3-3, 3+√3, -1-√3]

- The basis vector has to be zero padded to be the same size as the sub image.
- If an image is divided into 8 X 8 blocks :
LOWPASS : 1/4√2[ 1+√3, 3+√3, 3-√3, 1-√3, 0, 0, 0, 0]

HIGHPASS :1/4√2[ 1-√3, √3-3, 3+√3, -1-√3, 0, 0, 0, 0]

ORIGIN

- Because these are assumed circular convolution, we could zero pad equally on both ends.
LOWPASS : 1/4√2[0, 0, 1+√3, 3+√3, 3-√3, 1-√3, 0, 0]

HIGHPASS :1/4√2[0, 0, 1-√3, √3-3, 3+√3, -1-√3, 0, 0]

ORIGIN