SOME CV Question

Okay, here are five questions based on the “Standard Image Transforms in Image Processing.md” file you provided, covering both numerical calculations and conceptual understanding. Each question is worth 8 marks.

Question 8: General Image Transform and Separability

(a) (4 marks) Explain what the transform kernel, $K(x,y,u,v)$, represents in the general form of a 2D image transform. How does it relate the spatial domain to the transform domain? Give an example of a transform kernel and briefly describe what that transform does.

(b) (4 marks) What does it mean for an image transform to be “separable”? Explain the computational advantage of using a separable transform, showing the relevant equations for a forward transform to illustrate your point.

Solution 8:

(a) Transform Kernel Explanation:

• Representation: The transform kernel, $K(x,y,u,v)$, is a function that defines the specific image transformation being performed. It’s a mathematical rule that determines how each pixel in the original image, $f(x,y)$, contributes to each point in the transformed image, $T(u,v)$.
• Spatial to Transform Domain: The kernel acts as a “bridge” between the spatial domain (x, y) and the transform domain (u, v). For each (u, v) in the transform domain, the kernel is evaluated for every (x, y) in the spatial domain. The kernel’s value at a particular (x, y, u, v) determines the “weight” or contribution of the pixel $f(x,y)$ to the transform coefficient $T(u,v)$. Essentially, the transform projects the image onto a set of basis functions defined by the kernel.
• Example: The Discrete Fourier Transform (DFT) kernel is: $K(x,y,u,v)=e^{-j2\pi (\frac{ux}{M}+\frac{vy}{N})}$, where $j=\sqrt{-1}$. This transform decomposes the image into its constituent spatial frequencies. The kernel represents complex exponentials (sinusoids) with different frequencies (u, v).

(b) Separability and Computational Advantage:

• Separability: A 2D transform is separable if its kernel, $K(x,y,u,v)$, can be expressed as the product of two 1D kernels: $K(x,y,u,v)=K_1(x,u)K_2(y,v)$. This means the 2D transform can be broken down into two separate 1D transformations.

• Computational Advantage: Separability significantly reduces computational cost. Instead of performing a 2D convolution (which has a complexity of $O(M^2{N}^2)$ for an $M\times N$ image), we can perform two sets of 1D convolutions:

  1. Along Rows: First, we compute an intermediate result by applying $K_1(x,u)$ to each row of the image: $I(u,y)={\sum }_{x=0}^{M-1}f(x,y)K_1(x,u)$ This step has a complexity of $O(M{N}^2)$
  2. Along Columns: Then, we apply $K_2(y,v)$ to each column of the intermediate result $I(u,y)$: $T(u,v)={\sum }_{y=0}^{N-1}I(u,y)K_2(y,v)={\sum }_{y=0}^{N-1}{K}_2(y,v)\left[{\sum }_{x=0}^{M-1}f(x,y)K_1(x,u)\right]$ This step has a complexity of $O(M^{2}N)$

  The overall complexity becomes $O(M^{2}N)+O(M{N}^2)$, which, for a square image ($M=N$), simplifies to $O(2N^3)$ or simply $O(N^3)$. This is a significant reduction compared to $O(N^4)$ for a non-separable 2D transform. We perform $N$ 1D transforms of length $N$, twice (once for rows, once for columns).

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Question 9: DFT Calculation (Numerical)

(a) (5 marks) Given a 1D signal $f(x)=[1,2,3,4]$, where $x=[0,1,2,3]$, calculate the Discrete Fourier Transform (DFT), $F(u)$, for $u=[0,1,2,3]$. Show all steps, including the complex exponential calculations. You can leave the result in terms of complex numbers.

(b) (3 marks) What do the real and imaginary parts of the DFT coefficients represent? What does the magnitude, $|F(u)|$, represent?

Solution 9:

(a) 1D DFT Calculation:

The 1D DFT is given by:

$F(u)={\sum }_{x=0}^{N-1}f(x)e^{-j2\pi \frac{ux}{N}}$

Where $N=4$ in this case. $j=\sqrt{-1}$

• F(0): $F(0)={\sum }_{x=0}^{3}f(x)e^{-j2\pi \frac{0\cdot x}{4}}=1\cdot e^0+2\cdot e^0+3\cdot e^0+4\cdot e^0=1+2+3+4=10$

• F(1): $F(1)={\sum }_{x=0}^{3}f(x)e^{-j2\pi \frac{1\cdot x}{4}}=1\cdot e^0+2\cdot e^{-j\pi /2}+3\cdot e^{-j\pi }+4\cdot e^{-j3\pi /2}$ $F(1)=1+2(-j)+3(-1)+4(j)=1-2j-3+4j=-2+2j$

• F(2): $F(2)={\sum }_{x=0}^{3}f(x)e^{-j2\pi \frac{2\cdot x}{4}}=1\cdot e^0+2\cdot e^{-j\pi }+3\cdot e^{-j2\pi }+4\cdot e^{-j3\pi }$ $F(2)=1+2(-1)+3(1)+4(-1)=1-2+3-4=-2$

• F(3): $F(3)={\sum }_{x=0}^{3}f(x)e^{-j2\pi \frac{3\cdot x}{4}}=1\cdot e^0+2\cdot e^{-j3\pi /2}+3\cdot e^{-j3\pi }+4\cdot e^{-j9\pi /2}$ $F(3)=1+2(j)+3(-1)+4(-j)=1+2j-3-4j=-2-2j$

Therefore, $F(u)=[10,-2+2j,-2,-2-2j]$.

(b) Real, Imaginary, and Magnitude:

• Real Part: Represents the cosine component of the corresponding frequency. It indicates how much the signal aligns with a cosine wave of that frequency.
• Imaginary Part: Represents the sine component of the corresponding frequency. It indicates how much the signal aligns with a sine wave of that frequency.
• Magnitude, |F(u)|: Represents the amplitude or strength of that frequency component in the signal. It’s calculated as $|F(u)|=\sqrt{(\text{Real Part}{)}^2+(\text{Imaginary Part}{)}^2}$. It combines the sine and cosine contributions to give an overall measure of the presence of that frequency, regardless of phase.

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Question 10: DCT and Spatial Frequency

(a) (4 marks) Explain why the Discrete Cosine Transform (DCT) is particularly well-suited for image compression, as used in JPEG. Focus on the concept of “energy compaction.”

(b) (4 marks) In the context of the 2D Discrete Fourier Transform (DFT) of an image, what does a point $F(u,v)$ at the center of the transformed image (i.e., $u=0$, $v=0$) represent? What about points far from the center? How does this relate to the concept of spatial frequency?

Solution 10:

(a) DCT and Energy Compaction:

The DCT is well-suited for image compression due to its excellent energy compaction property. This means that for typical images, the DCT tends to concentrate a large portion of the total signal energy (information) into a small number of coefficients, specifically the low-frequency coefficients.

• Natural Images: Most natural images have smooth variations and relatively few sharp edges. This translates to a dominance of low-frequency components.
• DCT Basis Functions: The DCT basis functions are cosine waves of different frequencies. The DCT effectively decomposes the image into a weighted sum of these cosine waves.
• Concentration of Energy: Because natural images are dominated by low frequencies, the DCT coefficients corresponding to low frequencies (small u and v values) tend to have large magnitudes. The high-frequency coefficients (large u and v), representing fine details and noise, tend to have small magnitudes.
• Compression: This energy compaction allows for efficient compression. We can discard or coarsely quantize the high-frequency coefficients (which have low energy) with minimal impact on the visual quality of the reconstructed image. This significantly reduces the amount of data needed to represent the image. JPEG uses this by quantizing DCT coefficients, setting many high-frequency coefficients to zero.

(b) DFT Center and Spatial Frequency:

• Center (u=0, v=0): The point $F(0,0)$ at the center of the 2D DFT represents the DC component of the image. This is the average pixel intensity value of the entire image. It’s the zero-frequency component, representing a constant, uniform signal across the entire image. Mathematically, $F(0,0)={\sum }_{x=0}^{M-1}{\sum }_{y=0}^{N-1}f(x,y)$.

• Far from the Center: Points far from the center represent high spatial frequencies. The further away from the center a point $(u,v)$ is, the higher the spatial frequency it represents. These high frequencies correspond to rapid changes in pixel intensity, such as:

  • Sharp edges
  • Fine details
  • Texture
  • Noise

• Spatial Frequency Relationship: The DFT decomposes the image into a sum of complex sinusoids (basis functions). The coordinates (u, v) directly correspond to the frequencies of these sinusoids. The distance from the center, $\sqrt{u^2+v^2}$, represents the overall spatial frequency. The angle (relative to the u-axis) indicates the orientation of the sinusoidal pattern. For instance, points along the horizontal ‘u’ axis relate to vertical features and those on vertical ‘v’ axis represent horizontal features.

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Question 11: Inverse Transform and Basis Functions

(a) (4 marks) Write down the general discrete form of the inverse image transform. Explain how it reconstructs the original image from the transformed coefficients, $T(u,v)$. What role does the complex conjugate of the kernel play?

(b) (4 marks) Explain the concept of “basis functions” in the context of image transforms. How do the basis functions relate to the transform kernel? Give an example of basis functions for a specific transform.

Solution 11:

(a) Inverse Image Transform:

The general discrete form of the inverse image transform is:

$f(x,y)={\sum }_{u=0}^{M-1}{\sum }_{v=0}^{N-1}T(u,v)K^{\ast }(x,y,u,v)$

Where:

• $f(x,y)$: The reconstructed image in the spatial domain.
• $T(u,v)$: The transformed coefficients.
• $K^{\ast }(x,y,u,v)$: The complex conjugate of the transform kernel.

Reconstruction: The inverse transform reconstructs the original image by summing up the weighted contributions of all the transform coefficients. Each coefficient, $T(u,v)$, is multiplied by the complex conjugate of the kernel evaluated at $(x,y,u,v)$. This effectively “reverses” the projection performed by the forward transform.

Role of Complex Conjugate: The complex conjugate of the kernel is essential for ensuring that the inverse transform correctly reconstructs the original image. It’s particularly important for transforms involving complex numbers, like the DFT.

• In transforms with orthogonal/orthonormal basis function (such as DFT/DCT), inverse transform is equivalent to using complex conjugate of the kernel.
• In the case of the DFT, the forward transform uses $e^{-j2\pi (\frac{ux}{M}+\frac{vy}{N})}$, while the inverse transform uses $e^{j2\pi (\frac{ux}{M}+\frac{vy}{N})}$ (which is the complex conjugate). This ensures that the transform and its inverse are properly “matched” to decompose and reconstruct the signal accurately.

(b) Basis Functions:

• Concept: Basis functions are a set of fundamental “building blocks” that can be combined linearly (with different weights) to represent any image within the transform’s domain. They form a basis for the vector space of images.

• Relation to Kernel: The transform kernel, $K(x,y,u,v)$, implicitly defines the basis functions. For each specific pair of transform domain coordinates $(u,v)$, the kernel $K(x,y,u,v)$, when considered as a function of $x$ and $y$, represents a basis function. The forward transform projects the image onto these basis functions, and the inverse transform reconstructs the image as a linear combination of these basis functions, weighted by the transform coefficients.

• Example (DFT): For the Discrete Fourier Transform (DFT), the basis functions are complex exponentials (sinusoids). For each $(u,v)$, the basis function is: $K(x,y,u,v)=e^{-j2\pi (\frac{ux}{M}+\frac{vy}{N})}$ These are 2D sinusoids with different frequencies and orientations, determined by u and v. The DFT decomposes the image into a weighted sum of these sinusoidal basis functions. The inverse DFT reconstructs the image by summing these sinusoids, weighted by the corresponding DFT coefficients $F(u,v)$.

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Question 12: Separable 2D DCT Implementation (a)(5 marks) Given the following 2x2 image:

4   8
12  16

Calculate the 2x2 Type-II DCT of this image. Utilize the separable nature of the DCT kernel, perform first 1D vertical then horizontal operation. Use the formulas provided in the document.

(b) (3 marks) After performing the 2D DCT, you observe that the top-left coefficient, T(0,0), has a significantly larger magnitude than the other coefficients. Explain what this indicates about the original image, based on your understanding of spatial frequencies and the DCT.

Solution 12:

(a) 2x2 DCT Calculation:

The 1D DCT-II kernel is: $K(x,u)=\alpha (u)\cos \left[\frac{\pi (2x+1)u}{2M}\right]$

Where: $\alpha (u)=\{\begin{array}{ll}\sqrt{\frac{1}{M}},& u=0\\ \sqrt{\frac{2}{M}},& 1\le u\le M-1\end{array}$ Here M = N = 2

So, $\alpha (0)=\sqrt{1/2}$ and $\alpha (1)=\sqrt{2/2}=1$.

Step 1: 1D DCT along columns (Vertical): We calculate the intermediate I(x,v)

• For v=0: $I(0,0)=\alpha (0)\cdot [f(0,0)\cos (0)+f(0,1)\cos (0)]=\sqrt{1/2}\cdot [4\cdot 1+12\cdot 1]=\sqrt{1/2}\cdot 16=8\sqrt{2}$

  $I(1,0)=\alpha (0)\cdot [f(1,0)\cos (0)+f(1,1)\cos (0)]=\sqrt{1/2}\cdot [8\cdot 1+16\cdot 1]=\sqrt{1/2}\cdot 24=12\sqrt{2}$

• For v=1: $I(0,1)=\alpha (1)\cdot [f(0,0)\cos (\frac{\pi (2\cdot 0+1)}{4})+f(0,1)\cos (\frac{\pi (2\cdot 0+1)}{4})]=1\cdot [4\cos (\pi /4)+12\cos (\pi /4)]=16\cdot \frac{\sqrt{2}}{2}=8\sqrt{2}$

  $I(1,1)=\alpha (1)\cdot [f(1,0)\cos (\pi /4)+f(1,1)\cos (\pi /4)]=[8\cos (\pi /4)+16\cos (\pi /4)]=24\cdot \frac{\sqrt{2}}{2}=12\sqrt{2}$

Intermediate Matrix I(x,v) =

8√2   8√2
12√2  12√2

Step 2: 1D DCT along rows (Horizontal): We calculate the T(u,v)

• For u = 0: $T(0,0)=\alpha (0)\cdot [I(0,0)\cdot \cos (0)+I(1,0)\cdot \cos (0)]=\sqrt{1/2}\cdot [8\sqrt{2}\cdot 1+12\sqrt{2}\cdot 1]=\sqrt{1/2}\cdot 20\sqrt{2}=20$

  $T(0,1) = \alpha(0) \cdot [I(0,1)\cos(0) + I(1,1)\cos(0)] =  \sqrt{1/2} \cdot [8\sqrt{2} \cdot 1 + 12\sqrt{2}\cdot 1] =  \sqrt{1/2} \cdot (20\sqrt{2}) = 20$
  

• For u = 1: $T(1,0)=\alpha (1)\cdot [I(0,0)\cos (\pi /4)+I(1,0)\cos (\pi /4)]=1\cdot [8\sqrt{2}\cdot \frac{\sqrt{2}}{2}+12\sqrt{2}\cdot \frac{\sqrt{2}}{2}]=8+12=20$

  $T(1,1) =  \alpha(1) \cdot [I(0,1)\cos(\pi/4) + I(1,1)cos(\pi/4) ] = [8\sqrt{2}\cdot(\sqrt{2}/2) + 12\sqrt{2} \cdot (\sqrt{2}/2)] = 8 + 12 = 20$
  

Resulting DCT Matrix, T(u, v):

20  20
20  20

(b) Interpretation of T(0, 0):

The coefficient T(0, 0) in the DCT represents the DC component, which is proportional to the average pixel intensity of the original image. In this case, the large magnitude of T(0, 0) compared to other coefficients indicates:

1. Dominant Average Intensity: The original image has a significant average intensity value. It’s not an image that’s close to being all black. The larger the value of T(0,0) the brighter image we have.

2. Low-Frequency Dominance: Since T(0, 0) is a low-frequency coefficient (the lowest, in fact), its large magnitude signifies that the image is dominated by low-frequency components. This means there aren’t many rapid changes in intensity; the image is relatively smooth or uniform. If the other coefficients were zero (or very small), the original image would be a constant-intensity image (all pixels having the same value). The fact that, other coefficient are also the same, makes all the pixel of input image same, if we apply inverse transform.

In short, a large T(0, 0) indicates a significant average brightness and a predominance of smooth, gradual changes in intensity rather than sharp details or edges.