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# Lecture 6
## Continue on Light, eye/camera, and color
### BRDF (Bidirectional Reflectance Distribution Function)
$$
\rho(\theta_i,\phi_i,\theta_o,\phi_o)
$$
#### Diffuse Reflection
- Dull, matte surface like chalk or latex paint
- Most often used in computer vision
- Brightness _does_ depend on direction of illumination
Diffuse reflection governed by Lambert's law: $I_d = k_d N\cdot L I_i$
- $N$: surface normal
- $L$: light direction
- $I_i$: incident light intensity
- $k_d$: albedo
$$
\rho(\theta_i,\phi_i,\theta_o,\phi_o)=k_d \cos\theta_i
$$
#### Photometric Stereo
Suppose there are three light sources, $L_1, L_2, L_3$, and we have the following measurements:
$$
I_1 = k_d N\cdot L_1
$$
$$
I_2 = k_d N\cdot L_2
$$
$$
I_3 = k_d N\cdot L_3
$$
We can solve for $N$ by taking the dot product of $N$ and each light direction and then solving the system of equations.
Will not do this in the lecture.
#### Specular Reflection
- Mirror-like surface
$$
I_e=\begin{cases}
I_i & \text{if } V=R \\
0 & \text{if } V\neq R
\end{cases}
$$
- $V$: view direction
- $R$: reflection direction
- $\theta_i$: angle between the incident light and the surface normal
Near-perfect mirror have a high light around $R$.
common model:
$$
I_e=k_s (V\cdot R)^{n_s}I_i
$$
- $k_s$: specular reflection coefficient
- $n_s$: shininess (imperfection of the surface)
- $I_i$: incident light intensity
#### Phong illumination model
- Phong approximation of surface reflectance
- Assume reflectance is modeled by three compoents
- Diffuse reflection
- Specular reflection
- Ambient reflection
$$
I_e=k_a I_a + I_i \left[k_d (N\cdot L) + k_s (V\cdot R)^{n_s}\right]
$$
- $k_a$: ambient reflection coefficient
- $I_a$: ambient light intensity
- $k_d$: diffuse reflection coefficient
- $k_s$: specular reflection coefficient
- $n_s$: shininess
- $I_i$: incident light intensity
Many other models.
#### Measuring BRDF
Use Gonioreflectometer.
- Device for measuring the reflectance of a surface as a function of the incident and reflected angles.
- Can be used to measure the BRDF of a surface.
BRDF dataset:
- MERL dataset
- CURET dataset
### Camera/Eye
#### DSLR Camera
- Pinhole camera model
- Lens
- Aperture (the pinhole)
- Sensor
- ...
#### Digital Camera block diagram
![Digital Camera block diagram](https://static.notenextra.trance-0.com/images/CSE559A/DigitalCameraBlockDiagram.png)
Scanning protocols:
- Global shutter: all pixels are exposed at the same time
- Interlaced: odd and even lines are exposed at different times
- Rolling shutter: each line is exposed as it is read out

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# Lecture 6
## Review
### Linear Fractional Transformations
Transformations of the form $f(z)=\frac{az+b}{cz+d}$,$a,b,c,d\in\mathbb{C}$ and $ad-bc\neq 0$ are called linear fractional transformations.
#### Theorem 3.8 Preservation of clircles
We defined clircle to be a circle or a line.
The circle equation is:
Let $\zeta=u+iv$ be the center of the circle, $r$ be the radius of the circle.
$$
circle=\{z\in\mathbb{C}:|\zeta-c|=r\}
$$
This is:
$$
|\zeta|^2-c\overline{\zeta}-\overline{c}\zeta+|c|^2-r^2=0
$$
If $\phi$ is a non-constant linear fractional transformation, then $\phi$ maps clircles to clircles.
We claim that a map is circle preserving if and only if for some $\alpha,\beta,\gamma,\delta\in\mathbb{R}$.
$$
\alpha|\zeta|^2+\beta Re(\zeta)+\gamma Im(\zeta)+\delta=0
$$
when $\alpha=0$, it is a line.
when $\alpha\neq 0$, it is a circle.
Proof:
Let $w=u+iv=\frac{1}{\zeta}$, so $\frac{1}{w}=\frac{u}{u^2+v^2}-i\frac{v}{u^2+v^2}$.
Then the original equation becomes:
$$
\alpha\left(\frac{u}{u^2+v^2}\right)^2+\beta\left(\frac{u}{u^2+v^2}\right)+\gamma\left(-\frac{v}{u^2+v^2}\right)+\delta=0
$$
Which is in the form of circle equation.
EOP
## Chapter 4 Elements of functions
> $e^t=\sum_{n=0}^{\infty}\frac{t^n}{n!}$
So, following the definition of $e^\zeta$, we have:
$$
\begin{aligned}
e^{x+iy}&=e^xe^{iy} \\
&=e^x\left(\sum_{n=0}^{\infty}\frac{(iy)^n}{n!}\right) \\
&=e^x\left(\sum_{n=0}^{\infty}\frac{(-1)^ny^n}{n!}\right) \\
&=e^x(\cos y+i\sin y)
\end{aligned}
$$
### $e^\zeta$
The exponential of $e^\zeta=x+iy$ is defined as:
$$
e^\zeta=exp(\zeta)=e^x(\cos y+i\sin y)
$$
So,
$$
|e^\zeta|=|e^x||\cos y+i\sin y|=e^x
$$
#### Theorem 4.3 $e^\zeta$ is holomorphic
$e^\zeta$ is holomorphic on $\mathbb{C}$.
Proof:
$$
\begin{aligned}
\frac{\partial}{\partial\zeta}e^\zeta&=\frac{1}{2}\left(\frac{\partial}{\partial x}+\frac{i}{\partial y}\right)e^x(\cos y+i\sin y) \\
&=\frac{1}{2}e^x(\cos y+i\sin y)+ie^x(-\sin y+i\cos y) \\
&=0
\end{aligned}
$$
EOP
#### Theorem 4.4 $e^\zeta$ is periodic
$e^\zeta$ is periodic with period $2\pi i$.
Proof:
$$
e^{\zeta+2\pi i}=e^\zeta e^{2\pi i}=e^\zeta\cdot 1=e^\zeta
$$
EOP
#### Theorem 4.5 $e^\zeta$ as a map
$e^\zeta$ is a map from $\mathbb{C}$ to $\mathbb{C}$ with period $2\pi i$.
$$
e^{\pi i}+1=0
$$
This is a map from cartesian coordinates to polar coordinates, where $e^x$ is the radius and $y$ is the angle.
This map attains every value in $\mathbb{C}\setminus\{0\}$.
#### Definition 4.6-8 $\cos\zeta$ and $\sin\zeta$
$$
\cos\zeta=\frac{1}{2}(e^{i\zeta}+e^{-i\zeta})
$$
$$
\sin\zeta=\frac{1}{2i}(e^{i\zeta}-e^{-i\zeta})
$$
$$
\cosh\zeta=\frac{1}{2}(e^\zeta+e^{-\zeta})
$$
$$
\sinh\zeta=\frac{1}{2}(e^\zeta-e^{-\zeta})
$$
From this definition, we can see that $\cos\zeta$ and $\sin\zeta$ are no longer bounded.
And this definition is still compatible with the previous definition of $\cos$ and $\sin$ when $\zeta$ is real.
Moreover,
$$
\cosh(i\zeta)=\cos\zeta
$$
$$
\sinh(i\zeta)=i\sin\zeta
$$
### Logarithm
#### Definition 4.9 Logarithm
A logarithm of $a$ is any $b$ such that $e^b=a$.
If $a=0$, then no logarithm exists.
If $a\neq 0$, then there exists infinitely many logarithms of $a$.
Let $a=re^{i\theta}$, $b=x+iy$ be a logarithm of $a$.
Then,
$$
e^{x+iy}=re^{i\theta}
$$
Since logarithm is not unique, we can always add $2k\pi i$ to the angle.
If $y\in(-\pi,\pi]$, then $\log a=b$ means $e^b=a$ and $Im(b)\in(-\pi,\pi]$.
If $a=re^{i\theta}$, then $\log a=\log r+i(\theta_0+2k\pi)$.
#### Definition 4.10
Let $G$ be an open connected subset of $\mathbb{C}\setminus\{0\}$.
A branch of $\arg(\zeta)$ in $G$ is a continuous function $\alpha$, such that $\alpha(\zeta)$ is a value of $\arg(\zeta)$.
A branch of $\log(\zeta)$ in $G$ is a continuous function $\beta$, such that $e^{\beta(\zeta)}=\zeta$.
Note: $G$ has a branch of $\arg(\zeta)$ if and only if it has a branch of $\log(\zeta)$.
If $G=\mathbb{C}\setminus\{0\}$, then not branch of $\arg(\zeta)$ exists.
Suppose $\alpha_1$ and $\alpha_2$ are two branches of $\arg(\zeta)$ in $G$.
Then,
$$
\alpha_1(\zeta)-\alpha_2(\zeta)=2k\pi i
$$
for some $k\in\mathbb{Z}$.
#### Theorem 4.11
$\log(\zeta)$ is holomorphic on $\mathbb{C}\setminus\{0\}$.
Proof:
Method 1: Use polar coordinates. (See in homework)
Method 2: Use the fact that $\log(\zeta)$ is the inverse of $e^\zeta$.
Suppose $h=s+it$, $e^h=e^s(\cos t+i\sin t)$, $e^h-1=e^s(\cos t-1)+i\sin t$. So
$$
\begin{aligned}
\frac{e^h-1}{h}&=\frac{(s+it)e^s(\cos t-1)+i\sin t}{s^2+t^2} \\
&=\frac{e^s(\cos t-1)}{s^2+t^2}+i\frac{\sin t}{s^2+t^2}
\end{aligned}
$$
Continue next time.

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@@ -8,4 +8,5 @@ export default {
Math416_L3: "Complex Variables (Lecture 3)",
Math416_L4: "Complex Variables (Lecture 4)",
Math416_L5: "Complex Variables (Lecture 5)",
Math416_L6: "Complex Variables (Lecture 6)",
}

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