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43 changes: 43 additions & 0 deletions .github/workflows/deploy.yml
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name: Deploy Quartz site to GitHub Pages

on:
push:
branches:
- main

permissions:
contents: read
pages: write
id-token: write

concurrency:
group: "pages"
cancel-in-progress: false

jobs:
build:
runs-on: ubuntu-22.04
steps:
- uses: actions/checkout@v4
with:
fetch-depth: 0 # Fetch all history for git info
- uses: actions/setup-node@v4
- name: Install Dependencies
run: npm ci
- name: Build Quartz
run: npx quartz build
- name: Upload artifact
uses: actions/upload-pages-artifact@v3
with:
path: public

deploy:
needs: build
environment:
name: github-pages
url: ${{ steps.deployment.outputs.page_url }}
runs-on: ubuntu-latest
steps:
- name: Deploy to GitHub Pages
id: deployment
uses: actions/deploy-pages@v4
3 changes: 3 additions & 0 deletions README.md
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# GitHub Pages
- ***Site***: [https://uomocosa.github.io/UNISI-Sensors-and-Microsystems-Obsidian-Quartz-Publish](https://uomocosa.github.io/UNISI-Sensors-and-Microsystems-Obsidian-Quartz-Publish)
----
# Quartz v4

> “[One] who works with the door open gets all kinds of interruptions, but [they] also occasionally gets clues as to what the world is and what might be important.” — Richard Hamming
Expand Down
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52 changes: 52 additions & 0 deletions content/Notes & Images/Bode Plot.md
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- ***Source***: [Internet Page](https://lpsa.swarthmore.edu/Bode/BodeReviewRules.html) | [[Rules for Drawing Bode Diagrams 1.pdf|Downladed PDF]]
- ***Amplitude***:
- **Real zero**: $+20 * \text{zero's multiplicity} \kern3px {\text{dB}\over\text{decade}}$
- **Real pole**: $-20 * \text{pole's multiplicity} \kern3px {\text{dB}\over\text{decade}}$
- **Imaginary zero**:
- $+40 * \text{zero's multiplicity} \kern3px {\text{dB}\over\text{decade}}$
- Peak at $w = w_0 \sqrt{1 - 2 \zeta^2}$
- **Imaginary pole**:
- $-40 * \text{pole's multiplicity} \kern3px {\text{dB}\over\text{decade}}$
- Peak at $w = w_0 \sqrt{1 - 2 \zeta^2}$

- ***Phase***:
- **Starting Phase***:
- $+0°$ if $K \gt 0$ ($K$ is the gain)
- $\pm180°$ if $K \lt 0$ ($K$ is the gain)
- **Real zero**: $+90° * \text{zero's multiplicity}$
- **Real pole**: $-90° * \text{zero's multiplicity}$
- **Imaginary zero**: $+180° * \text{zero's multiplicity}$
- **Imaginary pole**: $-180° * \text{pole's multiplicity}$

- ***Frequency Domain Response (Bode Form)***:
- For a single-pole system, the transfer function in frequency domain is represented as: $H(j\omega) = \frac{A\_{DC}}{1 + j\frac{\omega}{\omega\_T}}$, where $A_{DC}$ is the DC gain and $\omega_T$ is the cutoff frequency.
- Cutoff frequency ($f_T$) or $-3 , \text{dB}$ frequency is related to the time constant $\tau$ as: $\omega_T = \frac{1}{\tau}$.

- ***Bode Plots***:
- The module of $H(j\omega)$ is represented in dB as $20 \log |H(j\omega)|$.
- The DC gain ($A_{DC}$) is the value of $H(j\omega)$ when $\omega = 0$.
- The cutoff frequency ($f_T$) corresponds to $|H(j\omega_T)| = \frac{1}{\sqrt{2}}$ in dB.

---
#### Gain
- A **real zero** means $+20 \cdot \mu \kern5px \frac{\text{dB}}{\text{dec}}$, where $\mu$ is its **multiplicity**
- A **real pole** means $-20 \cdot \mu \kern5px \frac{\text{dB}}{\text{dec}}$, where $\mu$ is its **multiplicity**
- An **imaginary zero**
- means $+40 \cdot \mu \kern5px \frac{\text{dB}}{\text{dec}}$, where $\mu$ is its **multiplicity**
- In the case of an imaginary zero there is a also peak is at $w = w_0 \sqrt{1 - 2 \kern2px \zeta^2}$
- An **imaginary pole**:
- means $+40 \cdot \mu \kern5px \frac{\text{dB}}{\text{dec}}$, where $\mu$ is its **multiplicity**
- There is a also peak is at $w = w_0 \sqrt{1 - 2 \kern2px \zeta^2}$
- ~Ex.:<br>![[Pasted image 20230418162842.png]]

----
#### Phase
- **Starting Phase***:
- $+0°$ if $K \gt 0$ ($K$ is the gain)
- $\pm180°$ if $K \lt 0$ ($K$ is the gain)
- **Real zero**: $+90° \cdot \mu \kern5px \frac{\text{dB}}{\text{dec}}$, where $\mu$ is its **multiplicity**
- **Real pole**: $-90° \cdot \mu \kern5px \frac{\text{dB}}{\text{dec}}$, where $\mu$ is its **multiplicity**
- **Couple of conjugate complex zero**: $+180° \cdot \mu \kern5px \frac{\text{dB}}{\text{dec}}$, where $\mu$ is its **multiplicity**
- **Couple of conjugate complex pole**: $-180° \cdot \mu \kern5px \frac{\text{dB}}{\text{dec}}$, where $\mu$ is its **multiplicity**

---
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#TODO Pasted from ChatGPT

To change variables in an integral, you can use a technique called variable substitution or change of variables. The idea is to replace the original variable of integration with a new variable to simplify the integral. Here are the general steps:

1. **Choose a New Variable**: Select a new variable to replace the original variable in the integral. The choice of the new variable should be based on making the integral easier to work with. Common choices include \(u\), \(t\), \(s\), or any other letter you prefer.

2. **Determine the Relationship**: Find a relationship between the new variable and the original variable. This relationship is often expressed as \(u = g(x)\), where \(u\) is the new variable, and \(g(x)\) is a function of the original variable \(x\). This function should be chosen so that it simplifies the integral.

3. **Calculate the Differentials**: Compute the differential \(du\) in terms of the new variable \(u\). This is done using the chain rule:

\[
du = g'(x) \, dx
\]

4. **Change the Limits**: If you are dealing with definite integrals, also change the limits of integration from the original variable to the new variable. This step is crucial when changing variables.

5. **Substitute**: Substitute the new variable \(u\) and its corresponding differential \(du\) into the integral.

6. **Simplify and Integrate**: After substitution, you should have a new integral in terms of the new variable \(u\). Simplify the integrand as much as possible. Then, integrate with respect to \(u\).

7. **Change Back (if necessary)**: If you want the final result in terms of the original variable, use the relationship \(u = g(x)\) to change back to the original variable after integrating.

Here's a simple example to illustrate the process:

Original Integral: \(\int x e^{x^2} \, dx\)

1. Choose a New Variable: Let \(u = x^2\).

2. Determine the Relationship: \(u = g(x) = x^2\).

3. Calculate the Differentials: \(du = 2x \, dx\).

4. Change the Limits: Not applicable for this indefinite integral.

5. Substitute: Replace \(x^2\) with \(u\) and \(dx\) with \(\frac{du}{2x}\):

\(\int e^u \frac{1}{2x} \, du\)

6. Simplify and Integrate: The integral becomes \(\frac{1}{2} \int \frac{e^u}{x} \, du\).

7. Change Back (if necessary): If you want the result in terms of \(x\), you can use the original relationship \(u = x^2\) to replace \(u\) with \(x^2\) in the final answer.

Remember that the choice of the new variable and the relationship \(u = g(x)\) is crucial for simplifying the integral effectively.
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- [ffmpeg](https://ffmpeg.org)
- Convert a video (`.mp4` file) to an only audio (`.mp3` file)
- Open the Window's Commands Promt, type:
- `cd C:\...`
- `ffmpeg -i "video_name.mp4" "video_name.mp3"`
- Go to [hugging-face: RASMUS](https://huggingface.co/spaces/RASMUS/Whisper-youtube-crosslingual-subtitles) drop the audio and create subtitles.
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- [Install VLC](https://www.videolan.org/vlc/) or [ffmpeg](https://ffmpeg.org) (better the latter one)
- Convert a video (`.mp4` file) to an only audio (`.mp3` file)
- Open VLC and go to `Media >> Converti/Salva`
- Press `+ Aggiungi` and select the video you want to create subtitles for
- Press `Converti/Salva`
- The dot `Converti` must be selected
- Change the `Profilo` to `Audio - MP3`
- Select a new `Destinazione`
- Press `Avvia`
- Go to [replicate.com](https://replicate.com/m1guelpf/whisper-subtitles) drop the audio into the `audio path` and press `submit`
- If you want to create ***NOTES*** for them (***IT DOES NOT WORK!***): %% fold %%
1. Go to [promt splitter](https://chatgpt-prompt-splitter.jjdiaz.dev) and pass the whole subtitle text
2. Pass each "split" to ChatGPT.
3. Ask him somenting like: ***The parts I passed you previously are subtitles taken from a single lecture of "Sensor and Microsystems", can you make COMPREHENSIVE and COMPLETE notes for me, given those subtitles and your extensive knowledge on the subject***
4. If there are formulas you can ask for a Latex formatting in this way: ***If you write any equation do it in this format `$$ <latex code here> $$`***
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The formula for the average value of a function ($f(x)$) over an interval $[a, b]$ is given by:
$$
\overline{f} = \frac{1}{b - a} \int_{a}^{b} f(x) \, dx
$$

In this formula:
- $a$ and $b$ are the lower and upper bounds of the interval over which you want to find the average of $f(x)$.
- $f(x)$ is the function whose average you want to calculate.
- $\int_{a}^{b} f(x) \, dx$ represents the definite integral of $f(x)$ over the interval $[a, b]$.
- $\frac{1}{b - a}$ scales the integral result to find the average value over the interval.

This formula essentially calculates the mean value of the function $f(x)$ over the specified interval.
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The nabla symbol, denoted as $\nabla$, represents a vector differential operator commonly used in vector calculus.
It is used to describe various operations involving vectors, such as gradients, divergences, and curls.

Here are some key operations associated with the nabla operator:
1. **Gradient ($\nabla f$):**
- The gradient of a **scalar function** $f$, results in a **vector**:$$\nabla f = \frac{\partial f}{\partial x}\hat{i} + \frac{\partial f}{\partial y}\hat{j} +\frac{\partial f}{\partial z}\hat{k}$$*~Example: If $f(x, y, z) = x^2 + y^2 + z^2$, then $\nabla f = \, <2x ,\ 2y\mathbf ,\ 2z>$*.

2. **Divergence ($\nabla \cdot \vec F$):**
- The divergence of a **vector** field $\vec F = (F_x, F_y, F_z)$ is given by $$\nabla \cdot \vec F = {\partial F_x \over \partial x} + {\partial F_y \over \partial y} + {\partial F_z \over \partial z}$$*~Example: If $F = (2x, y^2, z)$, then $\nabla \cdot \vec F = 2 + 2y + 1 = 2y + 3$*.

3. **Curl ($\nabla \times \vec F$):**
- The curl of a vector field $\vec F = (F_x, F_y, F_z)$ is given by:$$\nabla \times \vec F = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \right)\hat{i} + \left( \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right)\hat{j} + \left( \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right)\hat{k}$$*~Example: If $F = (2, \, y, \, 3z + 2x)$, then $\nabla \times F = (0-0)\hat{i} + (0-1)\hat{j} + (0-0)\hat{k} = -2\hat{i}$.

4. **Laplacian ($\nabla^2 f$):**
- The Laplacian of a **scalar function** $f$ is given by:$$ \nabla^2 f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2}$$*~Example:*
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