add ecse-1010-poc-lab03

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content/en/posts/ecse-1010/lab02/index.md

@@ -10,7 +10,7 @@ author:
   link: https/www.jamesflare.com
   email:
   avatar: /site-logo.avif
-description: This blog post discusses a detailed lab assignment focusing on proving various electrical concepts using resistors, diodes, op-amps, and nodal analysis. The experiments aim to validate Ohm's Law, non-linear IV curves for LEDs, differential resistance in diode IV curves, nodal voltage solving with Kirchhoff’s Laws, the function of an op amp comparator, mathematical op amp functionality, and two-channel audio mixer transfer functions.
+description: The experiments aim to validate Ohm's Law, non-linear IV curves for LEDs, differential resistance in diode IV curves, nodal voltage solving with Kirchhoff’s Laws, the function of an op amp comparator, mathematical op amp functionality, and two-channel audio mixer transfer functions.
 keywords: ["Electrical Engineering","Ohm's Law","IV curve","Nodal Analysis","Op-Amp"]
 license:
 comment: true
@@ -28,7 +28,7 @@ hiddenFromHomePage: false
 hiddenFromSearch: false
 hiddenFromRss: false
 hiddenFromRelated: false
-summary: This blog post discusses a detailed lab assignment focusing on proving various electrical concepts using resistors, diodes, op-amps, and nodal analysis. The experiments aim to validate Ohm's Law, non-linear IV curves for LEDs, differential resistance in diode IV curves, nodal voltage solving with Kirchhoff’s Laws, the function of an op amp comparator, mathematical op amp functionality, and two-channel audio mixer transfer functions.
+summary: The experiments aim to validate Ohm's Law, non-linear IV curves for LEDs, differential resistance in diode IV curves, nodal voltage solving with Kirchhoff’s Laws, the function of an op amp comparator, mathematical op amp functionality, and two-channel audio mixer transfer functions.
 resources:
   - name: featured-image
     src: featured-image.avif

content/en/posts/ecse-1010/lab03/index.md

@@ -0,0 +1,348 @@
+---
+title: ECSE 1010 Proof of Concepts - Alpha Lab03
+subtitle:
+date: 2024-12-18T01:07:47-05:00
+lastmod: 2024-12-18T01:07:47-05:00
+slug: ecse-1010-poc-lab03
+draft: false
+author:
+  name: James
+  link: https://www.jamesflare.com
+  email:
+  avatar: /site-logo.avif
+description: This blog post outlines a series of lab experiments and analyses involving Fourier analysis, signal reconstruction, filter design, and frequency manipulation using MATLAB and other tools. It includes discussions on theoretical concepts, simulations, and practical applications.
+keywords: ["Fourier Analysis","Signal Processing","Filter Design"]
+license:
+comment: true
+weight: 0
+tags:
+  - ECSE 1010
+  - Lab
+  - Electrical Engineering
+  - RPI
+categories:
+  - Electrical Engineering
+collections:
+  - ECSE 1010
+hiddenFromHomePage: false
+hiddenFromSearch: false
+hiddenFromRss: false
+hiddenFromRelated: false
+summary: This blog post outlines a series of lab experiments and analyses involving Fourier analysis, signal reconstruction, filter design, and frequency manipulation using MATLAB and other tools. It includes discussions on theoretical concepts, simulations, and practical applications.
+resources:
+  - name: featured-image
+    src: featured-image.jpg
+  - name: featured-image-preview
+    src: featured-image-preview.jpg
+toc: true
+math: true
+lightgallery: true
+password:
+message:
+repost:
+  enable: false
+  url:
+
+# See details front matter: https://fixit.lruihao.cn/documentation/content-management/introduction/#front-matter
+---
+
+<!--more-->
+
+## 0. Lab Document
+
+<div style="width: 100%; max-width: 600px; margin: 0 auto; display: block;">
+  <embed src="Lab03.pdf" type="application/pdf" width="100%" height="500px">
+</div>
+
+## 1. Prove the Result of Summing 3 Sinusoidal Waves in the Time and Frequency Domains
+
+### Description
+
+We are going to sum 3 basic type of sine wave and check its spectrum of frequency domain. See if that matches our expectation.
+
+### Analysis
+
+To determine the time domain signal and frequency spectrum of summing 3 sinusoidal waves. We can use the properties of sinusoidal waves in time and frequency domains.
+
+For example, Sine Wave $250Hz$, $500Hz$ and $750Hz$ should have a result of time domain like
+
+{{< image src="P1-1-b.avif" caption="P1-1-b" width=600px >}}
+
+For the frequency domain, it should be like
+
+{{< image src="P1-1-c.avif" caption="P1-1-c" width=600px >}}
+
+Since the sum is made by 3 single sine wave.
+
+### Simulation
+
+{{< image src="P1-2-a.avif" caption="P1-2-a" width=600px >}}
+
+{{< image src="P1-1-b.avif" caption="P1-1-b" width=600px >}}
+
+{{< image src="P1-1-a.avif" caption="P1-1-a" width=600px >}}
+
+### Discussion
+
+Both our sketch and the simulation of frequency domain has 3 positive identical peaks. Which proves our analysis. The reason that they are not fully matched is due to Discrete Fourier Transform.
+
+The sine wave used in this simulation is discrete sine wave. Unlike normal Fourier Transform, how many data point leads to how many frequency it went through. But the value in $dBm$ is extremely low. So, it makes no different.
+
+## 2. Prove the Concept of Fourier Analysis by Determining Which Two Basic Types of Signals were Summed to Create This Mystery Signal
+
+### Description
+
+We are going to find out two basic types of signals of a summed up signal by Fourier Analysis. To do that we are going to use the peak finder in spectrum analyzer in MATLAB.
+
+### Analysis
+
+We can pick the most strong signal in frequency spectrum. That should be the two basic types of signals were summed to create this mystery signal.
+
+### Simulation
+
+Let's load [Mystery Signal Audio File](lab03_mystery_signal.wav)
+
+{{< image src="P2-2-a.avif" caption="P2-2-a" width=600px >}}
+
+The time domain looks like
+
+{{< image src="P2-1-a.avif" caption="P2-1-a" width=600px >}}
+
+And the frequency domain looks like
+
+{{< image src="P2-1-b.avif" caption="P2-1-b" width=600px >}}
+
+### Discussion
+
+Use the peak finder, the top 2 positive frequency are
+
+| Peak Number | Frequency ($Hz$) | Magnitude ($dBm$) |
+|:---:|:---:|:---:|
+| 1 | $101.5050$ | $25.2965$ |
+| 2 | $749.5751$ | $19.6980$ |
+
+So, the two basic types of signals were summed to create this mystery signal are
+
+- Sine Wave $101.5050Hz$
+- Sine Wave $749.5751Hz$
+
+## 3. Prove the Concept of Fourier Synthesis by Reconstructing the Mystery Signal in the Time Domain
+
+### Description
+
+After find out the top 2 basic type of wave inside the mystery signal. We are going to reconstruct it and see if that matches the original work. We may use more than top 2 signal to get a better fitting.
+
+### Analysis
+
+A signal composed of two sine waves with frequencies of $101.5050 Hz$ and $749.5751 Hz$ can be mathematically expressed as follows:
+
+$$
+x(t) = \sin(2\pi \times 101.5050 \, t) + \sin(2\pi \times 749.5751 \, t)
+$$
+
+- **$t$**: Represents time in seconds.
+- **$\sin$**: The sine function, which generates the oscillating waveform.
+- **$2\pi \times f \, t$**: Converts the frequency $f$ from Hertz (cycles per second) to radians per second, which is necessary for the sine function argument.
+
+a more general form can be written as
+
+$$
+x(t) = A_1 \sin(2\pi f_1 t + \phi_1) + A_2 \sin(2\pi f_2 t + \phi_2)
+$$
+
+Where
+
+- $A_1$ and $A_2$ are the amplitudes of the sine waves.
+- $f_1 = 101.5050 Hz$ and $f_2 = 749.5751 Hz$ are the frequencies.
+- $\phi_1$ and $\phi_2$ are the phase shifts.
+
+Assuming both sine waves have an amplitude of 1 and no phase shift, the expression simplifies to:
+
+$$
+x(t) = \sin(2\pi \times 101.5050 \, t) + \sin(2\pi \times 749.5751 \, t)
+$$
+
+### Simulation & Discussion
+
+{{< image src="P3-2-a.avif" caption="P3-2-a" width=600px >}}
+
+Use the peak finder, the top 5 positive frequency are
+
+| Peak Number | Frequency ($Hz$) | Magnitude ($dBm$) |
+|:---:|:---:|:---:|
+| 1 | $101.5$ | $25.2965$ |
+| 2 | $749.6$ | $19.6980$ |
+| 3 | $148.4$ | $-3.2694$ |
+| 4 | $694.9$ | $-13.9588$ |
+| 5 | $1350.8$ | $-14.5331$ |
+
+Then, build a layout with 5 basic wave instead of 2 for better result
+
+{{< image src="P3-2-b.avif" caption="P3-2-b" width=600px >}}
+
+Let's check the result
+
+{{< image src="P3-2-c.avif" caption="P3-2-c" width=600px >}}
+
+The power is not exactly matched, since they were adjusted by hand. But they are close, and get a result reconstructing signal of time domain like
+
+{{< image src="P3-2-d.avif" caption="P3-2-d" width=600px >}}
+
+Compare our reconstructing signal to the original signal, they share the same period and shape.
+
+{{< image src="P2-1-a.avif" caption="P2-1-a" width=600px >}}
+
+The different in detail may due to leak of more basic type of waves and different power of basic type of waves.
+
+## 4. Prove the Concept of an Analog High-Pass Filter Using an Inductor and a Resistor and Could I Make a Low-Pass Filter Using the Same Components?
+
+### Circuit Schematic
+
+For low-pass filter
+
+{{< image src="P4-1-a.avif" caption="P4-1-a" width=600px >}}
+
+For high-pass filter
+
+{{< image src="P4-1-b.avif" caption="P4-1-b" width=600px >}}
+
+### Description
+
+We will make HF and LF filter using an inductor and a resistor. And check the result of frequency respond using network analyzer function in Analog Discovery 3.
+
+### Analysis
+
+The cutoff frequencies for both the LR Low-Pass and LR High-Pass filters using component values
+
+- **Resistor ($R$)**: $510 \Omega$
+- **Inductor ($L$)**: $1 mH = 0.001 H$
+
+The cutoff frequency for both filters is given by:
+
+$$
+f_c = \frac{R}{2\pi L}
+$$
+
+Plugging in the Values
+
+$$
+\begin{align*}
+  f_c &= \frac{510}{2 \times \pi \times 0.001} \\\
+  f_c &= \frac{510}{0.00628318} \approx 81,000 \text{ Hz}
+\end{align*}
+$$
+
+### Simulation
+
+For low-pass filter
+
+{{< image src="P4-3-a.avif" caption="P4-3-a" width=600px >}}
+
+We got a cut off frequency around $82.37kHz$
+
+{{< image src="P4-3-a-b.avif" caption="P4-3-a-b" width=600px >}}
+
+For high-pass filter
+
+{{< image src="P4-3-b.avif" caption="P4-3-b" width=600px >}}
+
+We got a cut off frequency around $80.69kHz$
+
+{{< image src="P4-3-b-b.avif" caption="P4-3-b-b" width=600px >}}
+
+### Measurement
+
+I am using the network analyzer to draw a frequency respond diagram. For low-pass filter, we had a setup like
+
+{{< image src="P4-4-a-b.avif" caption="P4-4-a-b" width=600px >}}
+
+We got a cut off frequency around $79.48kHz$
+
+{{< image src="P4-4-a.avif" caption="P4-4-a" width=600px >}}
+
+For high-pass filter
+
+{{< image src="P4-4-a-b.avif" caption="P4-4-a-b" width=600px >}}
+
+We got a cut off frequency around $80.98kHz$
+
+{{< image src="P4-4-b.avif" caption="P4-4-b" width=600px >}}
+
+### Discussion
+
+Our Simulation matches our Analysis. The Measurement also matches our Simulation and Analysis. Both of them has a cut off frequency around $80kHz$ with mark of $-3dBm$.
+
+## 5. Prove the concept of using Fourier Analysis to associate audible features of an audio signal with specific frequency ranges in its frequency spectrum
+
+### Description
+
+We are going to find out the leading signal of the given signal and try to reconstruct it. After reconstruction, we are going to compare some key feature to the original signal.
+
+### Analysis
+
+{{< image src="P5-1-a.avif" caption="P5-1-a" width=600px >}}
+
+{{< image src="P5-1-b.avif" caption="P5-1-b" width=600px >}}
+
+We used high-pass and low-pass filter to split two frequencies within a signal up. We ran these two frequencies, the original signal, and the two frequencies summed into the spectrum analyzer. We see the sum of the two filtered frequencies is fit to the original.
+
+Then, we point out the main frequency from [splay](splay.zip)
+
+{{< image src="P5-2-a.avif" caption="P5-2-a" width=600px >}}
+
+which are
+
+- $440.5Hz$
+- $480.1Hz$
+
+### Simulation & Discussion
+
+Then, we reconstructed it.
+
+{{< image src="P5-2-b.avif" caption="P5-2-b" width=600px >}}
+
+They have identical shape and frequency domain.
+
+{{< image src="P5-2-d.avif" caption="P5-2-d" width=600px >}}
+
+{{< image src="P5-2-d.avif" caption="P5-2-d" width=600px >}}
+
+## 6. Prove the Concept of Filtering Out or Isolating a Particular Range of Frequencies of a Signal
+
+### Description
+
+We are going to simulate the LP filter and using Simulink to create a mid-pass filter. After this, we are going to apply this to a song and see what can it change.
+
+### Analysis
+
+We simulated a low-pass filter. Red is the raw signal, and the black is the filtered signal. As we can see, part of signal got removed.
+
+{{< image src="P6-2-a.avif" caption="P6-2-a" width=600px >}}
+
+With the formula of
+
+$$
+F_{low} = \frac{1}{2 \pi RC}
+$$
+
+we can get the capacitance of the target LP filter.
+
+$$
+\begin{align*}
+  300 &= \frac{1}{2 \pi \cdot  1K \cdot C} \\\
+  C &= 5.6 \times 10^{-7}
+\end{align*}
+$$
+
+### Simulation & Discussion
+
+Our audio file was that of Mary Had a Little Lamb. We wanted to hear a higher pitched version of it.
+
+{{< image src="P6-1-a.avif" caption="P6-1-a" width=600px >}}
+
+by applying a Mid-pass filter from $800Hz$ to $1700Hz$, we are able to remove part of the signal and able to hear about the sound. We isolate a specific, higher range of the song.
+We end up with a higher pitched version of the song and some high pitched screechy sound accompanying it. The higher pitched version is also less loud.
+
+{{< image src="P6-1-b.avif" caption="P6-1-b" width=600px >}}
+
+{{< image src="P6-1-c.avif" caption="P6-1-c" width=600px >}}

content/zh-cn/posts/ecse-1010/lab02/index.md

@@ -10,7 +10,7 @@ author:
   link: https/www.jamesflare.com
   email:
   avatar: /site-logo.avif
-description: 本文讨论了一项详细的实验室作业，重点是通过使用电阻器、二极管、运算放大器和节点分析来验证各种电气概念。实验旨在验证欧姆定律、LED的非线性IV曲线、二极管IV曲线中的差分电阻、基尔霍夫定律下的节点电压求解、运算放大器比较器的功能、数学运算放大器功能以及双通道音频混音器的传递函数。
+description: 本实验旨在验证欧姆定律、LED的非线性IV曲线、二极管IV曲线中的差分电阻、基尔霍夫定律下的节点电压求解、运算放大器比较器的功能、数学运算放大器功能以及双通道音频混音器的传递函数。
 keywords: ["电气工程","欧姆定律","电流-电压曲线","节点分析法","运算放大器"]
 license:
 comment: true
@@ -28,7 +28,7 @@ hiddenFromHomePage: false
 hiddenFromSearch: false
 hiddenFromRss: false
 hiddenFromRelated: false
-summary: 本文讨论了一项详细的实验室作业，重点是通过使用电阻器、二极管、运算放大器和节点分析来验证各种电气概念。实验旨在验证欧姆定律、LED的非线性IV曲线、二极管IV曲线中的差分电阻、基尔霍夫定律下的节点电压求解、运算放大器比较器的功能、数学运算放大器功能以及双通道音频混音器的传递函数。
+summary: 本实验旨在验证欧姆定律、LED的非线性IV曲线、二极管IV曲线中的差分电阻、基尔霍夫定律下的节点电压求解、运算放大器比较器的功能、数学运算放大器功能以及双通道音频混音器的传递函数。
 resources:
   - name: featured-image
     src: featured-image.avif

content/zh-cn/posts/ecse-1010/lab03/index.md

@@ -0,0 +1,346 @@
+---
+title: ECSE 1010 概念验证 - Alpha Lab03
+subtitle:
+date: 2024-12-18T01:07:47-05:00
+lastmod: 2024-12-18T01:07:47-05:00
+slug: ecse-1010-poc-lab03
+draft: false
+author:
+  name: James
+  link: https://www.jamesflare.com
+  email:
+  avatar: /site-logo.avif
+description: 这篇博客文章概述了一系列涉及傅里叶分析、信号重建、滤波器设计和频率处理的实验和分析，使用MATLAB及其他工具。内容包括理论概念的讨论、模拟以及实际应用。
+keywords: ["傅里叶分析", "信号处理", "滤波器设计"]
+license:
+comment: true
+weight: 0
+tags:
+  - ECSE 1010
+  - Lab
+  - Electrical Engineering
+  - RPI
+categories:
+  - Electrical Engineering
+collections:
+  - ECSE 1010
+hiddenFromHomePage: false
+hiddenFromSearch: false
+hiddenFromRss: false
+hiddenFromRelated: false
+summary: 这篇博客文章概述了一系列涉及傅里叶分析、信号重建、滤波器设计和频率处理的实验和分析，使用MATLAB及其他工具。内容包括理论概念的讨论、模拟以及实际应用。
+resources:
+  - name: featured-image
+    src: featured-image.jpg
+  - name: featured-image-preview
+    src: featured-image-preview.jpg
+toc: true
+math: true
+lightgallery: true
+password:
+message:
+repost:
+  enable: false
+  url:
+
+# See details front matter: https://fixit.lruihao.cn/documentation/content-management/introduction/#front-matter
+---
+
+<!--more-->
+
+## 0. 参考文档
+
+<div style="width: 100%; max-width: 600px; margin: 0 auto; display: block;">
+  <embed src="Lab03.pdf" type="application/pdf" width="100%" height="500px">
+</div>
+
+## 1. 在时域和频域证明三个正弦波叠加后的结果
+
+### 描述
+
+我们将对三种基本类型的正弦波进行求和，并检查其频谱。看看是否符合我们的预期。
+
+### 分析
+
+为了确定三个正弦波叠加后的时域信号和频率谱，我们可以利用正弦波在时域和频域的性质。
+
+例如，250Hz、500Hz 和 750Hz 的正弦波求和后，在时域中应该如下所示：
+
+{{< image src="P1-1-b.avif" caption="P1-1-b" width=600px >}}
+
+在频域中应为
+
+{{< image src="P1-1-c.avif" caption="P1-1-c" width=600px >}}
+
+因为叠加波是由三个单一正弦波组成的。
+
+### 模拟
+
+{{< image src="P1-2-a.avif" caption="P1-2-a" width=600px >}}
+
+{{< image src="P1-1-b.avif" caption="P1-1-b" width=600px >}}
+
+{{< image src="P1-1-a.avif" caption="P1-1-a" width=600px >}}
+
+### 讨论
+
+我们的草图和频域模拟都有三个相同的正峰值，这证明了我们的分析。它们不完全匹配的原因是离散傅里叶变换。
+
+在这个模拟中使用的正弦波是离散的正弦波。与正常的傅里叶变换不同，数据点的数量决定了通过多少频率。但分贝值非常低，所以没有区别。
+
+## 2. 使用傅里叶分析确定创建神秘信号所用的基本类型信号
+
+### 描述
+
+我们将使用傅里叶分析来找出一个叠加信号中的两种基本类型的信号。为此，我们将在 MATLAB 中使用频谱仪的峰值查找器。
+
+### 分析
+
+我们可以选择频率谱中最强的信号作为两个基本类型的信号，它们组合形成了这个神秘信号。
+
+### 模拟
+
+让我们加载 [神秘信号音频文件](lab03_mystery_signal.wav)
+
+{{< image src="P2-2-a.avif" caption="P2-2-a" width=600px >}}
+
+时域看起来像
+
+{{< image src="P2-1-a.avif" caption="P2-1-a" width=600px >}}
+
+频域看起来像
+
+{{< image src="P2-1-b.avif" caption="P2-1-b" width=600px >}}
+
+### 讨论
+
+使用峰值查找器，前两个正频率为
+
+| 峰值编号 | 频率（Hz） | 幅度（dBm） |
+|:---:|:---:|:---:|
+| 1 | $101.5050$ | $25.2965$ |
+| 2 | $749.5751$ | $19.6980$ |
+
+因此，组合形成神秘信号的两个基本类型信号是
+
+- 101.5050Hz 正弦波
+- 749.5751Hz 正弦波
+
+## 3. 使用傅里叶合成重构神秘信号的时间域表示
+
+### 描述
+
+在找出神秘信号内的前两种基本类型的波之后，我们将对其进行重建，并看看是否与原始信号匹配。我们可能需要使用超过前两个信号来获得更好的拟合。
+
+### 分析
+
+一个由频率为 $101.5050 Hz$ 和 $749.5751 Hz$ 的两个正弦波组成的信号可以数学表达如下：
+
+$$
+x(t) = \sin(2\pi \times 101.5050 \, t) + \sin(2\pi \times 749.5751 \, t)
+$$
+
+- **$t$**：表示时间（秒）。
+- **$\sin$**：正弦函数，生成振荡波形。
+- **$2\pi \times f \, t$**：将频率 $f$ 从赫兹（每秒周期数）转换为弧度/秒，这是正弦函数参数所必需的。
+
+更一般的形式可以写成
+
+$$
+x(t) = A_1 \sin(2\pi f_1 t + \phi_1) + A_2 \sin(2\pi f_2 t + \phi_2)
+$$
+
+其中
+
+- $A_1$ 和 $A_2$ 是正弦波的振幅。
+- $f_1 = 101.5050 Hz$ 和 $f_2 = 749.5751 Hz$ 是频率。
+- $\phi_1$ 和 $\phi_2$ 是相位偏移。
+
+假设两个正弦波的幅度为 1，没有相位偏移，则表达式简化为：
+
+$$
+x(t) = \sin(2\pi \times 101.5050 \, t) + \sin(2\pi \times 749.5751 \, t)
+$$
+
+### 模拟与讨论
+
+{{< image src="P3-2-a.avif" caption="P3-2-a" width=600px >}}
+
+使用峰值查找器，前五个正频率为
+
+| 峰值编号 | 频率（Hz） | 幅度（dBm） |
+|:---:|:---:|:---:|
+| 1 | $101.5$ | $25.2965$ |
+| 2 | $749.6$ | $19.6980$ |
+| 3 | $148.4$ | $-3.2694$ |
+| 4 | $694.9$ | $-13.9588$ |
+| 5 | $1350.8$ | $-14.5331$ |
+
+然后，使用五个基本波形而不是两个来获得更好的结果
+
+{{< image src="P3-2-b.avif" caption="P3-2-b" width=600px >}}
+
+让我们检查一下结果
+
+{{< image src="P3-2-c.avif" caption="P3-2-c" width=600px >}}
+
+功率并不完全匹配，因为它们是手动调整的。但它们很接近，并且得到了一个时间域信号重构的结果如下：
+
+{{< image src="P3-2-d.avif" caption="P3-2-d" width=600px >}}
+
+将我们的重构信号与原始信号进行比较，它们具有相同的周期和形状。
+
+{{< image src="P2-1-a.avif" caption="P2-1-a" width=600px >}}
+
+细节上的不同可能是由于缺少更多基本类型的波形以及基本类型波形的功率差异造成的。
+
+## 4. 使用电感器和电阻证明模拟高通滤波器的概念，并且我能否使用相同的组件制作低通滤波器？
+
+### 电路图
+
+对于低通滤波器
+
+{{< image src="P4-1-a.avif" caption="P4-1-a" width=600px >}}
+
+对于高通滤波器
+
+{{< image src="P4-1-b.avif" caption="P4-1-b" width=600px >}}
+
+### 描述
+
+我们将使用电感器和电阻制作高频（HF）和低频（LF）滤波器，并使用 Analog Discovery 3 的网络分析仪功能检查频率响应结果。
+
+### 分析
+
+对于 LR 低通滤波器和 LR 高通滤波器，截止频率由以下公式给出：
+
+$$
+f_c = \frac{R}{2\pi L}
+$$
+
+其中
+
+- **电阻（$R$）**：510 欧姆
+- **电感（$L$）**：1 mH = 0.001 H
+
+代入数值计算截止频率：
+
+$$
+\begin{align*}
+f_c &= \frac{510}{2 \times \pi \times 0.001} \\
+f_c &= \frac{510}{0.00628318} \approx 81,000 \text{ Hz}
+\end{align*}
+$$
+
+### 模拟
+
+对于低通滤波器
+
+{{< image src="P4-3-a.avif" caption="P4-3-a" width=600px >}}
+
+我们得到了一个大约为 82.37 kHz 的截止频率。
+
+{{< image src="P4-3-a-b.avif" caption="P4-3-a-b" width=600px >}}
+
+对于高通滤波器
+
+{{< image src="P4-3-b.avif" caption="P4-3-b" width=600px >}}
+
+我们得到了一个大约为 80.69 kHz 的截止频率。
+
+{{< image src="P4-3-b-b.avif" caption="P4-3-b-b" width=600px >}}
+
+### 测量
+
+我使用网络分析仪绘制了频率响应图。对于低通滤波器，我们的设置如下：
+
+{{< image src="P4-4-a-b.avif" caption="P4-4-a-b" width=600px >}}
+
+我们得到了一个大约为 79.48 kHz 的截止频率。
+
+{{< image src="P4-4-a.avif" caption="P4-4-a" width=600px >}}
+
+对于高通滤波器
+
+{{< image src="P4-4-a-b.avif" caption="P4-4-a-b" width=600px >}}
+
+我们得到了一个大约为 80.98 kHz 的截止频率。
+
+{{< image src="P4-4-b.avif" caption="P4-4-b" width=600px >}}
+
+### 讨论
+
+我们的模拟与分析结果一致。测量也与模拟和分析结果一致，它们的截止频率都在 80kHz 左右，并且标记为 -3dBm。
+
+## 5. 使用傅里叶分析将音频信号的可听特征与其频谱中的特定频率范围关联起来的概念
+
+### 描述
+
+我们将找出给定信号的主要成分并尝试重构它。在重构之后，我们将比较一些关键特性与原始信号进行对比。
+
+### 分析
+
+{{< image src="P5-1-a.avif" caption="P5-1-a" width=600px >}}
+
+{{< image src="P5-1-b.avif" caption="P5-1-b" width=600px >}}
+
+我们使用高通和低通滤波器将信号中的两个频率分开。我们将这两个频率、原始信号以及它们的求和输入频谱分析仪，可以看到两个过滤后的频率之和与原始信号匹配。
+
+然后，我们用 [splay](splay.zip) 指出了主要频率
+
+{{< image src="P5-2-a.avif" caption="P5-2-a" width=600px >}}
+
+它们是
+
+- 440.5Hz
+- 480.1Hz
+
+### 模拟与讨论
+
+然后，我们进行了重构。
+
+{{< image src="P5-2-b.avif" caption="P5-2-b" width=600px >}}
+
+它们具有相同的形状和频域表示。
+
+{{< image src="P5-2-d.avif" caption="P5-2-d" width=600px >}}
+
+## 6. 过滤或隔离信号中特定频率范围的概念
+
+### 描述
+
+我们将模拟低通滤波器，并使用 Simulink 创建一个带通滤波器。之后，我们将将其应用于一首歌曲并看看它会带来什么变化。
+
+### 分析
+
+我们模拟了一个低通滤波器。红色是原始信号，黑色是经过过滤的信号。如图所示，部分信号被移除。
+
+{{< image src="P6-2-a.avif" caption="P6-2-a" width=600px >}}
+
+根据公式
+
+$$
+F_{low} = \frac{1}{2 \pi RC}
+$$
+
+我们可以得到目标低通滤波器的电容值。
+
+$$
+\begin{align*}
+300 &= \frac{1}{2 \pi \cdot 1K \cdot C} \\
+C &= 5.6 \times 10^{-7}
+\end{align*}
+$$
+
+### 模拟与讨论
+
+我们的音频文件是《玛丽有只小羊羔》。我们希望听到一个更高音调的版本。
+
+{{< image src="P6-1-a.avif" caption="P6-1-a" width=600px >}}
+
+通过应用从 800Hz 到 1700Hz 的带通滤波器，我们可以移除部分信号并听到声音。我们隔离了一个特定的、更高音调范围的声音。
+最终得到了一个更高音调版本的歌曲，并且伴随着一些高音刺耳声。更高的音调版本也更轻。
+
+{{< image src="P6-1-b.avif" caption="P6-1-b" width=600px >}}
+
+{{< image src="P6-1-c.avif" caption="P6-1-c" width=600px >}}