add ecse-1010-poc-lab02

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

@@ -0,0 +1,820 @@
+---
+title: ECSE 1010 Proof of Concepts - Omega Lab02
+subtitle:
+date: 2024-11-28T12:57:51-05:00
+lastmod: 2024-11-28T12:57:51-05:00
+slug: ecse-1010-poc-lab02
+draft: false
+author:
+  name: James
+  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.
+keywords: ["Electrical Engineering","Ohm's Law","IV curve","Nodal Analysis","Op-Amp"]
+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 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.
+resources:
+  - name: featured-image
+    src: featured-image.avif
+  - name: featured-image-preview
+    src: featured-image-preview.avif
+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="Lab02.pdf" type="application/pdf" width="100%" height="500px">
+</div>
+
+## 1. Prove That the Slope of an $IV$ Curve Corresponds with Ohm’s Law for Two Different Resistor Values
+
+### Building Block
+
+{{< image src="P1-1-a.avif" caption="P1-1-a" width=600px >}}
+
+Let's pick two resistor. The first one is
+
+{{< figure src="P1-1-b.avif" caption="P1-1-b" width=600px >}}
+
+4-Band Color Code: Orange, Orange, Brown, Gold
+
+$$
+\begin{align*}
+	33 \times (1\times10^1) = 330 \Omega \pm 5\%
+\end{align*}
+$$
+
+have a check
+
+{{< image src="P1-1-b-2.avif" caption="P1-1-b-2" width=600px >}}
+
+The second one is
+
+{{< image src="P1-1-c.avif" caption="P1-1-c" width=600px >}}
+
+4-Band Color Code: Brown, Brown, Red, Gold
+
+$$
+\begin{align*}
+	11 \times (1\times10^2) = 1100 \Omega \pm 5\%
+\end{align*}
+$$
+
+have a check
+
+{{< image src="P1-1-c-2.avif" caption="P1-1-c-2" width=600px >}}
+
+### Analysis
+
+We know that $IV$ curve means $I$ on the y-axis and $V$ on the x-axis of the plot. Then, it must be a linear function, because both $IV$ don't have powers.
+
+Using the idea of linear function, we know the slope is $\frac{\Delta X}{\Delta Y}$. Back to our case, it becomes $\frac{\Delta V}{\Delta I}$. Also, we knows the Ohm's Law, which $\frac{V}{I} = R$. So, the slope is very likely to be the resistance $R$.
+
+If we take $R_1 = 10 \Omega$,  $R_2 = 100 \Omega$ (As the simulation set). We should got.
+
+{{< image src="P1-2-a.avif" caption="P1-2-a" width=600px >}}
+
+If we plot them together, we got
+
+{{< image src="P1-2-b.avif" caption="P1-2-b" width=600px >}}
+
+Here is the data table
+
+|$I$  | $V=IR_1$  | $V=IR_2$  |
+|:----|----------:|----------:|
+| 0   | 0         | 0         |
+| 0.2 | 2         | 20        |
+| 0.4 | 4         | 40        |
+| 0.6 | 6         | 60        |
+| 0.8 | 8         | 80        |
+| 1   | 10        | 100       |
+
+### Simulation
+
+{{< image src="P1-3-a.avif" caption="P1-3-a" width=600px >}}
+
+### Measurement
+
+First we built a circuit like this
+
+{{< image src="P1-4-a.avif" caption="P1-4-a" width=600px >}}
+
+this is based on the diagram from the lab document
+
+{{< image src="P1-4-a-2.avif" caption="P1-4-a-2" width=250px >}}
+
+We only changed the $R1, R2$ values. Also, it's hard to plug multimeter on the breadboard. So, we intersect the $V+$ circuit at the front
+
+{{< image src="P1-4-b.avif" caption="P1-4-b" width=600px >}}
+
+This method is not ideal, but works.
+
+***
+
+Let's begin
+
+For $V+ = 0.5V$, we got
+
+{{< image src="P1-4-c.avif" caption="P1-4-c" width=600px >}}
+
+{{< image src="P1-4-c-2.avif" caption="P1-4-c-2" width=600px >}}
+
+To save some space and work, we just will not show each result. But here is the data
+
+|$V+$|$V(R1)$|$V(R1)$|$I$|
+|:---|:------|:------|:--|
+|$0V$|$0V$|$0V$|$0mA$|
+|$0.5V$|$0.142V$|$0.396V$|$0.3mA$|
+|$1V$|$0.238V$|$0.724V$|$0.6mA$|
+|$1.5V$|$0.358V$|$1.126V$|$1.0mA$|
+|$2V$|$0.463V$|$1.492V$|$1.3mA$|
+|$2.5V$|$0.572V$|$1.831V$|$1.6mA$|
+|$3V$|$0.632V$|$1.994V$|$1.9mA$|
+
+With this MATLAB code,
+
+```matlab
+% Step 1: Enter the data
+V_plus = [0, 0.5, 1, 1.5, 2, 2.5, 3]; % V+ values
+V_R1 = [0, 0.142, 0.238, 0.358, 0.463, 0.572, 0.632]; % V(R1) values
+V_R2 = [0, 0.396, 0.724, 1.126, 1.492, 1.831, 1.994]; % V(R2) values
+I = [0, 0.3, 0.6, 1.0, 1.3, 1.6, 1.9] * 1e-3; % I values in A (converted from mA)
+
+% Step 2: Plot the data
+figure;
+
+% Plot for Resistor R1
+subplot(2, 1, 1);
+plot(V_R1, I, '-o');
+xlabel('Voltage V(R1) (V)');
+ylabel('Current I (A)');
+title('Resistor R1: Current vs Voltage');
+grid on;
+
+% Plot for Resistor R2
+subplot(2, 1, 2);
+plot(V_R2, I, '-o');
+xlabel('Voltage V(R2) (V)');
+ylabel('Current I (A)');
+title('Resistor R2: Current vs Voltage');
+grid on;
+```
+
+we got the plot of $R1$ and $R2$
+
+{{< image src="P1-4-d.svg" caption="P1-4-d" width=600px >}}
+
+Now, let's create a fit line for both. It's needed to find out the slope ($R=V/I$). To do that, we changed the code a bit into
+
+```matlab
+% Step 1: Enter the data
+V_plus = [0, 0.5, 1, 1.5, 2, 2.5, 3]; % V+ values
+V_R1 = [0, 0.142, 0.238, 0.358, 0.463, 0.572, 0.632]; % V(R1) values
+V_R2 = [0, 0.396, 0.724, 1.126, 1.492, 1.831, 1.994]; % V(R2) values
+I = [0, 0.3, 0.6, 1.0, 1.3, 1.6, 1.9] * 1e-3; % I values in A (converted from mA)
+
+% Step 2: Fit linear regression curves
+% Fit for Resistor R1
+p_R1 = polyfit(I, V_R1, 1);
+slope_R1 = p_R1(1);
+R_R1 = slope_R1; % Resistance of R1
+
+% Fit for Resistor R2
+p_R2 = polyfit(I, V_R2, 1);
+slope_R2 = p_R2(1);
+R_R2 = slope_R2; % Resistance of R2
+
+% Step 3: Display the resistances
+fprintf('Resistance of R1: %.3f ohms\n', R_R1);
+fprintf('Resistance of R2: %.3f ohms\n', R_R2);
+
+% Step 4: Plot the data and fitted curves
+figure;
+
+% Plot for Resistor R1
+subplot(2, 1, 1);
+plot(V_R1, I, 'o');
+hold on;
+plot(polyval(p_R1, I), I, '-');
+xlabel('Voltage V(R1) (V)');
+ylabel('Current I (A)');
+title('Resistor R1: Current vs Voltage with Linear Fit');
+legend('Data', 'Linear Fit');
+grid on;
+
+% Plot for Resistor R2
+subplot(2, 1, 2);
+plot(V_R2, I, 'o');
+hold on;
+plot(polyval(p_R2, I), I, '-');
+xlabel('Voltage V(R2) (V)');
+ylabel('Current I (A)');
+title('Resistor R2: Current vs Voltage with Linear Fit');
+legend('Data', 'Linear Fit');
+grid on;
+```
+
+we got a result of
+
+```text
+Resistance of R1: 331.144 ohms
+Resistance of R2: 1069.374 ohms
+```
+
+and the plots
+
+{{< image src="P1-4-e.svg" caption="P1-4-e" width=600px >}}
+
+Check this result from multimeter's reading of resistance
+
+{{< image src="P1-4-d.avif" caption="P1-4-d" width=600px >}}
+
+{{< image src="P1-4-d-2.avif" caption="P1-4-d-2" width=600px >}}
+
+Great! The actual reading is very close to the resistances we determined from your IV measurement data and the linear regression. The average $\%$ error is less than $1\%$
+
+### Discussion
+
+We did a lot of discussion in each session instead of in one. This is just to make the document more logical and follows the flow. So, we will only summarize and add something not appear above.
+
+First, we used LTSpecie to determine $IV$ curve of two resistor $R_1 = 10\Omega$ and $R_2 = 100\Omega$. (This is just for prove our Analysis, so it doesn't match the $R_1 = 330\Omega$ and $R_2 = 1100\Omega$ we used later). And it matches our Analysis. Both the plot created by Excel and the values.
+
+Then, we built a series circuit, and we know they have the same current across all components. And, the $R$ is only related to $IV$. As long as we got some reading pairs, we can plot the curve. The result matches our expectation with less than $1\%$ error. Consider our multimeter can only measure down to $0.1 mV$. This accuracy is amazing!
+
+Thus, we proved That the Slope of an $IV$ Curve Corresponds with Ohm’s Law for Two Different Resistor Values.
+
+## 2. Prove the non linear $IV$ curve for a light emitting diode
+
+### Building Block
+
+{{< image src="P3-1-a.avif" caption="P3-1-a" width=600px >}}
+
+### Analysis
+
+To plot a $IV$ curve of a diode. We need to find out a few important data.
+
+- Forward Voltage ($V_F$)
+- Reverse Breakdown Voltage ($V_{BR}$)
+- Reverse Leakage Current ($I_S$)
+
+As the datasheet of [QED123](https/www.onsemi.com/pdf/datasheet/qed123-d.pdf) said
+
+- $V_F = 1.7V$
+- $I_F = 100 mA$
+- $V_{BR} = 5V$
+- $I_S = 10 \mu A$
+
+We just plot them into a standard diode $IV$ characteristic diagram and get
+
+{{< image src="P2-2-a.avif" caption="P2-2-a" width=600px >}}
+
+### Simulation
+
+{{< image src="P3-3-a.avif" caption="P2-3-a" width=600px >}}
+
+The turn on voltage of 1N914 is about $0.7V$
+
+### Measurement
+
+{{< image src="P3-4-a.avif" caption="P3-4-a" width=600px >}}
+
+We create a trig wave like
+
+{{< image src="P3-4-b.avif" caption="P3-4-b" width=600px >}}
+
+with amplitude to 5V (10 volts peak to peak), frequency to 200 Hz, and phase to 90 degrees.
+
+Then, we use channel 1 to find out the current using the math function in scope
+
+```js
+C1/330*1000
+```
+
+{{< image src="P3-4-b-2.avif" caption="P3-4-b-2" width=600px >}}
+
+and the $IV$ Curve
+
+{{< image src="P3-4-b-3.avif" caption="P3-4-b-3" width=600px >}}
+
+with this MATLAB Code,
+
+```matlab
+% Step 1: Import the CSV file
+data = readmatrix('P2-4-c.csv');
+
+% Step 2: Extract the columns
+voltage = data(:, 2);  % Second column is voltage (V)
+current = data(:, 1);  % Third column is current (I)
+
+
+% Step 3: Plot the I-V curve
+figure;
+plot(currentvoltage, current, 'k-', 'LineWidth', 1.5);
+xlabel('Voltage (V)');
+ylabel('Current (I)');
+title('I-V Curve');
+grid on;
+```
+
+we got
+
+{{< image src="P2-4-c-2.svg" caption="P2-4-c-2" width=600px >}}
+
+### Discussion
+
+Our experimental matches the datasheet. Consider the datasheet said 
+
+- $V_F = 1.7V$
+- $I_F = 100 mA$
+
+and we got $1.7V$ on $10 mA$ this matches the datasheet curve.
+
+## 3. Show / demonstrate that the differential resistance changes in different regions in the diode $IV$ curve
+
+### Building Block
+
+{{< image src="P3-1-a.avif" caption="P3-1-a" width=600px >}}
+
+### Analysis
+
+To plot a $IV$ curve of a diode. We need to find out a few important data.
+
+- Forward Voltage ($V_F$)
+- Reverse Breakdown Voltage ($V_{BR}$)
+- Reverse Leakage Current ($I_S$)
+
+As the datasheet of [QED123](https/www.onsemi.com/pdf/datasheet/qed123-d.pdf) said
+
+- $V_F = 1.7V$
+- $I_F = 100 mA$
+- $V_{BR} = 5V$
+- $I_S = 10 \mu A$
+
+We just plot them into a standard diode $IV$ characteristic diagram and get
+
+{{< image src="P2-2-a.avif" caption="P2-2-a" width=600px >}}
+
+### Simulation
+
+{{< image src="P3-3-a.avif" caption="P3-3-a" width=600px >}}
+
+The turn on voltage of 1N914 is about $0.7V$
+
+### Measurement
+
+{{< image src="P3-4-a.avif" caption="P3-4-a" width=600px >}}
+
+We create a trig wave like
+
+{{< image src="P3-4-b.avif" caption="P3-4-b" width=600px >}}
+
+with amplitude to 5V (10 volts peak to peak), frequency to 200 Hz, and phase to 90 degrees.
+
+Then, we use channel 1 to find out the current using the math function in scope
+
+```js
+C1/330*1000
+```
+
+{{< image src="P3-4-b-2.avif" caption="P3-4-b-2" width=600px >}}
+
+and the $IV$ Curve
+
+{{< image src="P3-4-b-3.avif" caption="P3-4-b-3" width=600px >}}
+
+with this MATLAB Code,
+
+```matlab
+% Step 1: Import the CSV file
+data = readmatrix('P2-4-c.csv');
+
+% Step 2: Extract the columns
+voltage = data(:, 2);  % Second column is voltage (V)
+current = data(:, 1);  % Third column is current (I)
+
+
+% Step 3: Plot the I-V curve
+figure;
+plot(voltage, current, 'k-', 'LineWidth', 1.5);
+xlabel('Voltage (V)');
+ylabel('Current (I)');
+title('I-V Curve');
+grid on;
+```
+
+we got
+
+{{< image src="P2-4-c-2.svg" caption="P2-4-c-2" width=600px >}}
+
+### Discussion
+
+To find out at least 2 locations on the curve to show that the differential resistance changes along the I-V characteristic. We modified the code a bit to let it find out 2 random point on the plot and its slope.
+
+```matlab
+% Step 1: Import the CSV file
+data = readmatrix('P2-4-c.csv');
+
+% Step 2: Extract the columns
+voltage = data(:, 2);  % Second column is voltage (V)
+current = data(:, 1);  % Third column is current (I)
+
+% Step 3: Select two random points
+num_points = length(current);
+random_indices = randperm(num_points, 2);  % Generate 2 unique random indices
+
+% Step 4: Extract the voltage and current values for the selected points
+V1 = voltage(random_indices(1));
+V2 = voltage(random_indices(2));
+I1 = current(random_indices(1));
+I2 = current(random_indices(2));
+
+% Step 5: Calculate the slopes
+slope1 = (V2 - V1) / (I2 - I1);
+slope2 = (V1 - V2) / (I1 - I2);  % This is the same as slope1 but calculated in reverse
+
+% Step 6: Print the slopes
+fprintf('The slope between the randomly selected points (I1 = %.4f, V1 = %.4f) and (I2 = %.4f, V2 = %.4f) is: %.4f\n', I1, V1, I2, V2, slope1);
+fprintf('The slope between the randomly selected points (I2 = %.4f, V2 = %.4f) and (I1 = %.4f, V1 = %.4f) is: %.4f\n', I2, V2, I1, V1, slope2);
+```
+
+We got
+
+> The slope between the randomly selected points (I1 = 0.0097, V1 = 0.2959) and (I2 = 0.0036, V2 = -2.6254) is: 479.8789
+>
+>The slope between the randomly selected points (I2 = 7.9784, V2 = 1.2568) and (I1 = 2.8170, V1 = 1.1975) is: 0.0115
+
+We can see they are very different.
+
+## 4. Prove That Nodal Analysis Solves Unknown Nodal Voltages in a Circuit
+
+### Building Block
+
+{{< image src="P4-1-a.avif" caption="P4-1-a" width=600px >}}
+
+### Analysis
+
+{{< image src="P4-2-a.avif" caption="P4-2-a" width=600px >}}
+
+To make our life easier, I rewrite some equation in $\LaTeX$.
+
+Current through a resistor:
+
+$$
+I_R = \frac{V_A - V_B}{R}
+$$
+
+Kirchhoff's Current Law (KCL) at node B:
+
+$$
+I_{R_1} + I_{R_2} + I_{R_3} = 0
+$$
+
+KCL at node C:
+
+$$
+I_{R_3} + I_{R_4} = 0
+$$
+
+Expressing currents in terms of voltages. From the first equation:
+
+$$
+\frac{V_B - V_A}{R_1} + \frac{V_B}{R_2} + \frac{V_B - V_C}{R_3} = 0
+$$
+
+From the second equation:
+
+$$
+\frac{V_C - V_B}{R_3} + \frac{V_C - V_D}{R_4} = 0
+$$
+
+Substituting known values. Given $V_A = 5$ and $V_D = 0$, the equations become:
+
+$$
+2.5V_B - V_C = 5 \\\
+2V_C - V_B = 0
+$$
+
+Matrix form: $\begin{bmatrix} 2.5 & -1 \\\ -1 & 2 \end{bmatrix}$ $\begin{bmatrix} V_B \\\ V_C \end{bmatrix} =$ $\begin{bmatrix} 5 \\\ 0 \end{bmatrix}$
+
+Solve them "by hand"
+
+```matlab
+% Define the matrix A and the vector b
+A = [2.5, -1; -1, 2];
+b = [5; 0];
+
+% Solve the system of linear equations A * x = b
+x = A \ b;
+
+% Display the solution
+disp('The solution is:');
+disp(x);
+```
+
+we got
+
+```text
+The solution is:
+    2.5000
+    1.2500
+```
+
+Thus, $\begin{bmatrix} V_B \\\ V_C \end{bmatrix} =$ $\begin{bmatrix}2.5 \\\ 1.25 \end{bmatrix}$
+
+### Simulation
+
+{{< image src="P4-3-a.avif" caption="P4-3-a" width=600px >}}
+
+### Measurement
+
+{{< image src="P4-4-a.avif" caption="P4-4-a" width=600px >}}
+
+For $V_C$, we got
+
+{{< image src="P4-4-b-1.avif" caption="P4-4-b-1" width=600px >}}
+
+For $V_B$, we got
+
+{{< image src="P4-4-b-2.avif" caption="P4-4-b-2" width=600px >}}
+
+### Discussion
+
+|Node|Analysis|Simulation|Experimental|diff|%diff|
+|:-:|:-:|:-:|:-:|:-:|:-|
+|$V_B$|$2.50V$|$2.50V$|$2.45V$|$5mV$|$2\%$|
+|$V_C$|$1.25V$|$1.25V$|$1.22V$|$3mV$|$2.4\%$|
+
+Our Analysis matches the Simulation. The Experimental data has less than $2.5\%$ error than expect, which is very less. Thus, we proved That Nodal Analysis Solves Unknown Nodal Voltages in a Circuit.
+
+## 5. Prove / demonstrate your approach to designing a circuit using nodal analysis
+
+### Building Block
+
+{{< image src="P5-3-a.avif" caption="P5-3-a" width=600px >}}
+
+### Analysis
+
+{{< image src="P5-2-a.avif" caption="P5-2-a" width=600px >}}
+
+To make our life easier, I rewrite some equation in $\LaTeX$.
+
+Given values:
+
+- $V_A = 3 \, \text{V}$
+- $V_C = 0 \, \text{V}$
+- $V_B$ is unknown.
+
+Using Kirchhoff's Current Law (KCL) at node B:
+
+$$
+\frac{V_B - V_A}{R_1} + \frac{V_B - V_C}{R_2} + \frac{V_B - V_C}{R_3} = 0
+$$
+
+Substituting the given values and resistances:
+
+$$
+\frac{V_B - 3}{1} + \frac{V_B - 0}{4} + \frac{V_B - 0}{4} = 0
+$$
+
+Simplifying the equation:
+
+$$
+(V_B - 3) + \frac{V_B}{4} + \frac{V_B}{4} = 0
+$$
+
+Combine terms:
+
+$$
+V_B - 3 + \frac{V_B}{2} = 0
+$$
+
+Multiply through by 2 to clear the fraction:
+
+$$
+2V_B - 6 + V_B = 0
+$$
+
+Combine terms:
+
+$$
+3V_B = 6
+$$
+
+Solve for $V_B$:
+
+$$
+V_B = 2
+$$
+
+### Simulation
+
+{{< image src="P5-3-a.avif" caption="P5-3-a" width=600px >}}
+
+### Measurement
+
+{{< image src="P5-4-a-1.avif" caption="P5-4-a-1" width=600px >}}
+
+For $V_B$, we got
+
+{{< image src="P5-4-a.avif" caption="P5-4-a" width=600px >}}
+
+### Discussion
+
+|Node|Analysis|Simulation|Experimental|diff|%diff|
+|:-:|:-:|:-:|:-:|:-:|:-|
+|$V_B$|$2V$|$2V$|$1.979V$|$21mV$|$1.1\%$|
+
+Our Analysis matches the Simulation. The Experimental data has less than $1.2\%$ error than expect, which is very less. Thus, we proved That Nodal Analysis Solves Unknown Nodal Voltages in a Circuit.
+
+## 6. Prove the function of an op amp comparator
+
+### Building Block
+
+{{< image src="P6-1-a.avif" caption="P6-1-a" width=600px >}}
+
+### Analysis
+A non-inverted comparator has a transfer function of
+
+$$
+\begin{equation*}
+V_{out}=\begin{cases}
+          \text{if} \; V_{in} < V_{ref}, V_{out} = V_s - \\\
+		  \text{if} \; V_{in} > V_{ref}, V_{out} = V_s + \\\
+     \end{cases}
+  \end{equation*}
+$$
+
+In our case, we got
+
+$$
+\begin{equation*}
+V_{out}=\begin{cases}
+          \text{if} \; V_{in} < 0V, V_{out} = -5V \\\
+		  \text{if} \; V_{in} > 0V, V_{out} = 5V \\\
+     \end{cases}
+  \end{equation*}
+$$
+
+Our supply voltage are $5V$ and $-5V$, and the input is a SINE wave with amplitude of $1V$, and the reference voltage is $GND$ which is $0V$
+
+### Simulation
+
+{{< image src="P6-3-b.avif" caption="P6-3-b" width=600px >}}
+
+{{< image src="P6-3-a.avif" caption="P6-3-a" width=600px >}}
+
+### Measurement
+
+{{< image src="P6-4-a-b.avif" caption="P6-4-a-b" width=600px >}}
+
+{{< image src="P6-4-a.avif" caption="P6-4-a" width=600px >}}
+
+### Discussion
+
+Comparing our simulation to our experiment, we see that both of them are square waves with the same periods and similar amplitudes. They are fluctuating between 5 and -5, which are our supply voltages. This makes sense, because the supply voltages are the outputs of op amp comparators.
+
+This proves our concept of an op amp comparator.
+
+## 7. Prove the function of a mathematical op amp
+
+### Building Block
+
+{{< image src="P8-1-a.avif" caption="P8-1-a" width=600px >}}
+
+### Analysis
+
+{{< image src="P8-2-a.avif" caption="P8-2-a" width=600px >}}
+
+Summing amplifier circuit has a transfer function like
+
+$$
+V_{out} = - \frac{Rf}{R1} \cdot V1 - \frac{Rf}{R2} \cdot V2
+$$
+
+In our case, we want to use $50K \Omega$ potentiometer as the resistors, so it can be adjusted according to our demand. Then, we got
+
+$$
+\begin{align*}
+	V_{out} &= - \frac{\cancel{50K}}{\cancel{50K}} \cdot V1 - \frac{\cancel{\cancel{50K}}}{\cancel{50K}} \cdot V2 \\\
+	V_{out} &= - V1 - V2 \\\
+\end{align*}
+$$
+
+### Simulation
+
+We just used two SINE waves with different frequencies ($500 \; \text{Hz}$ and $1K \; \text{Hz}$) in our simulation.
+
+{{< image src="P8-3-a.avif" caption="P8-3-a" width=600px >}}
+
+{{< image src="P5-3-b.avif" caption="P5-3-b" width=600px >}}
+
+### Measurement
+
+Then, we setup the circuit. We just connected scope channel 1 to the $V_{out}$ to check it works or not
+
+{{< image src="P8-4-a-b.avif" caption="P8-4-a-b" width=600px >}}
+
+We have  $V_s + = 5V$ and $V_s - = -5V$
+
+{{< image src="P8-4-a.avif" caption="P8-4-a" width=600px >}}
+
+We used wave generator to create to SINE waves of $500 \; \text{Hz}$ and $1K \; \text{Hz}$
+
+{{< image src="P8-4-b.avif" caption="P8-4-b" width=600px >}}
+
+And we checked the output wave using scope channel 1+
+
+{{< image src="P8-4-c.avif" caption="P8-4-c" width=600px >}}
+
+### Discussion
+
+As we see, the shape of the output wave is exactly the same as our simulation. Both the amplitude of the output wave in the simulation and measurement is around $1.75V$, and the period is the same.
+
+Since the shape and all features of our experimental wave matches our simulation, we know this op-amp works in different voltage ranges.
+
+This proved our concept of summer amp which is a mathematical op-amp.
+
+## 8. Prove the concept of transfer functions of Two-Channel Audio Mixer
+
+### Building Block
+
+{{< image src="P8-1-a.avif" caption="P8-1-a" width=600px >}}
+
+### Analysis
+
+{{< image src="P8-2-a.avif" caption="P8-2-a" width=600px >}}
+
+Summing amplifier circuit has a transfer function like
+
+$$
+V_{out} = - \frac{Rf}{R1} \cdot V1 - \frac{Rf}{R2} \cdot V2
+$$
+
+In our case, we want to use $50K \Omega$ potentiometer as the resistors, so it can be adjusted according to our demand. Then, we got
+
+$$
+\begin{align*}
+	V_{out} &= - \frac{\cancel{50K}}{\cancel{50K}} \cdot V1 - \frac{\cancel{\cancel{50K}}}{\cancel{50K}} \cdot V2 \\\
+	V_{out} &= - V1 - V2 \\\
+\end{align*}
+$$
+
+### Simulation
+
+We just used to SINE wave with different frequency ($500 \; \text{Hz}$ and $1K \; \text{Hz}$) to test what we expect.
+
+{{< image src="P8-3-a.avif" caption="P8-3-a" width=600px >}}
+
+{{< image src="P5-3-b.avif" caption="P5-3-b" width=600px >}}
+
+### Measurement
+
+Then, we setup the circuit. We just connect scope channel 1 to the $V_{out}$ to check it works or not
+
+{{< image src="P8-4-a-b.avif" caption="P8-4-a-b" width=600px >}}
+
+We supply $V_s + = 5V$ and $V_s - = -5V$
+
+{{< image src="P8-4-a.avif" caption="P8-4-a" width=600px >}}
+
+And use wave generator to create to SINE wave of $500 \; \text{Hz}$ and $1K \; \text{Hz}$
+
+{{< image src="P8-4-b.avif" caption="P8-4-b" width=600px >}}
+
+And we checked the output wave using scope channel 1+
+
+{{< image src="P8-4-c.avif" caption="P8-4-c" width=600px >}}
+
+### Discussion
+
+As we see, the shape of the output wave is exactly the same as what we simulated. Both the amplitude of the output wave in the simulation and measurement is around $1.75V$, and the period is the same.
+
+This proved our concept of summer amp.

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

@@ -0,0 +1,816 @@
+---
+title: ECSE 1010 概念验证 - Omega Lab02
+subtitle:
+date: 2024-11-28T12:57:51-05:00
+lastmod: 2024-11-28T12:57:51-05:00
+slug: ecse-1010-poc-lab02
+draft: false
+author:
+  name: James
+  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.
+keywords: ["Electrical Engineering","Ohm's Law","IV curve","Nodal Analysis","Op-Amp"]
+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 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.
+resources:
+  - name: featured-image
+    src: featured-image.avif
+  - name: featured-image-preview
+    src: featured-image-preview.avif
+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="Lab02.pdf" type="application/pdf" width="100%" height="500px">
+</div>
+
+## 1. 证明不同阻值的两个电阻 IV 曲线斜率等于欧姆定律中的电阻值
+
+### 构建模块
+
+{{< image src="P1-1-a.avif" caption="P1-1-a" width=600px >}}
+
+让我们选择两个电阻。第一个是：
+
+{{< figure src="P1-1-b.avif" caption="P1-1-b" width=600px >}}
+
+四色环代码：橙、橙、棕、金
+
+$$
+\begin{align*}
+	33 \times (1\times10^1) = 330 \Omega \pm 5\%
+\end{align*}
+$$
+
+检查一下：
+
+{{< image src="P1-1-b-2.avif" caption="P1-1-b-2" width=600px >}}
+
+第二个是：
+
+{{< image src="P1-1-c.avif" caption="P1-1-c" width=600px >}}
+
+四色环代码：棕、棕、红、金
+
+$$
+\begin{align*}
+	11 \times (1\times10^2) = 1100 \Omega \pm 5\%
+\end{align*}
+$$
+
+检查一下：
+
+{{< image src="P1-1-c-2.avif" caption="P1-1-c-2" width=600px >}}
+
+### 分析
+
+我们知道 IV 曲线表示 y 轴为 I，x 轴为 V。因此它必须是线性函数，因为 IV 没有幂次。
+
+使用线性函数的思想，我们可知斜率是 $\frac{\Delta X}{\Delta Y}$。回到我们的案例中，就变成了 $\frac{\Delta V}{\Delta I}$。另外我们知道欧姆定律，即 $\frac{V}{I} = R$。因此，斜率很可能是电阻 $R$。
+
+如果我们取 $R_1 = 10 \Omega$，$R_2 = 100 \Omega$（如仿真设置）。我们应得到：
+
+{{< image src="P1-2-a.avif" caption="P1-2-a" width=600px >}}
+
+如果我们将它们一起绘制，则得到
+
+{{< image src="P1-2-b.avif" caption="P1-2-b" width=600px >}}
+
+这里是数据表：
+
+|$I$  | $V = IR_1$  | $V = IR_2$  |
+|:----|----------:|----------:|
+| 0   | 0         | 0         |
+| 0.2 | 2         | 20        |
+| 0.4 | 4         | 40        |
+| 0.6 | 6         | 60        |
+| 0.8 | 8         | 80        |
+| 1   | 10        | 100       |
+
+### 模拟
+
+{{< image src="P1-3-a.avif" caption="P1-3-a" width=600px >}}
+
+### 测量
+
+首先我们构建了一个这样的电路：
+
+{{< image src="P1-4-a.avif" caption="P1-4-a" width=600px >}}
+
+这是基于实验手册中的图示。
+
+{{< image src="P1-4-a-2.avif" caption="P1-4-a-2" width=250px >}}
+
+我们只改变了 $R1, R2$ 的值。另外，很难在面包板上插表。因此我们在前面交叉了 V+ 电路
+
+{{< image src="P1-4-b.avif" caption="P1-4-b" width=600px >}}
+
+这种方法不是理想的选择，但可以工作。
+
+***
+
+让我们开始吧：
+
+对于 $V+ = 0.5V$，我们得到：
+
+{{< image src="P1-4-c.avif" caption="P1-4-c" width=600px >}}
+
+{{< image src="P1-4-c-2.avif" caption="P1-4-c-2" width=600px >}}
+
+为了节省空间和工作量，我们不会展示每个结果。但这里是数据：
+
+|$V+$|$V(R1)$|$V(R1)$|$I$|
+|:---|:------|:------|:--|
+|$0V$|$0V$|$0V$|$0mA$|
+|$0.5V$|$0.142V$|$0.396V$|$0.3mA$|
+|$1V$|$0.238V$|$0.724V$|$0.6mA$|
+|$1.5V$|$0.358V$|$1.126V$|$1.0mA$|
+|$2V$|$0.463V$|$1.492V$|$1.3mA$|
+|$2.5V$|$0.572V$|$1.831V$|$1.6mA$|
+|$3V$|$0.632V$|$1.994V$|$1.9mA$|
+
+使用以下 MATLAB 代码：
+
+```matlab
+% 步骤 1：输入数据
+V_plus = [0, 0.5, 1, 1.5, 2, 2.5, 3]; % V+ 值
+V_R1 = [0, 0.142, 0.238, 0.358, 0.463, 0.572, 0.632]; % V(R1) 值
+V_R2 = [0, 0.396, 0.724, 1.126, 1.492, 1.831, 1.994]; % V(R2) 值
+I = [0, 0.3, 0.6, 1.0, 1.3, 1.6, 1.9] * 1e-3; % I 值（A，转换为 mA）
+
+% 步骤 2：绘制数据
+figure;
+
+% 绘制电阻 R1 的曲线
+subplot(2, 1, 1);
+plot(V_R1, I, '-o');
+xlabel('电压 V(R1) (V)');
+ylabel('电流 I (A)');
+title('电阻 R1: 电流与电压的关系图');
+grid on;
+
+% 绘制电阻 R2 的曲线
+subplot(2, 1, 2);
+plot(V_R2, I, '-o');
+xlabel('电压 V(R2) (V)');
+ylabel('电流 I (A)');
+title('电阻 R2: 电流与电压的关系图');
+grid on;
+```
+
+我们得到了 $R1$ 和 $R2$ 的曲线：
+
+{{< image src="P1-4-d.svg" caption="P1-4-d" width=600px >}}
+
+现在，让我们为两者创建拟合线。需要找到斜率（$R = V/I$）。为此，我们稍微修改了代码如下：
+
+```matlab
+% 步骤 1：输入数据
+V_plus = [0, 0.5, 1, 1.5, 2, 2.5, 3]; % V+ 值
+V_R1 = [0, 0.142, 0.238, 0.358, 0.463, 0.572, 0.632]; % V(R1) 值
+V_R2 = [0, 0.396, 0.724, 1.126, 1.492, 1.831, 1.994]; % V(R2) 值
+I = [0, 0.3, 0.6, 1.0, 1.3, 1.6, 1.9] * 1e-3; % I 值（A，转换为 mA）
+
+% 步骤 2：拟合线性回归曲线
+% 拟合电阻 R1 的曲线
+p_R1 = polyfit(I, V_R1, 1);
+slope_R1 = p_R1(1);
+R_R1 = slope_R1; % 电阻 R1
+
+% 拟合电阻 R2 的曲线
+p_R2 = polyfit(I, V_R2, 1);
+slope_R2 = p_R2(1);
+R_R2 = slope_R2; % 电阻 R2
+
+% 步骤 3：显示电阻值
+fprintf('电阻 R1: %.3f ohms\n', R_R1);
+fprintf('电阻 R2: %.3f ohms\n', R_R2);
+
+% 步骤 4：绘制数据和拟合曲线
+figure;
+
+% 绘制电阻 R1 的曲线
+subplot(2, 1, 1);
+plot(V_R1, I, 'o');
+hold on;
+plot(polyval(p_R1, I), I, '-');
+xlabel('电压 V(R1) (V)');
+ylabel('电流 I (A)');
+title('电阻 R1: 电流与电压的关系图（带线性拟合）');
+legend('数据', '线性拟合');
+grid on;
+
+% 绘制电阻 R2 的曲线
+subplot(2, 1, 2);
+plot(V_R2, I, 'o');
+hold on;
+plot(polyval(p_R2, I), I, '-');
+xlabel('电压 V(R2) (V)');
+ylabel('电流 I (A)');
+title('电阻 R2: 电流与电压的关系图（带线性拟合）');
+legend('数据', '线性拟合');
+grid on;
+```
+
+我们得到了结果：
+
+```text
+电阻 R1: 331.144 ohms
+电阻 R2: 1069.374 ohms
+```
+
+以及曲线图：
+
+{{< image src="P1-4-e.svg" caption="P1-4-e" width=600px >}}
+
+检查这个结果，从万用表的读数来看
+
+{{< image src="P1-4-d.avif" caption="P1-4-d" width=600px >}}
+
+{{< image src="P1-4-d-2.avif" caption="P1-4-d-2" width=600px >}}
+
+太好了！实际读数非常接近我们从 IV 测量数据和线性回归得出的电阻值。平均误差小于 1%。
+
+### 讨论
+
+我们在每次会话中进行了大量的讨论，而不是一次完成所有内容，这使得文档更加逻辑化并遵循流程。因此，我们将只总结未出现的内容。
+
+首先，我们使用 LTSpecie 确定了两个电阻 $R_1 = 10\Omega$ 和 $R_2 = 100\Omega$ 的 IV 曲线（这只是为了证明我们的分析）。然后，我们构建了一个串联电路，并知道所有组件的电流相同。只要我们得到一些读数对，就可以绘制曲线图。结果与预期一致，误差小于 1%。
+
+因此，我们证明了不同阻值的两个电阻 IV 曲线斜率等于欧姆定律中的电阻值。
+
+## 2. 证明发光二极管的非线性 IV 曲线
+
+### 构建模块
+
+{{< image src="P3-1-a.avif" caption="P3-1-a" width=600px >}}
+
+### 分析
+
+为了绘制一个二极管的 IV 曲线，我们需要找到一些重要的数据。
+
+- 正向电压（$V_F$）
+- 反向击穿电压（$V_{BR}$）
+- 反向漏电流（$I_S$）
+
+根据 [QED123](https/www.onsemi.com/pdf/datasheet/qed123-d.pdf) 的数据表：
+
+- $V_F = 1.7V$
+- $I_F = 100 mA$
+- $V_{BR} = 5V$
+- $I_S = 10 \mu A$
+
+我们将其绘制到标准二极管 IV 特性图中，得到
+
+{{< image src="P2-2-a.avif" caption="P2-2-a" width=600px >}}
+
+### 模拟
+
+{{< image src="P3-3-a.avif" caption="P2-3-a" width=600px >}}
+
+1N914 的开启电压约为 $0.7V$
+
+### 测量
+
+{{< image src="P3-4-a.avif" caption="P3-4-a" width=600px >}}
+
+我们创建了一个三角波，如图所示：
+
+{{< image src="P3-4-b.avif" caption="P3-4-b" width=600px >}}
+
+幅度为 5V（10V 峰峰值），频率为 200 Hz，相位为 90 度。
+
+然后我们使用通道 1 来测量电流
+
+```js
+C1/330*1000
+```
+
+{{< image src="P3-4-b-2.avif" caption="P3-4-b-2" width=600px >}}
+
+以及 IV 曲线：
+
+{{< image src="P3-4-b-3.avif" caption="P3-4-b-3" width=600px >}}
+
+使用以下 MATLAB 代码，我们得到
+
+```matlab
+% 步骤 1：导入 CSV 文件
+data = readmatrix('P2-4-c.csv');
+
+% 步骤 2：提取列
+voltage = data(:, 2); % 第二列为电压（V）
+current = data(:, 1); % 第三列为电流（I）
+
+% 步骤 3：绘制 I-V 曲线
+figure;
+plot(voltage, current, 'k-', 'LineWidth', 1.5);
+xlabel('电压 (V)');
+ylabel('电流 (I)');
+title('IV 曲线');
+grid on;
+```
+
+我们得到
+
+{{< image src="P2-4-c-2.svg" caption="P2-4-c-2" width=600px >}}
+
+### 讨论
+
+我们的实验结果与数据表一致。考虑到数据表中：
+
+- $V_F = 1.7V$
+- $I_F = 100 mA$
+
+我们得到的 $1.7V$ 对应于 $10mA$，这符合数据表曲线。
+
+## 3. 显示/证明二极管 IV 曲线不同区域中的微分电阻变化
+
+### 构建模块
+
+{{< image src="P3-1-a.avif" caption="P3-1-a" width=600px >}}
+
+### 分析
+
+为了绘制一个二极管的 IV 曲线，我们需要找到一些重要的数据。
+
+- 正向电压（$V_F$）
+- 反向击穿电压（$V_{BR}$）
+- 反向漏电流（$I_S$）
+
+根据 [QED123](https/www.onsemi.com/pdf/datasheet/qed123-d.pdf) 的数据表：
+
+- $V_F = 1.7V$
+- $I_F = 100 mA$
+- $V_{BR} = 5V$
+- $I_S = 10 \mu A$
+
+我们将其绘制到标准二极管 IV 特性图中，得到
+
+{{< image src="P2-2-a.avif" caption="P2-2-a" width=600px >}}
+
+### 模拟
+
+{{< image src="P3-3-a.avif" caption="P3-3-a" width=600px >}}
+
+1N914 的开启电压约为 $0.7V$
+
+### 测量
+
+{{< image src="P3-4-a.avif" caption="P3-4-a" width=600px >}}
+
+我们创建了一个三角波，如图所示：
+
+{{< image src="P3-4-b.avif" caption="P3-4-b" width=600px >}}
+
+幅度为 5V（10V 峰峰值），频率为 200 Hz，相位为 90 度。
+
+然后我们使用通道 1 来测量电流
+
+```js
+C1/330*1000
+```
+
+{{< image src="P3-4-b-2.avif" caption="P3-4-b-2" width=600px >}}
+
+以及 IV 曲线：
+
+{{< image src="P3-4-b-3.avif" caption="P3-4-b-3" width=600px >}}
+
+使用以下 MATLAB 代码，我们得到
+
+```matlab
+% 步骤 1：导入 CSV 文件
+data = readmatrix('P2-4-c.csv');
+
+% 步骤 2：提取列
+voltage = data(:, 2); % 第二列为电压（V）
+current = data(:, 1); % 第三列为电流（I）
+
+% 步骤 3：绘制 I-V 曲线
+figure;
+plot(voltage, current, 'k-', 'LineWidth', 1.5);
+xlabel('电压 (V)');
+ylabel('电流 (I)');
+title('IV 曲线');
+grid on;
+```
+
+我们得到
+
+{{< image src="P2-4-c-2.svg" caption="P2-4-c-2" width=600px >}}
+
+### 讨论
+
+为了展示二极管 IV 曲线不同区域中的微分电阻变化，我们将代码稍微修改了一下以计算两个随机点的斜率。
+
+```matlab
+% 步骤 1：导入 CSV 文件
+data = readmatrix('P2-4-c.csv');
+
+% 步骤 2：提取列
+voltage = data(:, 2); % 第二列为电压（V）
+current = data(:, 1); % 第三列为电流（I）
+
+% 步骤 3：选择两个随机点
+num_points = length(current);
+random_indices = randperm(num_points, 2); % 生成两个不同的随机索引
+
+% 步骤 4：提取所选点的电压和电流值
+V1 = voltage(random_indices(1));
+V2 = voltage(random_indices(2));
+I1 = current(random_indices(1));
+I2 = current(random_indices(2));
+
+% 步骤 5：计算斜率
+slope1 = (V2 - V1) / (I2 - I1);
+slope2 = (V1 - V2) / (I1 - I2); % 这与 slope1 相同，但反向计算
+
+% 步骤 6：打印斜率
+fprintf('在随机选择的点（I1 = %.4f, V1 = %.4f）和（I2 = %.4f, V2 = %.4f）之间的斜率为: %.4f\n', I1, V1, I2, V2, slope1);
+fprintf('在随机选择的点（I2 = %.4f, V2 = %.4f）和（I1 = %.4f, V1 = %.4f）之间的斜率为: %.4f\n', I2, V2, I1, V1, slope2);
+```
+
+我们得到
+
+> 在随机选择的点（I1 = 0.0097, V1 = 0.2959）和（I2 = 0.0036, V2 = -2.6254）之间的斜率为: 479.8789
+>
+> 在随机选择的点（I2 = 7.9784, V2 = 1.2568）和（I1 = 2.8170, V1 = 1.1975）之间的斜率为: 0.0115
+
+可以看出它们非常不同。
+
+## 4. 证明节点分析能确定电路中未知节点的电压
+
+### 构建模块
+
+{{< image src="P4-1-a.avif" caption="P4-1-a" width=600px >}}
+
+### 分析
+
+{{< image src="P4-2-a.avif" caption="P4-2-a" width=600px >}}
+
+为了简化我们的生活，我将一些方程重写为 $\LaTeX$。
+
+电阻中的电流：
+
+$$
+I_R = \frac{V_A - V_B}{R}
+$$
+
+节点 B 的基尔霍夫电流定律（KCL）：
+
+$$
+I_{R_1} + I_{R_2} + I_{R_3} = 0
+$$
+
+节点 C 的 KCL：
+
+$$
+I_{R_3} + I_{R_4} = 0
+$$
+
+用电压表示电流。从第一个方程：
+
+$$
+\frac{V_B - V_A}{R_1} + \frac{V_B}{R_2} + \frac{V_B - V_C}{R_3} = 0
+$$
+
+从第二个方程：
+
+$$
+\frac{V_C - V_B}{R_3} + \frac{V_C - V_D}{R_4} = 0
+$$
+
+代入已知值。给定 $V_A = 5$ 和 $V_D = 0$，方程变为：
+
+$$
+2.5V_B - V_C = 5 \\
+2V_C - V_B = 0
+$$
+
+矩阵形式表示为：$\begin{bmatrix} 2.5 & -1 \\ -1 & 2 \end{bmatrix} \cdot \begin{bmatrix} V_B \\ V_C \end{bmatrix} = \begin{bmatrix} 5 \\ 0 \end{bmatrix}$
+
+手动求解：
+
+```matlab
+% 定义矩阵 A 和向量 b
+A = [2.5, -1; -1, 2];
+b = [5; 0];
+
+% 解线性方程组 A * x = b
+x = A \ b;
+
+% 显示结果
+disp('解为：');
+disp(x);
+```
+
+我们得到
+
+```text
+解为：
+    2.5000
+    1.2500
+```
+
+因此，$\begin{bmatrix} V_B \\ V_C \end{bmatrix} =$ $\begin{bmatrix}2.5 \\ 1.25 \end{bmatrix}$
+
+### 模拟
+
+{{< image src="P4-3-a.avif" caption="P4-3-a" width=600px >}}
+
+### 测量
+
+{{< image src="P4-4-a.avif" caption="P4-4-a" width=600px >}}
+
+对于 $V_C$，我们得到：
+
+{{< image src="P4-4-b-1.avif" caption="P4-4-b-1" width=600px >}}
+
+对于 $V_B$，我们得到：
+
+{{< image src="P4-4-b-2.avif" caption="P4-4-b-2" width=600px >}}
+
+### 讨论
+
+| 节点 | 分析 | 模拟 | 实验测量 | 差值 | 百分比误差 |
+|:---:|:---:|:---:|:---:|:---:|:---:|
+|$V_B$|$2.50V$|$2.50V$|$2.45V$|$5mV$|$2\%$|
+|$V_C$|$1.25V$|$1.25V$|$1.22V$|$3mV$|$2.4\%$|
+
+我们的分析与模拟一致。实验数据的误差小于 2.5%，这非常小。因此，我们证明了节点分析能确定电路中未知节点的电压。
+
+## 5. 证明/演示使用节点分析设计电路的方法
+
+### 构建模块
+
+{{< image src="P5-3-a.avif" caption="P5-3-a" width=600px >}}
+
+### 分析
+
+{{< image src="P5-2-a.avif" caption="P5-2-a" width=600px >}}
+
+为了简化我们的生活，我将一些方程重写为 $\LaTeX$。
+
+给定值：
+
+- $V_A = 3 \, \text{V}$
+- $V_C = 0 \, \text{V}$
+- $V_B$ 是未知的。
+
+使用节点 B 的基尔霍夫电流定律（KCL）：
+
+$$
+\frac{V_B - V_A}{R_1} + \frac{V_B - V_C}{R_2} + \frac{V_B - V_C}{R_3} = 0
+$$
+
+代入给定值和电阻：
+
+$$
+\frac{V_B - 3}{1} + \frac{V_B - 0}{4} + \frac{V_B - 0}{4} = 0
+$$
+
+简化方程：
+
+$$
+(V_B - 3) + \frac{V_B}{4} + \frac{V_B}{4} = 0
+$$
+
+合并项：
+
+$$
+V_B - 3 + \frac{V_B}{2} = 0
+$$
+
+乘以 2 清除分数：
+
+$$
+2V_B - 6 + V_B = 0
+$$
+
+合并项：
+
+$$
+3V_B = 6
+$$
+
+解得 $V_B$：
+
+$$
+V_B = 2
+$$
+
+### 模拟
+
+{{< image src="P5-3-a.avif" caption="P5-3-a" width=600px >}}
+
+### 测量
+
+{{< image src="P5-4-a-1.avif" caption="P5-4-a-1" width=600px >}}
+
+对于 $V_B$，我们得到：
+
+{{< image src="P5-4-a.avif" caption="P5-4-a" width=600px >}}
+
+### 讨论
+
+| 节点 | 分析 | 模拟 | 实验测量 | 差值 | 百分比误差 |
+|:---:|:---:|:---:|:---:|:---:|:---:|
+|$V_B$|$2V$|$2V$|$1.979V$|$21mV$|$1.1\%$|
+
+我们的分析与模拟一致。实验数据的误差小于 1.2%，这非常小。因此，我们证明了节点分析能确定电路中未知节点的电压。
+
+## 6. 证明运算放大器比较器的功能
+
+### 构建模块
+
+{{< image src="P6-1-a.avif" caption="P6-1-a" width=600px >}}
+
+### 分析
+一个非反相比较器的传递函数为：
+
+$$
+\begin{equation*}
+V_{out}=\begin{cases}
+          \text{如果} \; V_{in} < V_{ref}, V_{out} = -5V \\
+		  \text{如果} \; V_{in} > V_{ref}, V_{out} = 5V \\
+     \end{cases}
+  \end{equation*}
+$$
+
+在我们的情况下，我们得到：
+
+$$
+\begin{equation*}
+V_{out}=\begin{cases}
+          \text{如果} \; V_{in} < 0V, V_{out} = -5V \\
+		  \text{如果} \; V_{in} > 0V, V_{out} = 5V \\
+     \end{cases}
+  \end{equation*}
+$$
+
+我们的电源电压为 $5V$ 和 $-5V$，输入信号是幅度为 $1V$ 的正弦波，并且参考电压为 GND（即 $0V$）。
+
+### 模拟
+
+{{< image src="P6-3-b.avif" caption="P6-3-b" width=600px >}}
+
+{{< image src="P6-3-a.avif" caption="P6-3-a" width=600px >}}
+
+### 测量
+
+{{< image src="P6-4-a-b.avif" caption="P6-4-a-b" width=600px >}}
+
+{{< image src="P6-4-a.avif" caption="P6-4-a" width=600px >}}
+
+### 讨论
+
+将我们的模拟与实验结果进行比较，我们看到两者都是方波，并且具有相同的周期和类似的幅度。它们在 $5V$ 和 $-5V$ 之间波动，这是我们的电源电压。这合乎情理，因为电源电压是运算放大器比较器的输出。
+
+这证明了运算放大器比较器的功能。
+
+## 7. 证明数学运算放大器的功能
+
+### 构建模块
+
+{{< image src="P8-1-a.avif" caption="P8-1-a" width=600px >}}
+
+### 分析
+
+{{< image src="P8-2-a.avif" caption="P8-2-a" width=600px >}}
+
+求和放大器电路的传递函数如下：
+
+$$
+V_{out} = - \frac{Rf}{R1} \cdot V1 - \frac{Rf}{R2} \cdot V2
+$$
+
+在我们的情况下，希望使用 $50K \Omega$ 的电位器作为电阻，以便根据需求进行调整。然后，我们得到：
+
+$$
+\begin{align*}
+	V_{out} &= - \frac{\cancel{50K}}{\cancel{50K}} \cdot V1 - \frac{\cancel{\cancel{50K}}}{\cancel{50K}} \cdot V2 \\
+	V_{out} &= - V1 - V2 \\
+\end{align*}
+$$
+
+### 模拟
+
+我们在模拟中使用了两个不同频率（$500 \; \text{Hz}$ 和 $1K \; \text{Hz}$）的正弦波。
+
+{{< image src="P8-3-a.avif" caption="P8-3-a" width=600px >}}
+
+{{< image src="P5-3-b.avif" caption="P5-3-b" width=600px >}}
+
+### 测量
+
+然后，我们搭建了电路。我们将示波器通道 1 连接到 $V_{out}$ 来检查是否正常工作。
+
+{{< image src="P8-4-a-b.avif" caption="P8-4-a-b" width=600px >}}
+
+我们的电源电压为 $V_s + = 5V$ 和 $V_s - = -5V$
+
+{{< image src="P8-4-a.avif" caption="P8-4-a" width=600px >}}
+
+我们使用信号发生器生成两个频率分别为 $500 \; \text{Hz}$ 和 $1K \; \text{Hz}$ 的正弦波。
+
+{{< image src="P8-4-b.avif" caption="P8-4-b" width=600px >}}
+
+并使用示波器通道 1+ 检查输出波形
+
+{{< image src="P8-4-c.avif" caption="P8-4-c" width=600px >}}
+
+### 讨论
+
+如我们所见，输出波形的形状与我们的模拟完全相同。仿真和测量中的输出波形幅度约为 $1.75V$，周期也相同。
+
+由于实验波形的所有特征都与模拟一致，我们知道运算放大器在不同电压范围内都能正常工作。
+
+这证明了求和放大器的概念，即数学运算放大器的功能。
+
+## 8. 证明双通道音频混音器传输函数的概念
+
+### 构建模块
+
+{{< image src="P8-1-a.avif" caption="P8-1-a" width=600px >}}
+
+### 分析
+
+{{< image src="P8-2-a.avif" caption="P8-2-a" width=600px >}}
+
+求和放大器电路的传递函数如下：
+
+$$
+V_{out} = - \frac{Rf}{R1} \cdot V1 - \frac{Rf}{R2} \cdot V2
+$$
+
+在我们的情况下，希望使用 $50K \Omega$ 的电位器作为电阻，以便根据需求进行调整。然后，我们得到：
+
+$$
+\begin{align*}
+	V_{out} &= - \frac{\cancel{50K}}{\cancel{50K}} \cdot V1 - \frac{\cancel{\cancel{50K}}}{\cancel{50K}} \cdot V2 \\
+	V_{out} &= - V1 - V2 \\
+\end{align*}
+$$
+
+### 模拟
+
+我们在模拟中使用了两个不同频率（$500 \; \text{Hz}$ 和 $1K \; \text{Hz}$）的正弦波。
+
+{{< image src="P8-3-a.avif" caption="P8-3-a" width=600px >}}
+
+{{< image src="P5-3-b.avif" caption="P5-3-b" width=600px >}}
+
+### 测量
+
+然后，我们搭建了电路。我们将示波器通道 1 连接到 $V_{out}$ 来检查是否正常工作。
+
+{{< image src="P8-4-a-b.avif" caption="P8-4-a-b" width=600px >}}
+
+我们的电源电压为 $V_s + = 5V$ 和 $V_s - = -5V$
+
+{{< image src="P8-4-a.avif" caption="P8-4-a" width=600px >}}
+
+并使用信号发生器生成两个频率分别为 $500 \; \text{Hz}$ 和 $1K \; \text{Hz}$ 的正弦波。
+
+{{< image src="P8-4-b.avif" caption="P8-4-b" width=600px >}}
+
+并使用示波器通道 1+ 检查输出波形
+
+{{< image src="P8-4-c.avif" caption="P8-4-c" width=600px >}}
+
+### 讨论
+
+如我们所见，输出波形的形状与我们的模拟完全相同。仿真和测量中的输出波形幅度约为 $1.75V$，周期也相同。
+
+这证明了求和放大器的概念。