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fix LaTeX passthrough

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JamesFlare1212
2025-05-04 15:35:08 -04:00
parent 68ffcf4956
commit 6a996f1c09
5 changed files with 177 additions and 172 deletions
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22
content/en/posts/ecse-1010/lab02/index.md
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@@ -518,11 +518,11 @@ $$
Substituting known values. Given $V_A = 5$ and $V_D = 0$, the equations become:
$$
2.5V_B - V_C = 5 \\\
2.5V_B - V_C = 5 \\
2V_C - V_B = 0
$$
Matrix form: $\begin{bmatrix} 2.5 & -1 \\\ -1 & 2 \end{bmatrix}$ $\begin{bmatrix} V_B \\\ V_C \end{bmatrix} =$ $\begin{bmatrix} 5 \\\ 0 \end{bmatrix}$
Matrix form: $\begin{bmatrix} 2.5 & -1 \\ -1 & 2 \end{bmatrix}$ $\begin{bmatrix} V_B \\ V_C \end{bmatrix} =$ $\begin{bmatrix} 5 \\ 0 \end{bmatrix}$
Solve them "by hand"
@@ -547,7 +547,7 @@ The solution is:
1.2500
```
Thus, $\begin{bmatrix} V_B \\\ V_C \end{bmatrix} =$ $\begin{bmatrix}2.5 \\\ 1.25 \end{bmatrix}$
Thus, $\begin{bmatrix} V_B \\ V_C \end{bmatrix} =$ $\begin{bmatrix}2.5 \\ 1.25 \end{bmatrix}$
### Simulation
@@ -666,8 +666,8 @@ A non-inverted comparator has a transfer function of
$$
\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 + \\\
\text{if} \; V_{in} < V_{ref}, V_{out} = V_s - \\
\text{if} \; V_{in} > V_{ref}, V_{out} = V_s + \\
\end{cases}
\end{equation*}
$$
@@ -677,8 +677,8 @@ In our case, we got
$$
\begin{equation*}
V_{out}=\begin{cases}
\text{if} \; V_{in} < 0V, V_{out} = -5V \\\
\text{if} \; V_{in} > 0V, V_{out} = 5V \\\
\text{if} \; V_{in} < 0V, V_{out} = -5V \\
\text{if} \; V_{in} > 0V, V_{out} = 5V \\
\end{cases}
\end{equation*}
$$
@@ -723,8 +723,8 @@ In our case, we want to use $50K \Omega$ potentiometer as the resistors, so it c
$$
\begin{align*}
V_{out} &= - \frac{\cancel{50K}}{\cancel{50K}} \cdot V1 - \frac{\cancel{\cancel{50K}}}{\cancel{50K}} \cdot V2 \\\
V_{out} &= - V1 - V2 \\\
V_{out} &= - \frac{\cancel{50K}}{\cancel{50K}} \cdot V1 - \frac{\cancel{\cancel{50K}}}{\cancel{50K}} \cdot V2 \\
V_{out} &= - V1 - V2 \\
\end{align*}
$$
@@ -782,8 +782,8 @@ In our case, we want to use $50K \Omega$ potentiometer as the resistors, so it c
$$
\begin{align*}
V_{out} &= - \frac{\cancel{50K}}{\cancel{50K}} \cdot V1 - \frac{\cancel{\cancel{50K}}}{\cancel{50K}} \cdot V2 \\\
V_{out} &= - V1 - V2 \\\
V_{out} &= - \frac{\cancel{50K}}{\cancel{50K}} \cdot V1 - \frac{\cancel{\cancel{50K}}}{\cancel{50K}} \cdot V2 \\
V_{out} &= - V1 - V2 \\
\end{align*}
$$