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+---
+title: ENGR 2350 - Quiz 2
+date: 2025-02-13T12:26:20-05:00
+lastmod: 2025-02-13T12:26:20-05:00
+slug: engr-2350-quiz-02
+draft: false
+author:
+  name: James
+  link: https://www.jamesflare.com
+  email:
+  avatar: /site-logo.avif
+description: This post shows the back Quiz 2 of ENGR 2350 on Spring 2025. It includes the question and answer with explains.
+keywords: ["C","Programming","RPI","ENGR 2350","Quiz"]
+license:
+comment: true
+weight: 0
+tags:
+  - C
+  - Programming
+  - RPI
+  - ENGR 2350
+  - Quiz
+categories:
+  - Programming
+  - Electrical Engineering
+collections:
+  - ENGR 2350
+hiddenFromHomePage: false
+hiddenFromSearch: false
+hiddenFromRss: false
+hiddenFromRelated: false
+summary: This post shows the back Quiz 2 of ENGR 2350 on Spring 2025. It includes the question and answer with explains.
+resources:
+  - name: featured-image
+    src: featured-image.jpg
+  - name: featured-image-preview
+    src: featured-image-preview.jpg
+toc: true
+math: true
+lightgallery: false
+password:
+message:
+repost:
+  enable: false
+  url:
+
+# See details front matter: https://fixit.lruihao.cn/documentation/content-management/introduction/#front-matter
+---
+
+<!--more-->
+
+## Q1 Timer Calculations
+
+> For a Timer\_A module operating with a system clock of **12 MHz** and in "up mode"...
+
+### Q1.1
+
+> ...what must be the timer's period (in number of counts) be such that the overflow period is **8 ms**? Assume that the timer's clock divider is set to **32**.
+
+$$\text{Timer clock} = \frac{12\text{ MHz}}{32} = 375000 \text{ Hz} \quad (375 \text{ kHz})$$
+$$N = 375000 \times 0.008 \text{ s} = \boxed{3000} \text{ ticks}$$
+
+### Q1.2
+
+> What is the smallest divider possible that could still allow the timer to produce the same overflow period? Assume, of course, that the timer's period (in counts) can also change.
+
+If we want the smallest divider possible, we’d ideally choose a divider of $\boxed{1}$
+
+## Q2 Basic GPIO
+
+> Answer the following questions about GPIO functionality and usage considering the circuit as provided below.
+> {{< image src="q2-gpio.avif" width="480px" caption="Q2 Basic GPIO Pin Out" >}}
+
+### Q2.2
+
+> Initialize the GPIO pins in the circuit diagram above using the **DriverLib**. Do not modify any other pins in the port.
+
+```c
+GPIO_setAsInputPin(GPIO_PORT_P6, GPIO_PIN1 | GPIO_PIN6);
+GPIO_setAsOutputPin(GPIO_PORT_P6, GPIO_PIN4);
+```
+
+### Q2.3
+
+> Using **Registers** or the **DriverLib**, turn on LED1 only if PB1 is pressed and PB2 is not. The LED should be set off otherwise.
+
+```c
+PB1 = GPIO_getInputPinValue(GPIO_PORT_P6, GPIO_PIN1);
+PB2 = GPIO_getInputPinValue(GPIO_PORT_P6, GPIO_PIN6);
+
+if (!PB1 && !PB2){ 
+    GPIO_setOutputHighOnPin(GPIO_PORT_P3, GPIO_PIN6);
+} else { 
+    GPIO_setOutputLowOnPIN(GPIO_PORT_P3, GPIO_PIN6);
+}
+```
+
+## Q3
+
+> Convert the flow chart to an equivalent segment of code.
+>
+> ```mermaid
+>  flowchart TB
+>     A([Start]) --> B{"Are a and b\nboth zero?"}
+>     B -- Yes --> C([Done])
+>     B -- No --> D[Divide a by 2<br/>and save back into a]
+>     D --> E[Multiply b by a<br/>and save back into b]
+>     E --> B
+> ```
+
+```c
+while (a == 0 && b == 0) {
+    a = a / 2;
+    b = b * a;
+}
+```
+
+## Q4 Timer / Interrupt Code
+
+> Given the complete program below and knowing that SMCLK is **12 MHz**, answer the following questions.
+>
+> ```c
+> void IncA(),IncB(),IncC();
+> uint8_t A = 0,B = 0,C = 0;
+> Timer_A_UpModeConfig tim_config;
+> uint32_t timer = XXXXXXXXX; // Some valid value
+> 
+> void main(){
+>   SysInit();
+>   TimerInit();
+>   while(1){
+>         // To fill in
+>   }
+> }
+> 
+> void TimerInit(){
+>   tim_config.clockSource = TIMER_A_CLOCKSOURCE_SMCLK;
+>   tim_config.clockSourceDivider = TIMER_A_CLOCKSOURCE_DIVIDER_32;
+>   tim_config.timerPeriod = 12345;
+>   tim_config.timerClear = TIMER_A_DO_CLEAR;
+>   tim_config.timerInterruptEnable_TAIE = TIMER_A_TAIE_INTERRUPT_ENABLE;
+>   Timer_A_configureUpMode(timer,&tim_config);
+>   Timer_A_registerInterrupt(timer,TIMER_A_CCRX_AND_OVERFLOW_INTERRUPT,IncC);
+>   Timer_A_startCounter(timer,TIMER_A_UP_MODE);
+> }
+>
+> void IncA(){
+>   Timer_A_clearInterruptFlag(TIMER_A1_BASE);
+>   A++;
+> }
+> 
+> void IncB(){
+>   Timer_A_clearInterruptFlag(TIMER_A2_BASE);
+>   B++;
+> }
+> 
+> void IncC(){
+>   Timer_A_clearInterruptFlag(TIMER_A3_BASE);
+>  C++;
+> }
+> ```
+
+### Q4.1
+
+> What function given in this code is and will be called as an **Interrupt Service Routine**? Give the function name.
+
+```c
+IncC()
+```
+
+Because the line `Timer_A_registerInterrupt(timer, TIMER_A_CCRX_AND_OVERFLOW_INTERRUPT, IncC);` It sets up `IncC()` as the ISR for the timer interrupt.
+
+### Q4.2
+
+> How often is this function triggered by the hardware? Give your answer in milliseconds.
+
+$$\text{Timer Clock} = \frac{12\,\text{MHz}}{32} = 375\,\text{kHz}$$
+$$\text{Tick Period} = \frac{1}{375\,\text{kHz}} \approx 2.667\,\mu\text{s}$$
+$$\text{Interrupt Period} = 12345 \times 2.667\,\mu\text{s} \approx \boxed{32.92}\,\text{ms}$$
+
+### Q4.3
+
+> Write a segment of code to be placed in the `while(1)` loop that would implement a **blocking** delay of approximately 5 s without using `__delay_cycles()` such that the message "`5 seconds`" is printed after each delay.
+
+$$\text{Period} \approx \frac{12345}{12\,\text{MHz}/32} \approx 32.92\text{ ms}$$
+$$\frac{5000\text{ ms}}{32.92\text{ ms}} \approx 152$$
+
+```c
+C = 0;
+while(C < 152) {}
+printf("5 seconds\n");
+```
+
+### Q4.4
+
+> Write a segment of code to be placed in the `while(1)` loop that would implement a **non-blocking** delay of approximately 5 s without using `__delay_cycles()` such that the message "`5 seconds`" is printed after each delay and "`not blocked`" is printed continuously.
+
+```c
+printf("not blocked");
+if (C >= 152) {
+    printf("5 seconds");
+    C = 0;
+}
+```
diff --git a/content/en/posts/engr-2350/quiz02/q2-gpio.avif b/content/en/posts/engr-2350/quiz02/q2-gpio.avif
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diff --git a/content/zh-cn/posts/engr-2350/quiz02/index.md b/content/zh-cn/posts/engr-2350/quiz02/index.md
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+---
+title: ENGR 2350 - Quiz 2
+date: 2025-02-13T12:26:20-05:00
+lastmod: 2025-02-13T12:26:20-05:00
+slug: engr-2350-quiz-02
+draft: false
+author:
+  name: James
+  link: https://www.jamesflare.com
+  email:
+  avatar: /site-logo.avif
+description: 这篇帖子展示了2025年春季ENGR 2350课程的Quiz 2。它包含问题、答案及解释。
+keywords: ["C语言","编程","RPI","ENGR 2350","Quiz"]
+license:
+comment: true
+weight: 0
+tags:
+  - C语言
+  - 编程
+  - RPI
+  - ENGR 2350
+  - Quiz
+categories:
+  - 编程语言
+  - Electrical Engineering
+collections:
+  - ENGR 2350
+hiddenFromHomePage: false
+hiddenFromSearch: false
+hiddenFromRss: false
+hiddenFromRelated: false
+summary: 这篇帖子展示了2025年春季ENGR 2350课程的Quiz 2。它包含问题、答案及解释。
+resources:
+  - name: featured-image
+    src: featured-image.jpg
+  - name: featured-image-preview
+    src: featured-image-preview.jpg
+toc: true
+math: true
+lightgallery: false
+password:
+message:
+repost:
+  enable: false
+  url:
+
+# See details front matter: https://fixit.lruihao.cn/documentation/content-management/introduction/#front-matter
+---
+
+<!--more-->
+
+#### Q1 定时器计算
+
+> 对于一个工作在 **12 MHz** 系统时钟下并且处于“向上模式”的 Timer\_A 模块...
+
+##### Q1.1
+
+> 要让定时器的溢出周期为 **8 ms**，需要设置其周期（以计数值表示）为多少？假设定时器的时钟分频器设置为 **32**。
+
+$$\text{Timer clock} = \frac{12\,\text{MHz}}{32} = 375000 \text{ Hz} \quad (375 \text{ kHz})$$
+$$N = 375000 \times 0.008 \text{ s} = \boxed{3000} \text{ ticks}$$
+
+##### Q1.2
+
+> 要使定时器仍能产生相同的溢出周期，最小的分频器应该是多少？假设定时器的周期（以计数值表示）可以调整。
+
+如果我们想要最小的分频器设置为 1，则可以满足要求。因此，最小的分频器设置为 $\boxed{1}$
+
+#### Q2 基本 GPIO
+
+> 请回答以下关于 GPIO 功能和使用的问题，并考虑提供的电路图。
+>
+> {{< image src="q2-gpio.avif" width="480px" caption="Q2 Basic GPIO Pin Out" >}}
+
+##### Q2.2
+
+> 使用 **DriverLib** 初始化电路图中的 GPIO 引脚。不要修改端口的其他引脚。
+
+```c
+GPIO_setAsInputPin(GPIO_PORT_P6, GPIO_PIN1 | GPIO_PIN6);
+GPIO_setAsOutputPin(GPIO_PORT_P6, GPIO_PIN4);
+```
+
+##### Q2.3
+
+> 使用 **寄存器** 或 **DriverLib**，当 PB1 被按下且 PB2 未被按下时，点亮 LED1；否则关闭 LED。
+
+```c
+PB1 = GPIO_getInputPinValue(GPIO_PORT_P6, GPIO_PIN1);
+PB2 = GPIO_getInputPinValue(GPIO_PORT_P6, GPIO_PIN6);
+
+if (!PB1 && !PB2){ 
+    GPIO_setOutputHighOnPin(GPIO_PORT_P3, GPIO_PIN6);
+} else { 
+    GPIO_setOutputLowOnPIN(GPIO_PORT_P3, GPIO_PIN6);
+}
+```
+
+#### Q3
+
+> 将流程图转换为等效的代码段。
+>
+> ```mermaid
+>  flowchart TB
+>     A([Start]) --> B{"Are a and b\nboth zero?"}
+>     B -- Yes --> C([Done])
+>     B -- No --> D[Divide a by 2<br/>and save back into a]
+>     D --> E[Multiply b by a<br/>and save back into b]
+>     E --> B
+> ```
+
+```c
+while (a == 0 && b == 0) {
+    a = a / 2;
+    b = b * a;
+}
+```
+
+#### Q4 定时器/中断代码
+
+> 给定下面的完整程序，并且知道 SMCLK 是 **12 MHz**，回答以下问题。
+>
+> ```c
+> void IncA(),IncB(),IncC();
+> uint8_t A = 0,B = 0,C = 0;
+> Timer_A_UpModeConfig tim_config;
+> uint32_t timer = XXXXXXXXX; // Some valid value
+> 
+> void main(){
+>   SysInit();
+>   TimerInit();
+>   while(1){
+>         // To fill in
+>   }
+> }
+> 
+> void TimerInit(){
+>   tim_config.clockSource = TIMER_A_CLOCKSOURCE_SMCLK;
+>   tim_config.clockSourceDivider = TIMER_A_CLOCKSOURCE_DIVIDER_32;
+>   tim_config.timerPeriod = 12345;
+>   tim_config.timerClear = TIMER_A_DO_CLEAR;
+>   tim_config.timerInterruptEnable_TAIE = TIMER_A_TAIE_INTERRUPT_ENABLE;
+>   Timer_A_configureUpMode(timer,&tim_config);
+>   Timer_A_registerInterrupt(timer,TIMER_A_CCRX_AND_OVERFLOW_INTERRUPT,IncC);
+>   Timer_A_startCounter(timer,TIMER_A_UP_MODE);
+> }
+>
+> void IncA(){
+>   Timer_A_clearInterruptFlag(TIMER_A1_BASE);
+>   A++;
+> }
+> 
+> void IncB(){
+>   Timer_A_clearInterruptFlag(TIMER_A2_BASE);
+>   B++;
+> }
+> 
+> void IncC(){
+>   Timer_A_clearInterruptFlag(TIMER_A3_BASE);
+>  C++;
+> }
+> ```
+
+##### Q4.1
+
+> 在给定的代码中，哪个函数会被作为 **中断服务例程** 调用？给出函数名。
+
+```c
+IncC()
+```
+
+因为 `Timer_A_registerInterrupt(timer, TIMER_A_CCRX_AND_OVERFLOW_INTERRUPT, IncC);` 这一行将 `IncC()` 设置为定时器中断的 ISR。
+
+##### Q4.2
+
+> 这个函数被硬件触发的频率是多少？请给出答案并用毫秒为单位。
+
+$$\text{Timer Clock} = \frac{12\,\text{MHz}}{32} = 375\,\text{kHz}$$
+$$\text{Tick Period} = \frac{1}{375\,\text{kHz}} \approx 2.667\,\mu\text{s}$$
+$$\text{Interrupt Period} = 12345 \times 2.667\,\mu\text{s} \approx \boxed{32.92}\,\text{ms}$$
+
+##### Q4.3
+
+> 编写一段代码，放在 `while(1)` 循环中，实现一个 **阻塞** 的延迟，大约为 5 秒钟，并且不使用 `__delay_cycles()`，使得每次延迟后打印出消息 "`5 seconds`"。
+
+$$\text{Period} \approx \frac{12345}{12\,\text{MHz}/32} \approx 32.92\text{ ms}$$
+$$\frac{5000\text{ ms}}{32.92\text{ ms}} \approx 152$$
+
+```c
+C = 0;
+while(C < 152) {}
+printf("5 seconds\n");
+```
+
+##### Q4.4
+
+> 编写一段代码，放在 `while(1)` 循环中，实现一个 **非阻塞** 的延迟，大约为 5 秒钟，并且不使用 `__delay_cycles()`，使得每次延迟后打印出消息 "`5 seconds`" 并且连续打印出 "`not blocked`"。
+
+```c
+printf("not blocked");
+if (C >= 152) {
+    printf("5 seconds");
+    C = 0;
+}
+```
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