206 lines
5.4 KiB
Markdown
206 lines
5.4 KiB
Markdown
---
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title: ENGR 2350 - Quiz 2
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date: 2025-02-13T12:26:20-05:00
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lastmod: 2025-02-13T12:26:20-05:00
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slug: engr-2350-quiz-02
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draft: false
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author:
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name: James
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link: https://www.jamesflare.com
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email:
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avatar: /site-logo.avif
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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。
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keywords: ["C语言","编程","RPI","ENGR 2350","Quiz"]
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license:
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comment: true
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weight: 0
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tags:
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- C语言
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- 编程
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- RPI
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- ENGR 2350
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- Quiz
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categories:
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- 编程语言
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- Electrical Engineering
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collections:
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- ENGR 2350
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hiddenFromHomePage: false
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hiddenFromSearch: false
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hiddenFromRss: false
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hiddenFromRelated: false
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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。
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resources:
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- name: featured-image
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src: featured-image.jpg
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- name: featured-image-preview
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src: featured-image-preview.jpg
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toc: true
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math: true
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lightgallery: false
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password:
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message:
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repost:
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enable: false
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url:
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# See details front matter: https://fixit.lruihao.cn/documentation/content-management/introduction/#front-matter
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---
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<!--more-->
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#### Q1 定时器计算
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> 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...
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##### Q1.1
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> 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**。
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$$\text{Timer clock} = \frac{12\,\text{MHz}}{32} = 375000 \text{ Hz} \quad (375 \text{ kHz})$$
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$$N = 375000 \times 0.008 \text{ s} = \boxed{3000} \text{ ticks}$$
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##### Q1.2
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> 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。
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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 $\boxed{1}$
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#### Q2 基本 GPIO
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> 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。
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>
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> {{< image src="q2-gpio.avif" width="480px" caption="Q2 Basic GPIO Pin Out" >}}
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##### Q2.2
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> 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。
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```c
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GPIO_setAsInputPin(GPIO_PORT_P6, GPIO_PIN1 | GPIO_PIN6);
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GPIO_setAsOutputPin(GPIO_PORT_P6, GPIO_PIN4);
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```
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##### Q2.3
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> 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。
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```c
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PB1 = GPIO_getInputPinValue(GPIO_PORT_P6, GPIO_PIN1);
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PB2 = GPIO_getInputPinValue(GPIO_PORT_P6, GPIO_PIN6);
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if (!PB1 && !PB2){
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GPIO_setOutputHighOnPin(GPIO_PORT_P3, GPIO_PIN6);
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} else {
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GPIO_setOutputLowOnPIN(GPIO_PORT_P3, GPIO_PIN6);
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}
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```
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#### Q3
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> 将流程图转换为等效的代码段。
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>
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> ```mermaid
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> flowchart TB
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> A([Start]) --> B{"Are a and b\nboth zero?"}
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> B -- Yes --> C([Done])
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> B -- No --> D[Divide a by 2<br/>and save back into a]
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> D --> E[Multiply b by a<br/>and save back into b]
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> E --> B
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> ```
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```c
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while (a == 0 && b == 0) {
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a = a / 2;
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b = b * a;
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}
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```
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#### Q4 定时器/中断代码
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> 给定下面的完整程序，并且知道 SMCLK 是 **12 MHz**，回答以下问题。
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>
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> ```c
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> void IncA(),IncB(),IncC();
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> uint8_t A = 0,B = 0,C = 0;
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> Timer_A_UpModeConfig tim_config;
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> uint32_t timer = XXXXXXXXX; // Some valid value
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>
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> void main(){
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> SysInit();
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> TimerInit();
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> while(1){
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> // To fill in
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> }
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> }
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>
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> void TimerInit(){
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> tim_config.clockSource = TIMER_A_CLOCKSOURCE_SMCLK;
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> tim_config.clockSourceDivider = TIMER_A_CLOCKSOURCE_DIVIDER_32;
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> tim_config.timerPeriod = 12345;
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> tim_config.timerClear = TIMER_A_DO_CLEAR;
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> tim_config.timerInterruptEnable_TAIE = TIMER_A_TAIE_INTERRUPT_ENABLE;
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> Timer_A_configureUpMode(timer,&tim_config);
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> Timer_A_registerInterrupt(timer,TIMER_A_CCRX_AND_OVERFLOW_INTERRUPT,IncC);
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> Timer_A_startCounter(timer,TIMER_A_UP_MODE);
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> }
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>
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> void IncA(){
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> Timer_A_clearInterruptFlag(TIMER_A1_BASE);
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> A++;
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> }
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>
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> void IncB(){
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> Timer_A_clearInterruptFlag(TIMER_A2_BASE);
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> B++;
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> }
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>
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> void IncC(){
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> Timer_A_clearInterruptFlag(TIMER_A3_BASE);
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> C++;
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> }
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> ```
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##### Q4.1
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> 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。
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```c
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IncC()
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```
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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。
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##### Q4.2
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> 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。
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$$\text{Timer Clock} = \frac{12\,\text{MHz}}{32} = 375\,\text{kHz}$$
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$$\text{Tick Period} = \frac{1}{375\,\text{kHz}} \approx 2.667\,\mu\text{s}$$
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$$\text{Interrupt Period} = 12345 \times 2.667\,\mu\text{s} \approx \boxed{32.92}\,\text{ms}$$
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##### Q4.3
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> 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`"。
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$$\text{Period} \approx \frac{12345}{12\,\text{MHz}/32} \approx 32.92\text{ ms}$$
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$$\frac{5000\text{ ms}}{32.92\text{ ms}} \approx 152$$
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```c
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C = 0;
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while(C < 152) {}
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printf("5 seconds\n");
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```
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##### Q4.4
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> 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`"。
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```c
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printf("not blocked");
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if (C >= 152) {
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printf("5 seconds");
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C = 0;
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}
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```
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