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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
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  name: James
  link: https://www.jamesflare.com
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 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"]
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  - C
  - Programming
  - RPI
  - ENGR 2350
  - Quiz
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  - Programming
  - Electrical Engineering
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  - ENGR 2350
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 summary: This post shows the back Quiz 2 of ENGR 2350 on Spring 2025. It includes the question and answer with explains.
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    src: featured-image-preview.jpg
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 # 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;
 }
 ```
--- a/content/en/posts/engr-2350/quiz02/q2-gpio.avif
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--- a/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"]
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 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。
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    src: featured-image.jpg
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    src: featured-image-preview.jpg
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 math: true
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 # 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;
 }
 ```
--- a/content/zh-cn/posts/engr-2350/quiz02/q2-gpio.avif
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