205 lines
5.5 KiB
Markdown
205 lines
5.5 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: 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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keywords: ["C","Programming","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
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- Programming
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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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- Programming
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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: 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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resources:
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- name: featured-image
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src: featured-image.jpg
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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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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 Timer Calculations
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> For a Timer\_A module operating with a system clock of **12 MHz** and in "up mode"...
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### Q1.1
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> ...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**.
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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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> 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.
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If we want the smallest divider possible, we’d ideally choose a divider of $\boxed{1}$
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## Q2 Basic GPIO
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> Answer the following questions about GPIO functionality and usage considering the circuit as provided below.
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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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> Initialize the GPIO pins in the circuit diagram above using the **DriverLib**. Do not modify any other pins in the port.
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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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> Using **Registers** or the **DriverLib**, turn on LED1 only if PB1 is pressed and PB2 is not. The LED should be set off otherwise.
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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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> Convert the flow chart to an equivalent segment of code.
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>
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> ```mermaid
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> flowchart LR
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> A([Start]) --> B{Are a and b<br/>both 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 Timer / Interrupt Code
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> Given the complete program below and knowing that SMCLK is **12 MHz**, answer the following questions.
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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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> What function given in this code is and will be called as an **Interrupt Service Routine**? Give the function name.
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```c
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IncC()
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```
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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.
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### Q4.2
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> How often is this function triggered by the hardware? Give your answer in milliseconds.
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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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> 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.
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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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> 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.
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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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