FINAL EXAM. last name first name

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1 55:036 Embedded Systems and Systems Software Spring, 2011 FINAL EXAM last name first name Notes: This is an open-book, open-notes exam Be sure to write legibly and clearly mark your answers Show your work for partial credit Total = 131 points Maximum score = 125 points

2 Question 1. Complete the following glossary of terminology encountered in this course. Write a single phrase/sentence rather than detailed explanations. (1 point each) ALU Arithmetic Logic Unit PID UART USART Dekker s Algorithm Mutex Atomic Operation RTOS FIFO RMA RTC SecureCRT TWI SOC (Batteries) MISO 1

3 Question 2. (6 points) Explain what is the difference between synchronous and asynchronous serial communication. Address the issue of which transfer is inherently faster. Question 3. (3 points) In the table below, fill in asynchronous or synchronous as appropriate. RS232 SPI I2C Type of Communication Asynchronous Synchronous Synchronous Question 4. (1 point) Explain briefly (one/two sentences) the terms mark and space with respect to RS232 communications. These terms refer to the state of the communication line. Mark is or depending the RS232 variant. Space is + or +. RS232 uses negative logic: logic 1 is a mark on the Rx/Tx line and a logic 0 is a space on the Rx/Tx line. The idle state is mark. 2

4 Question 5. (5 points) An ATmega88PA controller uses an external MHz clock that is internally divided (using the CLKDIV option) by 8. What values should one load into UBRR0H:UBRR0L for proper operation of a N1, serial communications link? Answer From the ATmega88PA documentation: ( ) The effective clock frequency is ( ) Hz. Thus which means one should load 0:2 into UBRR0H:UBRR0L. Question 6. (5 points) Consider the serial (RS232) communication routine below, which uses the ATmega88 s USART hardware and polling to transmit a NULL-terminated string. The routine works reliably in an engineer s embedded application, until she turns on interrupts to service the INT0 interrupt. Explain what is going on, and provide a solution. void usart_print(const char *ptr){ } // Send NULL-terminated data from SRAM. // Uses polling (and it blocks). while(*ptr) { while (!( UCSR0A & (1<<UDRE0))) ; UDR0 = *(ptr++); } The statement: UDR0 = *(ptr++); is causing the problem. While this is a single C statement, it will compile to 10 or more assembly language statements. If the INT0 (or any interrupt) fires during the sequence, then the bit timing of the characters that the USART is generating, may be compromised,which could confuse the serial receiver. The other C statements will also compile to several assembly language statements, but delays here are not problematic. A solution is to bracket the statement with cli and sei instructions: : cli(); // Disable off interrupts UDR0 = *(ptr++); sei(); // Enable interrupts : 3

5 Question 7. (2 points) Provide a C (gcc) code snippet that declares and initializes a string msg with the contents Slide Card. Make sure the string is in Flash memory. const char msg[] PROGMEM = Slide Card ; Question 8. ( 1 point) What is the duration of a bit in RS232 serial communication with the following parameters: 19,200 7N1? The bit duration is Question 9. (2 points) Explain in a short paragraph what is a NACK when discussing I2C. 4

6 Question 10. (10 points) An embedded system engineer s supervisor assigns him the following task: explore two options to reduce the power consumption of a 5-V, 10-MHz embedded system by 45% or more. The first option is reducing the clock to 4 MHz, since this will still leave enough cycles available for the system to perform properly. The second option is to convert the system to a 3.3 V system. Explore each of these options, estimating the possible power savings, and comment on the downside/pitfalls of each option. Hint: start with the simple mathematical model for power consumption covered in the lectures. Solution A model for the power consumption of a microcontroller is where is the operating frequency and is the operating voltage, represents the total capacitance of all the microcontroller s gates, and the faction of active gates. Since not other information is available, we will assume is fixed. Option 1 If the embedded system is converted to a 3.3-V system while keeping the 10 MHz clock, the new system could conceivably consume ( ) of the power of the 5-V system. The downside of switching from a 5-V system to a 3.3 V system is that one may need additional hardware to interface with peripherals (i.e., logic and analog level translators), as well as a new 3.3-V power supply. Option 2 Converting the embedded system to a 4-MHz will give a 40% saving in power, since power consumption is proportional to the clock frequency. However, even though the controller has enough cycles to perform properly, the code may contain cycle wasting (loops, nops, ) timing sections. Reducing the clock to 4 MHz may cause the system to malfunction because of these sections. These failures may be very subtle and hard to detect and one should carefully scrutinize the code. 5

7 Question 11. (5 points) An embedded systems programmer finds that the following ISR is extremely slow. Explain in 3 5 sentences what the cause is. my_isr() { save_registers(); disable_interrupts(); y = getadc(1); /* measure voltage on ADC channel 1 */ printf( %3d,y); /* send value to PC over serial link */ } enable_interrupts() restore_registers(); On the surface, the ISR seems short and robust. The reason it is slow, is the printf statement. The printf statement is a large, complex, and slow function. Consequently, the single printf statement represents a large number of assembly language statements that takes many cycles to complete. Question 12. (5 points) Explain, by writing a sentence or two and providing a code snippet, how one can use WinAVR/gcc s stdio libraries to implement a routine that will convert a string to its corresponding number. For example, convert the string to the number What are two disadvantages of this method? A simple method is to use the C language string scan function, namely sscanf. For example, assuming the string is in a variable str, the following will convert and place the result in a variable n: float n; : sscanf(str, %f,&n); The sscanf function is large and complex, which leads to the two advantages of this method, namely, it is slow, and the amount of code the compiler will include will be large. 6

8 Question 13. (10 points) Consider a battery-operated consumer electronics device that uses an embedded microcontroller that will accept V power. The device can also be powered from a power supply that plugs into a mains outlet. Consumers expectation is that the switchover between battery- and mains power is transparent. That is, the instant a user inserts the power supply connector, the device switches to the power supply, and the instant the user unplugs the device, it switches to battery power. Draw a block diagram/schematic that shows how to implement this functionality using diode(s), linear regulator(s), battery, etc. Explain how the circuit works. Provide as much details as you can. For example, specify the type of diodes, indicate voltages on the diagram, include critical capacitors, and so on. You can assume that unregulated 9 V dc power is available. 7

9 Question 14. The table below shows the periods and worst-case execution times for two periodic real-time tasks. Task Period Duration (a) Determine the processor utilization. (b) Using the result from (a), explain whether one can or cannot properly schedule these tasks. (3 points each) Part (a) The processor utilization is Part (b) The processor utilization is less than 100% but this is a necessary but not sufficient condition for scheduling. Without knowledge of the task deadlines one cannot determine if one can schedule the tasks. 8

10 Question 15. (20 points) Consider a set of three periodic real-time tasks with period and duration as shown below. (Assume that the deadlines are equal to the periods.) These are to be preemptively scheduled using rate-monotonic scheduling. Task Period (ms) Duration (ms) A 6 2 B 20 3 C 9 2 On the diagram below, show a possible execution schedule for the three tasks using ratemonotonic scheduling over a 40 ms interval. Note that task A will execute 7 times during this interval, task B will execute twice, and task C will execute 5 times. Denote the task executions as A1 A7, B1 B2, and C1 C5. Be sure to show accurate timings on your diagram and clearly indicate where preemptions (if any) occur. Hint: use different styles of hatching and shading to indicate the different tasks, for example: Answer 9

11 Question 16. (10 points) A preemptive task switcher on an embedded controller switches between task1 and task2 below. The routines access a shared variable N that has initial value N = 3. shared int N = 3; task1() { N = N + 1; print N; } task2() { N = N + 1; print N; } Depending on when context switches occur, one scenario is that task1 and then task2 increments N. Alternatively, task2 can increment first, followed by task1. These are labeled as A and B in the table below, along with the expected result, namely N = 5, since each task increments N. The table also shows cases C and D, where despite the fact that the scheduler runs both tasks, N s value is 4 rather than 5. Explain how this can happen. Note, this is a 10-point question, so you must provide sufficient amount of detail. Order Result A task1(), task2() 5 B task2(), task1() 5 C task1(), task2() 4 D task1(), task2() 4 The key statement that is causing the undesired behavior is N = N + 1. Even though this is a single statement, the compiler generates several assembly language statement and the C statement is NOT atomic. 10

12 Question 17. (5 points) Assume a 1-byte variable called FLAG has been defined. Specific bits of FLAG are used to pass information from any one of several ISRs back to the mainline code to indicate that its interrupt event has occurred and that the mainline code can take action accordingly, and then clear the specific FLAG bit. Multiple ISRs and the mainline code therefore access and change FLAG. Why do the accesses and changes in the mainline code of this shared resource NOT constitute a critical region? Answer Thinking of FLAG as an array of 8 bits it is clear that individual bits are not shared between ISRs and are therefore not a shared resource. The following code will provide the same functionality unsigned char FLAG[8] where each of FLAG[0], FLAG[1],, FLAG[7] are accessed by a single ISR. However this would be less space-efficient compared to treating FLAG as an bit array. 11

13 Question 18. (5 points) Explain, by giving an example, why critical damping (that is, no overshoot) is an essential requirement in many industrial control systems. Expanding on your example, explain how you would guard against overshoot. Question 19. (6 points) An engineer designs an AVR-based PID controller for maintaining the water level in an open tank. Unfortunately, the engineer does not have a plant model, so she had to make assumptions about reasonable values for the P, I, and D gain constants. (a) After deploying her controller, she found the water level in the tank was too low. How should she adjust P, or I, or D (or their combinations) to address the problem? (b) On the other hand, if she found that the water level in the tank was too high, how should she adjust P, or I, or D (or their combinations) to address this problem? Either case (water level too high and water level too low) is a tracking error. She should increase the I and P constants. If this leads to oscillation, she can increase I only. If this still leads to oscillation, she could decrease P and D. 12

14 Question 20. (10 points) Consider a PID controller that uses the previous 5 error values for the I and the 2 previous error values for the, and has P = 0.1, I = 0.5, D = 0.1. Use the table below and determine the next actuator output (u 27 ). t e t ? For (discrete) PID control that use the 5 previous error values for and the two previous error values for, we have ( ) ( ) Thus, with and the next actuator output at is ( )( ) ( ) ( )( ) 13

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