EECS 473 Midterm Exam

Transcription

1 EECS 473 Midterm Exam Fall 2016 Name: KEY unique name: Sign the honor code: I have neither given nor received aid on this exam nor observed anyone else doing so. NOTES: 1. Closed book and Closed notes 2. There are 12 pages total for the exam as well as a handout. The last page of the exam can be removed and used as reference for the last problem. 3. Calculators are allowed, but no PDAs, Portables, Cell phones, etc. Using a calculator to store notes is not allowed nor is a calculator with any type of wireless capability. 4. You have about 120 minutes for the exam. 5. Be sure to show work and explain what you ve done when asked to do so. That may be very significant in the grading of this exam. Page 1 of 12

2 1) Circle the best answer. [10 points, -2 per wrong or blank answer, minimum 0] a. When powering a device that is very rarely used over a period of years, we generally would prefer to power it using alkaline batteries / lipo batteries / lead-acid batteries / nearly any secondary cell. b. Which of the following is not an advantage of creating a standard interface (API) for hardware devices? It should allow the programmer to not worry about the details (such as MMIO locations) for the hardware device. It should make the code itself more efficient (in terms of the number of instructions executed). It hopefully limits the number of people who need to have a detailed understanding of the hardware. It hopefully will make it easier to move existing code from one processor to another. c. Which of the following is an advantage of RM scheduling over EDF scheduling? RMS can schedule certain sets of periodic tasks EDF cannot. When RMS fails to schedule, it will always fail to schedule the task with the shortest period, while EDF could fail to schedule other tasks. When RMS fails to schedule, it will always fail to schedule the task with the largest CPU needs, while EDF could fail to schedule other tasks. RMS doesn t require dynamic priorities for periodic tasks, while EDF generally does. d. LDO is a type of linear regulator / switching regulator / primary cell / secondary cell. In general an LDO has a lower device order / direct output /maximum switching speed / minimum voltage drop compared to others of that same type. e. The command mknod /drill/memory c does which of the following: Creates a drill node in memory location c by 60 by 10. Creates a file named /drill/memory which is a block file with a minor number of 60 and a major number of 10. Creates a read-only character node named /drill/memory of 600 bytes. Creates a drill hole for a via in a Gerber file. Creates a file named /drill/memory which is a character device with a major number of 60 and a minor number of 10. Creates a non-plated (i.e. memoryless) drill hole for a node in an Excellon file. Page 2 of 12

3 2) Say you have the following groups of tasks. For each group find the CPU utilization and identify which groups are RM and which are EDF schedulable. Indicate if you needed to do the critical instant analysis. If needed, clearly show that analysis. The following equation may prove useful. [9 points] Group T1 Execution Time T1 Period T2 Execution Time T2 Period T3 Execution Time T3 Period % Utilization (Total) A % B % C % Group EDF Schedulable? RM Schedulable? Did you need to examine the critical instance? A Yes Yes Yes B No No No C Yes Yes No Page 3 of 12

4 3) Consider a battery with the following discharge characteristics [8 points] a) What is the (approximate) capacity of this battery in mah? Provide a brief justification.[3] Max capacity at slowest drain 3.0A * 20h = 60Ah = 60,000mAh b) Say your application requires 55A at 11V. i. For about how long would a single battery of this type run your application? [1] 20 minutes ii. If your application used 4 of these batteries in parallel, about how long could your application run? Briefly justify your answer. [4] Roughly 1.75 hours. 55A/4 = 13.75A so we use the 14.3A curve Page 4 of 12

5 4) Say we have a 6V 200mAh Lipo battery. We are using it through an LDO which outputs 4V and has a quiescent current of 1mA. You determine that the load draws a constant current and that the battery lasts 10 hours under that constant load. [7 points] a. What is the effective resistance of load? Show your work. [3] Draw 20mA since the battery lasts 10 hours. 19mA go to the load as there is 1mA lost to quiescent current. 4V / 19mA = 210 Ohms b. What percent of the battery s capacity is used by the LDO? You may assume that the battery keeps its voltage constant at 6V the whole time. Show your work. [4] (20mA * 6V 19mA * 4V) / 20mA * 6V) * 100% = 36.67% lost 5) Some variation of round robin is generally the default scheduling algorithm used by operating systems (including FreeRTOS and Linux). Why is round robin the default rather than EDF or RMS? [5 points] The operating system doesn t always know the period and/or runtime of a task. Round robin allows the tasks to run equally in the hope that every task can meet it s deadline. Page 5 of 12

6 6) PCB traces [10 points] a) Say your PCB has a metal layer that is 1 Oz/ft 2 thick operates in an ambient temperature of 20 C the trace needs to be kept below 30 C If you wish to drive a motor that uses up to 60 Watts at 12 Volts over a trace that is 4in long, what is your minimum trace width according to the figure above? Draw lines on the chart (much like the two examples) to show your work and write your answer below. [6] 0.30 inches b) Explain why necking-down a 40 thou trace to 20 thou can allow you to lower the overall resistance of a trace. [4] Necking-down allows you to fit a trace through a tight area, potentially reducing the overall length of a trace as the route can be more direct between two components. This in turn reduces the overall resistance of a trace. Page 6 of 12

7 7) Decoupling capacitors [10 points] a) The above graph shows the frequency vs. impedance for a given capacitor. Redraw the graph showing the same information for 10 of these capacitors in parallel. [5] b) Explain why we often need a variety of different sizes of capacitors when addressing power integrity. [5] Each size has a different response for a given frequency. So you need to have different capacitances to cover the whole range. This is because larger capacitors generally have higher parasitics (ESL and ESR) so a larger capacitor won t be as useful at higher frequencies. Page 7 of 12

8 8) Short answers [11 points] a. Does GPLv3 require that voters be able to modify GPL-licensed software running in a voting machine? Briefly explain your answer. [4] No. Companies distributing devices that include software under GPLv3 are at most required to provide the source and Installation Information for the software to people who possess a copy of the object code. The voter who uses a voting machine (like any other kiosk) doesn't get possession of it, not even temporarily, so the voter also does not get possession of the binary software in it. b. What is a deferred interrupt and why do we use them? [3] A deferred interrupt is when there s a task that must be run after the event that caused the interrupt occurs but it does not need to have a high priority. To prevent the interrupt from taking up processing time we create a task with a lower priority to run the desired outcome of the interrupt. When the interrupt occurs, the interrupt signals the task in some way such as giving a semaphore and this allows the task to run. c. A loadable kernel module is loadable because it can be attached or removed from the kernel as the kernel is running. Why is this useful? [4] Loadable kernel modules are useful because if they did not exist every time you want to add a driver you would be required to rebuild the entire kernel. Additionally, if all the drivers had to be built into the kernel it would create a very large image, which is not ideal (it would take up memory space and could slow things down quite a bit in the extreme case). Page 8 of 12

9 9) The EECS department is looking for a method to dim the lights in the EECS 473/373 lab when the sun is set in order to give students a better idea of what time of day it is. You ve been assigned the task to create a prototype to demonstrate to the department. For the prototype, you are to just have an LED turn on when it is light outside, otherwise have the LED be off (This corresponds to when the lights should be bright and dim). To demonstrate this idea, you have selected the TSL2561 light-to-digital convertor (also called a light sensor ), connected to an Arduino Uno, which controls the LED. Your job is to design a circuit diagram (devices shown below) and to write a sketch to accomplish this. Be sure to read the entire question and briefly look over the various documents available to you before you start. [31 points] a. Answer the following questions: [9 points] i. What voltage range can you use for the light sensor? [1] 2.7V 3.6V ii. What is the I 2 C slave address if the ADDR SEL pin is connected to GND? [2] iii. What address would you read to get DATA0LOW"? [2] 0xC iv. How much power would you expect the light sensor to draw when it s active? [1] 3V * 0.24 ma = 0.72 mw or 0.75mW v. How much power would you expect the light sensor to draw when it s not active (powered down)? Show your work. [2] 3V * 3.2uA = 9.6uW vi. When power is first applied, does the device start in active mode or powered down? [1] Powered down Page 9 of 12

10 GND b. Below are a number of components: an LDO, the light-to-digital convertor, an Arduino Uno, an LED and a 9V battery. No devices have power unless you supply it. Indicate, by drawing wires, how you will connect the components, and draw any other components needed. (Note on the Uno pin A4 = SDA and A5= SCL). Also, indicate the voltage you d want out of the LDO. [7 points] +3.3V GND GND 0.1uF A4 A5 1K Ohm 1K Ohm +3.3V +3.3V GND +9V 1K Ohm GND +3.3V LDO +3.3V GND 9V The LDO should output 3.3 V GND Page 10 of 12

11 c. Write an Arduino sketch that lights the LED if the Lux value is above 0.5. You should sample the sensor approximately every 30 seconds. We ve provided a sample sketch (just to remind you of the syntax and basic functions of the Arduino environment) and some functions for this code as the last page of this exam. Feel free to (carefully) remove it from the exam to use it as a reference. Be sure to read them before proceeding. [15 points] int outpin = 8; void setup() { pinmode(outpin, OUTPUT); initi2c(); writei2c(0x29, 0x00, 0x3); //set to active mode } void loop() { delay(30000); //30 seconds unsigned int ch0, ch1; unsigned char lower = readi2c(0x29,0xc); unsigned char upper = readi2c(0x29,0xd); ch0 = 256 * upper + lower; lower = readi2c(0x29,0xe); upper = readi2c(0x29,0xf); ch1 = 256 * upper + lower; } if ( findlux(ch0, ch1) < 0.5) { digitalwrite(outpin, HIGH): } else { digitalwrite(outpin,low); } Page 11 of 12

12 Reference sheet. (You can rip this out.) Sample Arduino sketch: int outpin = 8; void setup() { pinmode(outpin, OUTPUT); } void loop() { digitalwrite(outpin, HIGH); delay(1000); // delay 1000ms digitalwrite(outpin, LOW); delay(1000); } I2C functions Note: The functions listed below are not actually part of Arduino but for the sake of this exam please treat them as built-in functions. You may not use the actual I2C library or otherwise talk to the I2C devices. See page 12 of the specification. void initi2c(); Description Initializes the I2C connection. Must be called once before other functions are used. char readi2c(char address, unsigned char command_code) Description Reads a single byte from the 7-bit address specified using the 4-bit command code. Returns The character read void writei2c(char address, unsigned char command_code, char data) Description Writes a single byte (data) from the 7-bit address specified using the 4-bit command code. LUX function float findlux(int ch0, int ch1); Description Computes the LUX value given the 16-bit values read by channel 0 and channel 1. It assumes the TIMING register is set to the default value, 02h. (In other words, don t change the TIMING register) Returns A float between 0 and 1 (inclusive) that is the LUX value. Page 12 of 12