Model Solution and marking scheme for Examination Paper EEE305J1: Microcontroller Systems 2004/5 General Observations

Transcription

1 Model Solution and marking scheme for Examination Paper EEE305J1: Microcontroller Systems 2004/5 General Observations Design questions like A1 below are extremely difficult to mark, not least because there are literally an infinity of answers and the required information may pop up in several places in your script. The student should be careful to maximise his/her marks by clearly setting out all required information and assumptions, and by using aids to clarification, such as task lists, flowcharts and program comments. Time is of the essence with a question like this. Taking the primary mark (ex. bonus) of 60% gives us a total of 110 minutes (just under two hours). Thus a 10 mark section should be completed in around 18 minutes, or 1.8 minutes per mark! Do not write a two-page essay for six marks; it isn t cost effective. Question A1 A dedicated PIC16F84-based microcontroller is to be designed as an intruder alarm to continually monitor the state of four sensors at various points in a building. Each of these acts as a 2-state device, giving a high (logic 1) or low (logic 0) voltage as defined below. The system output is to be a 4-LED array with flashing lights showing which sensor is active and a single siren if any movement sensor is active. You are asked to design the hardware and software which will: Read the bank of four sensors. Activate the appropriate LEDs if sensors are active. Activate the siren alarm for a minimum period of 30 seconds. Repeat the above continually. Pertinent hardware details are: The intruder sensors output a logic 1 when activated. Each LED requires a nominal 10 ma to illuminate. The siren is activated via a relay activated directly from +5V. The PIC16F84 is clocked using a 1 MHz crystal and is powered using a +5V supply. You are required to design both input and output interface for the sensors, LEDs and buzzer/relay and software to implement the scheme. Answers (1) through (3) should be illustrated with a diagram and any necessary Parallel port configuration shown at that point in your answer. 1

2 1. The input interface for the four sensors. 9 minutes...[5 marks] 2. The output interface for the four LEDs. Determine the value of the current limiting resistors and state if your connection illuminates with a logic 0 or logic 1 (low or high). 14 minutes...[8 marks] 3. A single output line to drive the buzzer. 4 minutes...[2 marks] 4. A flow chart or task list and assembly-level coding implementing the software specified below. You can assume that all locations between File h 20 and File h 2F are available for general storage. (a) A main routine to: i. initialize the system; ii. check the state of the sensors continually until an intruder is sensed; iii. flash the appropriate LED(s) on and off at a nominal 1 Hz rate for a total duration of 30 seconds whilst sounding the siren; iv. disable the warning devices and repeat indefinitely; 58 minutes...[32 marks] (b) A subroutine to give a nominal fixed delay of 0.5 s based on the 1 MHz crystal. You must show your calculations for full marks. 23 minutes...[13 marks] For bonus marks: 5. How might you go about expanding the system, say, to cope with up to eight sensors in a zone and up to four zones? 13 minutes...[6 marks] 6. Rather than using discrete LEDs to indicate active sensors, how might you display the sensor number as a decimal digit? What hardware and software changes would have to be made? 7 minutes...[4 marks] Overall Specifications Be sure to understand the question! There are four LEDs and one buzzer to be activated by the PIC and four digital sensors to be read. The most convenient way to do this is to use Port A set to output for the indicators and Port B to read the sensors. Some students used Ports C or D which lost a mark as the PIC16F84 has only a 5-bit Port A and 8-bit Port B! Basically all the software has to do is to continually scan the four sensors connected to Port B. If these do not all read as logic 0 then a loop has to be entered, counting 30 passes. 2

3 In each pass the appropriate LED is turned on for 30 seconds and then off for 30 seconds. After these 30 passes, the sensors are again tested and the process repeated forever. Worked Solution A possible overall circuit is shown in Fig. 1. S3 RA3 S2 RA2 S1 RA1 S0 Intruder sensors RA0 PIC16F84 RB7 Siren RB2 LED3 R RB2 LED2 RB1 LED1 RB0 LED0 Zone display Figure 1: Block diagram of system covering solutions 1.1, 1.2 and Key points are: Connection of sensors to appropriate port lines (Port B[3:0] in the diagram). 2 marks. Explicit statement that Port B is configured as input on Reset or shows TRISB[3:0] set to all 1s. Also shown are the settings of Port B for (2) below. 3 marks. bsf STATUS,RP0 ; Change to Bank 1 by setting bit RP0 in STATUS movlw b ; PortB pins 7 and 3 thru 0 Outputs movwf TRISB ; TRISB set-up movlw b ; PortA[4:0] set to Inputs movwf TRISA ; TRISA set-up bcf STATUS,RP0 ; Back to Bank 0 3

4 2. Key points are: Connection of the LEDs to Port B. LEDs shown with either the anode connected to the port (lit on a high at port pin) or else (one student did this) with cathode to the port pin in which case they will be lit on a zero 2 marks. Use of a series resistor to limit the current. 1 mark. To calculate the current limiting resistors we know that a LED drops around 2 V when conducting and that the PIC outputs are 5 V (actually it drops a little when it supplies 10 ma) when logic 1. Thus V = IR; R = 5 2 = 300Ω. 2 marks Setting up of Port B as shown in (1) above. 3 marks. 3. Key points are: Alarm buzzer connected to a port line Makes pin output (as shown in the listing of (1) above). 1 mark. 1 mark. 4. It is possible to get more than 1 of the entire paper s marks from the software just in 3 this one answer. Thus you ignore software at your peril! Although marks are not specifically allocated for task lists, flowcharts etc. it is strongly recommended that you spend some time putting your thoughts down on paper before doing any coding. Not only does this help clarify your deliberations, but it makes it easier for us to follow your program and give marks. (a) The routine starts by checking the sensors. Although each bit can be checked in turn (using in total four btfsc instructions), the easiest way of doing this is simply to load the entire Port A pattern and clear the upper four bits. If all four lower bits are zero (no intruder) then overall there will be all 0s and the Z flag will be set. If it finds a non-zero situation, then it launches into the flash-light sequence, otherwise it continues on testing again indefinitely. The loop begins by turning on the siren and then setting up a number 30 in a File, which is decremented on each pass through the loop; that is there will be 30 flashes of the appropriate LED(s). The easiest way of illuminating the LEDs corresponding to the non-zero sensors is simply to copy the state of the sensors attached to Port B (which is sitting in the Working register after being ANDed) straight out to Port B. Then bit 7 needs to be set to 1 to turn on the buzzer. Nobody got this, and indeed it depends on making the connections of the sensors to Port B mirror the LEDs connected to Port A. The more general way would be to zero all LEDS and then check each sensor in turn, turning on the appropriate LED if the sensor was 1. After the first 30 seconds, all LEDS (but not the buzzer) are turned off and another 30-second delay called. This is repeated 30 times in all, before the complete main routine is repeated. Here is a possible implementation: 4

5 Initialize ports, etc LEDs and Siren off Sensors? All zero Not zero Siren on Count = 30 Activate LEDS Delay 0.5 seconds Turn off LEDs Delay 0.5 seconds Decrement Count Yes == 0? No Figure 2: Flow chart of a possible intruder alarm. 5

6 cblock h 20 ; Variables COUNT:1, SENSOR:1 endc include "p16f84.inc" ; First initialize ports (QA1(1..3)) ALARM bsf STATUS,RP0 ; Change to Bank 1 movlw b ; RB[3:0] and RB0 Outputs movwf PORTB ; RB[6:4] Inputs movlw b ; Make Port A pins Inputs movwf TRISA ; Do it bcf STATUS,RP0 ; Back to Bank 0 ; Turn everything off MAIN clrf PORTB ; All LEDs and siren off ; INTRUDER movf PORTA,w ; Get state of the sensors andlw b ; Isolate bottom four sensors movwf SENSOR ; and save in memory btfsc STATUS,Z ; Skip if non zero goto INTRUDER ; ELSE try again ; Pass here if intruder detected movlw d 30 ; Set up a loop of 30 seconds movwf COUNT ; Flash LEDs for 30s FLASH movf SENSOR,w ; Get the sensor pattern movwf PORTB ; Activate LEDs bsf PORTB,7 ; and buzzer ; call DELAY_500MS ; for 0.5 s ; clrf PORTB ; Turn off LEDs bsf PORTB,7 ; but not siren ; call DELAY_500MS ; for 0.5s ; decfsz COUNT,f ; Count decremented goto FLASH ; and repeat forever end Key points in the marking scheme were: Program structure/algorithm. Naming variables/header file. Testing for intruder (non-zero bits) at start of main loop. Activating LEDs from sensor pattern. Activating siren throughout 30 s period. Calling up the delay subroutine and hence flashing LEDs. Endless loop structure 8 marks. 1 mark. 4 marks. 4 marks. 4 marks. 7 marks. 4 marks. 6

7 Note: Other possibilities would be to test each sensor in turn and activate LEDs as appropriate or even to place the various activation patterns in a look-up table. (b) A nominal 500 ms delay subroutine can be implemented by using a 2-loop structure, with the inner loop being a 2 ms delay at 1 MHz. The next loop counts 250 times around this inner loop. As the crystal is 1 MHz, then each cycle is 4 µs and so an inner 2 ms delay is possible with the standard structure. ; ***************************************************************** ; * FUNCTION: Delays 500ms with a 1MHz crystal * ; * ENTRY : None * ; * EXIT : DELAY_COUNT = 0, W = 0 * ; ***************************************************************** cblock DELAY_COUNT endc DELAY_500MS movlw d 250 ; Outer loop of 250 passes movwf DELAY_COUNT ; DELAY_2MS movlw K ; Delay constant 1 D_LOOP addlw -1 ; Decrement K btfss STATUS,Z ; to zero K+1 goto D_LOOP ; 2*(K-1) ; decfsz DELAY_COUNT,f ; One more 2ms delay goto DELAY_2MS ; Repeat until 250 passes return ; Core cycles are 1 + K + (K+1) + 2K - 2 = 4K ; With a 1MHz crystal one cycle is 4us, so delay is 4*4K = 16*K = 2000us (2ms) K equ d 125 Key points in the marking scheme were: Loop structure algorithm and implementation of inner core. 3 marks. Outer loop structure. 3 marks. Timing calculation of inner loop for 1 MHz crystal. 6 marks. Use of return for subroutine. 1 2 mark. Locates the File DELAY_COUNT (may be done as part of the main routine). 1 2 mark. As an alternative a triple-loop structure can be used. Bonus marks 5. Student should consider the problem of coping with larger numbers of I/O lines. The PIC16F84 could be replaced by a larger device, such as a PIC16F87. However, even this can only deal with 16 sensors/16 LEDs and one buzzer. 7

8 Use a more efficient display with one LED per zone and a LED for each of eight sensors within a zone (=12) handled by one PIC16F84 and one PIC16F877 to handle 32 sensors with a serial link between devices. 6. Both hardware and software considerations should be considered> Hardware: Use a 7-segment display plus the buzzer connected to Port B. Software: Use a 7-segment look-up table to encode to the appropriate 7-segment pattern. 8

9 Question B2 (a) Software targeted to microcontroller systems is coded either at assembly level or using a high-level language; typically C. Discuss the advantages and disadvantages of each approach. 6 minutes...[3 marks] (b) A certain petrol tank capacity detector uses a digital pressure transducer whose naturally coded 8-bit output can be read by a PIC MCU at Port B and varies linearly from zero (empty, b ) to 255 (full, b ). Four LEDs and a buzzer are connected to Port A to act as a petrol gauge according to the following scheme: Green (Port A, pin 0) : Over 1 2 full Amber (Port A, pin 1) : Over 1 4 and up to 1 2 full 1 Red (Port A, pin 2) : full or under 4 Buzzer (Port A, pin 3) : Under 1 8 full Design and implement a function in the C language that will set up Ports A and B appropriately and activate the indicators as listed. A CCS C compiler reference set is given at the end of the paper. 30 minutes...[17 marks] Solution (a) Key points are: i. Assembly code is a symbolic representation of the natural machine code relevant to the underlying hardware. Advantages: Closeness to hardware can allow efficient (short and fast) code to be written and ease of debugging software-hardware interaction. Easier to access features of processor. Cheaper to translate to machine code compared to a compiler. Disadvantages: Each language is unique to hardware; that is non portable. Not related to the problem; rather to the hardware. ii. High-level code is a language oriented to problem solving. Advantages: High level instructions; e.g. mathematics, which are closer to human thought patterns. Relatively independent of the underlying hardware; i.e. portable, means that skills are transferable between hardware platforms. Quicker to write (more productive), as a high-level instruction is worth several machine instructions. 9

10 Easier to document and thus less error prone and easier to debug problem algorithms. Disadvantages: More expensive to purchase a compiler compared to an assembler, to translate to machine code Code takes up more space and runs slower. Can be difficult to see and debug interaction with hardware. (b) A possible coding would be: #include <16f84.h> /* Always put in this header file */ #byte PORTA = 5 /* Tell the compiler PORTA is File 5 */ #byte FUEL = 6 /* Tells that FUEL can be read from PORTB */ #bit GREEN = PORTA.0 /* GREEN LED is connected to pin RA0 */ #bit AMBER = PORTA.1 /* AMBER LED is connected to pin RA1 */ #bit RED = PORTA.2 /* RED LED is connected to pin RA2 */ #bit BUZZER = PORTA.3 /* BUZZER is connected to pin RA3 */ main() { set_tris_a(0xf0); /* Lower four bits of Port A Outputs */ set_tris_b(0xff); /* Make Port B all bits Inputs */ /* Now do a loop that checks the fuel level and activates LEDS forever */ while(1) /* DO forever */ { if(fuel > 0x80) /* or (FUEL & 0x80) or bit_test(fuel,7) */ {GREEN=1, AMBER=RED=BUZZER=0;} /* Activate LEDS & buzzer as approp*/ else if(fuel > 0x40) {AMBER=1, GREEN=RED=BUZZER=0;} /* Activate LEDS & buzzer as approp*/ else if(fuel > 0x20) {RED=1, GREEN=AMBER=BUZZER=0;} /* Activate LEDS & buzzer as approp*/ else {RED=BUZZER=1; GREEN=AMBER=0;} } /* End of forever loop */ } /* End of main() */ Basically all you do is to check each level of Fuel (read in at Port B) in turn and activate the appropriate pins. As defined using #bit, simply equate the named bit as 0 (for Low pin) and 1 (for High pin). Marks were given for: Defining the Port locations and individual bits using #byte and #bit directives. 2 marks. Setting up the Port A and Port B pin directions using set_tris_x() function. 1 mark Endless loop. 2 marks Testing levels or bits using if, else if decision tree 8 marks. Activating correct LEDs and buzzer. 4 marks. 10

11 Question B3 (a) The ASCII code for the letter C is h 43. If this character is to be transmitted using the asynchronous serial protocol at 1200 baud with eight data bits, no parity and one stop bit, draw a waveform annotated clearly with time and voltage, that represents what you would observe monitoring the signal using an oscilloscope from (i) the transmit line from the UART, (ii) the RS232 transmit line. 25 minutes...[14 marks] (b) If a device receiving asynchronous serial data is incapable of processing the incoming data fast enough, several techniques can be employed to allow the receiver to slow down the transmission character rate; these involve either hardware or software handshake techniques. Discuss, with the aid of diagrams either of these methods. 11 minutes...[6 marks] Solution (a) Key points are: 5V 0V Idle Start Least-Significant Bit Most-Significant Bit Stop Idle Figure 3: Waveform transmitting the ASCII code for the letter C. Start bit going to logic 0 in an idling line. 1.5 marks. Least-Significant Bit (LSB) follows and then each bit upwards to Most-Significant Bit (MSB); that is, a reversal of the normal way of writing a number. 3.5 marks. The frame terminates with a Stop bit and then line idles. 1.5 marks. Each bit takes the same time, which in the case of this example is seconds, giving 833 µs per bit. 3.5 marks. There are ten bit minimum in all, so the character rate is 120 characters per second. 3.5 marks. (b) Key points are: 11

12 TX RX Logic levels Transmission path Logic levels 0/5V RS232 levels ±12V 0/5V +12V 0V -12V Idle Start Least-Significant Bit Time Most-Significant Bit Stop Idle Figure 4: RS-232 transmission for the ASCII code for the letter C. Uses bipolar voltage levels at transmitter between ±12 V through ±15 V with respect to local ground. At receiver the signals should be at least ±3 V with respect to local ground voltage. 2 marks Voltage polarity are V for logic 1 and +V for logic 0 (that is voltage reversal). 2 marks. (c) Hardware handshaking uses additional lines to the data line(s) to check that the Start transmission Assert RTS line No CTS line active? Yes Send a character More characters? Yes No Deassert RTS line End of transmission Figure 5: Hardware RS-232 handshake. 12

13 modem or receiver device is ready to perform it s duties. In the simplest case two lines are used. RTS is activated by the transmitter to tell the device at the far end that that it is Ready To Transmit. When the receiver is ready to accept data, it asserts its CTS Clear To Send line. Note: Both handshake are active on logic 0 and in a RS-232 system this means a positive voltage for active and negative voltage for inactive. A task list would be: i. Transmitter asserts its RTS line. ii. Wait until the receiver s CTS line is active. iii. Send one frame. iv. If transmission is not finished then go to ii. v. Transmitter de-asserts its RTS line. Alternatively where the link is duplex; that is the transmitter can send data and receive data from the far end then software handshaking may be used. In this case two control characters are used by the receiver to tell the transmitter whether it is ready to accept its data or not. XON ([CONTROL-S] from the keyboard or ASCII h 11 ) means "I am ready to receive" and XOFF ([CONTROL-Q] from the keyboard or ASCII h 13 ) means "I am not ready". Start transmission Read character from receiver Read character from receiver Yes Was it XOFF? No No Was it XON? Yes Send a character More characters? Yes No End of transmission Figure 6: Software RS-232 handshake. A task list for the transmitter would be: 13

14 i. Have I received an XOFF character? IF not THEN goto iii. ii. Keep listening for an XON character. iii. Send a data character to the receiver. iv. If transmission is not finished THEN go to i. v. End of transmission. Hardware is much more reliable and faster, but needs two special lines for the handshake. Question B4 (a) Digital to analog convertor (DAC)s are used to convert a digital word to an analogue equivalent. Thus an 8-bit DAC can transform a digital byte to one of 256 possible levels. Describe one method of conversion. 11 minutes...[6 marks] (b) An 8-bit DAC is to be used to convert varying analog signals to their digital numerical equivalent. This analog input is to be connected to one input of an analog comparator whilst the other is derived from Port B of a PIC MCU via an 8-bit DAC. The output of the comparator is connected to pin RA0 of Port A. Illustrate how this circuit may be used to convert from analog to digital in eight steps. Show how you could implement this successive approximation algorithm as a subroutine returning with an 8-bit digitized equivalent of the input analog voltage. 25 minutes...[14 marks] Solution (a) A natural binary word is weighted in powers of two. That is, amplifying each bit by a relevant factor 2 n will convert a natural binary pattern into its analog equivalent. For instance: b 7 b 6 b 5 b 4 b 3 b 2 b 1 b 0 R 2R 4R 8R 16R 32R 64R 128R - + R V out Figure 7: An 8-bit weighted-resistor digital-to-digital converter. 14

15 In Fig. 7 the gain of each of the eight inputs b n is R 2 n R. Each value of R in doubles going up from b 7 (gain is 1) to b 0 (gain 1 ). All amplified inputs are added 128 together and thus b 7 is worth 128 times that of b 0. Marks given for concept of weighting, summation and correct values. (b) The concept of successive approximation as a directed trial and error process, starting with half scale and working downwards. DAC RB7 RB6 RB5 RB4 RB3 RB2 RB1 RB0 - + V in RA0 Figure 8: An 8-bit successive approximation analog-to-digital converter. Figure 8 shows the hardware configuration. A digital port generates the series of trial digital patterns, starting with b This is connected to a digital-toanalog converter, which transforms the pattern to its analog equivalent. This is compared to the analog input voltage and the outcome (higher or lower) is read from the comparator (logic 1 if V in > trial, else logic 0). Starting from the MS bit, each bit is set in software in turn, working down in significance. If the converted trial voltage is higher than V in, as read at pin RA0, then the last trial bit is zeroed else it is kept. A possible subroutine to do this trial sequence would be: CONVERT movlw b ; Initial walking bit pattern movwf PATTERN ; Put away in a File called PATTERN clrf TRIAL ; TRIAL is where the digitized value is built up ; Now DO the eight tries and build up the trial digitized value bit by bit LOOP movf TRIAL,w ; Get partial value iorwf PATTERN,w ; Add walking bit to this partial value movwf PORTB ; Send out to the DAC to convert to analog nop ; Short delay to allow circuits to settle nop ; Now check out if resulting analog voltage is too high? 15

16 btfsc PORTA,0 ; Check comparator output. Skip if too high movwf TRIAL ; ELSE trail is too low, so keep new bit ; Now generate the next walking bit pattern; e.g > > bcf STATUS,C ; Shift pattern right, but first clear Carry flag rrf PATTERN,f ; and then shift ; If a 1 pops out, THEN the eight tests are complete btfss STATUS,C ; Now check if 1 pops out into the carry flag goto LOOP ; IF not THEN do the next estimate return ; All done; answer in File TRIAL Marks for: Description of the successive approximation process. Diagram showing the hardware (Fig. 8). Software implementation. 4 marks. 4 marks. 6 marks. eee305j2_first_solution_04.tex L A T E X April 27,