EE472 Mid Term SOLUTION KEY. Rules. Prof. Blake Hannaford Department of Electrical Engineering The University of Washington 31-Oct-2005 NAME

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1 EE472 Mid Term SOLUTION KEY Prof. Blake Hannaford Department of Electrical Engineering The University of Washington 31-Oct-2005 NAME Rules Individual Work Closed Book, Closed Notes, No Calculators 1-page 1-sided cheat sheet may be used. 1 hour and 10 minutes 60 points DO ANY 4 PROBLEMS No partial credit will be given for work on a 5th problem. 1

2 1 C Programming (15 pts) pts Write a C function to use bitwise logical operators to find out whether or not an unsigned int, x, isa power of two. In other words, if x == 2 n return 1, else return 0. Where n is an integer: 0 n 15. Solution Attached pts Extend your C function to use bitwise logical operators to find out whether or not an int (two s complement representation) is a positive OR negative power of two. In other words, if x == ±2 n return 1, else return 0. Where n is an integer: 0 n 15. Solution This problem relies on understanding of two s-complement representation of negative integers. Attached. 2

3 2 Logical Operators in C (15 pts) Evaluate the following C expressions: #define A #define B #define C 0x33 0x20 0xB7 unsigned char a,b,c,d,e; a = (A B) & C; b = ~(A & B); c = A^C; d = B<<3; e = C B ~A; Give answers in binary: Solution Attached. 3

4 3 Pointers (15 pts) (Similar to questions in a Microsoft job interview!) For the following declarations and statements, // declarations int i,j,k[10]=2,4,6,8,10,12,14,16,18,20; int *ip; char a,b,c[10]=11,12,13,14,15,16,17,18,19,20; char *cp; void *vp; typedef struct stst int i; char *p; st; // assignment statements st x; ip = &k[4]; vp = ip; cp = &c[0]; x.p = cp+3; x.i = 27; Evaluate the following expressions (if the expression is an error, indicate that with X ): Solution *(ip+2) + c[3] ; (*cp +1); *(ip) + 6; 16 vp = &x ; ((st*)vp)->p; *(ip-2) ; X (p is still a pointer, need another deref) 6 4

5 4 Schedulers (15 pts.) Consider a Pre-emptive, Priority Based Scheduler. Diagram the schedule for the following three tasks: A: (Priority 2) int i = 0; while(1) i++; compute(0.2333ms); if(i==3) OS_start(Task_C); OS_IO(device, buffer); B: (Priority 3) while(1) compute(0.2333ms); yield(); C: (Priority 1) while(1) compute(0.6566ms); OS_delay(3); The scheduler is pre-emptive and priority-based. Whenever it runs, this scheduler starts the highest priority task whos status is READY. A clock interrupt starts the scheduler every 1.0ms. Assume the scheduler always takes 0.1ms to operate. Priorities use low numbers for high priority (like golf). All tasks are initially READY, except task C which is INACTIVE. A function yield() is available which immediately starts the scheduler. Each task may call a function void OS_IO(int device, char *buffer ) which requests the OS to perform I/O on a certain device. All I/O requests take ms. Calling OS IO immediately sets the current task status to WAIT and calls the scheduler. A task may call a function void OS_start(Task *task) where *task is a pointer to a task struct. This function changes the state of task from INACTIVE to READY and starts the scheduler immediately. Completion of I/O delay does NOT invoke the scheduler immediately, but the task status becomes READY. Each task may call a function void OS delay(unsigned int n) which causes its status to be set to WAIT until n clock interrupts have occured. The scheduler starts immediately when this function is called. 5

6 Solution: 6

7 5 Real Time Systems (15 pts) Consider the following problem statement and then answer the questions below. This is deliberately an open-ended question. There may be more than one right answer. Select answers to these questions which illustrate that you understand how to match concepts we have covered in EE472 to the task requirements. Assume a processor with about the same capabilities as the 80188/TERN board in the lab. A photocopier controller must have the following functions: 1. To make a single copy, it must generate timing signals (pulses) to all the paper handling hardware so that a sheet of paper goes all the way through the machine, flash the copying light, etc. These pulses range from 10ms to 500ms in length and must have an accuracy of ±50μsec. The entire copy cycle takes 2.0 seconds. 2. It must update the LCD display on the control panel at least three times per second. Humans will notice any timing errors greater than 10ms. 3. It must scan the buttons on the control panel at least 100 times per second (to eliminate multiple signals due to button bounce ). (This only takes 100μsec. each time) 4. If copying is complete and no buttons are pressed for 20 seconds, a task must clear all settings and reset the zoom lens. This takes 2.0 seconds. 5. When the user pressess the big green button, it must initiate N copy cycles, where N is the number entered by the user using control panel buttons. 6. Each hour it must send the number of copies made over a serial port to a supervisory computer. 7. A sensor signal can give a measure of copy quality if it is Fourier transformed and certain frequency bands are examined. At least once per day, this analysis should be completed and if quality is low, a signal should be sent to the maintenance service. With no other tasks running, this task would take 20 seconds on the machines microcontroller pts What type of scheduler would you use? Explain how your answer meets the task requirements. Solution: These tasks could probably be accomplished with several scheduler types. The timing requirements of the copier timing signals are quite strict (50μsec). This suggests that we need synchronization and priorities so that this task always gets the CPU when needed. The last task (quality measurement) is an ideal background task because it needs CPU time but has a very long deadline (24 hrs). The Fourier Transform is a complex calculation and it would be very nice if we don t have to figure out how to put a lot of scheduler yield() functions in it. This strongly suggests we need a preemptive scheduler. 7

8 5.2 8pts List the tasks you would design as parts of the real-time software system. For each task, give its type (either Periodic, Complex, Intermittent, or Background); list its inputs and outputs; if applicable, its period, P, and deadline, D. If it is not specified in the functional description, assume that each task will take 500μsec per cycle. Solution: Task Type Inputs/Outputs P C D CopyCycle Periodic timing/pulses 1ms 10μ?? 50μ LCD Periodic data/lcd 333ms.500ms 10ms Bscan Periodic keypad/data 10ms 100μ 5ms? Reset Complex Timer/lens Copy Complex Green Button / start CopyCycle Report Periodic Timer 1hr.500ms 1min? QCtl Background Timer 24hr 20sec. 1hr? Notes: C was not specified for the CopyCycle task but it clearly cannot be 500μsec because of the 50μsec requirement. From experience with the TERN boards, we know that it only takes a couple of μsec to set or clear a pulse, so with proper coding this task might take as little as 10μsec each time. Also, we have to add the constraint that pulses must be an integral multiple of 1ms because otherwise we need a 50μsec scheduler cycle which is beyond the capacity of our 40Mhz chip. There is some guesswork here on Deadlines, not all of them are clear from the requirements. 8

9 Solution to Problems 1-3 EE 472 University of Washington * Midterm Exam 31-Oct-2005 * Solutions to programming problems void prob1_1(); void prob1_2(); void prob1_2a(); void prob2(); void prob3(); void bin_display(unsigned int x); main() printf("ee 472 A2005\n MIDTERM SOLUTIONS\n\n"); prob1_1(); prob1_2(); prob1_2a(); prob2(); prob3(); void prob1_1() * a positive number is a power of 2 if just one bit is set. * int j; unsigned int x[3] = 2, 7, 16; printf("\nproblem 1.1\n"); for (j=0;j<3;j++) if(power2(x[j])) printf("%d is a power of two\n",x[j]); else printf("%d is NOT a power of two\n",x[j]); #define TRUE 1 #define FALSE 0 #define SIGN_BIT 0x8000 // bit 15 int power2(unsigned int x) unsigned int j, mask, bitcount; mask = 1; bitcount = 0; while(mask) // check all bits if(x & mask) bitcount++; mask = mask << 1; if(bitcount == 1) return(true); return(false); 9

10 void prob1_2() int j; #define NUM_TESTS 6 int x[num_tests] = -2, -7, -16, 13, 128, -256; printf("\nproblem 1.2\n"); for (j=0;j<num_tests;j++) if(power22(x[j])) printf("%d is a power of two\n",x[j]); else printf("%d is NOT a power of two\n",x[j]); int power22(int x) * a positive number is a power of 2 if just one bit is set. * * a negative number is a power of 2 if all bits with position * >= n are 1 and positions <n are 0. Example: * -64 = int j,mask,bitcount; int bit1_0 = 0, bit0_1 = 0, prev_bit=0; if(!(x & SIGN_BIT)) // positive numbers printf("%d is positive\n",x); mask = 1; bitcount = 0; while(!(mask & SIGN_BIT)) // don t check the sign bit if(x & mask) bitcount++; mask = mask << 1; if(bitcount ==1) return(true); else // negative numbers printf("%d is negative\n",x); mask = 1; bitcount = 0; prev_bit = 0; for(j=0;j<16;j++) // don t check the sign bit if(!(x & mask) && prev_bit!= 0) bit0_1++; if( (x & mask) && prev_bit == 0) bit1_0++; prev_bit = x & mask; mask = mask << 1; if(bit1_0 == 1 && bit0_1 == 0) return(true); return(false); void prob1_2a() * a positive number is a power of 2 if just one bit is set. * int j; 10

11 #define NUM_TESTS 6 int x[num_tests] = -2, -7, -16, 13, 128, -256; printf("\nproblem 1.2a (alternate solution to 1.2)\n"); for (j=0;j<num_tests;j++) if(power22a(x[j])) printf("%d is a power of two\n",x[j]); else printf("%d is NOT a power of two\n",x[j]); int power22a(int x) * a positive number is a power of 2 if just one bit is set. * * a negative number is a power of 2 if all bits with position * >= n are 1 and positions <n are 0. Example: * -64 = int j,mask,bitcount; int bit1_0 = 0, bit0_1 = 0, prev_bit=0; * This method is easier and OK if((x & SIGN_BIT)) x = ~x + 1 ; // better than x *= -1 because it s bitwise! printf("+-%d is either positive or negative\n",x); mask = 1; bitcount = 0; while(!(mask & SIGN_BIT)) // don t check the sign bit if(x & mask) bitcount++; mask = mask << 1; if(bitcount ==1) return(true); else return(false); * Problem 2 void prob2() #define A 0x33 #define B 0x20 #define C 0xB7 unsigned char a,b,c,d,e; printf("\nproblem 2\n"); a = (A B) & C; b = ~(A & B); c = A^C; d = B<<3; e = C B ~A; printf ("\n a = Ox%x = ",a); bin_display(a); 11

12 printf ("\n b = Ox%x = ",b); bin_display(b); printf ("\n c = Ox%x = ",c); bin_display(c); printf ("\n d = Ox%x = ",d); bin_display(d); printf ("\n e = Ox%x = ",e); bin_display(e); printf("\n"); * bin_display * * Display the individual bits of an integer void bin_display(unsigned int x) unsigned int i,j; for(i=0;i<16;i++) j = 1<<(15-i); if(j & x) printf("1"); else printf("0"); * Problem 3 - pointers void prob3() // declarations int i,j,k[10]=2,4,6,8,10,12,14,16,18,20; int *ip; char a,b,c[10]=11,12,13,14,15,16,17,18,19,20; char *cp; void *vp; typedef struct stst int i; char *p; st; printf("\nproblem 3\n"); // assignment statements st x; ip = &k[4]; vp = ip; cp = &c[0]; x.p = cp+3; x.i = 27; printf ("*(ip+2) + c[3] = %d\n",*(ip+2) + c[3]); printf ("(*cp +1) = %d\n", (*cp +1)); printf ("*(ip) + 6 = %d\n", *(ip) +6); vp = &x ; printf ("vp = &x ; ((st*)vp)->p; = %d\n", ((st*)vp)->p ); printf ("*(ip-2) = %d\n", *(ip-2)); 12

13 ************************************************************** Results: code]$ gcc mt_a05.c mt_a05.c: In function prob2 : mt_a05.c:190: warning: large integer implicitly truncated to unsigned type [blake@beagle code]$./a.out EE 472 A2005 MIDTERM SOLUTIONS Problem is a power of two 7 is NOT a power of two 16 is a power of two Problem is negative -2 is a power of two -7 is negative -7 is NOT a power of two -16 is negative -16 is a power of two 13 is positive 13 is NOT a power of two 128 is positive 128 is a power of two -256 is negative -256 is a power of two Problem 1.2a (alternate solution to 1.2) +-2 is either positive or negative -2 is a power of two +-7 is either positive or negative -7 is NOT a power of two +-16 is either positive or negative -16 is a power of two +-13 is either positive or negative 13 is NOT a power of two is either positive or negative 128 is a power of two is either positive or negative -256 is a power of two Problem 2 a = Ox33 = b = Oxdf = c = Ox84 = d = Ox0 = e = Oxff = Problem 3 *(ip+2) + c[3] = 28 (*cp +1) = 12 *(ip) + 6 = 16 vp = &x ; ((st*)vp)->p; = *(ip-2) = 6 [blake@beagle code]$ // this should be answered "X" 13