Chapter 1. Solutions Solution 1.1. Most RAMs are volatile, meaning the information is lost if power is removed then restored.

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1 Chapter 1. Solutions Solution 1.1. Most RAMs are volatile, meaning the information is lost if power is removed then restored. Solution 1.2. The bus signals of the MC9S12C32 are AD15-0, R/W, LSTRB and E. The bus signals of the MC9S12DP512 are XADDR19-16, AD15-0, ECS, R/W, LSTRB and E. Solution 1.3. This table shows the function of devices for the two types of CPU bus cycles. CPU read cycle CPU write cycle can drive the address bus a) Processor b) processor can drive the data bus c) RAM, ROM, and I/O d) processor can drive the R/W bus e) Processor f) processor can receive info from data bus g) Processor h) RAM, and I/O Solution 1.4. The 74LS I IH is 0.02 ma, and the I IL is 0.4 ma The 9S12 I OH is 10 ma, and the I OL is 10 ma; the worst case is 10 ma/0.4 ma, which is 25. Solution 1.5. The supply current (I CC ) is smaller for CMOS devices. The CMOS devices also have smaller output currents. Because of the complementary nature (one PNP and one NPN) the total current will be small, meaning it operates at lower power. Solution 1.6. A 14-bit ADC has 2 14 = alternatives Solution 1.7. A 12-bit ADC has 2 12 = 4096 alternatives. This is about 4000 alternatives, which equals 3¾ decimal digits. Solution 1.8. A precision of 3½ decimal digits equals 2000 alternatives. The next higher power of 2 is 2048 alternatives, which equals 11 bits. Solution 1.9. A 16-bit ADC has 2 16 = alternatives Solution An 11-bit ADC has 2 11 = 2048 alternatives. This is about 2000 alternatives, which equals 3½ decimal digits. Solution A precision of 2¾ decimal digits equals 400 alternatives. The next higher power of 2 is 512 alternatives, which equals 9 bits. Solution A precision of 4¾ decimal digits equals alternatives. The next higher power of 2 is alternatives, which equals 16 bits. A 15-bit ADC would have alternatives, which is less than the desired amount of Solution is = 0001, is = 0011, is = 0111, is = 1100,1000 1

2 Solution is = 0001, is = 0011, is = 1000, is = 1111,1110 Solution $25 is 2*16+5 = 37 $63 is 6*16+3 = 99 $A3 is 10*16+3 = 163 $FE is 15*16+14 = 254 Solution ,0000,0110, is $206A = 2*4096+6*16+10 = 8298 Solution $1234 = 1*4096+2*256+3*16+4 = 4660 Solution = 4*256+13*16+2 = $04D2 Solution = 2*4096+7*256+1*16 = $2710 = 0010,0111,0001, Solution $1234 = 1*4096+2*256+3*16+4 = 4660 Solution $ABCD =1010,1011,1100, = = Solution = 4*256+13*16+2 = $04D2 Solution = =1101,1000,1111, = $D8F0 Solution The 9S12 has accumulators A,B or D, index registers X,Y, stack pointer SP, program counter PC and a condition code register CCR. Register A and B together comprise Register D. Solution The MC9S12C32 has 0x3800-0x3FFF 2 kibibytes of RAM 0x4000-0x7FFF 16 kibibytes Flash EEPROM 0xC000-0x7FFF 16 kibibytes Flash EEPROM The MC9S12DP512 has 0x0800-0x3FFF 14 kibibytes of RAM 0x4000-0xFFFF 512 kibibytes Flash EEPROM Solution Condition Code Register : S X H I N Z V C S : Stop disable X : X interrupt mask H : half-carry ( from bit 3 ) I : I interrupt mask N : negative Z : zero V : overflow C : carry 2

3 Solution MC9S12C32 Ports AD, A, B, E(bits 7-2), J(bits 7-6), M(bits5-0), P, S(bits 3-0), T MC9S12DP512 Ports A, B, E(bits 7-2), H, J(bits 7-6, 1-0), K (bits 7, 5-0), M, P S T Solution MC9S12C32 Port E(bits 1-0) MC9S12DP512 Ports AD1 AD0 E (bits 1-0) Solution A direction register specifies whether an I/O pin is input (read only) or output (read and write). Solution Direction registers make the pin more flexible, allowing for more uses of the pin. Each application can specify the functionality of the pin for its own purposes V 300Ω Solution D1.31. The 7405 can deliver the 5 ma output current required. Assume the V OL of the 7405 is 0.5 V. To get a diode current of 5 ma, we use a ( )/0.005 = 300- resistor. Solution D1.32. Because 1 ma LED current is so low, it can be interfaced directly to the microcontroller. Assume the V OL of the microcontroller is 0.5 V. To get a diode current of 1 ma, we use a ( )/0.001 = resistor. There are two ways to connect the LED to the microcomputer output. +5V microcontroller 2000Ω output 2000Ω microcontroller output Solution D.33. The I IH of the 74LS04 is 0.02 ma and V IH is 2 V. The pullup resistor must be smaller than (5-2 V)/0.02 ma = 150 k. This means the circuit will work as is, with the 10-k resistor. Solution D1.34. ldaa #$AA ;7,5,3,1 output staa DDRT ;6,4,2,0 input // 9S12 C 6,4,2,0 inputs DDRT = 0xAA; // 7,5,3,1 outputs Solution D1.35. // 9S12 C 7,6,3,2,1,0 ldaa #$30 ; 5,4 output DDRT = 0x30; // 5,4 output staa DDRT ; 7,6,3,2,1,0 Solution D1.36. This approach will work for all I/O ports with separate data and direction registers. The basic idea is to create a Virtual Data Direction Register (permanently stored in a global LogicalDDRT). Notice that for all three cases the data port can be zero. So the code PTT=0x00; is executed once in the ritual. There are three possibilities for each bit 1) Virtual Data Direction Register is an input, Make actual DDR an input 2) Virtual Data Direction Register is an output and the data out is 1, Make actual DDR an input (this makes it float as desired) 3

4 3) Virtual Data Direction Register is an output and the data out is 0, Make actual DDR an output (this activates the output transistors) Make the data bit 0 unsigned char LogicalDDRT; // simulated or virtual 6811 direction register void Pinit(unsigned char direction){ DDRT = 0x00; // 1 means output=0, 0 means input or output=1 PTT = 0x00; // we will toggle the DDRT to make output change LogicalDDRT = direction; // 1 means open collector output, 0 means input Pout(0xFF); // (optional) output pins are already HiZ void Pout(unsigned char data){ // Pout does not affect DDRC bits of pins which are inputs // for LogicalDDRT&data pins (high level=hiz outputs) make DDRT=0 DDRT &= ~( LogicalDDRT&data ); // for LogicalDDRC&(~data) pins (low level=0 outputs) make DDRT=1 DDRT = ( LogicalDDRT&(~data) ); +5V unsigned char Pin(void){ return PTT; // unchanged R Extra Question Design a circuit that interfaces a 2.4-V, 13-mA LED to YourComputer. Extra Solution The 7405 can deliver the 13 ma output current required. Assume the V OL of the 7405 is 0.5 V. To get a diode current of 13 ma, we use R= ( )/0.013 = 162- resistor Extra Question Design a circuit that interfaces a 2.3-V, 14-mA LED to YourComputer. Extra Solution The 7405 can deliver the 14 ma output current required. Assume the V OL of the 7405 is 0.5 V. To get a diode current of 14 ma, we use R= ( )/0.014 = 157- resistor. Extra Question Design a circuit that interfaces a 2.2-V, 11-mA LED to YourComputer. Extra Solution The 7405 can deliver the 11 ma output current required. Assume the V OL of the 7405 is 0.5 V. To get a diode current of 11 ma, we use R= ( )/0.011 = 209- resistor. Extra Question You are given the 9S12 voltage and current parameters for PT0: V OH, V OL, V IH, V IL, I OH, I OL, I IH, and I IL. The desired LED voltage is V D and the desired current is I D. Assume the LED current is small enough, so the LED can be interfaced directly to the 9S12 PT0 pin as shown. Give the equation to calculate the resistance R in terms of V D, I D, V OH, V OL, V IH, V IL, I OH, I OL, I IH, and I IL. Extra Solution When active the output voltage of PT0 will be V OH, and the LED voltage will be V D. This means (V OH - V D ) will develop across the resistor. To generate the desired LED current of I D, the resistor should be R = (V OH -V D )/ I D Extra Question You are given the 9S12 voltage and current parameters for PT1: V OH, V OL, V IH, V IL, I OH, I OL, I IH, and I IL. The desired LED voltage is V D and the desired current is I D. Assume the LED current is small enough, so the LED can be interfaced directly to the 9S12 PT1 pin as shown. Give the equation to calculate the resistance R in terms of V D, I D, V OH, V OL, V IH, V IL, I OH, I OL, I IH, and I IL. Extra Solution When active the output voltage of PT1 will be V OL, and the LED voltage will be V D. This means (5- V OL - V D ) will develop across the resistor. To generate the desired LED current of I D, the resistor should be R = (5 - V OL - V D )/ I D PT0 PT1 R R +5V 4

5 Extra Question If a system uses a 15-bit ADC, about how many decimal digits will it have? Extra Solution A 15-bit ADC has 2 15 = alternatives. This is more than alternatives, which equals 4½ decimal digits. Extra Question If a system uses a 17-bit ADC, about how many decimal digits will it have? Extra Solution A 17-bit ADC has 2 17 = alternatives. This is more than alternatives, which equals 5 decimal digits. It is enough less than 200,000 that we would not claim 5½ decimal digits. Extra Question If a system requires 2¾ decimal digits of precision, what is the smallest number of bits the ADC needs to have? Extra Solution A precision of 2¾ decimal digits equals 400 alternatives. The next higher power of 2 is 512 alternatives, which equals 9 bits. Extra Question The resolution of an 8-bit unsigned binary fixed-point number is 2 3, which equals 1/8. What is the value of the number if the integer stored in memory is 83? Extra Solution value = integer*resolution = 83/8 = Extra Question The resolution of an 8-bit unsigned binary fixed-point number is 2 7, which equals 1/128. What is the value of the number if the integer stored in memory is 49? Extra Solution value = integer*resolution = 49/128 = Extra Question The resolution of an 8-bit unsigned binary fixed-point number is 2 2, which equals 1/4. What is the value of the number if the integer stored in memory is 50? Extra Solution value = integer*resolution = 50/4 = 12.5 Extra Question Assume you are going to use a 16-bit signed decimal fixed-point number system to represent the values from to What is the smallest fixed-point resolution you could use? Extra Solution value = integer*resolution. The range of 16-bit signed integers is to The smallest resolution is 0.1, which would create a value range of to Extra Question Assume you are going to use a 16-bit unsigned decimal fixed-point number system to represent the values from 0 to 100. What is the smallest fixed-point resolution you could use? Extra Solution value = integer*resolution. The range of 16-bit unsigned integers is 0 to The smallest resolution is 0.01, which would create a value range of 0 to Extra Question Assume you are going to use a 16-bit signed binary fixed-point number system to represent the values from to What is the smallest fixed-point resolution you could use? Extra Solution value = integer*resolution. The range of 16-bit signed integers is to The smallest resolution is 1/32, which would create a value range of /32 to /32, which is to Extra Question A signed fixed point system has a range of values from -50 to +50 with a resolution of 2 8. Note: 2 8 equals 1/256. With which of the following data types should the software variables be allocated? When more than one answer is possible choose the most space efficient type. A) char B) unsigned char C) float D) short E) unsigned short F) double G) long H) unsigned long Extra Solution value = integer*resolution. The range of integers will be 50*256 to +5*256, which is to These integers will fit into D) short Extra Question A signed fixed point system has a range of values from -100 to +100 with a resolution of 2 9. Note: 2 9 equals 1/512. With which of the following data types should the software variables be allocated? When more than one answer is possible choose the most space efficient type. A) char B) unsigned char C) float D) short E) unsigned short F) double 5

6 G) long H) unsigned long Extra Solution value = integer*resolution. The range of integers will be 100*512 to +100*512, which is to These integers will fit into G) long Extra Question An unsigned fixed point system has a range of values from 0 to +100 with a resolution of 2 9. Note: 2 9 equals 1/512. With which of the following data types should the software variables be allocated? When more than one answer is possible choose the most space efficient type. A) char B) unsigned char C) float D) short E) unsigned short F) double G) long H) unsigned long Extra Solution value = integer*resolution. The range of integers will be 0 to 100*512, which is 0 to These integers will fit into E) unsigned short Extra Question Write software that initializes 9S12 Port T, so pins 3 and 4 are output without affecting the other 6 bits. Extra Solution bset DDRT,#$18 ; 4,3 output // 9S12 C DDRT = 0x18; // 4,3 output Extra Question Write software that initializes 9S12 Port T, so pins 6 and 1 are inputs without affecting the other 6 bits. Extra Solution bclr DDRT,#$42 ; 6,1 inputs // 9S12 C DDRT &= ~0x42; // 6,1 inputs Extra Question Write software that initializes 9S12 Port H, so all the pins are output, then outputs the repeating sequence $E0, $3C, $07. This software outputs the sequence over and over without stopping. Extra Solution main movb #$FF,DDRH loop movb #$E0,PTH movb #$3C,PTH movb #$07,PTH bra loop ;7-0 outputs // 9S12 C DDRH = 0xFF; // 7-0 outputs while(1){ PTH = 0xE0; PTH = 0x3C; PTH = 0x07; Extra Question Write software that initializes 9S12 Port M, so all the pins are output, then outputs the repeating sequence $01, $02, $04, $08. This software outputs the sequence over and over without stopping. Extra Solution main movb #$FF,DDRM loop movb #$01,PTM movb #$02,PTM movb #$04,PTM movb #$08,PTM bra loop ;7-0 outputs // 9S12 C DDRM = 0xFF; // 7-0 outputs while(1){ PTM = 0x01; PTM = 0x02; PTM = 0x04; PTM = 0x08; Extra Question Does the associative principle hold for signed integer multiply and divide? In particular do these two C calculations always achieve identical outputs? If no, give an example. Out1 = (A*B)/C; Out2 = A*(B/C); Extra Solution No (4*7)/2 = 14 4*(7/2) = 12 6

7 Extra Question Does the associative principle hold for signed integer addition and subtraction? In particular do these two C calculations always achieve identical outputs? If no, give an example. Out3 = (A+B)-C; Out4 = A+(B-C); Extra Solution Yes Extra Question In this question the input parameter is the temperature, T C, which is a 16-bit signed binary fixed-point number with a resolution of 1/8 ºC. The integer portion of this parameter is stored in memory as the variable Ic. The output of this function will be the temperature, T F, which is a 16-bit signed binary fixed-point number with a resolution of 1/8 ºF. Write C code that implements the conversion from ºC to ºF using fixed-point math. In particular, calculate T F =1.8*T C +32º. The function operates on the integer portion of the fixed point numbers. You may assume the input temperature range is bounded -50 < T C < +100 ºC. The prototype is short Convert(short Ic); // convert Ic to If Extra Solution =256/8. The largest input occurs at 100 ºC, yielding an integer value of 800. Overflow during the *9 using 16-bit math can t happen because 800*9=7200, which is less than short Convert(short Ic){ short If; // convert Ic to If If = (Ic*9)/ ; return If; Extra Question In this question the input parameter is the temperature, T C, which is a 16-bit signed binary fixed-point number with a resolution of 1/16 ºC. The integer portion of this parameter is stored in memory as the variable Ic. The output of this function will be the temperature, T F, which is a 16-bit signed binary fixed-point number with a resolution of 1/16 ºF. Write C code that implements the conversion from ºC to ºF using fixed-point math. In particular, calculate T F =1.8*T C +32º. The function operates on the integer portion of the fixed point numbers. You may assume the input temperature range is bounded -20 < T C < +50 ºC. The prototype is short Convert(short Ic); // convert Ic to If Extra Solution =512/16. The largest input occurs at 50 ºC, yielding an integer value of 800. Overflow during the *9 using 16-bit math can t happen because 800*9=7200, which is less than short Convert(short Ic){ short If; // convert Ic to If If = (Ic*9)/ ; return If; Extra Question In this question the input parameter is the temperature, T F, which is a 16-bit signed binary fixed-point number with a resolution of 1/8 ºF. The integer portion of this parameter is stored in memory as the variable If. The output of this function will be the temperature, T C, which is a 16-bit signed binary fixed-point number with a resolution of 1/8 ºC. Write C code that implements the conversion from ºF to ºC using fixed-point math. In particular, calculate T C =(5/9)*(T f -32º). The function operates on the integer portion of the fixed point numbers. You may assume the input temperature range is bounded -40 < T F < +120 ºF. The prototype is short Convert(short If); // convert If to Ic Extra Solution =256/8. The largest input occurs at 120 ºF, yielding an integer value of 960. The smallest input occurs at -40 ºF, yielding an integer value of Overflow during the *9 using 16-bit math can t happen because ( )*5= 3520, which is less than Similarly ( )*5= short Convert(short If){ short Ic; // convert Ic to If Ic = ((Ic-256)*5)/9; return Ic; Extra Question In this question the input parameter is the temperature, T F, which is a 16-bit signed binary fixed-point number with a resolution of 1/16 ºF. The integer portion of this parameter is stored in memory as the variable If. The output of this function will be the temperature, T C, which is a 16-bit signed binary fixed-point number with a resolution of 1/16 ºC. Write C code that implements the conversion from ºF to ºC using fixed-point math. In particular, calculate T C =(5/9)*(T f -32º). The function operates on the integer portion of the fixed point numbers. You may assume the input temperature range is bounded -40 < T F < +120 ºF. The prototype is short Convert(short If); // convert If to Ic 7

8 Extra Solution =512/16. The largest input occurs at 120 ºF, yielding an integer value of The smallest input occurs at -40 ºF, yielding an integer value of Overflow during the *9 using 16-bit math can t happen because ( )*5= 7040, which is less than Similarly ( )*5= short Convert(short If){ short Ic; // convert Ic to If Ic = ((Ic-512)*5)/9; return Ic; Dan, are you pondering what I m pondering? Yah. Ben, who plugged this typerwriter into our TV? J V w 8

EE319 K Lecture 3. Introduction to the 9S12 Lab 1 Discussion Using the TExaS simulator. University of Texas ECE

EE319 K Lecture 3. Introduction to the 9S12 Lab 1 Discussion Using the TExaS simulator. University of Texas ECE EE319 K Lecture 3 Introduction to the 9S12 Lab 1 Discussion Using the TExaS simulator University of Texas ECE Introduction (von Neumann architecture) processor Bus Memory Mapped I/O System Input Devices