Physics 116B Winter 2006: Problem Assignment 8

Transcription

1 Physics 116B Winter 2006: Problem Assignment 8 Due Wednesday, March You may use a calculator for the following: (a) Convert to decimal: $9F03. (b) Express in binary: $3CFA. (c) Convert to hexadecimal: 343. (d) Convert to a 16 bit signed integer in 2 s complement notation: -27. (e) Convert from 2 s complement signed integer to decimal: $FCA3. 2. Consider the MAS M68000 Mac program on the next page. The lines are numbered in the left column. (a) The MAS code uses certain Macintosh conventions to allow instruction code and data segments to be relocated in memory without needing to change the instruction codes. i. Which two M68000 registers are used to achieve this capability? ii. Give an example of an M68000 addressing mode which may be used for the instruction area to achieve this. Give an example for the data area. (Refer to Sec. 2.2 of the M68000 Programmer s Manual.) iii. In writing your own code, what precaution must be taken regarding use of the address registers because of this? iv. Which (if any) of the following lines of code are unaffected by this convention: 12, 14, 15, 20, 21? Explain briefly. (b) Suppose the jsr on line 14 was replaced with jmp. Would the program go to line 29 anyway? If so, what would go wrong later? Explain in some detail. (c) Do any instructions make use of numerical constants stored in the program code segment? If so, give one example and explain briefly. (d) On line 41, suppose D0 ($40) initially. What is the HEX number (and the corresponding ASCII character) in D0 after the operation on line 41? (e) The program modifies characters so it only has to search in the range ASCII a to f (lower case only). Show how to change the program to modify characters such that the search is in the range A to F (upper case only) and return the proper HEX value. Just list the lines which you change. 1

2 Sum_of_Chars_Mod_MASn.a Printed: Thursday, March 17, :31:27 PM Page 1 of 1 1 ; Program to read nchar ASCII characters and sum up any HEX digits (0-F) 2 ; HEX digits can also be lower case 3 ; This is the MAS version 4 ; Just uses a pre-defined string of characters starting at msg. Sum is in D2 at end. 5 ; Output result to screen when done 6 7 xref strout, decout, newline, stop 8 9 start: lea msg,a1 ; put address of msg in a1 10 move.w nchar,d1 ; number of bytes to read 11 clr.w d2 ; clear the register for the sum (specify.w) 12 jmp enter ; enter loop at end 13 loop: move.b (a1)+,d0 ; move the character from msg array to d0 14 jsr cnvhx ; subroutine to find valid HEX digits 15 add.w d0,d2 ; number or 0 returned in d0 16 enter: dbra d1,loop ; subtract 1 from d1 and see if done 17 ; all data processed, sum in d2 18 ; now output the information 19 jsr newline 20 lea msg5,a0 ; output sum message 21 move.w #31,d0 22 jsr strout 23 move.w d2,d0 ; output the sum 24 jsr decout 25 jsr newline 26 jsr stop ; end of program 27 ; Subroutine cnvhx tests ASCII characters to see if they represent HEX digits 28 ; and if so, returns value (.w) in d0. If not valid digit, returns 0 29 cnvhx: and.b #$7F,d0 ; mask off parity bit of character 30 cmp.b #$30,d0 ; see if it is less than $30 31 blt exit ; if so, exit subroutine with zero result 32 cmp.b #$39,d0 ; see if it is greater than $39 33 bgt skip ; if so, check for a-f 34 and.w #$000F,d0 ; return 16 bit pos. number in d0 35 rts ; all done if exit: clr.w d0 ; not a hex digit - return 0 as 16 bit word 37 rts 38 ; Look for hex digits ($41-$46 for A-F or $61-$66 for a-f) 39 ; Set bit 5 (lsb=0) so we only have to check lower case characters 40 ; When get valid digit, subtract $57 to convert to HEX value 41 skip: or.b #$20,d0 ; set bit to assure lower case 42 cmp.b #$61,d0 ; see if less than $61 (a) 43 blt exit ; if so exit with 0 result 44 cmp.b #$66,d0 ; see if greater than $66 (f) 45 bgt exit ; exit with 0 result 46 sub.b #$57,d0 ; found valid HEX digit offset by $57 from value 47 and.w #$000F,d0 ; clear high order bits (if any) for 16 bit word 48 rts 49 data 50 msg5: dc.b 'The sum of numbers entered is: ' 51 msg: dc.b 'abc89hijkl85973nb642abc89hijkl85973nb642' ; define character string 52 nchar: dc.w 40 ; number of characters to read 53 even 54 end 55 2

3 3. Consider the EASY68k program on the next page. The lines are numbered in the left column. This time, we are just summing any ASCII characters that happen to be decimal integers. The data are read from the keyboard and output to the screen using EASY68k I/O routines which rely on the trap #15 instruction. The use of these routines is explained in the included table from the EASY68k help commands. The trap instruction itself is discussed on p of the M68000 Microprocessor User s Manual. It is useful for implementing system program calls as is done here. The result is equivalent to a MAS system subroutine call although the mechanism is different. Notice the use of absolute addressing for the code and data via the org directive for the assembler. That is, the first instruction with the label start will be placed at location $1000. The storage area for msg will begin at $2000. (a) The instruction on line 8, lea are hex). msg, a1, is assembled as 43F (numbers i. Look up LEA in the M68000 Programmer s Reference Manual. Compare the binary pattern of the first word of this instruction and verify it is what you expect (explain). Check the register field and effective address field (mode, register) to determine the addressing mode for this operation. ii. Look up the behavior of this addressing mode in Sec. 2.2 of the manual. Verify that the last 4 bytes are what you expect (explain). iii. Would you expect this addressing mode to be used in the Mac program with the MAS system? Explain briefly. (b) The instruction jsr addit (line 18) is located at address $1022 and the next instruction (line 19) at $1028. Prior to executing line 18, the stack pointer contains $ Upon entry to the addit subroutine on line 32, answer the following. i. What is the contents of the stack pointer? ii. What is the 32 bit word contained in the memory address referenced by the stack pointer? (c) If the ASCII character for zero ( 0) is in d0 on entry to the addit subroutine on line 32, what is the contents (i) of the Z condition code and (ii) the N condition code after line 33 is executed? Note: You are strongly encouraged to get EASY68k to run or modify this program. A text file of the code is on the Physics 116 site. You can use the EASY68k simulator to watch the registers, memory and stack as you run the program. Breakpoints can be set to stop the execution if you want to look at registers. Also, there are two modes of single-stepping. One steps over user subroutines (runs them but does not trace execution) while the other traces execution into the subroutine. 3

4 Sum_of_chars_wIO.txt Printed: Thursday, March 9, :09:38 PM Page 1 of 1 1 ; Program to read, store and echo string of 20 ASCII characters and sum up any integers 2 ; Modified for Easy68K simulator - uses Easy68k trap instructions for I/O 3 ; Sum in D2 at end, display on screen 4 5 start: org $ move.w #2,d0 ; set up registers for trap instruction to read string 7 move.w nchar,d1 ; number of characters to read in d1 (80 max) 8 lea msg,a1 ; load the address of msg in a1 (must use LEA) 9 trap #15 ; do EASY68k I/O operation 10 move.w #0,d0 ; display string on screen 11 trap #15 ; do EASY68k I/O operation ; begin processing clr d2 ; clear the register for the sum 16 jmp enter ; enter loop at end 17 loop: move.b (a1)+,d0 ; move the character from msg array to d0 18 jsr addit ; subroutine to test and add integers 19 enter: dbra d1,loop ; subtract 1 from d1 and see if done ; all data processed, sum in d2 - now display number on screen move.l d2,d1 ; put result in d1 for EASY68k I/O 24 move.w #3,d0 ; display number on screen 25 trap #15 ; do EASY68k I/O operation STOP #$2000 ; end of program ; Subroutine addit tests ASCII characters to see if they represent numbers 30 ; and if so, adds them to the sum in d addit: and.b #$7F,d0 ; mask off parity bit of character 33 cmp.b #$30,d0 ; see if it is less than $30 34 blt skip ; if so, skip to return statement 35 cmp.b #$39,d0 ; see if it is greater than $39 36 bgt skip ; if so, skip to return statement 37 and.w #$000F,d0 ; get the number 38 add.w d0,d2 ; add to sum in d2 39 skip: rts ; return from subroutine org $2000 ; data storage area 42 msg: ds.b 80 ; storage for string of ASCII characters 43 nchar: dc.w 20 ; number of characters to read 44 END START 4

5 4. Refer to the posted Week 11 lecture notes for the following: (a) A block diagram of a successive approximation ADC is shown above. Bit 1 is the least significant bit of the output binary number. The voltage to be measured is v in. i. Consider a 3 bit ADC as an example (n=3 in the diagram above). What is the first binary number sent to the DAC by the successive approximation control register? ii. If the comparator output is low, what 3 bit number is sent next? iii. As a result, if the comparator output is now high, what number is sent next? iv. Now the comparator output is low. What is the 3 bit ADC value for this input voltage? v. What is the main advantage of this ADC over a simple counting (or counter-ramp) ADC? (b) The high frequency cutoff of ordinary telephone signals is 3400 Hz. Could such signals be digitized at a sampling rate of 8000 Hz and reproduced accurately? Explain briefly. (c) Would the presence of a 5000 Hz signal cause a problem with an 8000 Hz sampling rate? Explain briefly. 5