|
ECEN 3213 Fall 2002 Exam 2 7 November 2002 |
Name______SOLUTION______ |
The
questions in this exam refer to the 68HC11A8 single chip, unless otherwise
noted in the question. This is an OPEN
Book Open notes exam. Read the
instructions carefully before completing the exam. All answers must be shown on these pages – no attached pages will
be graded.
1. (4 Pts) Draw a State Diagram
that:
-
has 4 states;
-
has 1 binary input value;
-
has 1 binary output value;
-
stays in the initial state
while the input is 0;
-
returns to the initial state
whenever the input is 0 two or more times in a row;
-
outputs a 0 whenever the
previous inputs is the same as the current input values. Outputs a 1 whenever the input is the
opposite of the previous input value.
The following True / False
questions are worth 2 points each.
Circle T for True or F for False for your answer.
F 2. All digital computer systems have the
CPU and some of the Memory on the same chip.
T 3. All digital computer systems have I/O.
T 4. The 6811 we use in this class (the single
chip version) has all three blocks of a digital computer system on one chip.
T 5. A system
using a 6811 could have extra memory added of the chip that has the CPU.
T 6. Extended
addressing is required to access external memory in a 6811 system.
T 7. The stack
pointer is a 16 bit register.
T 8. The stack
pointer can point to ROM.
F 9. The stack can be
in either RAM or ROM but not both.
F 10. Binary numbers
are always greater in value than their decimal equivalents.
11.(21pts) Fill in the following
table with hex values for the given values and operations.
|
Value A |
Value B |
Logical OR of Value A and Value
B |
Logical AND of Value A and Value
B |
Logical Exclusive-OR of Value A
and Value B |
|
$00 |
$97 |
$97 |
$00 |
$97 |
|
$FF |
$97 |
$FF |
$97 |
$68 |
|
$F0 |
$97 |
$F7 |
$90 |
$67 |
|
$0F |
$97 |
$9F |
$07 |
$98 |
|
$AA |
$97 |
$BF |
$82 |
$3D |
|
$CC |
$97 |
$DF |
$84 |
$5B |
|
$33 |
$97 |
$B7 |
$13 |
$A4 |
The following multiple-choice
questions will be scored as 2 pts for each correct
answer, -1 for an incorrect answer and 0 for unanswered
questions.
12. For a state
diagram, how many edges enter each state for a diagram with 1 binary input? ___g___
a. 0
b. 1 c. 2 d. 3 e. 4 f. 8 g. Can’t tell or varies
13. How many bytes are available for a stack in a 6811A8 without
external memory chips ?___g__
a. 0
b. 1 c. 2 d. 3 e. 4 f. 8 g. 256
14. If the stack is used only for storing subroutine return
addresses, how deep can the subroutine nesting be for a 9 byte stack?___e__
a. 0
b. 1 c. 2 d. 3 e. 4 f. 8 g. None
of these choices
15. For the same clock frequency, How much faster is a parallel 1
byte port than a 1 bit serial port with no start or stop bits?______f________
a. 0 b. 1 c. 2 d. 3 e. 4 f. 8 g. None of these choices
16. For the same clock frequency, How much faster is a parallel 1
byte port than a 1 bit serial port with 2 start bits and 1stop bits?_____c_________
a. 0 b. 10 c.
11 d. 12 e. 4 f.8 g. None
of these choices
17. What does it mean when the PC
register has the value 35? __b___
a. The next instruction to be fetched is in memory location
60735.
b. Something is probably wrong because 35 addresses RAM and
instructions are usually in ROM.
c. This is a trick question because a 16-bit PC register
cannot hold a value of less than 1024.
Consider the assembly program
listing on the last two pages of this exam.
18. (15pts) For each of the addressing modes: give an
example instruction, its address from the assembly listing, and the location of
the operand:
|
Addressing mode |
Example instruction |
Address of example instruction |
Location of operand for example
instruction |
|
Inherent |
inx |
$E11A |
x
register |
|
Immediate |
lds
#$f0 |
$E102 |
instruction
EEPROM |
|
Direct |
staa
oldpta |
$E108 |
mem
location $0005 |
|
Extended |
ldab
pte |
$E117 |
$100A |
|
Indexed |
stab
List, x |
$E11B |
mem
loca 17 + [reg x] |
19. (14 pts) Complete the following
RAM memory Map for the Assembly listing on the last two pages. (There are more
blanks than answers, put an XXX in unused spaces, which could help with
question on memory usage).
|
RAM Memory Address |
Name |
Size (number of Bytes) |
|
$00-$04 |
XXX |
5 unused bytes |
|
$05 |
oldpta |
1 byte |
|
$06 |
count |
2 bytes |
|
$08-$19 |
XXX |
17 unused bytes |
|
$20 |
lastled |
1 byte |
|
$21 to $60 |
XXX |
63 unused bytes |
|
$61 to $80 |
List |
31 bytes
for storing values |
|
$81 to $E9 |
XXX |
115 unused bytes(or 0 if
stack is 112) |
|
$f0 down to $EA |
Stack |
6 bytes (or 122 bytes
available) |
|
$f1 to $FF |
XXX |
14 unused bytes |
|
|
|
|
|
|
|
|
|
$FF |
XXX |
unused |
Consider the program listing on
the last two pages. Answer the
following memory usage questions.
20. (4pts)How many bytes of EPROM
are used to store the program instructions?__ 94 = $5E (accept 96=$60)_______
21. (4pts)How many bytes of the
RAM are NOT used?_215 (or 99)_____________
22. (4pts)What is in EPROM
location $E005?_____Don’t Know_________
23. (4pts) What is in EPROM
location $EE00? ______Don’t Know ________
$1004 ptb equ $1004 ; Port B
output bits for LED indicators
$100A pte equ $100A ; Port E
input digits
$0005 oldpta equ $5 ; place to store old value
$0061 List equ 97
$0006 count equ $06 ; store the count of
values
$0020 lastled equ $20 ;store last value for leds.
$E100 org $E100
$E100 8600 [ 2](
0){pp }start ldaa #$00
$E102
8E00F0 [ 3]( 2){ppp } lds #$f0
$E105
BDE300 [ 6]( 5){ppprss } jsr
subgreen
$E108 9705 [
3]( 11){ppw } staa oldpta
$E10A 9706 [ 3]( 14){ppw } staa count
$E10C
CE0000 [ 3]( 17){ppp }
ldx #$00
$E10F
BDE300 [ 6]( 20){ppprss } jsr subgreen
$E112 8601 [ 2]( 26){pp }begin ldaa #$01
$E114
BDE300 [ 6]( 28){ppprss } jsr subgreen
$E117
F6100A [ 4]( 34){pppr }reade ldab pte ;load value
$E11A 08 [ 3]( 38){pfn } inx
$E11B E761 [ 4]( 41){ppnw } stab List,x
$E11D DF06 [ 4]( 45){ppww } stx count
$E11F CF [ 2]( 49){pf } stop
$E120
BDE410 [ 6]( 51){ppprss } jsr subaddy
$E123
8C001F [ 4]( 57){pppn }
cpx #$1F
$E126 26EA [ 3]( 61){pfn } bne begin
$E128
BDE400 [ 6]( 64){ppprss } jsr subzero
$E12B CF [ 2]( 70){pf } stop
$E12C Þ20D2 [ 3]( 72){pfn }
bra start
$E200 org $E200
$E200 36 [ 3](
0){pfs }subred psha
$E201 8601 [ 2](
3){pp } ldaa #$01
$E203
B71004 [ 4]( 5){pppw } staa ptb
$E206 9720 [ 3](
9){ppw } sta lastled
$E208 32 [
4]( 12){pfxu pula
$E209 39 [ 5]( 16){pfxuu }
rts
$E300 org $E300
$E300 36 [ 3](
0){pfs }subgreen psha
$E301 8602 [ 2](
3){pp } ldaa #$02
$E303
B71004 [ 4]( 5){pppw } staa ptb
$E306 9720 [ 3](
9){ppw } staa lastled
$E308 32 [ 4]( 12){pfxu } pula
$E309 39 [ 5]( 16){pfxuu }
rts
$E400 org $E400
$E400 36 [ 3](
0){pfs }subzero psha
$E401
BDE200 [ 6]( 3){ppprss } jsr subred
$E404
BDE300 [ 6]( 9){ppprss } jsr
subgreen
$E407 8600 [ 2]( 15){pp } ldaa #$00
$E409
B71004 [ 4]( 17){pppw }
staa ptb
$E40C 9720 [ 3]( 21){ppw } staa lastled
$E40E 32 [ 4]( 24){pfxu } pula
$E40F 39 [ 5]( 28){pfxuu }
rts
$E410 36 [ 3]( 33){pfs }subaddy psha
$E411 9620 [ 3]( 36){ppr } ldaa lastled
$E413 8A04 [ 2]( 39){pp } oraa #$04
$E415
B71004 [ 4]( 41){pppw }
staa ptb
$E418 9720 [ 3]( 45){ppw } staa lastled
$E41A 32 [ 4]( 48){pfxu } pula
$E41B 39 [ 5]( 52){pfxuu }
rts
$FFFE org $FFFE
$FFFE E100 dc.w start
$0000 end
***************Symbol
Table*********************
List $0061
begin $E112
count $0006
lastled $0020
oldpta $0005
ptb $1004
pte $100A
reade $E117
start $E100
subaddy $E410
subgreen $E300
subred $E200
subzero $E400
Assembly
successful