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Lecture 3 9/7/11

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Lab 202a: Door combo 1 4 2 3 <enter>

- Do homework problems 2.25, 2.31-2.35 (we will go over in class) for Friday 9/9/11

- Floating point: overhead too high for many embedded apps – hundreds of operations to do FP ops
- Fixed point: 2 part representation
- Explicit variable integer I
- Implicit (i.e. not stored, determined and remembered by programmer!) fixed constant
- Represented number F=I

- Precision: number of distinguishable patterns, governed by number of bits in I
- Resolution: smallest difference that can be represented, by definition

- Money! Integral pennies: =0.01
- Decimal-based implicit constant makes sense here – “Decimal fixed-point” though will be represented in binary of course

- Voltmeter with resolution of 0.001V (i.e. integral mV) also makes sense: =0.001
- Decimal fixed point useful for managing displays to humans, binary fixed point where =2m is easier to computer with!

- Voltmeter on 6811/12
- Built-in 8-bit ADC takes 0-5V input
- 256 bit patterns to represent the entire range
- Output N of the ADC is such that
- Vin = 5*N/255 = 0.019607843*N

- Smallest change we can detect is about 20mV
- This one won’t discriminate between 3.005 and 3.008 V, or even 3.01 and 3.02 V

- Design decisions
- We now need to represent the voltage in a format convenient for display

- Built-in 8-bit ADC takes 0-5V input

- Since a decimal display of voltage is the end game, we choose a decimal fixed point display with =0.01V
- Why not =0.1V?
- Why not =0.001V?

- Since a decimal display of voltage is the end game, we choose a decimal fixed point display with =0.01V
- Why not =0.1V?
- Why not =0.001V?
- Essential point: this resolution is a bit better than the raw ADC resolution (500 vs 256 patterns): we don’t want to lose resolution as an artifact of display format! Thermometers!!!
- Complication: doesn’t fit in 8 bits! Need 916

- Main reason we use it: to add/subtract Fixed point numbers with same , just do integer add/subtract!
- If numbers have different values, we have to convert to a common basis first.

- we want z=x+y with
x=I2n

y=J2m

z=K2p

- algebraic manipulation gives
K=I2n-p + J2m-p

- alot like equalizing exponents in FP math

- multiplication, division more challenging
- reasons similar to excess-bias problem in FP math
- if you multiply two numbers representing dollar values stored as pennies (=0.01), straight multiplication gives the answer in terms of 1/10,000 dollars (1/100 of a penny)
- need to multiply by 100 to get back to pennies representation

- if you multiply two numbers representing dollar values stored as pennies (=0.01), straight multiplication gives the answer in terms of 1/10,000 dollars (1/100 of a penny)

- reasons similar to excess-bias problem in FP math

- Arithmetic, logical operaions
- full complement of logical bitwise operators
- AND, OR, NOT, XOR (EOR on Freescale)
- details of instruction variants will follow in ch3
- 10 variants of AND instruction, for instance!

- shift instructions
- ASR, LSR; ASL, LSL; ROR, ROL
- difference ASR,LSR?
- difference ASL,LSL?

- ASR, LSR; ASL, LSL; ROR, ROL

- full complement of logical bitwise operators

- There are integer instructions for + −×÷
- Challenge: programmer needs to check for anomalies such as overflow explicitly
- Aid: CCR: Condition Code Register
- unlike MIPS but like most other processors
- 4 flags of interest to us now (there are more)
- N negative: result of last computation was negative
- Z zero: result of last computation was zero
- V overflow: result of last computation was signed overflow (i.e. 2’s complement)
- C carry: result of last computation was unsigned overflow

- Each operation sets appropriate flag bits, programmer then checks them EVERY TIME!

- For example, knowing we just did an unsigned add (bcc = Branch if C flag Clear)
ldaaA8get first input

addaB8regA = A8 + B8

bccOK1if C=0 then no error

so skip to end

ldaa #255overflow

OK1 staaR8 R8= (smaller of A8+B8, 255)

- Strategy here is to round an unsigned overflow back down to largest possible unsigned integer
- Why not just abort?
- What if V set?

- similar idea: doing R8=A8-B8, set R8 to 0 on unsigned overflow (i.e. negative result)

- The -128 – +127 problem complicates life a bit
- several typos in your book on page 53 (at least in some printings) also complicate understanding
- nb BMI is Branch if Minus, i.e. N=1.
- BPL is Branch if Plus, i.e. N=0.
- BVC is Branch if V flag Clear

ldaaA8get first input

addaB8regA = A8 + B8

bvc ok3 if V=0 no error so

skip to end

err3 bmi over3if V=1 and N=1, it

was overflow

ldaa#-128 if V=1 and N=0,

it was underflow

bra ok3

over3 ldaa#127overflow

ok3 staaR8WAS MISSING IN BOOK