# Lecture 3 9/7/11 - PowerPoint PPT Presentation

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Lecture 3 9/7/11. 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 . Fixed Point Arithmetic. Floating point: overhead too high for many embedded apps – hundreds of operations to do FP ops Fixed point: 2 part representation

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Lecture 3 9/7/11

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### Lecture 3 9/7/11

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

### Fixed Point Arithmetic

• 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

### Examples

• 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!

### Examples

• 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

### Voltmeter design, cont.

• 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?

### Voltmeter design, cont.

• 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 916

### Fixed point arithmetic

• 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.

### Fixed point arithmetic

• we want z=x+y with

x=I2n

y=J2m

z=K2p

• algebraic manipulation gives

K=I2n-p + J2m-p

• alot like equalizing exponents in FP math

### Fixed point arithmetic

• 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

### Freescale instruction set

• 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?

### Arithmetic operations

• 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

bccOK1if C=0 then no error

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?

### Unsigned subtraction

• 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

bvc ok3 if V=0 no error so

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