How to approximate float function by integer numbers?

I need to spare same space in 2 kB Flash MCU to finish with the program to control servos where x = 0.0° to 90.0°

How to approximate the function float y = x / 90 * 1250 + 3750 by an integer function, preferably using uint16_t

The integer divisor should be a power of 2.

Answer 1

[Les Beckham]

I'm assuming you want to replace the floating point calculation with an integer one? I'm not sure how that is going to "spare some space" unless you can't do floating point calculations and you don't have room for floating point emulation in software.

There are a lot of possible ways to do that depending on details of your application which you didn't provide.

This example comes pretty close.

x = 0:0.1:90;
y = x./90*1250 + 3750;
x2 = uint16(0:90 * 111); % scale your x values by 111 
y2 = x2/8 + 3750; % replace /8 with right-shift 3 places
plot(x, y, x2/111, y2)
xlabel 'x'
ylabel 'y'
legend('Float calculation', 'Integer calculation', 'Location', 'southeast')
grid on

[Dan Richter]

Perfect, thank you very much. I am working with Microchip MCU ATtiny202 (128B SRAM, 2k Flash). I was able to adopt Arduino Cat laser toy DIY for my cat ;-).

[Les Beckham]

You are quite welcome.

Answer 2

[Walter Roberson]

General exact process:

This can be done more compactly using bitand() and bitget() and similar, but sometimes it is easier to think in terms of streams of bits.

format long g
R = (1250/90)

R = 
          13.8888888888889

U64 = typecast(double(R), 'uint64');
B64 = dec2bin(U64, 64);

numerator = int64(bin2dec(['1', B64(end-51:end)]))

numerator = int64
   7818749353073778

denominator = 2^(510 + 52 - bin2dec(B64(2:11)))

denominator = 
           562949953421312

sign = 2 * (B64(1)=='0') - 1

sign = 
     1

reconstructed = sign * double(numerator) / double(denominator)

reconstructed = 
          13.8888888888889

R - reconstructed

ans = 
     0

Approximating with a 16 bit denominator would take more work. Or perhaps less...

[Walter Roberson]

format long g
R = (1250/90)

R = 
          13.8888888888889

D = 15 - ceil(log2(abs(R)));
denominator = uint16(2^D)

denominator = uint16
   2048

sign = 1; if R < 0; sign = -1; end
numerator = sign * int16(floor(abs(R) * denominator))

numerator = int16
   28444

reconstructed = double(numerator) / double(denominator)

reconstructed = 
              13.888671875

R - reconstructed

ans = 
      0.000217013888889284

When you look at those, at first it looks as if it would be plausible that you could gain another bit of accuracy by using a numerator one bit different from twice as large as the existing one, so 56888 +/- 1. But if you do that then you lose the room for the numerator to be negative.

This code will not work properly for input values less than 1.