Linear interpolation works on Windows but not on my PIC18F, trouble with 8-bit arch?-CodePudding

I've implemented linear interpolation on my PIC18F device where I'm interpolating ADC values with intervals stored in EEPROM. I tested the code and it doesn't calculate properly. So I put it into my computer and the calculations works well so I don't really know where the problem might come from, as the two functions I use right now for testing purposes are same on both platforms.

These are the functions same for both platforms:

uint16_t interpolate(uint16_t a, uint16_t b, uint16_t c, uint16_t d, uint16_t raw) {
    // b (x-a)*(d-b)/(c-a)
    return b   (raw - a) * (d - b) / (c - a);
}

uint16_t mapRawToCal(uint16_t raw) {
    uint16_t cal = 0;

    if (raw < 150) {
        cal = 0;
    }
    else if (raw >= 150 && raw <= 457) {
        cal = interpolate(150, 0, 457, 800, raw);
    }
    else if (raw > 457 && raw <= 1235) {
        cal = interpolate(457, 800, 1235, 1000, raw);
    }
    else if (raw > 1235 && raw <= 1988) {
        cal = interpolate(1235, 1000, 1988, 1988, raw);
    }
    else if (raw > 1988 && raw <= 2700) {
        cal = interpolate(1988, 1988, 2700, 3500, raw);
    }
    else if (raw > 2700 && raw <= 3560) {
        cal = interpolate(2700, 3500, 3560, 3300, raw);
    }
    else if (raw > 3560 && raw <= 3987) {
        cal = interpolate(3560, 3300, 3987, 4096, raw);
    }
    else if (raw > 3987) {
        cal = 4096;
    }

    return cal;
}

This is the PIC18F device main function:

int main(void) {
    // Setup
 hardware_init();
   
 ADC_Init();
    
    // Loop
    while(1) {
        uint16_t ch1Data;
        ch1Data = ADC_Read(1);
        
        uint16_t *buf = usb_get_in_buffer(1);
        buf[0] = ch1Data; // Rx Axis
        buf[1] = mapRawToCal(ch1Data); // Ry Axis
           
        usb_send_in_buffer(1, 4);
           
        __delay_ms(200); //Delay

        #ifndef USB_USE_INTERRUPTS
        usb_service();
        #endif
    }
    
 return 0;
}

And this is the main function on Windows:

int main() {
    for (uint16_t raw = 0; raw <= 4096; raw  ) {
        printf("input=%d - output=%d\n", raw, mapRawToCal(raw));
    }

    return 0;
}

Here is chunk of the C Windows program result in console (that is accurate):

input=129 - output=0
input=130 - output=0
input=131 - output=0
input=132 - output=0
input=133 - output=0
input=134 - output=0
input=135 - output=0
input=136 - output=0
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input=162 - output=31
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input=170 - output=52
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input=452 - output=786
input=453 - output=789
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input=456 - output=797
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input=458 - output=800
input=459 - output=800
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input=491 - output=808
input=492 - output=808
input=493 - output=809
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input=495 - output=809
input=496 - output=810
input=497 - output=810
input=498 - output=810
input=499 - output=810
input=500 - output=811
input=501 - output=811
input=502 - output=811
input=503 - output=811
input=504 - output=812
input=505 - output=812
input=506 - output=812

And here is the video what my PIC outputs, the top chart is the raw data from ADC, the bottom one is with the calculated/interpolated data: https://youtu.be/Dxc60d8wDHw

Any help is highly appreciated! Thank you!

CodePudding user response：

You are a "victim" of the integer promotion, and that integers have different sizes on the PIC and on the PC.

All integer calculation is done with int because of integer promotion.

Just an example with raw = 2488:

a is 1988
b is 1988
c is 2700
d is 3500

(raw - a) gives 500 and (d - b) gives 1512.

The product of (raw - a) by (d - b) should be 756000.

On the PC you get this result, because its int uses 32 bits and has therefore a range of -2147483648 to 2147483647.

But your PIC uses only 16 bits for int and so the result overflows the range of -32768 to 32767. The calculated result is -30.432, and gives you a wrong interpolation.

The solution is to use an intermediate result with a type of int32_t or uint32_t that provides a bigger range.

CodePudding user response：

In function interpolate(), this multiplication:

(raw - a) * (d - b)

produces a result of type int or unsigned int, depending on the width of those types. For an 8-bit arch, they are probably 16 bits wide (the minimum), and the result will then have type unsigned int. Again assuming 16-bit width, that result will overflow for some of your input values.

On Windows, you will find that int and unsigned int are both 32 bits wide. The result of the same multiplication will have type int (though it doesn't make a difference in this case) and it will not overflow.

There are a couple of ways you could handle this. Supposing that you want to produce the same numeric results, you can cast one of the operands to unsigned long:

    return b   (unsigned long)(raw - a) * (d - b) / (c - a);

That will probably come at a small computational cost.

Alternatively, if you were willing to accept a loss of precision then you could perform the division before the multiplication, which I would write this way for improved clarity:

    return b   (d - b) * ((raw - a) / (c - a));