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Let's say I have a 12-bit Analog to Digital Converter (4096 bins). And let's say I have a signal from 0 to 5 Volts.
What is the proper conversion formula to convert ADC bins into Volts?
V = ADC / 4096 * 5
or
V = ADC / 4095 * 5
Do I divide by 4096 because there are 4096 bins in the ADC?
Or do I divide by 4095 because that is the highest value that the ADC returns?
V = ADC / 4096 * 5
is the correct formula for converting the digital value back to (an approximation of) the analog voltage.
This is according to The Data Conversion Handbook, edited by Walt Kester (Newnes, 2005),
and available at (as of 2018/10/18) at:
https://www.analog.com/en/education/education-library/data-conversion-handbook.html
See in particular Figures 2.4 and 2.5 in Chapter 2:
In your case, FS would be 5 V. (And of course you're using a 12-bit ADC, not a 3-bit one.) Note that even if the ADC value is the maximum possible value (4095 in your case), the corresponding analog voltage will be slightly less than the "full-scale" voltage (5 V in your case).
Brian's suggestion about checking ADC datasheet is ideal. BUT! Assuming your maximum voltage (5V) equates to the maximum ADC input (12-bits = 4095), the following conversion should work for you:
const float maxAdcBits = 4095.0f; // Using Float for clarity
const float maxVolts = 5.0f; // Using Float for clarity
const float voltsPerBit = (maxVolts / maxAdcBits);
float yourVoltage = ADCReading * voltsPerBit;
A quick inspection of the math with Excel leads me to believe this is correct.
How picky do you want to get? If you want to really get picky, then you should also consider that each "bin" represents a small range of values (about 1.2 mV in your case). So when you convert to a voltage value, do you want to return the voltage value of the middle of the bin, or the lower edge of the bin? That is, do you want to effectively "truncate" or "round" in the value you report?
Also, the ADC's steps are probably even (linear), but take care as to what the ADC does with the bins at the two extremities of the range. Those bins may be possibly half the width of the others. It depends on the ADC, so check the spec.
Whether this concern matters at all depends on your application.
The spec for the ADC should identify how 5V is represented in terms of your 12 bits.
I would suspect that 4095 corresponds to 5V and thus your second solution is correct. Otherwise you would never be able to identify a signal of 5V correctly.
For a 12-bit value the maximum representable value is 4095, but of course there are 4096 values total (including zero). Assuming that your ADC is linear then yes, 4095 is equivalent to full scale. This is not necessarily 5V but whatever your reference voltage is equivalent to OR a value exceeding that voltage (of course).
Related
Why do some numbers lose accuracy when stored as floating point numbers?
For example, the decimal number 9.2 can be expressed exactly as a ratio of two decimal integers (92/10), both of which can be expressed exactly in binary (0b1011100/0b1010). However, the same ratio stored as a floating point number is never exactly equal to 9.2:
32-bit "single precision" float: 9.19999980926513671875
64-bit "double precision" float: 9.199999999999999289457264239899814128875732421875
How can such an apparently simple number be "too big" to express in 64 bits of memory?
In most programming languages, floating point numbers are represented a lot like scientific notation: with an exponent and a mantissa (also called the significand). A very simple number, say 9.2, is actually this fraction:
5179139571476070 * 2 -49
Where the exponent is -49 and the mantissa is 5179139571476070. The reason it is impossible to represent some decimal numbers this way is that both the exponent and the mantissa must be integers. In other words, all floats must be an integer multiplied by an integer power of 2.
9.2 may be simply 92/10, but 10 cannot be expressed as 2n if n is limited to integer values.
Seeing the Data
First, a few functions to see the components that make a 32- and 64-bit float. Gloss over these if you only care about the output (example in Python):
def float_to_bin_parts(number, bits=64):
if bits == 32: # single precision
int_pack = 'I'
float_pack = 'f'
exponent_bits = 8
mantissa_bits = 23
exponent_bias = 127
elif bits == 64: # double precision. all python floats are this
int_pack = 'Q'
float_pack = 'd'
exponent_bits = 11
mantissa_bits = 52
exponent_bias = 1023
else:
raise ValueError, 'bits argument must be 32 or 64'
bin_iter = iter(bin(struct.unpack(int_pack, struct.pack(float_pack, number))[0])[2:].rjust(bits, '0'))
return [''.join(islice(bin_iter, x)) for x in (1, exponent_bits, mantissa_bits)]
There's a lot of complexity behind that function, and it'd be quite the tangent to explain, but if you're interested, the important resource for our purposes is the struct module.
Python's float is a 64-bit, double-precision number. In other languages such as C, C++, Java and C#, double-precision has a separate type double, which is often implemented as 64 bits.
When we call that function with our example, 9.2, here's what we get:
>>> float_to_bin_parts(9.2)
['0', '10000000010', '0010011001100110011001100110011001100110011001100110']
Interpreting the Data
You'll see I've split the return value into three components. These components are:
Sign
Exponent
Mantissa (also called Significand, or Fraction)
Sign
The sign is stored in the first component as a single bit. It's easy to explain: 0 means the float is a positive number; 1 means it's negative. Because 9.2 is positive, our sign value is 0.
Exponent
The exponent is stored in the middle component as 11 bits. In our case, 0b10000000010. In decimal, that represents the value 1026. A quirk of this component is that you must subtract a number equal to 2(# of bits) - 1 - 1 to get the true exponent; in our case, that means subtracting 0b1111111111 (decimal number 1023) to get the true exponent, 0b00000000011 (decimal number 3).
Mantissa
The mantissa is stored in the third component as 52 bits. However, there's a quirk to this component as well. To understand this quirk, consider a number in scientific notation, like this:
6.0221413x1023
The mantissa would be the 6.0221413. Recall that the mantissa in scientific notation always begins with a single non-zero digit. The same holds true for binary, except that binary only has two digits: 0 and 1. So the binary mantissa always starts with 1! When a float is stored, the 1 at the front of the binary mantissa is omitted to save space; we have to place it back at the front of our third element to get the true mantissa:
1.0010011001100110011001100110011001100110011001100110
This involves more than just a simple addition, because the bits stored in our third component actually represent the fractional part of the mantissa, to the right of the radix point.
When dealing with decimal numbers, we "move the decimal point" by multiplying or dividing by powers of 10. In binary, we can do the same thing by multiplying or dividing by powers of 2. Since our third element has 52 bits, we divide it by 252 to move it 52 places to the right:
0.0010011001100110011001100110011001100110011001100110
In decimal notation, that's the same as dividing 675539944105574 by 4503599627370496 to get 0.1499999999999999. (This is one example of a ratio that can be expressed exactly in binary, but only approximately in decimal; for more detail, see: 675539944105574 / 4503599627370496.)
Now that we've transformed the third component into a fractional number, adding 1 gives the true mantissa.
Recapping the Components
Sign (first component): 0 for positive, 1 for negative
Exponent (middle component): Subtract 2(# of bits) - 1 - 1 to get the true exponent
Mantissa (last component): Divide by 2(# of bits) and add 1 to get the true mantissa
Calculating the Number
Putting all three parts together, we're given this binary number:
1.0010011001100110011001100110011001100110011001100110 x 1011
Which we can then convert from binary to decimal:
1.1499999999999999 x 23 (inexact!)
And multiply to reveal the final representation of the number we started with (9.2) after being stored as a floating point value:
9.1999999999999993
Representing as a Fraction
9.2
Now that we've built the number, it's possible to reconstruct it into a simple fraction:
1.0010011001100110011001100110011001100110011001100110 x 1011
Shift mantissa to a whole number:
10010011001100110011001100110011001100110011001100110 x 1011-110100
Convert to decimal:
5179139571476070 x 23-52
Subtract the exponent:
5179139571476070 x 2-49
Turn negative exponent into division:
5179139571476070 / 249
Multiply exponent:
5179139571476070 / 562949953421312
Which equals:
9.1999999999999993
9.5
>>> float_to_bin_parts(9.5)
['0', '10000000010', '0011000000000000000000000000000000000000000000000000']
Already you can see the mantissa is only 4 digits followed by a whole lot of zeroes. But let's go through the paces.
Assemble the binary scientific notation:
1.0011 x 1011
Shift the decimal point:
10011 x 1011-100
Subtract the exponent:
10011 x 10-1
Binary to decimal:
19 x 2-1
Negative exponent to division:
19 / 21
Multiply exponent:
19 / 2
Equals:
9.5
Further reading
The Floating-Point Guide: What Every Programmer Should Know About Floating-Point Arithmetic, or, Why don’t my numbers add up? (floating-point-gui.de)
What Every Computer Scientist Should Know About Floating-Point Arithmetic (Goldberg 1991)
IEEE Double-precision floating-point format (Wikipedia)
Floating Point Arithmetic: Issues and Limitations (docs.python.org)
Floating Point Binary
This isn't a full answer (mhlester already covered a lot of good ground I won't duplicate), but I would like to stress how much the representation of a number depends on the base you are working in.
Consider the fraction 2/3
In good-ol' base 10, we typically write it out as something like
0.666...
0.666
0.667
When we look at those representations, we tend to associate each of them with the fraction 2/3, even though only the first representation is mathematically equal to the fraction. The second and third representations/approximations have an error on the order of 0.001, which is actually much worse than the error between 9.2 and 9.1999999999999993. In fact, the second representation isn't even rounded correctly! Nevertheless, we don't have a problem with 0.666 as an approximation of the number 2/3, so we shouldn't really have a problem with how 9.2 is approximated in most programs. (Yes, in some programs it matters.)
Number bases
So here's where number bases are crucial. If we were trying to represent 2/3 in base 3, then
(2/3)10 = 0.23
In other words, we have an exact, finite representation for the same number by switching bases! The take-away is that even though you can convert any number to any base, all rational numbers have exact finite representations in some bases but not in others.
To drive this point home, let's look at 1/2. It might surprise you that even though this perfectly simple number has an exact representation in base 10 and 2, it requires a repeating representation in base 3.
(1/2)10 = 0.510 = 0.12 = 0.1111...3
Why are floating point numbers inaccurate?
Because often-times, they are approximating rationals that cannot be represented finitely in base 2 (the digits repeat), and in general they are approximating real (possibly irrational) numbers which may not be representable in finitely many digits in any base.
While all of the other answers are good there is still one thing missing:
It is impossible to represent irrational numbers (e.g. π, sqrt(2), log(3), etc.) precisely!
And that actually is why they are called irrational. No amount of bit storage in the world would be enough to hold even one of them. Only symbolic arithmetic is able to preserve their precision.
Although if you would limit your math needs to rational numbers only the problem of precision becomes manageable. You would need to store a pair of (possibly very big) integers a and b to hold the number represented by the fraction a/b. All your arithmetic would have to be done on fractions just like in highschool math (e.g. a/b * c/d = ac/bd).
But of course you would still run into the same kind of trouble when pi, sqrt, log, sin, etc. are involved.
TL;DR
For hardware accelerated arithmetic only a limited amount of rational numbers can be represented. Every not-representable number is approximated. Some numbers (i.e. irrational) can never be represented no matter the system.
There are infinitely many real numbers (so many that you can't enumerate them), and there are infinitely many rational numbers (it is possible to enumerate them).
The floating-point representation is a finite one (like anything in a computer) so unavoidably many many many numbers are impossible to represent. In particular, 64 bits only allow you to distinguish among only 18,446,744,073,709,551,616 different values (which is nothing compared to infinity). With the standard convention, 9.2 is not one of them. Those that can are of the form m.2^e for some integers m and e.
You might come up with a different numeration system, 10 based for instance, where 9.2 would have an exact representation. But other numbers, say 1/3, would still be impossible to represent.
Also note that double-precision floating-points numbers are extremely accurate. They can represent any number in a very wide range with as much as 15 exact digits. For daily life computations, 4 or 5 digits are more than enough. You will never really need those 15, unless you want to count every millisecond of your lifetime.
Why can we not represent 9.2 in binary floating point?
Floating point numbers are (simplifying slightly) a positional numbering system with a restricted number of digits and a movable radix point.
A fraction can only be expressed exactly using a finite number of digits in a positional numbering system if the prime factors of the denominator (when the fraction is expressed in it's lowest terms) are factors of the base.
The prime factors of 10 are 5 and 2, so in base 10 we can represent any fraction of the form a/(2b5c).
On the other hand the only prime factor of 2 is 2, so in base 2 we can only represent fractions of the form a/(2b)
Why do computers use this representation?
Because it's a simple format to work with and it is sufficiently accurate for most purposes. Basically the same reason scientists use "scientific notation" and round their results to a reasonable number of digits at each step.
It would certainly be possible to define a fraction format, with (for example) a 32-bit numerator and a 32-bit denominator. It would be able to represent numbers that IEEE double precision floating point could not, but equally there would be many numbers that can be represented in double precision floating point that could not be represented in such a fixed-size fraction format.
However the big problem is that such a format is a pain to do calculations on. For two reasons.
If you want to have exactly one representation of each number then after each calculation you need to reduce the fraction to it's lowest terms. That means that for every operation you basically need to do a greatest common divisor calculation.
If after your calculation you end up with an unrepresentable result because the numerator or denominator you need to find the closest representable result. This is non-trivil.
Some Languages do offer fraction types, but usually they do it in combination with arbitary precision, this avoids needing to worry about approximating fractions but it creates it's own problem, when a number passes through a large number of calculation steps the size of the denominator and hence the storage needed for the fraction can explode.
Some languages also offer decimal floating point types, these are mainly used in scenarios where it is imporant that the results the computer gets match pre-existing rounding rules that were written with humans in mind (chiefly financial calculations). These are slightly more difficult to work with than binary floating point, but the biggest problem is that most computers don't offer hardware support for them.
This question already has answers here:
BMI for generating masks with AVX512
(2 answers)
Closed 3 years ago.
One of the benefits of Intel's AVX-512 extension is that nearly all operations can be masked by providing in addition to the vector register a kreg which specifies a mask to apply to the operation: elements excluded by the mask may either be set to zero or retain their previous value.
A particularly common use of the kreg is to create a mask that excludes N contiguous elements at the beginning or end of a vector, e.g., as the first or final iteration in a vectorized loop where less than a full vector would be processed. E.g., for a loop over 121 int32_t values, the first 112 elements could be handled by 7 full 512-bit vectors, but that leaves 9 elements left over which could be handled by masked operations which operate only on the first 9 elements.
So the question is, given a (runtime valued) integer r which is some value in the range 0 - 16 representing remaining elements, what's the most efficient way to load a 16-bit kreg such that the low r bits are set and the remaining bits unset? KSHIFTLW seems unsuitable for the purpose because it only takes an immediate.
BMI2 bzhi does exactly what you want: Zero High Bits Starting with Specified Bit Position. Every CPU with AVX512 so far has BMI2.
__mmask16 k = _bzhi_u32(-1UL, r);
This costs 2 instructions, both single-uop: mov-immediate and bzhi. It's even single-cycle latency. (Or 3 cycles on KNL)
For r=0, it zeros all the bits giving 0.
For r=1, it leaves only the low bit (bit #0) giving 1
For r=12, it zeros bit #12 and higher, leaving 0x0FFF (12 bits set)
For r>=32 BZHI leaves all 32 bits set (and sets CF)
The INDEX is specified by bits 7:0 of the second source operand
If you had a single-vector-at-a-time cleanup loop that runs after an unrolled vector loop, you could even use this every loop iterations, counting the remaining length down towards zero, instead of a separate last-vector cleanup. It leaves all bits set for high lengths. But this costs 2 uops inside the loop, including port 5 kmovw, and means your main loop would have to use masked instructions. This only works for r<=255 because it only looks at the low byte, not the full integer index. But the mov reg, -1 can be hoisted because bzhi doesn't destroy it.
PS. Normally I think you'd want to arrange your cleanup to handle 1..16 elements, (or 0..15 if you branch to maybe skip it). But the full 17-possibility 0..16 makes sense if this cleanup also handles small lengths that never enter the main loop at all, and len=0 is possible. (And your main loop exits with length remaining = 1..16 so the final iteration can be unconditional)
I'm writing a tone generator program for a microcontroller.
I use an hardware timer to trigger an interrupt and check if I need to set the signal to high or low in a particular moment for a given note.
I'm using pretty limited hardware, so the slower I run the timer the more time I have to do other stuff (serial communication, loading the next notes to generate, etc.).
I need to find the frequency at which I should run the timer to have an optimal result, which is, generate a frequency that is accurate enough and still have time to compute the other stuff.
To achieve this, I need to find an approximate (within some percent value, as the higher are the frequencies the more they need to be imprecise in value for a human ear to notice the error) LCM of all the frequencies I need to play: this value will be the frequency at which to run the hardware timer.
Is there a simple enough algorithm to compute such number? (EDIT, I shall clarify "simple enough": fast enough to run in a time t << 1 sec. for less than 50 values on a 8 bit AVR microcontroller and implementable in a few dozens of lines at worst.)
LCM(a,b,c) = LCM(LCM(a,b),c)
Thus you can compute LCMs in a loop, bringing in frequencies one at a time.
Furthermore,
LCM(a,b) = a*b/GCD(a,b)
and GCDs are easily computed without any factoring by using the Euclidean algorithm.
To make this an algorithm for approximate LCMs, do something like round lower frequencies to multiples of 10 Hz and higher frequencies to multiples of 50 Hz. Another idea that is a bit more principled would be to first convert the frequency to an octave (I think that the formula is f maps to log(f/16)/log(2)) This will give you a number between 0 and 10 (or slightly higher --but anything above 10 is almost beyond human hearing so you could perhaps round down). You could break 0-10 into say 50 intervals 0.0, 0.2, 0.4, ... and for each number compute ahead of time the frequency corresponding to that octave (which would be f = 16*2^o where o is the octave). For each of these -- go through by hand once and for all and find a nearby round number that has a number of smallish prime factors. For example, if o = 5.4 then f = 675.58 -- round to 675; if o = 5.8 then f = 891.44 -- round to 890. Assemble these 50 numbers into a sorted array, using binary search to replace each of your frequencies by the closest frequency in the array.
An idea:
project the frequency range to a smaller interval
Let's say your frequency range is from 20 to 20000 and you aim for a 2% accurary, you'll calculate for a 1-50 range. It has to be a non-linear transformation to keep the accurary for lower frequencies. The goal is both to compute the result faster and to have a smaller LCM.
Use a prime factors table to easily compute the LCM on that reduced range
Store the pre-calculated prime factors powers in an array (size about 50x7 for range 1-50), and then use it for the LCM: the LCM of a number is the product of multiplying the highest power of each prime factor of the number together. It's easy to code and blazingly fast to run.
Do the first step in reverse to get the final number.
I have propably very stupid/simple question to GnuRadio users.
I have a Random Source as a source of bits [-1, 1]. And I want to multiply every bit with cosinus to make bpsk modulator.
Problem is that Bits are generated as fast as possible... (dont have enything in common with samp_rate). When I have 1 period of cosinus, there are generated many bits from Random Source.
Question is, how can I slow down bit rate generation ??
Thanks for any help
(I dont want to use DPSK Mod :))
Strictly speaking you can not delay the generation of bits. However you can increase the duration of each symbol. This can be done with the repeat block of GNU Radio. This block takes a parameter called interpolation that corresponds to the number of times an input item will be repeated at the output.
So you find the period of your cosine in samples, lets say p. For each random bit produced by the Random source block, you repeated it p times with the repeat block. With this way you increase the duration of your random symbol. Then you pass the resulting samples to the multiply block of your flowgraph.
This question comes from a homework assignment I was given. You can base your storage system off of one of the three following formats:
DD MM SS.S
DD MM.MMM
DD.DDDDD
You want to maximize the amount of data you can store by using as few bytes as possible.
My solution is based off the first format. I used 3 bytes for latitude: 8 bits for the DD (-90 to 90), 6 bits for the MM (0-59), and 10 bits for the SS.S (0-59.9). I then used 25 bits for the longitude: 9 bits for the DDD (-180 to 180), 6 bits for the MM, and 10 for the SS.S. This solution doesn't fit nicely on a byte border, but I figured the next reading can be stored immediately following the previous one, and 8 readings would use only 49 bytes.
I'm curious what methods others can come up. Is there a more efficient method to storing this data? As a note, I considered an offset based storage, but the problem gave no indication of how much the values may change between readings, so I'm assuming any change is possible.
Your suggested method is not optimal. You are using 10 bits (1024 possible values) to store a value in the range (0..599). This is a waste of space.
If you'll use 3 bytes for latitude, you should map the range [0, 2^24-1] to the range [-90, 90]. Hence each of the 2^24 values represents 180/2^24 degrees, which is 0.086 seconds.
If you want only 0.1 second accuracy, you'll need 23 bits for latitudes and 24 bits for longitudes (you'll get 0.077 seconds accuracy). That's 47 bit total instead of your 49 bits, with better accuracy.
Can we do even better?
The exact number of bits needed for 0.1 second accuracy is log2(180*60*60*10 * 360*60*60*10) < 46.256. Which means that you can use 46256 bits (5782 bytes) to store 1000 (lat,lon) pairs, but the mathematics involved will require dealing with very large integers.
Can we do even better?
It depends. If your data set has concentrations, you can store only some points and relative distances from these points, using less bits. Clustering algorithms should be used.
Sticking to existing technology:
If you used half precision floating point numbers to store only the DD.DDDDD data, you can be a lot more space-efficent, but you'd have to accept an exponent bias of 15, which means: The coordinates stored might not be exact, but at an offset from the original value.
This is due to the way floating point numbers are stored, essentially: A normalized significant is multiplied by an exponent to result in a number, instead of just storing a single value (as in integer numbers, the way you calculated the numbers for your solution).
The next highest commonly used floating point number mechanism uses 32 bits (the type "float" in many programming languages) - still efficient, but larger than your custom format.
If, however, you would design your own custom floating point type as well, and you gradually added more bits, your results would become more exact and it would STILL be more efficient than the solution you first found. Just play around with the number of bits used for significant and exponent, and find out how close your fp approximations come to the desired result in degrees!
Well, if this is for a large number of readings, then you may try a differential approach. Start with an absolute location, and then start saving incremental changes, which should ideally require less bits, depending on the nature of the changes. This is effectively compressing the stream. But somehow I don't think that's what this homework is about.