Could someone explain how one arrives at the equation below for high pass filtering of the accelerometer values? I don't need mathematical derivation, just an intuitive interpretation of it is enough.
#define kFilteringFactor 0.1
UIAccelerationValue rollingX, rollingY, rollingZ;
- (void)accelerometer:(UIAccelerometer *)accelerometer didAccelerate:(UIAcceleration *)acceleration {
// Subtract the low-pass value from the current value to get a simplified high-pass filter
rollingX = (acceleration.x * kFilteringFactor) + (rollingX * (1.0 - kFilteringFactor));
rollingY = (acceleration.y * kFilteringFactor) + (rollingY * (1.0 - kFilteringFactor));
rollingZ = (acceleration.z * kFilteringFactor) + (rollingZ * (1.0 - kFilteringFactor));
float accelX = acceleration.x - rollingX;
float accelY = acceleration.y - rollingY;
float accelZ = acceleration.z - rollingZ;
// Use the acceleration data.
}
While the other answers are correct, here is an simplistic explanation. With kFilteringFactor 0.1 you are taking 10% of the current value and adding 90% of the previous value. Therefore the value retains a 90% similarity to the previous value, which increases its resistance to sudden changes. This decreases noise but it also makes it less responsive to changes in the signal. To reduce noise and keep it responsive you would need non trivial filters, eg: Complementary, Kalman.
The rollingX, rollingY, rollingZ values are persistent across calls to the function. They should be initialised at some point prior to use. These "rolling" values are just low pass filtered versions of the input values (aka "moving averages") which are subtracted from the instantaneous values to give you a high pass filtered output, i.e. you are getting the current deviation from the moving average.
ADDITIONAL EXPLANATION
A moving average is just a crude low pass filter. In this case it's what is known as ARMA (auto-regressive moving average) rather than just a simple MA (moving average). In DSP terms this is a recursive (IIR) filter rather than a non-recursive (FIR) filter. Regardless of all the terminology though, yes, you can think of it as a smoothing function" - it's "smoothing out" all the high frequency energy and leaving you with a slowly varying estimate of the mean value of the signal. If you then subtract this smoothed signal from the instantaneous signal then the difference will be the content that you have filtered out, i.e. the high frequency stuff, hence you get a high pass filter. In other words: high_pass_filtered_signal = signal - smoothed_signal.
Okay, what that code is doing is calculating a low-pass signal and then substracting the current value.
Thing of a square wave that takes two values 5 and 10. In other words, it oscillates between 5 and 10. Then the low pass signal is trying to find the mean (7.5). The high-pass signal is then calculated as current value minus mean, i.e. 10 - 7.5 = 2.5, or 5 - 7.5 = -2.5.
The low-pass signal is computed by integrating over past values by adding a fraction of the current value to 90% of the past low-pass value.
Related
I am getting analog voltage data, in mV, from a fuel gauge. The calibration readings were taken for every 10% change in the fuel gauge as mentioned below :
0% - 2000mV
10% - 2100mV
20% - 3200mV
30% - 3645mV
40% - 3755mV
50% - 3922mV
60% - 4300mV
70% - 4500mv
80% - 5210mV
90% - 5400mV
100% - 5800mV
The tank capacity is 45L.
Post calibration, I am getting reading from adc as let's say, 3000mV. How to calculate the exact % of fuel left in the tank?
If you plot the transfer function of ADC reading agaist the percentage tank contents you get a graph like this
There appears to be a fair degree of non linearity in the relationship between the sensor and the measured quantity. This could be down to a measurement error that was made while performing the calibration or it could be a true non linear relationship between the sensor reading and the tank contents. Using these results will give fairly inaccurate estimates of tank contents due to the non linearity of the transfer function.
If the relationship is linear or can be described by another mathematical relationship then you can perform an interpolation between known points using this mathematical relationship.
If the relationship is not linear than you will need many more known points in your calibration data so that the errors due to the interpolation between points is minimised.
The percentage value corresponding to the ADC reading can be approximated by finding the entries in the calibration above and below the reading that has been taken - for the ADC reading example in the question these would be the 10% and 20% values
Interpolation_Proportion = (ADC - ADC_Below) / (ADC_Above - ADC_Below) ;
Percent = Percent_Below + (Interpolation_Proportion * (Percent_Above - Percent_Below)) ;
.
Interpolation proportion = (3000-2100)/(3200-2100)
= 900/1100
= 0.82
Percent = 10 + (0.82 * (20 - 10)
= 10 + 8.2
= 18.2%
Capacity = 45 * 18.2 / 100
= 8.19 litres
When plotted it appears that the data id broadly linear, with some outliers. It is likely that this is experimental error or possibly influenced by confounding factors such as electrical noise or temperature variation, or even just the the liquid slopping around! Without details of how the data was gathered and how carefully, it is not possible to determine, but I would ask how many samples were taken per measurement, whether these are averaged or instantaneous and whether the results are exactly repeatable over more than one experiment?
Assuming the results are "indicative" only, then it is probably wisest from the data you do have to assume that the transfer function is linear, and to perform a linear regression from the scatter plot of your test data. That can be most done easily using any spreadsheet charting "trendline" function:
From your date the transfer function is:
Fuel% = (0.0262 x SensormV) - 54.5
So for your example 3000mV, Fuel% = (0.0262 x 3000) - 54.5 = 24.1%
For your 45L tank that equates to about 10.8 Litres.
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 want an object to smoothly stop at a specified point. I have the initial velocity, V0, of the object, the end velocity (which is set to 0), V1, and the distance, d, required to reach the destination. V1 and V0 are measured in radians per frame, and d is also measured in radians.
I've tried using the following formula:
a = (V0*V0 - V1*V1) / (2.0 * d);
But it always seems to overshoot its target.
Edit:
Essentially, I have a wheel that starts spinning at an initial velocity of V0. The wheel has to do a certain number of spins before stopping at a specified location. The distance, d is the amount of radians required to do the specified amount of spins and stop at a specified location. The velocity is number of radians per frame.
We have enough information now for an educated guess.
The formula is correct, but only in the ideal case, where the length of an iteration time is very small. My guess is that in your loop you are updating position before velocity, so that for that time period the wheel can advance with undiminished velocity, and you overshoot. (If you updated velocity before position, you would undershoot.)
You can either make your frames shorter, which will make the overshoot less severe, or you can modify the formula to eliminate it:
a = (V0*V0) / (2.0 * d - V0*tdelta);
where tdelta is the length of time of a single frame. (I've assumed V1=0.)
To make animation independent of frame rate, is it necessary to multiply the delta value by both velocity and acceleration?
// Multiply both acceleration and velocity by delta?
vVelocity.x += vAcceleration.x * delta;
vVelocity.y += vAcceleration.y * delta;
position.x += vVelocity.x * delta;
position.y += vVelocity.y * delta;
Should I apply delta to the velocity only and not acceleration?
Assuming your "delta" is the amount of time passed since last update:
Short answer: yes.
Long answer:
One way to check this sort of thing is to see if the units work out. It's not guaranteed, but usually if your units work out, then you've figured things correctly.
Velocity measures distance per unit time, and delta is time. So velocity times delta is (picking arbitrary units meters and seconds) (m/s) * s = m. So you can see that velocity times delta does create a distance, so that appears reasonable for position.
Acceleration measures velocity per unit time, that is, with the same units (m/s)/s. So, acceleration times delta is ((m/s)/s) * s = m/s. Looks like a velocity to me. We're good!
Yes, it is necessary to involve delta with both the velocity and the acceleration. They're both properties that are defined with respect to time (m/s for one, m/s/s for the other - units may vary), so delta should be used whenever they have to change non-instantaneously.
I've tried the typical physics equations for this but none of them really work because the equations deal with constant acceleration and mine will need to change to work correctly. Basically I have a car that can be going at a large range of speeds and needs to slow down and stop over a given distance and time as it reaches the end of its path.
So, I have:
V0, or the current speed
Vf, or the speed I want to reach (typically 0)
t, or the amount of time I want to take to reach the end of my path
d, or the distance I want to go as I change from V0 to Vf
I want to calculate
a, or the acceleration needed to go from V0 to Vf
The reason this becomes a programming-specific question is because a needs to be recalculated every single timestep as the car keeps stopping. So, V0 constantly is changed to be V0 from last timestep plus the a that was calculated last timestep. So essentially it will start stopping slowly then will eventually stop more abruptly, sort of like a car in real life.
EDITS:
All right, thanks for the great responses. A lot of what I needed was just some help thinking about this. Let me be more specific now that I've got some more ideas from you all:
I have a car c that is 64 pixels from its destination, so d=64. It is driving at 2 pixels per timestep, where a timestep is 1/60 of a second. I want to find the acceleration a that will bring it to a speed of 0.2 pixels per timestep by the time it has traveled d.
d = 64 //distance
V0 = 2 //initial velocity (in ppt)
Vf = 0.2 //final velocity (in ppt)
Also because this happens in a game loop, a variable delta is passed through to each action, which is the multiple of 1/60s that the last timestep took. In other words, if it took 1/60s, then delta is 1.0, if it took 1/30s, then delta is 0.5. Before acceleration is actually applied, it is multiplied by this delta value. Similarly, before the car moves again its velocity is multiplied by the delta value. This is pretty standard stuff, but it might be what is causing problems with my calculations.
Linear acceleration a for a distance d going from a starting speed Vi to a final speed Vf:
a = (Vf*Vf - Vi*Vi)/(2 * d)
EDIT:
After your edit, let me try and gauge what you need...
If you take this formula and insert your numbers, you get a constant acceleration of -0,0309375. Now, let's keep calling this result 'a'.
What you need between timestamps (frames?) is not actually the acceleration, but new location of the vehicle, right? So you use the following formula:
Sd = Vi * t + 0.5 * t * t * a
where Sd is the current distance from the start position at current frame/moment/sum_of_deltas, Vi is the starting speed, and t is the time since the start.
With this, your decceleration is constant, but even if it is linear, your speed will accomodate to your constraints.
If you want a non-linear decceleration, you could find some non-linear interpolation method, and interpolate not acceleration, but simply position between two points.
location = non_linear_function(time);
The four constraints you give are one too many for a linear system (one with constant acceleration), where any three of the variables would suffice to compute the acceleration and thereby determine the fourth variables. However, the system is way under-specified for a completely general nonlinear system -- there may be uncountably infinite ways to change acceleration over time while satisfying all the constraints as given. Can you perhaps specify better along what kind of curve acceleration should change over time?
Using 0 index to mean "at the start", 1 to mean "at the end", and D for Delta to mean "variation", given a linearly changing acceleration
a(t) = a0 + t * (a1-a0)/Dt
where a0 and a1 are the two parameters we want to compute to satisfy all the various constraints, I compute (if there's been no misstep, as I did it all by hand):
DV = Dt * (a0+a1)/2
Ds = Dt * (V0 + ((a1-a0)/6 + a0/2) * Dt)
Given DV, Dt and Ds are all given, this leaves 2 linear equations in the unknowns a0 and a1 so you can solve for these (but I'm leaving things in this form to make it easier to double check on my derivations!!!).
If you're applying the proper formulas at every step to compute changes in space and velocity, it should make no difference whether you compute a0 and a1 once and for all or recompute them at every step based on the remaining Dt, Ds and DV.
If you're trying to simulate a time-dependent acceleration in your equations, it just means that you should assume that. You have to integrate F = ma along with the acceleration equations, that's all. If acceleration isn't constant, you just have to solve a system of equations instead of just one.
So now it's really three vector equations that you have to integrate simultaneously: one for each component of displacement, velocity, and acceleration, or nine equations in total. The force as a function of time will be an input for your problem.
If you're assuming 1D motion you're down to three simultaneous equations. The ones for velocity and displacement are both pretty easy.
In real life, a car's stopping ability depends on the pressure on the brake pedal, any engine braking that's going on, surface conditions, and such: also, there's that "grab" at the end when the car really stops. Modeling that is complicated, and you're unlikely to find good answers on a programming website. Find some automotive engineers.
Aside from that, I don't know what you're asking for. Are you trying to determine a braking schedule? As in there's a certain amount of deceleration while coasting, and then applying the brake? In real driving, the time is not usually considered in these maneuvers, but rather the distance.
As far as I can tell, your problem is that you aren't asking for anything specific, which suggests that you really haven't figured out what you actually want. If you'd provide a sample use for this, we could probably help you. As it is, you've provided the bare bones of a problem that is either overdetermined or way underconstrained, and there's really nothing we can do with that.
if you need to go from 10m/s to 0m/s in 1m with linear acceleration you need 2 equations.
first find the time (t) it takes to stop.
v0 = initial velocity
vf = final velocity
x0 = initial displacement
xf = final displacement
a = constant linear acceleration
(xf-x0)=.5*(v0-vf)*t
t=2*(xf-x0)/(v0-vf)
t=2*(1m-0m)/(10m/s-0m/s)
t=.2seconds
next to calculate the linear acceleration between x0 & xf
(xf-x0)=(v0-vf)*t+.5*a*t^2
(1m-0m)=(10m/s-0m/s)*(.2s)+.5*a*((.2s)^2)
1m=(10m/s)*(.2s)+.5*a*(.04s^2)
1m=2m+a*(.02s^2)
-1m=a*(.02s^2)
a=-1m/(.02s^2)
a=-50m/s^2
in terms of gravity (g's)
a=(-50m/s^2)/(9.8m/s^2)
a=5.1g over the .2 seconds from 0m to 10m
Problem is either overconstrained or underconstrained (a is not constant? is there a maximum a?) or ambiguous.
Simplest formula would be a=(Vf-V0)/t
Edit: if time is not constrained, and distance s is constrained, and acceleration is constant, then the relevant formulae are s = (Vf+V0)/2 * t, t=(Vf-V0)/a which simplifies to a = (Vf2 - V02) / (2s).