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Kalman, White Noise, Sensor Specification, Discretization? 
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#1
Feb1612, 12:40 PM

P: 3

Hi.
I have a few questions about sensor specifications and its implementation in a Kalman Filter and simulation of gyroscope/accelerometer output. Abbreviation used: d  discrete c  continuous Q1: From book: Aided Navigation  Farrell (you don't need the book to understand the question). Section 7.2 Methodology: Detailed Example: Here [itex]\sigma_1[/itex], [itex]\sigma_{bu}[/itex] and [itex]\sigma_{by}[/itex] are root PSD of continuous white noise. [itex]\sigma_2[/itex] is std. deviation of discrete white noise. [itex]Qc[/itex] = diag([[itex]\sigma_{bu}^2[/itex] [itex]\sigma_{by}^2[/itex] [itex]\sigma_1^2[/itex] ]) [itex]Qd[/itex] = function of [itex]Qc[/itex] and sampling time [itex]Rd[/itex] = [itex]\sigma_2^2[/itex] This I understand, but not section 7.4. Section 7.4 An Alternative Approach: Same sigma values as above, plus [itex] sigma_3[/itex] is root PSD of continuous white noise. Qc = diag([[itex]\sigma_{bu}^2[/itex] [itex] \sigma_{by}^2[/itex] [itex] \sigma_3^2[/itex]]) Qd = function of Qc and sampling time Rd =[itex] \sigma_1^2[/itex] and later Rd = diag([[itex]\sigma_2^2[/itex] [itex] \sigma_1^2[/itex]]) a) Can you just use [itex]\sigma[/itex] in Rd no matter if it's std. deviation for discrete white noise ([itex]\sigma_2[/itex]) or root PSD of continuous white noise([itex]\sigma_1[/itex])? b) Shouldn't [itex]\sigma_1[/itex] be "discretizised" or something? Q2: If you get noise specification for the sensor noise from PSD method, Allan Variance method or data sheet, it is for continuous white noise. Is this correct? Q3: How do you simulate the output of a sensor in e.g. Matlab/simulink if you have noise specification in continuous time? Q4: When we discretizise a stochastic linear system, we get: Qd = f(Qc, A, T) Rd = Rc (http://en.wikipedia.org/wiki/Discretization) I don't understand why Rd = Rc? Q5: In the book: Optimal State Estimation  Kalman, Hinf and Nonlinear Approaches  2006  Dan Simon, section 8.1 DISCRETETIME AND CONTINUOUSTIME WHITE NOISE, it says: "... discretetime white noise with covariance Qd in a system with a sample period of T, is equivalent to continuoustime white noise with covariance Qc*[itex]\delta[/itex] (t) (*[itex]\delta[/itex] (t): dirac delta function), where Qc = Qd / T." "... Rd = Rc / T ... This establishes the equivalence between white measurement noise in discrete time and continuous time. The effects of white measurement noise in discrete time and continuous time are the same if v(k) ~ (0,Rd) v(t) ~ (0,Rc) " How does this relate to my other questions? It seems to suggest that Rd [itex]\neq[/itex] Rc as Q4 suggests. Best Regard Jonas 


#2
Feb2412, 08:49 AM

P: 3

Any experts out there?



#3
Feb2512, 05:10 AM

Sci Advisor
P: 3,321

I'm not an expert in you topic, but I'd find it interesting to chat about it until one comes along.
My intuitive understanding of white noise goes like this. Suppose you wanted to simulate a white noise at 1 second time intervals by drawing a random number from some distribution. You do this an produce a graph. Then someone (say your boss) wants you to simulate "the same" noise but at a time step of 1/10 of a second. If you draw numbers from the same distribution at each 1/10 of a second, you get another jumpy graph. Maybe you think it looks OK. But suppose your boss is using the white noise in some other function that adds it up, like the simple sum of all the white noise jumps you provide. He probably won't think that you did a good job because in a time interval of a certain size , say 60 seconds, the graph of his function will look a more jumpy with your 1/10 th second noise that it did with the 1 second noise. Suppose you try to fix this by scaling the numbers you draw. For 1/10 a second, you draw random numbers from the same distribution that you were using and then you divide the numbers by 10. This won't please the boss either. The graph of his function won't look jumpy enough. (Intuitively, this is because the jumps he gets at 1 second intervals is now the average of 10 draws at 1/10 second intervals which has a smaller variation that one draw per second. The way to please your boss is to scale the numbers so their variance is 1/10 of the standard deviation of the original distribution. Then, according the law for finding the variance of a sum of independent random variables, the variation he gets a 1 second intervals is the sum of 10 terms, each of which has variance that is 1/10 of the variance he had at 1 second intervals. So he sees the same variability over a given time interval as he did with your 1 second noise. If I want to talk about the variance of white noise, in the sense of a process that takes places in continuous time rather than at discrete steps, I think of the units of measure as being somethingsquared per unit time. Only when both units are established (e.g. feet and seconds) is the variance of the noise well defined. I can't tell what the texts you presented are saying about Rc=Rd. An emprical measurment of noise at an arbitrary time step has the proper units for noise at a unit time step, e.g. 2.4 ft^2/ 2 secs = 1.2 ft^2/ 1 sec. But I don't know if that is what is being asserted. 


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