My problem is the following:
I have fit a surface to some xyz coordinate data to obtain a polynomial surface. That's a polynomial in the variables x and y giving a surface of z-values. I know how to compute 95 percent condidence intervals on the mean response and also new predicition intervals (the standard stuff). Having obtained the surface function it's also easy to determine the derivative at a certain point on the the surface however, what I'm looking for is:
How do I compute confidence levels on the derivative value dS/dx, the (partial) derivative in the x direction at a certain point?
Are there expressions for calculating such intervals similar to the intervals for the 95 percent confidence intervals on the mean response?
(Please don't bother to post an answer on the confidence interval for the slope of a simple regression line. That's not the answer to this question. Anyway, I think it isn't.)
Related
I was studying for a signals & systems project and I have come across this code on high and low pass filters for an audio signal on the internet. Now I have tested this code and it works but I really don't understand how it is doing the low/high pass action.
The logic is that a sound is read into MATLAB by using the audioread or wavread function and the audio is stored as an nx2 matrix. The n depends on the sampling rate and the 2 columns are due to the 2 sterio channels.
Now here is the code for the low pass;
[hootie,fs]=wavread('hootie.wav'); % loads Hootie
out=hootie;
for n=2:length(hootie)
out(n,1)=.9*out(n-1,1)+hootie(n,1); % left
out(n,2)=.9*out(n-1,2)+hootie(n,2); % right
end
And this is for the high pass;
out=hootie;
for n=2:length(hootie)
out(n,1)=hootie(n,1)-hootie(n-1,1); % left
out(n,2)=hootie(n,2)-hootie(n-1,2); % right
end
I would really like to know how this produces the filtering effect since this is making no sense to me yet it works. Also shouldn't there be any cutoff points in these filters ?
The frequency response for a filter can be roughly estimated using a pole-zero plot. How this works can be found on the internet, for example in this link. The filter can be for example be a so called Finite Impulse Response (FIR) filter, or an Infinite Impulse Response (IIR) filter. The FIR-filters properties is determined only from the input signal (no feedback, open loop), while the IIR-filter uses the previous signal output to control the current signal output (feedback loop or closed loop). The general equation can be written like,
a_0*y(n)+a_1*y(n-1)+... = b_0*x(n)+ b_1*x(n-1)+...
Applying the discrete fourier transform you may define a filter H(z) = X(z)/Y(Z) using the fact that it is possible to define a filter H(z) so that Y(Z)=H(Z)*X(Z). Note that I skip a lot of steps here to cut down this text to proper length.
The point of the discussion is that these discrete poles can be mapped in a pole-zero plot. The pole-zero plot for digital filters plots the poles and zeros in a diagram where the normalized frequencies, relative to the sampling frequencies are illustrated by the unit circle, where fs/2 is located at 180 degrees( eg. a frequency fs/8 will be defined as the polar coordinate (r, phi)=(1,pi/4) ). The "zeros" are then the nominator polynom A(z) and the poles are defined by the denominator polynom B(z). A frequency close to a zero will have an attenuation at that frequency. A frequency close to a pole will instead have a high amplifictation at that frequency instead. Further, frequencies far from a pole is attenuated and frequencies far from a zero is amplified.
For your highpass filter you have a polynom,
y(n)=x(n)-x(n-1),
for each channel. This is transformed and it is possble to create a filter,
H(z) = 1 - z^(-1)
For your lowpass filter the equation instead looks like this,
y(n) - y(n-1) = x(n),
which becomes the filter
H(z) = 1/( 1-0.9*z^(-1) ).
Placing these filters in the pole-zero plot you will have the zero in the highpass filter on the positive x-axis. This means that you will have high attenuation for low frequencies and high amplification for high frequencies. The pole in the lowpass filter will also be loccated on the positive x-axis and will thus amplify low frequencies and attenuate high frequencies.
This description is best illustrated with images, which is why I recommend you to follow my links. Good luck and please comment ask if anything is unclear.
I have performed a MLE fit to some data that I have using the fminunc function in matlab and estimated the parameter confidence intervals from the Hessian output. Does anyone have a method for generating 95% confidence bands around the fitted curve?
A crude way to do this is generate parameters grid on the intervals that you have obtained, and plot the curve for every point of this grid. This works only if the number of parameters is low. You should keep in mind that the resulting confidence interval for curve is lower than that of the parameters, i.e. if you have 2 parameters with 95% confidence, you will obtain 95%*95%=90% confidence interval
If the curve depends on the parameters monotonically at every point, you can generate the interval just by taking boundary values of parameter intervals.
I'm calculating displacement from the motion of an accelerometer and using a Kalman filter to improve the displacement accuracy. Please note that I am aware of ineffectiveness of using acceleration to obtain displacement in real scenarios, but in my case displacement is pretty small (like 10 cm over 2–3 seconds).
I am following this paper (PDF). In the paper there are two matrices, Q and R, for noise modeling and they are set such that displacement error is minimized. The authors tested the above with synthetic acceleration data of a known covariance to use the same in matrices Q and R.
I decided to vary the particular covariance and find its corresponding minimum error in displacement. But in my case there is no change in displacement at any value of covariance. Any help?
let us assume,
I have a vector t with the times in seconds of my samples. (These samples are not equally distributed on the time domain.
Also I have a vector data containing the samplevalues at the time t.
t and data have the same length.
If I plot the graph some sort of periodical signal is obtained.
now I could perform: abs(fft(data)) to get my spectrum, which is then plotted over the amount of data points on the x-axis.
How can I obtain my spectrum regarding the times in vector t and plot it?
I want to see which frequencies in 1/s or which period in s my signal contains.
Thanks for your help.
[Not the OP's intention]: FFT will give you the spectrum (global) for any number of input data points. You cannot have a specific data point (in time) associated with parts (or the full) spectrum.
What you can do instead is use spectrogram and obtain the Short-Time Fourier Transform (STFT). This will give you a NxM discrete grid of time-frequency FT values (N: FT frequency bins, M: signal time-windows).
By localizing the (overlapping) STFT windows on your data samples of interest you will get N frequency magnitude values, thus the distribution of short-term spectrum estimates as the signal changes in time.
See also the possibly relevant answer here: https://stackoverflow.com/a/12085728/651951
EDIT/UPDATE:
For unevenly spaced data you need to consider the Non-Uniform DFT (and Non-uniform FFT implementations). See the relevant question/answer here https://scicomp.stackexchange.com/q/593
The primary approaches for NFFT or NUFFT, are based on creating a uniform grid through local convolutions/interpolation, running FFT on this and undoing the convolutional effect of the interpolation filter.
You can read more:
A. Dutt and V. Rokhlin, Fast Fourier transforms for nonequispaced data, SIAM J. Sci. Comput., 14, 1993.
L. Greengard and J.-Y. Lee, Accelerating the Nonuniform Fast Fourier Transform, SIAM Review, 46 (3), 2004.
Pippig, M. und Potts, D., Particle Simulation Based on Nonequispaced Fast Fourier Transforms, in: Fast Methods for Long-Range Interactions in Complex Systems, 2011.
For an implementation (with an interface to MATLAB) try NFFT and possibly its parallelized version PNFFT. You may find a nice walk-through on how to set-up and use here.
You can resample or interpolate your sample points to get another set of sample points that are equally spaced in t. The chosen spacing or sample rate of the second set of equally spaced sample points will allow you to infer frequencies to the result of an FFT of that second set.
The results may be noisy or include aliasing unless the initial data set is bandlimited to a sufficiently low frequency to allow interpolation. If bandlimited, then you might try something like cubic splines as an interpolation method.
Although it may look like one can get a high FFT bin frequency resolution by resampling to a larger number of data points, the actual useful resolution accuracy will be more related to the original number of samples.
1) Besides the negative frequencies, which is the minimum frequency provided by the FFT function? Is it zero?
2) If it is zero how do we plot zero on a logarithmic scale?
3) The result is always symmetrical? Or it just appears to be symmetrical?
4) If I use abs(fft(y)) to compare 2 signals, may I lose some accuracy?
1) Besides the negative frequencies, which is the minimum frequency provided by the FFT function? Is it zero?
fft(y) returns a vector with the 0-th to (N-1)-th samples of the DFT of y, where y(t) should be thought of as defined on 0 ... N-1 (hence, the 'periodic repetition' of y(t) can be thought of as a periodic signal defined over Z).
The first sample of fft(y) corresponds to the frequency 0.
The Fourier transform of real, discrete-time, periodic signals has also discrete domain, and it is periodic and Hermitian (see below). Hence, the transform for negative frequencies is the conjugate of the corresponding samples for positive frequencies.
For example, if you interpret (the periodic repetition of) y as a periodic real signal defined over Z (sampling period == 1), then the domain of fft(y) should be interpreted as N equispaced points 0, 2π/N ... 2π(N-1)/N. The samples of the transform at the negative frequencies -π ... -π/N are the conjugates of the samples at frequencies π ... π/N, and are equal to the samples at frequencies
π ... 2π(N-1)/N.
2) If it is zero how do we plot zero on a logarithmic scale?
If you want to draw some sort of Bode plot you may plot the transform only for positive frequencies, ignoring the samples corresponding to the lowest frequencies (in particular 0).
3) The result is always symmetrical? Or it just appears to be symmetrical?
It has Hermitian symmetry if y is real: Its real part is symmetric, its imaginary part is anti-symmetric. Stated another way, its amplitude is symmetric and its phase anti-symmetric.
4) If I use abs(fft(y)) to compare 2 signals, may I lose some accuracy?
If you mean abs(fft(x - y)), this is OK and you can use it to get an idea of the frequency distribution of the difference (or error, if x is an estimate of y). If you mean abs(fft(x)) - abs(fft(y)) (???) you lose at least phase information.
Well, if you want to understand the Fast Fourier Transform, you want to go back to the basics and understand the DFT itself. But, that's not what you asked, so I'll just suggest you do that in your own time :)
But, in answer to your questions:
Yes, (excepting negatives, as you said) it is zero. The range is 0 to (N-1) for an N-point input.
In MATLAB? I'm not sure I understand your question - plot zero values as you would any other value... Though, as rightly pointed out by duffymo, there is no natural log of zero.
It's essentially similar to a sinc (sine cardinal) function. It won't necessarily be symmetrical, though.
You won't lose any accuracy, you'll just have the magnitude response (but I guess you knew that already).
Consulting "Numerical Recipes in C", Chapter 12 on "Fast Fourier Transform" says:
The frequency ranges from negative fc to positive fc, where fc is the Nyquist critical frequency, which is equal to 1/(2*delta), where delta is the sampling interval. So frequencies can certainly be negative.
You can't plot something that doesn't exist. There is no natural log of zero. You'll either plot frequency as the x-axis or choose a range that doesn't include zero for your semi-log axis.
The presence or lack of symmetry in the frequency range depends on the nature of the function in the time domain. You can have a plot in the frequency domain that is not symmetric about the y-axis.
I don't think that taking the absolute value like that is a good idea. You'll want to read a great deal more about convolution, correction, and signal processing to compare two signals.
result of fft can be 0. already answered by other people.
to plot 0 frequency, the trick is to set it to a very small positive number (I use exp(-15) for that purpose).
already answered by other people.
if you are only interested in the magnitude, yes you can do that. this is applicable, say, in many image processing problems.
Half your question:
3) The results of the FFT operation depend on the nature of the signal; hence there's nothing requiring that it be symmetrical, although if it is you may get some more information about the properties of the signal
4) That will compare the magnitudes of a pair of signals, but those being equal do no guarantee that the FFTs are identical (don't forget about phase). It may, however, be enough for your purposes, but you should be sure of that.