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...is two-fold: firstly, I need to test for periodicity, and secondly, I need to try to predict the temporal location of future peaks. Of course, the peaks might be occurring on unmeasured days. I have been looking at this model: wave.form <- deriv3( ~ sin(2*pi*((day-offset)/period + 0.25)) * amplitude + mean, c("period", "offset", "amplitude", "mean"), function(day, period, offset, amplitude, mean){}) curve(wave.form(x, period=7, offset=2, mean=5, amplitude=4), from=1, to=30) So, for my purposes, the mean and the amplitude seem t...

...disorders, head injuries, and/or vertigo. On the contrary, I am aware of the definition of a periodogram, and I know what the unnormalized periodogram in the data I presented looks like. Spec.pgram() is actually normalized too something, because it's discrete integral is not well above the SS amplitude of the signal it computed the periodogram for. In other words, the powers are not in units of around 4,000, which the peak would be if the units were merely the modulus squared of the Fourier coeficients of the data I presented. Alas, the modulus squared of the Fourier coeficients IS the TWO SIDED...

...: int 1 1 1 1 1 1 1 1 1 1 ... $ Reproducao : int 3 3 3 3 3 3 3 3 3 3 ... $ Estacao : int 2 2 2 2 2 2 2 2 2 2 ... $ X30cm : int 1 1 1 1 1 1 1 1 1 1 ... $ Dir.mov : int 0 0 0 0 0 0 0 0 0 0 ... $ Temp.media : num 12.0 11.3 12.1 12.3 12.4 ... $ Temp.amplitude : num 1.167 -0.750 0.875 0.125 ... $ Vel.media : num 0.479 0.514 0.517 0.445 0.468 ... $ Vel.amplitude : num -0.04865 0.03417 0.00312 0.02364 ... $ Caudal.medio : num 0.570 0.585 0.589 0.485 0.501 ... $ Caudal.amplitude: num -0.04323 0.01449 0.00405 0.01617 ... $ Prof....

Dear list, What on earth is spec.pgram() normalized too? If you would like to skip my proof as to why it's not normed too the mean squared or sum squared amplitude of the discrete function a[], feel free too skip the rest of the message. If it is, but you know why it's not exact in spec.pgram() when it should be, skip the rest of this message. The issue I refer herein refers only too a single univariate time series, and may not reflect the issues encount...

...############## patient <- data.frame( day = c(1, 3, 5, 8, 10, 12, 15, 17, 19, 22, 24, 26), protein = c(5, 3, 10, 7, 2, 8, 25, 22, 7, 10, 12, 5) ) plot(patient$day, patient$protein, type="b") # This is my model: wave.form <- deriv3( ~ sin(2*pi*((day-offset)/period + 0.25)) * amplitude + mean, c("period", "offset", "amplitude", "mean"), function(day, period, offset, amplitude, mean){}) curve(wave.form(x, period=7, offset=2, mean=5, amplitude=4), from=1, to=30) require(nlme) wave.1 <- gnls(protein ~ wave....

Dear list, I apologies first for my English, hope you will understand well my question. I am working on 1/2 hour piezometric data, time unit is second. They present daily oscillation when using the spectrum() function. What I am really interested in, is to find the amplitude corresponding to this oscillation. I work with a college using Matlab, and although we apply the same methodology, our results differs : we find the same frequencies, but not the same amplitudes. I put the code I use here to understand if the difference comes from the way that fft function a...

...block align, 16 bits/samp, 146966 data bytes sox: 73483 Samps/chans sox: Input file mysoundfile.wav: using sample rate 8000 size shorts, encoding signed (2's complement), 1 channel Samples read: 73482 Length (seconds): 9.185250 Scaled by: 2147483647.0 Maximum amplitude: 0.980347 Minimum amplitude: -0.980347 Midline amplitude: 0.000000 Mean norm: 0.190652 Mean amplitude: -0.004612 RMS amplitude: 0.294026 Maximum delta: 1.554443 Minimum delta: 0.000000 Mean delta: 0.096981 RMS delta: 0.163...

...ups. The seasonal variation can be best described by two harmonics, I called the frequencies omega and omega2. I now want to test whether (1) the seasonal pattern differs between the treatments (some kind of overall test). If this is the case, (2) I want to conduct tests to find out whether the amplitude of the fist harmonic, the amplitude of the second harmonic, the phase of the first harmonic and the phase of the second harmonic differs between the treatment groups. My attempt was to create a lme containing the two oscillations and the treatment group: omegaDay=omega*Day...

Hello. I'm from Federal University of Rio Grande do Sul - Brazil and I'm trying to adapt speex_echo.h (speex-1.2beta3-win32) to our video-conference software code. We are working on amplitude values, but I realised that in order to AEC works, one has to convert amplitude to frequency and filter the echo frequencies. So, my question is: Do I have to manually convert amplitude samples to frequency samples or is it enough to send amplitude parameters to speex_echo.h functions that this f...

...adius 1 (in terms of simplicity) but unfortunately I have the cartesian coordinates legends that do not help on reading my Figure. You can help me by executing the code below? require("rgls") degreeToRadian<-function(degree){? return ? (0.01745329252*degree)} turnPolarToX<-function(Amplitude,Coordinate){? return (Amplitude*cos(degreeToRadian(Coordinate)))} turnPolarToY<-function(Amplitude,Coordinate){? return (Amplitude*sin(degreeToRadian(Coordinate)))} # Putting the first circle on 3d space. Circle of radius 1X1<-turnPolarToX(1,1:360)Y1<-turnPolarToY(1,1:360)Z1<-rep(0,360...

...ather be able to stay with just the two-way (and one three-way) interactions that I know are of interest (I know this from previous logistic regression modelling of the dataset). For those who like to see the actual R code, here are two fits I've tried: gam(outcome ~ te(acl, dispersal) + te(amplitude, dispersal) + te(slope, curvature, amplitude), family=binomial, data=rla, method="REML") gam(outcome ~ te(slope) + te(curvature) + te(amplitude) + te(acl) + te(dispersal) + te(slope, curvature) + te(slope, amplitude) + te(curvature, amplitude) + te(acl, dispersal) + te(amplitude, disper...

...ue7, f2=some.value8, f3=some.value9 ) So far, I managed to create the "building blocks" I need but when it comes to defining their number I still haven't found a solution (see the following code). ## some fake starting values par.ini <- data.frame(period=1:5, invR=1:5, a0=1:5, Amplitude=1:5, Phase=1:5) n <- nrow(par.ini) - 1 # I'm not interested in the last harmonic inip <- par.ini[1:n, c(1, 3:5)] # neither in the second column amps <- paste("A", 1:n, sep="") phas <- paste("P", 1:n, sep=...

...plicity) but >unfortunately I have the cartesian coordinates legends that do not help >on reading my Figure. You can help me by executing the code below? > >require("rgls") >degreeToRadian<-function(degree){? return ? (0.01745329252*degree)} >turnPolarToX<-function(Amplitude,Coordinate){? return >(Amplitude*cos(degreeToRadian(Coordinate)))} >turnPolarToY<-function(Amplitude,Coordinate){? return >(Amplitude*sin(degreeToRadian(Coordinate)))} ># Putting the first circle on 3d space. Circle of radius >1X1<-turnPolarToX(1,1:360)Y1<-turnPolarToY(1,1:...

What I did was to plot your initial values, then plot the smoothed values and guess the constants. That is, I got an "eyeball" fit to the smoothed values. As I have described this as "gross cheating" in the past, you should either split your data, estimate on one subset and then test on another, or estimate on your data and test on a replication. If you get pretty much the same

...rge an ammount of the original signal at time offset j the >> echo canceller thinks should be removed from the current input frame. > > Careful when looking at W because of how the real and imaginary parts > are packed in the array. Err. Ok, as I got it, 'bin 0' has it's amplitude in W[0], bin 1 to N-1 has it's real part in W[i*2-1] and it's imag in W[i*2], and finally the nyquist amplitude is in W[N-1] I took this from how power_spectrum() computes, so I might be off :) >> Anyway, I did some proper testing. I took my headset, bent the microphone >>...

useR's, I want to create a "movie" of a sin function (from 0 to pi/2) using levelplot() in the lattice package. I basically want to create 20 or so plots of the sin function starting with an amplitude of 0 and ending at amplitude 1. By using a loop and plotting these in succession, it will have the appearance of a movie or animation. I believe I know how to do this part, but I am having trouble working with the colors. I want to use a range of colors starting with light blue (denoting zero) a...

...retization interval $w = [-\pi:.1*\pi: \pi]$? Then, how do I find the 'magnitude' of spectrum and 'phase' of spectrum in R? Is there an existing package/function in R? An example would be very helpful. -- View this message in context: http://r.789695.n4.nabble.com/calculate-phase-amplitude-of-fourier-transform-function-in-R-tp3303345p3303345.html Sent from the R help mailing list archive at Nabble.com.

You have the point. "I get very distorted speech when using libspeex in my application. What¡¯s wrong? .... If the input speech has an amplitude close to ¡¾2^15, it is possible that at decoding, the amplitude be a bit higher than that, causing clipping when saving as 16-bit PCM." This is one of FAQs in 'speex-manual' you wrote. I guess amplitudes close to 2^15 bother me. Do you have any idea to avoid it? I believe speex gi...

Kindly send a cc to me when replying to the list. I'm having trouble using lmer beyond a first step. My data: > some(exp1B) sub ba amplitude a b c d 2 1 1.00 1.5 65 63 4 8 41 4 1.15 0.0 92 41 3 4 43 4 1.15 3.0 88 48 2 2 63 6 1.00 3.0 50 72 9 9 77 8 1.15 0.0 112 25 2 1 89 10 1.15 0.0 37 33 36 34 126 13 1.15 1.5 80 50 6 4 140 14 1.15 4.5 1...

I have been attempting to estimate the periodic contribution of an effect to some data but have not been able to fit a sine wave within R. It would be nice to start by being able to fit a sine wave with an amplitude and frequency. x<-seq(0,20,by=0.5) y<-2*sin(2*pi*.5*x) #amplitude =2, frequency=0.5 # This failed to converge r<-nls(y ~ A*sin(2*pi*F*x), start=list(A = 1, F = 1), trace=T) # even this gave a max iteration error r<-nls(y ~ A*sin(2*pi*F*x), start=list(A = 1, F = .5), trace=T) I hav...