I have a signal of some length, say 1000 samples. I would like to extend this signal to 5000 samples, sampled at the same rate as the original (i.e., I want to predict what the signal would be if I continued to sample it for a longer period of time). The signal is composed of several sinusoidal components added together.

The method that first came to me was to take the entire FFT, and extend it, but this leaves a very strong discontinuity at frame 1001. I've also considered only using the part of the spectrum near the peaks, and while this seems to improve the signal somewhat, it doesn't seem to me that the phase is guaranteed to be correct. What is the best method for extending this signal?

Here's some MATLAB code showing an idealized method of what I want. Of course, I won't know beforehand that there are exactly 3 sinusoidal components, nor their exact phase and frequency. I want to make sure that the function is continuous, that there isn't a jump as we move to point 501,

vals = 1:50;
signal = 100+5*sin(vals/3.7+.3)+3*sin(vals/1.3+.1)+2*sin(vals/34.7+.7); % This is the measured signal
% Note, the real signal will have noise and not be known exactly.
output_vals = 1:200;
output_signal = 100+5*sin(output_vals/3.7+.3)+3*sin(output_vals/1.3+.1)+2*sin(output_vals/34.7+.7); % This is the output signal

hold all;

Basically, given the green line, I want to find the blue line. enter image description here

  • 3
    $\begingroup$ Is this in the context of tiling an image, doing a something like a sky extension, or something else? In other words, are there any other quality measures for the extension besides it being "smooth" at the edges? $\endgroup$
    – datageist
    Aug 25, 2011 at 4:55
  • 3
    $\begingroup$ Your question isn't clear. What do you mean by "extend"? Are you trying to duplicate and catenate copies of your signal (tiling) or do you want to upsample it by 5x or do you want to record it for 5x longer? Is your signal periodic? I think a more concrete and well defined question is necessary and a minimal example of what you're trying to achieve will definitely help. $\endgroup$ Aug 25, 2011 at 5:00
  • 1
    $\begingroup$ does extension mean interpolation? question is not very clear. $\endgroup$
    – Sriram
    Aug 25, 2011 at 11:41
  • $\begingroup$ Never realized it could mean so many things... Will try and make it clearer, basically I want to have a 1-D signal sampled at the same rate, but for a longer period of time. $\endgroup$ Aug 25, 2011 at 12:59
  • 1
    $\begingroup$ @endolith: Does that improve things? $\endgroup$ Aug 25, 2011 at 15:00

4 Answers 4


Depending on the source material, the DCT-based spectral interpolation method described in the following paper looks promising:

lk, H.G., Güler S. "Signal transformation and interpolation based on modified DCT synthesis", Digital Signal Processing, Article in Press, 2011.

Here's one of the figures from the paper showing an example of interpolation:

enter image description here

The applications of the technique to the recovery of lost segments (e.g. 4.2. Synthesis of lost pitch periods) are probably most relevant to extrapolation. In other words, grab a piece of the existing source material, pretend there's a lost segment between the edge and the arbitrary segment you selected, then "reconstruct" the "missing" portion (and perhaps discard the piece you used at the end). It appears an even simpler application of the technique would work for looping the source material seamlessly (e.g. 3.1. Interpolation of the spectral amplitude).

  • 1
    $\begingroup$ Sadly the link is dead. I could only find this payed download. $\endgroup$
    – knedlsepp
    Jan 6, 2015 at 10:48
  • $\begingroup$ @knedlsepp This link appears to give the paper without being behind a paywall. $\endgroup$
    – Peter K.
    May 25, 2017 at 11:35

I think linear predictive coding (otherwise known as an auto-regressive moving average) is what you are looking for. LPC extrapolates a time series by first fitting a linear model to the time series, in which each sample is assumed to be a linear combination of previous samples. After fitting this model to the existing time series, it can be run forward to extrapolate further values while maintaining a stationary(?) power spectrum.

Here is a little example in Matlab, using the lpc function to estimate the LPC coefficients.

N = 150;    % Order of LPC auto-regressive model
P = 500;    % Number of samples in the extrapolated time series
M = 150;    % Point at which to start predicting

t = 1:P;

x = 5*sin(t/3.7+.3)+3*sin(t/1.3+.1)+2*sin(t/34.7+.7); %This is the measured signal

a = lpc(x, N);

y = zeros(1, P);

% fill in the known part of the time series
y(1:M) = x(1:M);

% in reality, you would use `filter` instead of the for-loop
for ii=(M+1):P      
    y(ii) = -sum(a(2:end) .* y((ii-1):-1:(ii-N)));

plot(t, x, t, y);
l = line(M*[1 1], get(gca, 'ylim'));
set(l, 'color', [0,0,0]);
legend('actual signal', 'extrapolated signal', 'start of extrapolation');

Of course, in real code you would use filter to implement the extrapolation, by using the LPC coefficients a as an IIR filter and pre-loading the known timeseries values into the filter state; something like this:

% Run the initial timeseries through the filter to get the filter state 
[~, zf] = filter(-[0 a(2:end)], 1, x(1:M));     

% Now use the filter as an IIR to extrapolate
y((M+1):P) = filter([0 0], -a, zeros(1, P-M), zf); 

Here is the output:

LPC example

It does a reasonable job, though the prediction dies off with time for some reason.

I don't actually know much about AR models and would also be curious to learn more.


EDIT: @china and @Emre are right, the Burg method appears to work much better than LPC. Simply by changing lpc to arburg in the above code yields the following results:

Extrapolation using the Burg method

The code is available here: https://gist.github.com/2843661

  • $\begingroup$ LPC is actually AR, sans MA. The extrapolate signal "dies off" because the transfer function $H(z)=b(z)/a(z)$ of the extrapolator is causing attenuation. $\endgroup$
    – Emre
    May 18, 2012 at 23:31
  • $\begingroup$ @Emre Is there a way to improve the extrapolation? $\endgroup$
    – nibot
    May 21, 2012 at 7:52
  • $\begingroup$ As @chinnu says, the easy way is to feed the output into the input. $\endgroup$
    – Emre
    May 22, 2012 at 5:02
  • 2
    $\begingroup$ All right, you made me fire up MATLAB :P The problem can be avoided by using a different AR estimation algorithm; simply replace a=lpc(x,N) by a=arburg(x,N). For a (dry) discussion on AR algorithms see Why Yule-Walker Should Not be Used for Autoregressive Modelling $\endgroup$
    – Emre
    May 25, 2012 at 17:18
  • 1
    $\begingroup$ This is obviously a bit late, but there's a problem with your setup. You provide the full time series of x (P measurements) to lpc (or arburg) when estimating model coefficients. For extrapolation per the original question, you should really only base this estimation on the first M measurements. Providing fewer time points results in an inferior extrapolation, but still works reasonably well. $\endgroup$
    – Chris C
    Apr 25, 2016 at 19:11

1-D 'Extrapolation' is quite simple by using the BURG's method to estimate the LP coefficients. Once LP coefficients are available one can easily compute the time samples by applying filter. The samples which are predicted with Burg's are the next time samples of your input time segment.


If you're completely sure that there are only few frequency components to the signal, you can try the MUSIC algorithm to find out which frequencies are contained in your signal and try to work from there. I'm not completely sure that this can be made to work perfectly.

Additionally, because your data is completely deterministic, you can try to build some sort of a non-linear predictor, train it using your existing data set and let it extrapolate the rest.

In general this is an extrapolation problem, do you might want to Google something like Fourier extrapolation of price.

  • $\begingroup$ (Years later) another algorithm for finding a few frequency components is Harmonic inversion . $\endgroup$
    – denis
    Jan 31, 2016 at 15:32

Not the answer you're looking for? Browse other questions tagged or ask your own question.