I have tried to perform sinc interpolation (in 1D) with the following Matlab code:



%% Deduce sinc from Fourier domain

xp=[zeros(1,5) x zeros(1,5)];



%plot(door1D);hold on;plot(fftshift(abs(sinc1D)))

%% Interp with sinc in spatial domain
%plot(x,'b*');hold on;plot(x_up,'r*');

hold on;
hold on;

It seems to work (except for ripples due to the Gibbs phenomenon I guess?). However, I worked this out in a "deductive" manner and I dont completely understand the parameters (period/frequency) of the sinc. The approach was to extract the sinc from the fft of the door function. Then use that sinc as if I had computed it beforehand and convolve the upsampled (not yet interpolated) 1D signal with it.

Can someone help me understand it better? How would I built this sinc ? e.g. from the sinc() function of Matlab. And for it to work, I have convolved with abs of sinc cf. conv2(x_up,sinc1D,'same'); , but this seems strange... can someone explain/develop/correct this ?

REM.:also it is probably badly scaled, but that is another detail

  • $\begingroup$ please help! :) $\endgroup$
    – SheppLogan
    Aug 18 '19 at 15:15

The sinc function actually represents an ideal (brickwall) lowpass filter that's used to complete the interpolation process after the data has been expanded (zero stuffed) properly. So let me outline the time domain approach here:

Assume you have data samples $x[n]$ of length $N$, and you want to upsample this by the integer factor of $L$, yielding a new interpolated data $y[n]$ of length $M = L \times N $ samples.

The first stage is to zero stuff the input $x[n]$; i.e., expand it by $L$ by the expander block :

$$ x[n] \longrightarrow \boxed{ \uparrow L } \longrightarrow x_e[n] $$

where $x_e[n]$ is related to $x[n]$ by the following:

$$ x_e[n] = \begin{cases} { x[n/L] ~~~,~~~ n = 0, \pm L, \pm 2L... \\ ~~~ 0 ~~ ~~~,~~~\text{otherwise} } \end{cases} $$

Then, to complete the interpolation process and fill in the empty (zeroed) samples of $x_e[n]$, one has to lowpass filter $x_e[n]$ by an ideal lowpass filter $h[n]$ with the following frequency domain definition $H(\omega)$ :

$$ H(\omega) = \begin{cases} { ~ L ~ ~~~,~~~ |\omega| < \frac{\pi}{L} \\ ~ 0 ~ ~~~,~~~\text{otherwise} } \end{cases} $$

The impulse response $h[n]$ of this ideal filter is computed by the inverse discrete-time Fourier transform of $H(\omega)$ and is given by

$$ h[n] = L \frac{ \sin( \frac{\pi}{L} n) } {\pi n} $$

This is an infitely long and non-causal filter, and thus cannot be implemented in this form. (See Hilmar's comments) Practically it's truncated and weighted by a window function, for example by a Hamming or Kaiser window.

The following MATLAB / OCTAVE code represents designing the filter and applying it into data in time domain:

L = 5;      % interpolation factor
N = 500;    % data length
x = hamming(N)'.*randn(1,N);  % generate bandlimited data...

% expanded signal
xe = zeros(1,N*L);
xe(1:L:end) = x;          % generate th expanded signal 

% interpolation sinc filter
n = -32:32;               % timing index
h = L*sin(pi*n/L)./(pi*n); % ideal lowpass filter
h(33) = L;                % fill in the zero divison sample
h = hamming(65)'.*h;      % apply weighting window on h[n]

% interpolate:
y = filter(h,1,xe);     % y[n] is the interpolated signal
  • $\begingroup$ great answer, thanks. This is exactly what I was looking for, and I think it will be useful to many. $\endgroup$
    – SheppLogan
    Aug 18 '19 at 18:48
  • $\begingroup$ So the lowpass filter is needed because the "useful" spectrum of xis shrinked when comparing x and xe (from an img proc lecture), and aliases apprear around it. (not sure I express myself correctly?). What is exactly this shinking factor ? $\endgroup$
    – SheppLogan
    Aug 18 '19 at 18:57
  • $\begingroup$ also, why did you choose -32:32 for sinc? Is is arbitrary? $\endgroup$
    – SheppLogan
    Aug 18 '19 at 19:01
  • $\begingroup$ last question, also would it be possible for you to say intuitively why the sinc interpolation is the ideal one? ( I remember as you just mentioned that the ideal lowpass filter is a brickwall to cut all freq outside it, in Fourier domain, and therefore a convolution with sinc in time/spatial domain, but "intuitively" compared to nearest neighbor interp for example why is is so much better? $\endgroup$
    – SheppLogan
    Aug 18 '19 at 20:07
  • $\begingroup$ yes, yes&no and yes. 1- As a technical terminology, the spectrum of $x_e[n]$ contains images rather than aliases and the lowpass filter removes all the images outside the frequency band $|\omega| < \pi/L$; the shrinking factor is $L$. 2-Length of the filter is arbitrary but not so, to get a better approximation it must be long enough. 3- Sinc() interpolator is the ideal one because it corresponds to the impulse response of the perfect interpolation filter and derived based on inverse Fourier transform it. Nearest neighbor is not a perfect interpolator by definition. $\endgroup$
    – Fat32
    Aug 18 '19 at 20:59

Sinc() interpolation looks nice on paper or in text books but in practice it's rarely a good solution. The main problem is that the sinc() impulse response is infinitely long and it's not causal. Not only does it have infinite length, but it also decays only very slowly with time, so you typically need a large number of samples to get a decent accuracy.

This, in turn, results in long latency, high computational cost and fairly large "transition" areas at the beginning and end of the output.

  • 1
    $\begingroup$ i think, in practice, a Kaiser-windowed sinc() interpolation can be very good. nearly always, not rarely. nearly as good as Parks-McClellan (firpm()) or Least Squares (firls()) and sometimes better, if you compare apples to apples. also sometimes firpm() chokes for kernels bigger than 2048 and firls() chokes for kernels bigger than 4096. but the kernal size is the product of the number of phases (sometimes i like that as high as 512) and the number of FIR taps (like 32 or 64). but you can make it as big as you want with kaiser() and sinc(). $\endgroup$ Aug 18 '19 at 19:18
  • $\begingroup$ so the choice by @Fat32 range of -32 to 32 is arbitrary ? $\endgroup$
    – SheppLogan
    Aug 18 '19 at 19:25
  • 1
    $\begingroup$ not arbitrary, it's just that we know from FIR filter design that we will need about 32 taps (which means you're looking at 16 samples on the left and 16 samples on the right) to do a decent job of high-quality interpolation of audio. but to do a really good job of it, you might need 64 taps. the number of FIR taps times the number of phases in the polyphase FIR filter is the total length of the the FIR that you would need to design using tools like firpm() or firls() in MATLAB. $\endgroup$ Aug 18 '19 at 22:56

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