I implemented an interpolation method in C++ based on this equation

y(t) = x[n]*sinc((t-nT)/T)

n - index of input signal,
x - input signal buffer,
t - is an actual time, 
nT - time of input samples,
T - period of input signal.

I calculated it on 33 samples:1+2*16. I tested it on Matlab and C++ on VS 2010. It works great on close frequencies (eg. 24kHz to 20kHz) but when I tested situation where the input signal is 24kHz and has a peak on 8kHz and then I wanted to interpolated it to 700Hz it gave me a peak at 300Hz so it is totally wrong. I expected that the main couse of this situation is that I did not use the cutoff frequency for low pass filter in this equation above. Seriously, I do not know how to add low pass filtration to my method.

Thanks for all your suggestions or advise.

  • 1
    $\begingroup$ Welcome to DSP.SE! Your question is a little unclear for me. You have discrete samples $x[n]$ that you want to turn into a continuous-time function $y(t)$. If that's the case, I would have thought you'd do: $y(t) = \sum_{n=0}^{N-1} x[n] {\rm sinc}((t-nT)/T)$ ? And I don't understand what you mean by I calculated it on 33 samples:1+2*16. $\endgroup$
    – Peter K.
    Sep 29, 2015 at 13:22
  • 1
    $\begingroup$ so i calculated i on the 16 in the past, 16 in the future and the closest to the current timestamp, do the window has 33. The rest, you understood correct. $\endgroup$
    – eryk
    Sep 29, 2015 at 13:28
  • $\begingroup$ Next questions: you say T - period of input signal but I would have called T the sampling period. I'm not sure how you interpolated it to 700Hz? Does that mean you set T=1/700? $\endgroup$
    – Peter K.
    Sep 29, 2015 at 13:33
  • $\begingroup$ T is a period in nanoseconds and here we have 10^9/24000 = 41667 ns, but of input signal not a cycle time of output $\endgroup$
    – eryk
    Sep 29, 2015 at 13:38
  • $\begingroup$ I interpolate it to 700kHz because I have a time domain in 700 Hz so the time t is buffer {0, 1s/700, 2s/700, 3s/700.. } and so on. $\endgroup$
    – eryk
    Sep 29, 2015 at 13:48

2 Answers 2


When you use interpolation to decrease the sample rate, you need to first low pass filter the signal to a bandwidth below the new FsNew/2. You can do this by using a wider Sinc kernel related to the new lower sample rate. e.g. intead of T in the denominator on the Sinc parameter in your interpolation kernel, use 1/FsNew.

  • $\begingroup$ OK thank you very much for your answer! You really understand my problem. I used equation y(t) = x[n]*sinc((t-nT)/T) and calculated T as a period of input signal (for 24kHz it should be 41667ns). When I tried calculated this period from output time domain (for 700 it is 1428571ns)it is going extremely wrong and then calculated spectrum shows wrong results at all. $\endgroup$
    – eryk
    Oct 1, 2015 at 7:27

I'm still not quite sure I understand what you're trying to achieve, but here's an example I cooked up in scilab.

The input is 33 points of a sine wave:

fs = 100; // 10kHz sampling rate
Tin = 1/fs; // Input sampling period
N = 16;
tin = [-N:N]; // Times of samples
f0 = 5;

x = sin(2*%pi*f0/fs*tin + 1.123791823)


The sampling frequency is 100Hz, the frequency of the sine wave is 5 Hz.

Then, I try to interpolate this up to a sampling frequency of 700Hz using the sinc formula:

Tout = 1/700;
tout = -N:Tout:N;
y = zeros(1,length(tout))
for n = 1:length(tin)
    y =  y + x(n)*sinc(%pi*(tout-tin(n)))

title('Input (blue line) and interpolated (red dotted line)')

dx = splin(tin,x);
x_interp = interp(tout,tin,x,dx);

The results are as plotted below.

enter image description here

  • $\begingroup$ Just curious, does the 1.123791823 in the first code snippet mean anything? Or is it just to add some variety to the sine wave? $\endgroup$ Apr 2, 2021 at 0:17
  • 1
    $\begingroup$ @MarkLeMoine. It's just to add some variety. Otherwise the sinewave starts at (0,0), which can make things appear simpler than they really are. $\endgroup$
    – Peter K.
    Apr 2, 2021 at 21:28

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