interpolation data and FFT resolution

Im trying to interpolate time domain data to improve the FFT resolution. Im not sure what Im doing wrong but Im getting a larger amount of bins but not decreasing the frequency spacing between the bins. I think Im doing something wrong or otherwise have made a mistake in theory. It should be simple.

    %create a 1MHz sinewave

    Fs = 1/1.0000e-09                  % Sampling frequency

    T = 1/Fs;                     % Sample time

    L = 2500;                     % Length of signal

    t = (0:L-1)*T;                % Time vector

    x = 15*sin(2*pi*1000000*t)

    figure, plot (t, x)

    %%%%%%%

    %%%%%%%

    %%FFT before interpolation

    Fs = 1/( t(2) - t(1));

    L = size(t,2);

    NFFT = 2^nextpow2(L); % Next power of 2 from length of y

    f = Fs/2*linspace(0,1,NFFT/2+1);

    test= fft(x ,NFFT)/L;

    figure,plot(f,2*abs(test(1:NFFT/2+1)))

    ind =find (f>= 1000000,1)

    freq_res = (f(2)-f(1))

    %%%%%%%%%%

    %%%%%%%%

    %interpolate the sinewave by 100
    new_t = t(1):(t(2)-t(1))/100:max(t);

    new= interpn( t, x , new_t, 'bicubic');  %TX
    %%%%
    %%FFT before interpolation

    Fs2 = 1/( new_t(2) - new_t(1));

    L2 = size(new_t,2);

    NFFT2 = 2^nextpow2(L2); % Next power of 2 from length of y

    f2 = Fs2/2*linspace(0,1,NFFT2/2+1);

    test= fft(new ,NFFT2)/L2;

    figure,plot(f2,2*abs(test(1:NFFT2/2+1)));

    ind2 =find (f2>= 1000000,1)

    freq_res = (f2(2)-f2(1))

I was hoping by interpolating the 1MHz sinewave by 1000 I would get a frequency improvement of 1000. Yet, in both, the frequency resolution is approximately the same. Isnt the FFT resolution roughly equal to the sampling frequency divided by the number of samples. I know we are increasing the number of samples, but isn't interpolation similar to sampling at higher frequency?

Answer 1

[David Young]

The frequency resolution is always 1/T, where T is the time from the start to the end of the time series. The resolution is independent of the sampling frequency, so interpolation can't change it.

As you say, the frequency resolution equals f/N where f is the sampling frequency and N is the number of samples. (Since N = Tf, this is 1/T.) Interpolation increases f and N in proportion, so the frequency resolution stays the same.

What interpolation does is to increase the highest frequency represented in the DFT. This is N/(2T) or f/2. However, this high-frequency extension to the spectrum will not contain any new information, just as in the time domain interpolation doesn't add anything new.

It follows that sampling at a higher frequency does not increase the frequency resolution either - it also just increases the highest frequency represented, though in this case the high frequencies may carry useful information.

To increase the frequency resolution, you either need to sample your signal for a longer time, or you need to introduce some assumptions about the spectrum that allow you to use, for example, maximum entropy methods to measure peak positions.