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?