It seems that your data has been taken by sampling a sinusoid using a sampling frequency that is only two time the frequency of the sine. Unfortunately, if you want to respect the Nyquist's theorem, you have to respect the following equation.
Anyway, try to run the following example:
close all
N = 500;
fsin = 100;
fsample = 10*fsin;
tsin = 1/fsin;
t = 0:1/fsample:N/fsample;
s =sin(2*pi*t*fsin);
upsampling = 10;
figure; plot(s)
title('Original Signal')
faxis = -0.5*fsample:fsample/N:0.5*fsample-1/N;
sF = (1/length(s))*(fft(s));
sFPad = [sF(1:round(0.5*length(sF))) zeros(1,(upsampling-1)*length(sF)) sF(round(0.5*length(sF))+1:end) ];
ss = (length(sFPad))*(ifft((sFPad)));
tss = t(end)*linspace(0,1,length(ss));
figure; plot(ss)
title('Signal after upsampling')
As you can see, after the padding the anti-transformed signal it's the oversampled version of the original signal.
After the zero padding, there is more "distance" (in the frequency domain) between the two peak of the original tone. This is the same effect that you get if you try to increase the sampling frequency of the original signal.
Now you can get the same result if you keep the same sampling frequency and try to put a zero between every sample of the original signal (zero interleaving) and then try to do a filtering in time domain.
So upsampling in frequency domain (zero padding with filtering) <--> upsampling in time domain (zero interleaving with filtering).
You can find other useful information here: https://www.dsprelated.com/freebooks/sasp/Upsampling_Downsampling.html
