FFT of a signal with 0 added between samples I'm having a rough time doing the following question from a HW. Given a sequence of your choice, what happens to the DFT if we add a 0 between every sample? I.e. if we have x[n] = [A,B,C], we turn it into x[n] = [A,0,B,0,C,0].  

My initial thought was that the sampling frequency would double, similar to when we zero pad at the end of the signal. To do some analysis, I used the following MATLAB code.  

Fs = 300;            % Sampling frequency                    
T = 1/Fs;             % Sampling period       
L = 1500;             % Length of signal
t = (0:L-1)*T;        % Time vector

S = 0.7*sin(2*pi*50*t); %Signal - sine 50 Hz
z = zeros(1,3000);
z(1:2:end) = S; %Zero interp
z_fft = fft(z);
z_abs = abs(z_fft/(L*2));

P = z_abs(1:L*2/2+1);

P(2:end-1) = 2*P(2:end-1);
f = Fs*(0:(L*2/2))/(L*2);
plot(f,P);


Which yields the following plot:


That is different from the result when I add the same number of 0's but to the end of the sequence:


So I got confused, what is the result of adding the 0 between every sample? Does that doubles the frequency of my signal? What about that peak on f = 125Hz? Or am I making something wrong on my simulation?

Thanks for your time!
You can use expansion analysis to deduce what happens when a sequence is zero staffed in between its samples.

Let $x[n]$ be your original length $N$ sequence. Then its expansion by $2$ yields the new sequence $y[n]$ of length $2N$ :

$$ y[n] = \begin{cases} { x[n/2] ~~~,~~~n = 2m \\ ~~~~~ 0 ~~~,~~~ \text{otherwise} } \end{cases} $$

Then the DTFT relation between $Y(e^{j\omega})$ and $X(e^{j \omega})$ will be : $$Y(e^{j\omega}) = X(e^{j 2 \omega})$$

Then the DFT relation between N-point $X[k]$ and 2N-point $Y[k]$ will be

$$Y[k] = Y(e^{j \frac{2\pi}{2N} k}) = X(e^{ j 2 \frac{2\pi}{2N} k}) =  X[k] ~~~,~~~ k = 0,1,...,2N-1$$

In other words, $Y[k]$ will be two copies of $X[k]$