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However, I do know that the command X[k] = fftshift(fft(ifftshift(x[n])))*Ts works, and produces the correct thing, because I have shown it in the graphs in the OP, and so this command must be equivalent to Eq. (4).

It is now in a form that I can recognise Matlab's fft() expression in there (although in the term I have indicated the fft() is applied to $x[n]$, not the shifted version), but this is where I am stuck - how can I show that this last equation is indeed calculated by the command

However, I do know that the command X[k] = fftshift(fft(ifftshift(x[n])))*Ts works, and produces the correct thing, because I have shown it in the graphs in the OP. So, this command must be equivalent to Eq. (4), but I am not sure how to mathematically prove it.

It is now in a form that I can recognise Matlab's fft() expression in there   but this is where I am stuck - how can I show that this last equation is indeed calculated by the command

However, I do know that the command X[k] = fftshift(fft(ifftshift(x[n])))*Ts works, and produces the correct thing, because I have shown it in the graphs in the OP. Therefore, I must be able to show that

• $y[n]$ is an ifftshifted version of $x[n]$
• $X[k]$ is an fftshifted version of $Y[k]$.

In doing so, I should be able to express Eq. (4) in such a way that if ifftshift() is first applied to my signal $x[n]$, then Eq. (5) (the Matlab fft) is applied, and finally fftshift() is applied to the result, then I should obtain exactly the expression for $X[k]$, given by Eq. (4).

I am not even sure how to mathematically represent these two operations (which essentially just swap the two halves of the vectors around). As a sanity check so far, I can also verify that Eq.(6) is correct up to this point, by manually evaluating the two things for a simple test vector:

However, I do know that the command X[k] = fftshift(fft(ifftshift(x[n])))*Ts works, and produces the correct thing, because I have shown it in the graphs in the OP, and so this command must be equivalent to Eq. (4).

I have a feeling that there might be a way to use the shift theorem to show that if you circularly shift the vectors through ifftshift and fftshift then you can cancel the extra phase factors, and the fft can be applied. 

As a quick sanity check on Eq.  (6) so far, I can verify it by manually evaluating the two things for a simple test vector of six numbers:

I can put this sampled version into the CTFT formula (assuming I captured sufficiently the underlying signal) to obtain $$ \begin{align} X(f) &= \lim_{T_s\rightarrow0} \sum^{N}_{n=1} x[n] \exp \bigg( -2\pi i f \bigg[(n-1)\;T_s-\frac{T}{2}\bigg] \bigg)\cdot T_s\\ &\approx \sum^{N}_{n=1} x[n] \exp \bigg( -2\pi i f \bigg[(n-1)\;T_s-\frac{T}{2}\bigg] \bigg)\cdot T_s. \end{align} $$

I can write the CTFT formula as a Riemann sum, and insert this sampled signal into it (assuming I captured sufficiently the underlying signal) to obtain $$ \begin{align} X(f) &= \lim_{T_s\rightarrow0} \sum^{N}_{n=1} x[n] \exp \bigg( -2\pi i f \bigg[(n-1)\;T_s-\frac{T}{2}\bigg] \bigg)\cdot T_s\\ &\approx \sum^{N}_{n=1} x[n] \exp \bigg( -2\pi i f \bigg[(n-1)\;T_s-\frac{T}{2}\bigg] \bigg)\cdot T_s. \end{align} $$