How does this phase shift in x-space affect the position of a spectrum in k-space? I'm working on a new form of signal detection with which I hope to recover both the amplitude and phase of a very small signal. However, doing this requires the use of some Fourier maths that I don't fully understand - I've modelled it numerically and it seems to work, but I'm hoping that someone could shed some light on whether it is possible to solve the problem analytically. 
In essence, everything boils down to doing a Fourier transform that looks like this:
$$
\mathcal{F}(k) = \int A_s(x)e^{i\phi_s(x)}\;A_re^{ia/x}\; e^{-ikx}\;dx 
$$
where $A_s(x)e^{i\phi_s(x)}$ is the small signal signal that we're trying to recover (with amplitude $A_s(x)$ and phase $\phi_s(x)$). There's also a reference wave with constant amplitude $A_r$ and a phase which $inversely$ depends on $x$. The point of the reference wave is to introduce a shift in the spectral position of the signal so that it can be filtered out from other detected bits and pieces we don't want - more or less a holographic detection technique.
So my question is how, precisely, the phase of the reference (given by $a/x$) affects the spectral position of the signal? How does this phase shift in $x$-space shift the spectrum in $k$-space? Is this even something that can even be determined analytically?
Many thanks for your help!
 A: It's going to do more than just shift the spectrum. A pure shift will be caused by modulating with a constant amplitude and linear phase signal $e^{jk_0 x}$, which has infinitesimal bandwidth. Your modulating signal of $e^{ja/x}$, however, is not so simple. A common way to get a feel for what modulation by $e^{j\phi(x)}$ will do to your signal is to express $\phi(x)$ as a series: $\phi(x) \approx a + bx + cx^2 + \cdots$. That way, we when we multiply some signal of interest, say $z(x)$, by it, we get
$$
z(x) e^{j\phi(x)} \approx z(x) e^{ja} e^{jbx} e^{jcx^2} \cdots
\\
\Rightarrow \mathcal{F}(z(x) e^{j\phi(x)}) \approx Z(k) \ast \delta(k-b) \ast e^{j\pi f^2/c} \ast \cdots = Z(k-b) \ast e^{j\pi f^2/c} \ast \cdots
$$
We can see from the above that the bulk shift will be given by the linear term in the series representation of $\phi(x)$. The quadratic term is usually associated with a blurring, the cubic term with a skew, and so on. (I say usually because it of course comes down to the specific signal in question.)
The thing about the above is that for it to be accurate, we need the series representation $\phi(x)$ to be accurate and with quickly diminishing coefficients. For instance, if $dx^4$ plays a larger role in the series approximation than $bx$, then we cannot say that the above spectral shift of $Z(k-b)$ will be a good approximation.
So for your signal, we'd really need some bounds to get a better idea. (For starters, $x=0$ is not defined.) I'd have to assume that the actual integral is of the form:
$$
\int_{x_0-T/2}^{x_0+T/2} A_s(x) e^{i\phi_s(x)}\;A_re^{ia/x}\; e^{-ikx}\;\mathrm{d}x 
$$
The series representation of $1/x$ is a bit of a bugger when it comes to convergence, so depending on the value of $T$, you might be able to get an accurate approximation with a few terms. The series in the area of $x_0$ will look like
$$
a/x \vert _{x_0} \approx \sum_n (-1)^n {{(x-x_0)^n} \over {x_0^{n+1}} }
$$
You'll have to see if $T$ is small enough to get an accurate 2nd or 3rd order approximation. If so, then you can apply the above anlaysis to predict what $e^{ja/x}$ will do to your signal.
