In the context of single-pixel imaging, the following statement is given:
"A Fourier basis pattern $P_F (x,y) $ can be obtained by applying an inverse Fourier transform $\delta_F (u, v, \phi)$to a delta function": $$P_F (x,y) = \frac{1}{2} \left[ 1 +| F^{-1} \{\delta_F (u,v)\}|\right] \tag{1}$$ where $F^{-1}\{\}$ denotes an inverse Fourier transform and $$\delta_F (u,v, \phi) = \begin{cases} e^{i\phi} & u=u_0\,\,\mathrm{and}\,\,v=v_0 \tag{2}\\ 0 &\mathrm{otherwise}. \end{cases}$$
How do I calculate the inverse Fourier of a delta function as requested in (1)?
A colleague has stated to me that the inverse Fourier transform of delta functions at different positions can be calculated using eq. (3):
$$P_\phi (x,y) = \frac{1}{2} \text{real} \left[ 1 + F^{-1} \{\delta_F (u -u_0,v- v_0) e^{i\phi}\}\right] \tag{3}$$
Is this delta function any different from the one presented in the eq. (1)?
How does one calculate the Fourier inverse of a product of functions?
[1] Hadamard single-pixel imaging versus Fourier single-pixel imaging, Z. Zhang et al. Opt. Express 25, 19619-19639 ,(2017)
For a Dirac delta function $\delta(k-k_0)$ the inverse Fourier transform would simply be $$ F^{-1}\{\delta(k-k_0)\} = \int \exp(i k x)\delta(k-k_0)\ \frac{\text{d}k}{2\pi} = \frac{1}{2\pi} \exp(i k_0 x) . $$ In other words, just substitute the Dirac delta function into the inverse Fourier transform integral and use the standard rule to evaluate integrals with Dirac delta functions in them.
For the 2-dimensional case, the integral becomes $$ F^{-1}\{\delta(k_x-k_1,k_y-k_2)\} = \int \exp(i k_x x+i k_y y) \delta(k_x-k_1) \delta(k_y-k_2)\ \frac{\text{d}k_x}{2\pi}\ \frac{\text{d}k_y}{2\pi} . $$ I'm sure you can work out the result.