Appendix A: Proofs

Proof of proposition 1

Substitute the atomic signal:

wi(x,y)=Ae2iπ(xXi+yYi) (0x<M,0y<N)\begin{equation*} w_i\left(x,y\right) = A e^{2i\pi\left(\frac{x}{X_i} + \frac{y}{Y_i}\right)} \ \left(0 \leq x < M, 0 \leq y < N \right) \end{equation*}

into the discrete Fourier transform:

F(u,v)=1M×Nx=0M1y=0N1wi(x,y)e2iπ(xuM+yvN)\begin{equation*} F(u,v) = \frac{1}{M \times N} \sum\limits_{x = 0}^{M - 1} \sum\limits_{y = 0}^{N - 1} w_{i}(x,y) e^{2i\pi \left(\frac{x u}{M} + \frac{y v}{N}\right)} \end{equation*}

We have

F(u,v)=AM×Nx=0M1y=0N1e2iπ(xXi+yYi)e2iπ(xuM+yvN)\begin{equation*} F\left(u,v\right) = \frac{A}{M \times N} \sum\limits_{x = 0}^{M - 1} \sum\limits_{y = 0}^{N - 1} e^{2i\pi\left(\frac{x}{X_i} + \frac{y}{Y_i}\right)} e^{2i\pi \left(\frac{x u}{M} + \frac{y v}{N}\right)} \end{equation*}

Moreover

x=0M1y=0N1e2iπ(xXi+yYi)e2iπ(xuM+yvN)=x=0M1y=0N1e2iπx(1Xi+uM)e2iπy(1Yi+vN)=(x=0M1e2iπx(1Xi+uM))(x=0N1e2iπx(1Yi+vN))\begin{align*} &\sum\limits_{x = 0}^{M - 1} \sum\limits_{y = 0}^{N - 1} e^{2i\pi\left(\frac{x}{X_i} + \frac{y}{Y_i}\right)} e^{2i\pi \left(\frac{x u}{M} + \frac{y v}{N}\right)} \\ =&\sum\limits_{x = 0}^{M - 1} \sum\limits_{y = 0}^{N - 1} e^{2i\pi x \left(\frac{1}{X_i} + \frac{u}{M}\right)} e^{2i\pi y \left(\frac{1}{Y_i} + \frac{v}{N}\right)} \\ =&\left(\sum\limits_{x = 0}^{M - 1} e^{2i\pi x \left(\frac{1}{X_i} + \frac{u}{M}\right)}\right) \left(\sum\limits_{x = 0}^{N - 1} e^{2i\pi x \left(\frac{1}{Y_i} + \frac{v}{N}\right)}\right) \end{align*}

Using geometric sum formulae:

=1e2iπM(1Xi+uM)1e2iπ(1Xi+uM)1e2iπN(1Yi+vN)1e2iπ(1Yi+vN)=1e2iπMXi+2iπu1e2iπ(1Xi+uM)1e2iπNYi+2iπv1e2iπ(1Yi+vN)=1e2iπMXi1e2iπ(1Xi+uM)1e2iπNYi1e2iπ(1Yi+vN)\begin{align*} =& \frac{1 - e^{2i\pi M \left(\frac{1}{X_i} + \frac{u}{M}\right)}}{1 - e^{2i\pi \left(\frac{1}{X_i} + \frac{u}{M}\right)}} \frac{1 - e^{2i\pi N \left(\frac{1}{Y_i} + \frac{v}{N}\right)}}{1 - e^{2i\pi \left(\frac{1}{Y_i} + \frac{v}{N}\right)}} \\ =& \frac{1 - e^{2i\pi \frac{M}{X_i} + 2i\pi u}}{1 - e^{2i\pi \left(\frac{1}{X_i} + \frac{u}{M}\right)}} \frac{1 - e^{2i\pi\frac{N}{Y_i} + 2i\pi v}}{1 - e^{2i\pi \left(\frac{1}{Y_i} + \frac{v}{N}\right)}} \\ =& \frac{1 - e^{2i\pi \frac{M}{X_i}}}{1 - e^{2i\pi \left(\frac{1}{X_i} + \frac{u}{M}\right)}} \frac{1 - e^{2i\pi\frac{N}{Y_i}}}{1 - e^{2i\pi \left(\frac{1}{Y_i} + \frac{v}{N}\right)}} \end{align*}

Hence:

F(u,v)=AM×N1e2iπMXi1e2iπ(1Xi+uM)1e2iπNYi1e2iπ(1Yi+vN)\begin{equation*} F\left(u,v\right) = \frac{A}{M \times N} \frac{1 - e^{2i\pi \frac{M}{X_i}}}{1 - e^{2i\pi \left(\frac{1}{X_i} + \frac{u}{M}\right)}} \frac{1 - e^{2i\pi\frac{N}{Y_i}}}{1 - e^{2i\pi \left(\frac{1}{Y_i} + \frac{v}{N}\right)}} \end{equation*}

Last updated