Fourier transform of complex atomic signals
Substitute the atomic signal:
w i ( x , y ) = A e 2 i π ( x X i + y Y i ) ( 0 ≤ x < M , 0 ≤ y < 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*} w i ( x , y ) = A e 2 iπ ( X i x + Y i y ) ( 0 ≤ x < M , 0 ≤ y < N ) into the discrete Fourier transform:
F ( u , v ) = 1 M × N ∑ x = 0 M − 1 ∑ y = 0 N − 1 w i ( x , y ) e 2 i π ( x u M + y v N ) \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*} F ( u , v ) = M × N 1 x = 0 ∑ M − 1 y = 0 ∑ N − 1 w i ( x , y ) e 2 iπ ( M xu + N y v ) We have
F ( u , v ) = A M × N ∑ x = 0 M − 1 ∑ y = 0 N − 1 e 2 i π ( x X i + y Y i ) e 2 i π ( x u M + y v N ) \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*} F ( u , v ) = M × N A x = 0 ∑ M − 1 y = 0 ∑ N − 1 e 2 iπ ( X i x + Y i y ) e 2 iπ ( M xu + N y v ) Moreover
∑ x = 0 M − 1 ∑ y = 0 N − 1 e 2 i π ( x X i + y Y i ) e 2 i π ( x u M + y v N ) = ∑ x = 0 M − 1 ∑ y = 0 N − 1 e 2 i π x ( 1 X i + u M ) e 2 i π y ( 1 Y i + v N ) = ( ∑ x = 0 M − 1 e 2 i π x ( 1 X i + u M ) ) ( ∑ x = 0 N − 1 e 2 i π x ( 1 Y i + v N ) ) \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*} = = x = 0 ∑ M − 1 y = 0 ∑ N − 1 e 2 iπ ( X i x + Y i y ) e 2 iπ ( M xu + N y v ) x = 0 ∑ M − 1 y = 0 ∑ N − 1 e 2 iπ x ( X i 1 + M u ) e 2 iπ y ( Y i 1 + N v ) ( x = 0 ∑ M − 1 e 2 iπ x ( X i 1 + M u ) ) ( x = 0 ∑ N − 1 e 2 iπ x ( Y i 1 + N v ) ) Using geometric sum formulae:
Hence:
Convergence of energies
is continuous. We define the variable:
By the continuous mapping theorem:
= 1 − e 2 i π M ( 1 X i + u M ) 1 − e 2 i π ( 1 X i + u M ) 1 − e 2 i π N ( 1 Y i + v N ) 1 − e 2 i π ( 1 Y i + v N ) = 1 − e 2 i π M X i + 2 i π u 1 − e 2 i π ( 1 X i + u M ) 1 − e 2 i π N Y i + 2 i π v 1 − e 2 i π ( 1 Y i + v N ) = 1 − e 2 i π M X i 1 − e 2 i π ( 1 X i + u M ) 1 − e 2 i π N Y i 1 − e 2 i π ( 1 Y i + v N ) \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*} = = = 1 − e 2 iπ ( X i 1 + M u ) 1 − e 2 iπ M ( X i 1 + M u ) 1 − e 2 iπ ( Y i 1 + N v ) 1 − e 2 iπ N ( Y i 1 + N v ) 1 − e 2 iπ ( X i 1 + M u ) 1 − e 2 iπ X i M + 2 iπ u 1 − e 2 iπ ( Y i 1 + N v ) 1 − e 2 iπ Y i N + 2 iπ v 1 − e 2 iπ ( X i 1 + M u ) 1 − e 2 iπ X i M 1 − e 2 iπ ( Y i 1 + N v ) 1 − e 2 iπ Y i N F ( u , v ) = A M × N 1 − e 2 i π M X i 1 − e 2 i π ( 1 X i + u M ) 1 − e 2 i π N Y i 1 − e 2 i π ( 1 Y i + v N ) \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*} F ( u , v ) = M × N A 1 − e 2 iπ ( X i 1 + M u ) 1 − e 2 iπ X i M 1 − e 2 iπ ( Y i 1 + N v ) 1 − e 2 iπ Y i N The random variables X i , Y i ( i ∈ N ) \mathcal{X}_i, \mathcal{Y}_i \left(i \in \mathbb{N}\right) X i , Y i ( i ∈ N ) are independent then two variables:
X ‾ n = X 1 + … X n n and Y ‾ n = Y 1 + … Y n n \overline{\mathcal{X}}_n = \frac{\mathcal{X}_1 + \dots \mathcal{X}_n}{n} \quad \text{and} \quad \overline{\mathcal{Y}}_n = \frac{\mathcal{Y}_1 + \dots \mathcal{Y}_n}{n} X n = n X 1 + … X n and Y n = n Y 1 + … Y n are independent for all n ∈ N n \in \mathbb{N} n ∈ N . Since X i ( i ∈ N ) \mathcal{X}_i \left(i \in \mathbb{N}\right) X i ( i ∈ N ) are independent and identically distributed and X i ∼ N ( μ , σ 2 ) \mathcal{X}_i \sim \mathcal{N}\left(\mu, \sigma^2\right) X i ∼ N ( μ , σ 2 )
X ‾ n → n → ∞ a . s μ \overline{\mathcal{X}}_n \xrightarrow[n \to \infty]{a.s} \mu X n a . s n → ∞ μ by strong law of large numbers; similarly Y ‾ n → n → ∞ a . s μ \overline{\mathcal{Y}}_n \xrightarrow[n \to \infty]{a.s} \mu Y n a . s n → ∞ μ . Hence
[ X ‾ n Y ‾ n ] t → n → ∞ a . s [ X ‾ Y ‾ ] t {[\overline{\mathcal{X}}_n \quad \overline{\mathcal{Y}}_n]^t} \xrightarrow[n \to \infty]{a.s} {[\overline{\mathcal{X}} \quad \overline{\mathcal{Y}}]^t} [ X n Y n ] t a . s n → ∞ [ X Y ] t for some independent and identically distributed variables X ‾ ∼ Y ‾ ∼ μ \overline{\mathcal{X}} \sim \overline{\mathcal{Y}} \sim \mu X ∼ Y ∼ μ . Let us fix some 0 ≤ u < M 0 \leq u < M 0 ≤ u < M and 0 ≤ v < N 0 \leq v < N 0 ≤ v < N , then it is obvious that the function
F ( u , v ) ≜ ( x , y ) ↦ A M × N ( 1 − e 2 i π M x ) ( 1 − e 2 i π N y ) ( 1 − e 2 i π ( 1 x + u M ) ) ( 1 − e 2 i π ( 1 y + v N ) ) F_{\left(u,v\right)} \triangleq \left(x,y\right) \mapsto \frac{A}{M \times N} \frac{\left(1 - e^{2i\pi \frac{M}{x}}\right) \left(1 - e^{2i\pi \frac{N}{y}}\right)}{\left(1 - e^{2i\pi\left(\frac{1}{x} + \frac{u}{M}\right)}\right) \left(1 - e^{2i\pi\left(\frac{1}{y} + \frac{v}{N}\right)}\right)} F ( u , v ) ≜ ( x , y ) ↦ M × N A ( 1 − e 2 iπ ( x 1 + M u ) ) ( 1 − e 2 iπ ( y 1 + N v ) ) ( 1 − e 2 iπ x M ) ( 1 − e 2 iπ y N ) F ( u , v ) : Ω → R 2 ω ↦ F ( u , v ) ( X ( ω ) , Y ( ω ) ) \mathcal{F}_{\left(u,v\right)} \colon \Omega \to \mathbb{R}^2 \\ \omega \mapsto F_{\left(u,v\right)} \left(\mathcal{X} \left(\omega\right),\mathcal{Y}\left(\omega\right)\right) F ( u , v ) : Ω → R 2 ω ↦ F ( u , v ) ( X ( ω ) , Y ( ω ) ) for some random variables X \mathcal{X} X and Y \mathcal{Y} Y , then F ( u , v ) ( ω ) \mathcal{F}_{\left(u,v\right)}\left(\omega\right) F ( u , v ) ( ω ) is nothing but the Fourier transform of the atomic signal whose the horizontal and the vertical period are X ( ω ) \mathcal{X}\left(\omega\right) X ( ω ) and Y ( ω ) \mathcal{Y}\left(\omega\right) Y ( ω ) respectively. Let us consider the sequence ( F ( u , v ) n ) n ∈ N \left(\mathcal{F}_{\left(u,v\right)}^n\right)_{n \in \mathbb{N}} ( F ( u , v ) n ) n ∈ N where:
F ( u , v ) n ≜ ω ↦ F ( u , v ) ( X ‾ n ( ω ) , Y ‾ n ( ω ) ) \mathcal{F}_{\left(u,v\right)}^n \triangleq \omega \mapsto F_{\left(u,v\right)} \left(\overline{\mathcal{X}}_n \left(\omega\right),\overline{\mathcal{Y}}_n\left(\omega\right)\right) F ( u , v ) n ≜ ω ↦ F ( u , v ) ( X n ( ω ) , Y n ( ω ) ) F ( u , v ) n → n → ∞ a . s ω ↦ F ( u , v ) ( X ‾ ( ω ) , Y ‾ ( ω ) ) \mathcal{F}_{\left(u,v\right)}^n \xrightarrow[n \to \infty]{a.s} \omega \mapsto F_{\left(u,v\right)} \left(\overline{\mathcal{X}} \left(\omega\right),\overline{\mathcal{Y}}\left(\omega\right)\right) F ( u , v ) n a . s n → ∞ ω ↦ F ( u , v ) ( X ( ω ) , Y ( ω ) ) Since we have proved that X ‾ ∼ Y ‾ ∼ μ \overline{\mathcal{X}} \sim \overline{\mathcal{Y}} \sim \mu X ∼ Y ∼ μ then the limit of the sequence is the constant variable F ( u , v ) ( μ , μ ) F_{\left(u,v\right)} \left(\mu, \mu\right) F ( u , v ) ( μ , μ ) . Also, since F ( u , v ) F_{\left(u,v\right)} F ( u , v ) is continuous:
∣ F ( u , v ) ( X ‾ n , Y ‾ n ) − F ‾ ( u , v ) ( X n , Y n ) ∣ → n → ∞ 0 \lvert F_{\left(u,v\right)}\left(\overline{\mathcal{X}}_n, \overline{\mathcal{Y}}_n\right) - \overline{F}_{\left(u,v\right)}\left(\mathcal{X}_n, \mathcal{Y}_n\right) \rvert \xrightarrow[n \to \infty]{} 0 ∣ F ( u , v ) ( X n , Y n ) − F ( u , v ) ( X n , Y n ) ∣ n → ∞ 0 Hence F ‾ ( u , v ) ( X n , Y n ) → n → ∞ a . s F ( u , v ) ( μ , μ ) \overline{F}_{\left(u,v\right)}\left(\mathcal{X}_n, \mathcal{Y}_n\right) \xrightarrow[n \to \infty]{a.s} F_{\left(u,v\right)} \left(\mu, \mu\right) F ( u , v ) ( X n , Y n ) a . s n → ∞ F ( u , v ) ( μ , μ ) .
Remark. The proof does not require that X i X_i X i and Y i Y_i Y i have the same distribution. Indeed, if X i ∼ N ( μ x , ⋅ ) X_i \sim \mathcal{N}\left(\mu_{x},\cdot\right) X i ∼ N ( μ x , ⋅ ) and Y i ∼ N ( μ y , ⋅ ) Y_i \sim \mathcal{N}\left(\mu_{y},\cdot\right) Y i ∼ N ( μ y , ⋅ ) then F ‾ ( u , v ) ( X n , Y n ) → n → ∞ a . s F ( u , v ) ( μ x , μ y ) \overline{F}_{\left(u,v\right)}\left(\mathcal{X}_n, \mathcal{Y}_n\right) \xrightarrow[n \to \infty]{a.s} F_{\left(u,v\right)} \left(\mu_x, \mu_y\right) F ( u , v ) ( X n , Y n ) a . s n → ∞ F ( u , v ) ( μ x , μ y ) .