# On Beamforming with the Single-Sideband Transform
## Appendix A Properties of the Real Fourier Transform  
The Fourier Transform (FT) is defined as  
\[X_{\mathcal{F}}(f)=\int\limits_{-\infty}^{\infty}x(t)e^{-|2\pi ft}\mathrm{d}t \tag{1}\]  
with an inverse  
\[x(t)=\int\limits_{-\infty}^{\infty}X_{\mathcal{F}}(f)e^{|2\pi ft}\mathrm{d}t\,. \tag{2}\]  
The Real Fourier Transform (RFT) is defined as  
\[X_{\mathcal{R}}(f)=\sqrt{2}\mathrm{R}\left\{\int\limits_{-\infty}^{\infty}x(t )e^{-|2\pi ft+|3\frac{\pi}{4}}\mathrm{d}t\right\} \tag{3}\]with an inverse  
\[x(t)=\sqrt{2}\mathrm{R}\left\{\int\limits_{-\infty}^{\infty}X_{\mathcal{R}}(f)e^{ \mathrm{i}2\pi ft-\mathrm{i}\frac{3\pi}{4}}\mathrm{d}f\right\}. \tag{10}\]  
We assume that \(x(t)\) is a real-valued time-domain signal unless stated otherwise. The FT and RFT are bijective transformations of one another. Proof. : By manipulating Equation (11), we can obtain  
\[X_{\mathcal{R}}(f)=\sqrt{2}\mathrm{R}\left\{\left(\int\limits_{\infty}^{ \infty}x(t)e^{-\mathrm{i}2\pi ft}\mathrm{d}t\right)\mathrm{e}^{\mathrm{i}\frac {3\pi}{4}}\right\} \tag{11}\]  
where the term in parenthesis is trivially the FT of \(x(t)\) from Equation (10). Thus, we obtain that  
\[X_{\mathcal{R}}(f) =\mathrm{R}\left\{\left(X_{\mathcal{F}}^{\mathrm{R}}(f)+\mathrm{ i}X_{\mathcal{F}}^{\mathrm{I}}(f)\right)\cdot(-1+\mathrm{i})\right\} \tag{12}\] \[=-X_{\mathcal{F}}^{\mathrm{R}}(f)-X_{\mathcal{F}}^{\mathrm{I}}(f )\.\]  
Likewise, considering that the FT of a real signal is complex-conjugate, such that \(X_{\mathcal{F}}(f)=X_{\mathcal{F}}^{*}(-f)\), it is easy to see that  
\[X_{\mathcal{R}}(-f)=-X_{\mathcal{F}}^{\mathrm{R}}(f)+X_{\mathcal{F}}^{\mathrm{ I}}(f). \tag{13}\]  
With this, we have that  
\[\sqrt{2}X_{\mathcal{R}}(f)e^{-\mathrm{i}\frac{3\pi}{4}} =X_{\mathcal{F}}^{\mathrm{R}}(f)+X_{\mathcal{F}}^{\mathrm{I}}(f)+ \mathrm{i}X_{\mathcal{F}}^{\mathrm{R}}(f)+\mathrm{i}X_{\mathcal{F}}^{ \mathrm{R}}(f) \tag{14}\] \[\sqrt{2}X_{\mathcal{R}}(-f)e^{\mathrm{i}\frac{3\pi}{4}} =X_{\mathcal{F}}^{\mathrm{R}}(f)-X_{\mathcal{F}}^{\mathrm{I}}(f)- \mathrm{i}X_{\mathcal{F}}^{\mathrm{R}}(f)+\mathrm{i}X_{\mathcal{F}}^{ \mathrm{R}}(f)\]  
and, therefore, we have  
\[x(t)=\int\limits_{-\infty}^{\infty}X_{\mathcal{F}}(f)e^{|2\pi ft}\mathrm{d}t\,. \tag{2}\]0 \[x(t)=\int\limits_{-\infty}^{\infty}X_{\mathcal{F}}(f)e^{|2\pi ft}\mathrm{d}t\,. \tag{2}\]1  
Similarly, using Equation (11), we have that  
\[x(t)=\int\limits_{-\infty}^{\infty}X_{\mathcal{F}}(f)e^{|2\pi ft}\mathrm{d}t\,. \tag{2}\]2  
Using the property of complex numbers that \(\mathrm{R}\{a\}=\frac{a+a^{*}}{2}\), then  
\[x(t)=\int\limits_{-\infty}^{\infty}X_{\mathcal{F}}(f)e^{|2\pi ft}\mathrm{d}t\,. \tag{2}\]3  
Since for each \(X_{\mathcal{F}}(f)\) there exists one and only one \(X_{\mathcal{R}}(f)\), and vice-versa, they form a bijective relationship. The IRFT is the inverse of the RFT. Proof. : From Property 10, we have that  
\[x(t)=\int\limits_{-\infty}^{\infty}X_{\mathcal{F}}(f)e^{|2\pi ft}\mathrm{d}t\,. \tag{2}\]4Substituting this in Equation (A4), we have  
\[ \begin{split}\mathcal{R}^{-1}\{X_{\mathcal{R}}(f)\}(t)&= \sqrt{2}\text{R}\left\{\int\limits_{-\infty}^{\infty}\left(\frac{X_{\mathcal{F }}(f)e^{\text{i}\frac{2\pi}{T}}+X_{\mathcal{F}}(-f)e^{-\text{i}\frac{2\pi}{T} }}{\sqrt{2}}\right)e^{\text{i}2\pi f-\text{i}\frac{2\pi}{T}}\text{d}f\right\} \\ &=\text{R}\left\{\int\limits_{-\infty}^{\infty}\left(X_{ \mathcal{F}}(f)e^{\text{i}2\pi f\,t}+X_{\mathcal{F}}(-f)e^{\text{i}2\pi f\,t -\text{i}\frac{2\pi}{T}}\right)\text{d}f\right\}.\end{split} \] (A13)  
The first term expands to the inverse Fourier transform of \(X_{\mathcal{F}}(f)\), which is trivially \(x(t)\); the second term is the inverse Fourier transform of \(X_{\mathcal{F}}(-f)\), which is from the time reversal property and is \(x(-t)\). Therefore,  
\[ \begin{split}\mathcal{R}^{-1}\{X_{\mathcal{R}}(f)\}(t)& =\text{R}\left\{x(t)+x(\cdot+)e^{-\text{i}\frac{2\pi}{T}}\right\} \\ &=\text{R}\left\{x(t)+\text{i}x(\cdot)\right\}\\ &=x(t)\,.\end{split} \] (A14)  
Note that the invertibility of the RFT is a direct consequence of \(x(t)\) being a real signal.