# On Beamforming with the Single-Sideband Transform
## Appendix A Properties of the Real Fourier Transform
Assuming the contrary, then, obviously, \(\mathcal{R}^{-1}\{\mathcal{R}\{x(t)\}(f)\}(t)\
eq x(t)\), given that the inverse RFT involves taking the real part of the inverse, and, therefore, it maps the original signal to a different one. The FT convolution theorem does not apply for the RFT. Proof. : Let \(h(t)\) be the impulse response of an LTI system, with input \(x(t)\). It is trivial that the system's output, \(y(t)\), is given by  
\[y(t)=h(t)*x(t)\] (A15)  
with \(*\) being the convolution operator. For the Fourier transform, through the convolution theorem, it is trivial that  
\[Y_{\mathcal{F}}(f)=H_{\mathcal{F}}(f)X_{\mathcal{F}}(f)\,.\] (A16)  
Expanding these in terms of real and imaginary parts (omitting the frequency index for clarity),  
\[Y_{\mathcal{F}}=H_{\mathcal{F}}^{\text{R}}X_{\mathcal{F}}^{\text{R}}+\text{j} H_{\mathcal{F}}^{\text{R}}X_{\mathcal{F}}^{\text{I}}+\text{j}H_{\mathcal{F}}^{ \text{I}}X_{\mathcal{F}}^{\text{R}}-H_{\mathcal{F}}^{\text{I}}X_{\mathcal{F }}^{\text{I}}.\] (A17)  
Now in the RFT domain, with Equation (A6), we have that  
\[ \begin{split} X_{\mathcal{R}}(f)&=-X_{\mathcal{F}} ^{\text{R}}(f)-X_{\mathcal{F}}^{\text{I}}(f)\\ H_{\mathcal{R}}(f)&=-H_{\mathcal{F}}^{\text{R}}(f)- H_{\mathcal{F}}^{\text{I}}(f)\,.\end{split} \] (A18)  
Assuming the convolution theorem is true for the RFT,  
\[Y_{\mathcal{R}}=H_{\mathcal{F}}^{\text{R}}X_{\mathcal{F}}^{\text{R}}+H_{ \mathcal{F}}^{\text{R}}X_{\mathcal{F}}^{\text{I}}+H_{\mathcal{F}}^{\text{I}}X _{\mathcal{F}}^{\text{R}}+H_{\mathcal{F}}^{\text{I}}X_{\mathcal{F}}^{\text{I }}.\] (A19)  
Now, by applying Equation (A6) on Equation (A17), we have  
\[\tilde{Y}_{\mathcal{R}}=-H_{\mathcal{F}}^{\text{R}}X_{\mathcal{F}}^{\text{R}}- H_{\mathcal{F}}^{\text{R}}X_{\mathcal{F}}^{\text{I}}-H_{\mathcal{F}}^{\text{I}}X _{\mathcal{F}}^{\text{R}}+H_{\mathcal{F}}^{\text{I}}X_{\mathcal{F}}^{\text{I }}\] (A20)  
where it is explicit that \(Y_{\mathcal{R}}(f)\
eq\tilde{Y}_{\mathcal{R}}(f)\). Therefore, the RFT of the convolution (Equation (A20)) is not the product of the RFTs of the signals (Equation (A19)), and thus, the convolution theorem does not hold for the RFT. There is an equivalent of the convolution theorem for the RFT. Now, in the RFT domain, we take the correlation of a given frequency Equation (A6),  
\[\mathbb{E}\big{\{}X_{\mathcal{R}}(f)^{2}\big{\}}=\mathbb{E}\big{\{}X_{\mathcal{F} }^{\mathbf{R}}(f)^{2}+2X_{\mathcal{F}}^{\mathbf{R}}(f)X_{\mathcal{F}}^{\mathbf{ I}}(f)+X_{\mathcal{F}}^{\mathbf{I}}(f)^{2}\big{\}}\,.\] (A30)  
Using the fact that \(X_{\mathcal{F}}^{\mathbf{R}}(f)\) and \(X_{\mathcal{F}}^{\mathbf{I}}(f)\) are i.i.d. and zero-mean, the cross terms are zero, and thus,  
\[\mathbb{E}\big{\{}X_{\mathcal{R}}(f)^{2}\big{\}} =2\sigma_{f}^{2}\] (A31) \[=\mathbb{E}\big{\{}X_{\mathcal{F}}(f)X_{\mathcal{F}}^{*}(f)\big{\}}\,.\]  
It is trivial to see that the same applies for \(X_{\mathcal{R}}(-f)\). Conjugate frequencies in the RFT domain are independent. Proof. : We take the same assumptions as those in Property A5. Taking the complex correlation between the two conjugate frequencies,  
\[\mathbb{E}\big{\{}X_{\mathcal{F}}(f)X_{\mathcal{F}}^{*}(-f)\big{\}}=\mathbb{ E}\big{\{}X_{\mathcal{F}}^{\mathbf{R}}(f)^{2}+2\mathrm{j}X_{\mathcal{F}}^{ \mathbf{R}}(f)X_{\mathcal{F}}^{\mathbf{I}}(f)-X_{\mathcal{F}}^{\mathbf{I}}(f) ^{2}\big{\}}\,.\] (A32)  
Using the fact that the \(X_{\mathcal{F}}^{\mathbf{R}}(f)\) and \(X_{\mathcal{F}}^{\mathbf{I}}(f)\) are independent and zero-mean, the cross terms are zero, and with Equation (A28), then  
\[\mathbb{E}\big{\{}X_{\mathcal{F}}(f)X_{\mathcal{F}}^{*}(-f)\big{\}}=0\,.\] (A33)  
This result is known, but it is useful to show it since this same procedure is used for the RFT. We now consider Equations (A6) and (A7).