## 2 Mathematical Background  
The DFT of a discrete signal \(x(n)\) of size \(N\) is given by  
\[c(n)=\sum_{k=0}^{N-1}x(k)\exp\frac{-j2\pi nk}{N}=\sum_{k=0}^{N-1}x(k)\bigg{(} \cos\frac{2\pi nk}{N}-j\sin\frac{2\pi nk}{N}\bigg{)} \tag{1}\]  
for \(n=0,1,\ldots,N-1\), where \(j\) is the imaginary unit. Each coefficient \(c(n)\) of DFT, as a complex number, can be written as a sum  
\[c(n)=a(n)+jb(n) \tag{2}\]  
where \(a(n)\) and \(b(n)\) are the a real part and an imaginary part of \(c(n)\), respectively. For real signal \(x(n)\)  
\[a(n)=\sum_{k=0}^{N-1}x(k)\cos\frac{2\pi nk}{N} \tag{3}\]  
\[b(n)=-\sum_{k=0}^{N-1}x(k)\sin\frac{2\pi nk}{N} \tag{4}\]In this case, the DFT is redundant since it is conjugate-symmetric. Its real and imaginary parts meet the following properties:  
\[a(N-n)=\sum_{k=0}^{N-1}x(k)\cos\frac{2\pi(N-n)k}{N}=\sum_{k=0}^{N-1}x(k)\cos \frac{2\pi nk}{N}=a(n) \tag{5}\]  
\[b(N-n)=-\sum_{k=0}^{N-1}x(k)\sin\frac{2\pi(N-n)k}{N}=\sum_{k=0}^{N-1}x(k)\sin \frac{2\pi nk}{N}=-b(n) \tag{6}\]  
Taking into account the periodicity of the signal \(x(n)\) and their DFT with the period \(N\), we can write  
\[a(N-n)=a(-n)=a(n) \tag{7}\]  
so the real part of DFT is an even function. Since  
\[b(N-n)=b(-n)=-b(n) \tag{8}\]  
then the imaginary part of DFT is an odd function. From (8) for \(n=0\) we have \(b(-0)=b(0)=-b(0)\), so we obtain  
\[b(0)=0 \tag{9}\]  
and if \(N\) is an even number \(b(N/2)=b(N-N/2)=b(-N/2)=-b(N/2)\), so  
\[b\bigg{(}\frac{N}{2}\bigg{)}=0 \tag{10}\]  
In order to avoid redundancy and to use real transform for real signals the real discrete Fourier transform (RDFT) has been defined [25]  
\[c(n)=a(n)+jb(n) \tag{2}\]0  
where  
\[\theta(n)=\left\{ \begin{array}{ll}0&0\leq n\leq\lfloor\frac{N}{2}\rfloor\\ \frac{\pi}{2}&\lfloor\frac{N}{2}\rfloor<n\leq N-1\end{array} \right. \tag{12}\]  
and \(\lfloor\cdot\rfloor\) denotes the floor function. To see the relationship between the RDFT and the DFT of the signal \(x(n)\), it is better to write the RDFT in a slightly different form  
\[y(n)=\left\{ \begin{array}{ll}\sum_{k=0}^{N-1}x(k)\cos\frac{2\pi nk}{N}&0 \leq n\leq\lfloor\frac{N}{2}\rfloor\\ -\sum_{k=0}^{N-1}x(k)\sin\frac{2\pi nk}{N}&\lfloor\frac{N}{2}\rfloor<n\leq N- 1\end{array} \right. \tag{13}\]  
Now it is clear that  
\[y(n)=\left\{ \begin{array}{ll}a(n)&0\leq n\leq\lfloor\frac{N}{2}\rfloor\\ b(n)&\lfloor\frac{N}{2}\rfloor<n\leq N-1\end{array} \right. \tag{14}\]  
so the RDFT contains the not repeated and non-zero coefficients from the real and imaginary parts of the DFT. Knowing the RDFT coefficients of the signal and the relationships (7)-(10) and (14), it is easy to obtain the DFT coefficients. Namely, if \(N\) is an even number then  
\[ \begin{array}{ll}a(0)=y(0)&b(0)=0\\ a(1)=y(1)&b(1)=-b(N-1)=-y(N-1)\\ \vdots&\vdots\\ a(\frac{N}{2}-1)=y(\frac{N}{2}-1)&b(\frac{N}{2}-1)=-b(\frac{N}{2}+1)=-y(\frac{ N}{2}+1)\\ a(\frac{N}{2})=y(\frac{N}{2})&b(\frac{N}{2})=0\\ a(\frac{N}{2}+1)=a(\frac{N}{2}-1)=y(\frac{N}{2}-1)&b(\frac{N}{2}+1)=y(\frac{N}{ 2}+1)\\ \vdots&\vdots\\ a(N-1)=a(1)=y(1)&b(N-1)=y(N-1)\end{array}  \tag{15}\]  
If \(N\) is an odd number then  
\[ \begin{array}{ll}a(0)=y(0)&b(0)=0\\ a(1)=y(1)&b(1)=-b(N-1)=-y(N-1)\\ \vdots&\vdots\\ a(\frac{N-1}{2})=y(\frac{N-1}{2})&b(\frac{N-1}{2})=-b(\frac{N+1}{2})=-y(\frac{ N+1}{2})\\ a(\frac{N+1}{2})=a(\frac{N-1}{2})=y(\frac{N-1}{2})&b(\frac{N+1}{2})=y(\frac{N+1}{2})\\ \vdots&\vdots\\ a(N-1)=a(1)=y(1)&b(N-1)=y(N-1)\end{array}  \tag{16}\]  
Later in the article, the matrix-vector notation will be used, so the discrete real input signal \(x(n)\) of size \(N\) will be represented by a column vector \(\mathbf{x}_{N}=[x_{0},x_{1},\ldots,x_{N-1}]^{T}\), the \(N\)-by-\(N\) RDFT matrix will be denoted by \(\mathbf{R}_{N}\) and the output signal \(y(n)\)--by \(\mathbf{y}_{N}=[y_{0},y_{1},\ldots,y_{N-1}]^{T}\), where \([\cdot]^{T}\) means standard transposition operation.