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In an OFDM system, serial-to-parallel conversion for data is done, then the DFT is performed and then adding the cyclic prefix (CP).

My question is related to that step of adding a CP. As I know, adding the CP is My question, if $H$ is the vector of $y$ in frequency domain, "$H = Fy$", so Can we diagonalize $H$ by left-right multiplying by proper DFT? Which mean $A = F \, H \, F^H$ and $A$ is a diagonal matrix of $N \times N$ ?? Does that right in OFDM? And what's about after removing the CP when performing DFT operating in receiver side, will this propriety of circular convolution will be kept?

thank you

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  • $\begingroup$ i wonder whatta " $CP$ " is. $\endgroup$ – robert bristow-johnson Sep 18 '18 at 18:33
  • $\begingroup$ @robertbristow-johnson: cyclic prefix. $\endgroup$ – MBaz Sep 18 '18 at 22:27
  • $\begingroup$ +1 for showing efforts $\endgroup$ – Ahmad Bazzi Sep 19 '18 at 11:47
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Assuming you have $N$ symbols to transmit encoded in block $k$, \begin{equation} s(k) = \begin{bmatrix} s_1(k) \\ \vdots \\ s_N(k) \end{bmatrix} \end{equation} Performing $N-$FFT at the transmitter, you now have \begin{equation} x(k) = F^Hs(k) \end{equation} It is here that you insert a cyclic prefix of size $D > L - 1$, where $L$ is the number of taps in your FIR channel, i.e. at the receiver you have \begin{equation} r(k) =H_{ISI}\begin{bmatrix} \overbrace{x_{N-D+1}(k)} \\ \vdots \\ \underbrace{x_N(k)}_{\text{CP}} \\ x_1(k) \\ \vdots \\ x_N(k) \end{bmatrix} + H_{IBI} \begin{bmatrix} x_{N-D+1}(k-1) \\ \vdots \\ x_N(k-1) \\ x_1(k-1) \\ \vdots \\ x_N(k-1) \end{bmatrix} \end{equation} where $\lbrace c_k \rbrace_{k=0}^{L-1}$ are the channel taps. Note that the channel matrix is decomposed into two parts, ISI (intersymbol interference) and IBI (interblock interference) \begin{equation} H = H_{ISI} + H_{IBI} \end{equation} i.e. \begin{align} H_{ISI} &= \begin{bmatrix} c_0 & 0 & \ldots & 0\\ \vdots & \ddots & \ldots & \vdots \\ c_{L-1} & \ddots & & \vdots \\ \vdots & \ddots & & \vdots \\ 0 & \ldots & c_{L-1} & c_0 \end{bmatrix} \\ H_{IBI} &= \begin{bmatrix} 0 & \ldots & 0 & c_{L-1} & \ldots & \ldots & c_1 \\ \vdots & \ddots & & & \ddots & & \vdots \\ \vdots & \ddots & & & \ddots & & c_{L-1}\\ 0 & \ldots & & & \ldots & & 0\\ \vdots & \ddots & & & \ddots & & \vdots \\ 0 & \ldots & & & \ldots & & 0\\ \end{bmatrix} \end{align} If we remove the first $D$ samples of $r(k)$, we have \begin{equation} \tilde{r}(k) = \begin{bmatrix} r_{D+1}(k) \\ \vdots \\ r_{N+D}(k) \end{bmatrix} = \begin{bmatrix} 0 & \ldots & 0 & c_{L-1} \ldots & c_0 & 0 & \ldots & 0\\ \vdots & & & \ddots & & \ddots & & \vdots \\ 0 & \ldots & & \ldots & 0 & c_{L-1} & \ldots & c_0\\ \end{bmatrix} \begin{bmatrix} x_{N-D+1}(k) \\ \vdots \\ x_N(k) \\ x_1(k) \\ \vdots \\ x_N(k) \end{bmatrix} \end{equation} After a little bit of manipulation, you can see that \begin{equation} \tilde{r}(k) = C F s(k) \end{equation} where $C$ is a circular matrix that contains \begin{equation} \begin{bmatrix} c_0 & 0 & \ldots & c_{L-1} & \ldots c_1 \end{bmatrix} \end{equation} in its first row. Now, applying a DFT, we get \begin{equation} y(k) = F \tilde{r}(k) = F C F^H s(k) \end{equation} Since $C$ is circular, then $F C F^H$ is diagonalizable because any circular matrix is diagonal in a Fourier basis, hence you get \begin{equation} y(k) = D s(k) \end{equation} where $D$ is diagonal containing the eigenvalues of $C$.

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  • $\begingroup$ Thank you for your detailed answer, I have a question about decomposition of channel into $HISI$ and $HIBI$ .. you mean the $HIBI$ is corresponding to $CP$ length ?? $\endgroup$ – Fatima_Ali Sep 20 '18 at 9:51
  • $\begingroup$ Welcome $H_{IBI}$ refers to the part of the channel where there is InterBlock Interference, i.e. interference between consecutive OFDM blocks @Fatima_Ali $\endgroup$ – Ahmad Bazzi Sep 21 '18 at 2:35
  • $\begingroup$ Yes I see .. But I think that even that part of channel with $H_{IBI}$, it includes $H_{ISI}$ .. I mean both part included $ISI$ .. Is that right? $\endgroup$ – Fatima_Ali Sep 21 '18 at 7:38
  • $\begingroup$ Yes you are right .. even within the blocks there is inter symbol interference @Fatima_Ali $\endgroup$ – Ahmad Bazzi Sep 21 '18 at 18:29
  • $\begingroup$ Ok .. thank you so much .. Do you have any idea about this? dsp.stackexchange.com/questions/52102/… $\endgroup$ – Fatima_Ali Sep 22 '18 at 4:20

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