Jan 28

# How the Cyclic Prefix (CP) works in OFDM – Part II – The Theory.

This is a follow up to the Cyclic Prefix (CP) I did here.
There I gave a graphical explanation and here I will discuss how it happens theoretically. I recently came across the derivation in a text [1]. So here goes.

Considering an N-point FFT system, a block of data points
$\textbf{X}=[x_0 \quad x_1 \quad \ldots \quad x_{N-1}]$,
is to be transmitted during a symbol time. The data vector $\textbf{x}$ is sent to an IFFT module. The operation of the FFT, i.e. the discrete Fourier Transformation (DFT) is given by the following matrix:
$[\textbf{F}]_{k,m} = \frac{1}{\sqrt{N}} e^{-j2\pi\frac{km}{N}}, k,m=0,1,\ldots.N-1$
where $(k,m)$ denotes the $(k+1, m+1)$-th entry in the DFT matrix $\textbf{F}$.

Then the transmitted data vector after the IFFT operation is.
$\textbf{x}=\textbf{F}^H\textbf{X}$.
The normalization factor $\frac{1}{\sqrt{N}}$ takes care of the total bit energy as it is the same as the original vector $\textbf{x}$ because $\textbf{F}\textbf{F}^H=\textbf{F}^H\textbf{F}$.

Assuming the channel length (number of multipaths) is $L$, the OFDM symbol requires a CP of length $N_{CP}\geqslant (L-1)$. Then the total symbol length is $\bar{N}=N+N_{CP}$.

After the (linear) convolution through the channel and removal of the CP, the received signal (at the input to the FFT) can be written,

$\textbf{y}=\textbf{H}\textbf{x} + \textbf{n}$

where $\textbf{n}$ is the AWGN noise vector. The channel matrix $\textbf{H}$ is a circulant matrix due to the insertion of the CP in the transmitted data vector and can be written as,

$\textbf{H}=\begin{bmatrix} h_0 & h_{N-1} & \cdots & h_1 \\ h_1 & h_0 & \cdots &h_2 \\ \vdots & \vdots & \ddots & \vdots \\ h_{N-1} & h_{N-2} & \cdots & h_0 \end{bmatrix}$
Here $h_i$ are the channel impulse responses and $h_i=0$ for $L\leqslant i \leqslant (N-1)$.

Circulant matrices have an important property:

The eigenvectors of a circulant matrix of a given size are the columns of the discrete Fourier transform matrix of the same size [Wikipedia].

So we have,

$\textbf{H}=\textbf{F}^H \textbf{H}_{\mathsf{Eig}}\textbf{F}$

Thus,

$\textbf{F}\textbf{H}\textbf{F}^H = \textbf{H}_{\mathsf{Eig}} =\mathsf{diag}(H_0, H_1, ldots, H_{N-1})$

where,

$H_n =\sum_{l=0}^{L-1} h_l e^{-j2\pi\frac{nl}{N}}, n=0,1,\ldots.N-1$

Therefore, $H_n$ is just the frequency response of the $n$-th subcarrier and hence, the eigenvalues of $\textbf{H}$ are the frequency response of the channel.

Then, going back to the received signal vector,

$\textbf{y}=\textbf{H}\textbf{x} + \textbf{n}$

Sending this received data vector through the DFT,

$\begin{array} {lcl}\textbf{Y}&=&\textbf{F}\textbf{y} \\ & = &\textbf{F}(\textbf{H}\textbf{x} + \textbf{n}) \\ &=& \textbf{F}(\textbf{H}\textbf{F}^H\textbf{X}+ \textbf{n} \\ &=& \textbf{F}\textbf{H}\textbf{F}^H\textbf{X}+\textbf{F}\textbf{n} \\ &=& \overline{\textbf{H}}\textbf{X} + \tilde{\textbf{n}}\\ \end{array}$

Here $\overline{\textbf{H}}= \textbf{H}_{\mathsf{Eig}}$ and that the statistical properties of $\textbf{n}=\textbf{F}\textbf{n}=\tilde{\textbf{n}}$.

So you see how the insertion of the CP helps OFDM receivers easily decode the transmitted date vector by a simple inverse DFT operation (a single-tap equalization).

1. ##### ga

Thank you very much for this great explanations about cp. I have been looking for something like that in many books and web sites.
Which one is the reference [1] you mentioned?
Best regards

Welcome.
The reference is:
“Adaptive and Iterative Signal Processing in Communications” by Jinho Choi.

Reason I wanted to publish this was also because not many texts on OFDM discuss all the purposes of CP (they only mention ISI elimination) and this is the only book I came across the theoretical derivation for the circular convolution.

3. ##### kal

thanks for the post. it give me clear insight another perspective of cyclic prefix. may i know what the reference that u refer in previous and this post. thanks.

4. ##### Arun

Thanks, made my understanding a lot better!

Arun

5. ##### Vishnu

Hello sir, I am doing Mtech thesis on SCFDMA.In that I have to implement Tomlinson Harashima precoding , I got a paper for simulation,but in that they had not mention about channel impulsse response, which is very much important in this case, what can I do ?Is there any method for calculating h(n) for a multipath fading channel,please help me

6. ##### pościel

Bardzo mnie to zaciekawiło,będę tu zaglądać