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In ZP-OFDM, The convolution is linear. it's different from CP-OFDM which is circular convolution leading to have one tap equalizer in frequency domain.

if I have a ZP-OFDM system, and we have a signal to transmit $X$ of length $M$ x $1$, after adding the zero padding, the resultant transmitted signal will be $x$ of length $N$ x $1$. where the received signal is:

$y = h*x + v$, where $h$,$x$,$v$,$*$ denote multi-path channel of length $L$ x $1$, transmitted signal of length $N$x$1$, AWGN, and convolution operation, respectively.

to recover the transmitted signal from the received signal $y$, let the matrix $G$ to be the MMSE equalizer built based on the toeplitz matrix $H$ equivalent to $h$. so estimated signal $x$ = $Gy$.

The question which I couldn't get a clear answer for it, Is it better to built the matrix $G$ of size $M$x$N$ and multiply it directly with $y$ which is $N$x$1$, resulting the transmitted signal of size $M$x$1$. OR, I start discard the part equivalent to the zero padding from $y$ to have the length of $M$x$1$ and then build the matrix $G$ of size $M$x$M$ to be multiplied with $y$ after removing the part equivalent into the zero padding?

In simulation, the first option gives slightly better performance and more stable. Why ? what is the explanation of each case?

EDIT

1- The length of the received signal $y$ is considered to be $N$x $1$, it means that the delay of channel is already discarded.

2- We considered the length of the guard interval to be longer than the channel delay

3- The toeplitz matrix $H$ and $G$ was built as following,

C = [h;zeros(N-L,1)]; 
R = [h(1), zeros(1,N-1)];
H = toeplitz(C,R);  
G = inv((1/SNRv(i))*eye(N) + (H')*H) * (H'); 
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  • $\begingroup$ well, using more of the signal obviously gives you more information; using only a part of the signal that was "spread out" by the channel convolution loses energy and thus information. But I think you already know that, considering you understand how you need to work with the linear convolution, so where specifically does this question arise from? $\endgroup$ – Marcus Müller Aug 28 at 20:32
  • $\begingroup$ Hi Marcus, I couldn't understand what you mean "using more of signal obviously gives more information and using only a part of the signal that was spread out by the channel convolution loses energy", Is there detailed reason for that or mathematical expression? 2- the question was arise based on my notice on the MATLAB simulation results. $\endgroup$ – Fatima_Ali Aug 29 at 2:19
  • $\begingroup$ @Fatima_Ali As I don't know how you designed your matrix $G$, this comment is possibly wrong. By $y = h*x + v$, the output $y$ is of length $N+L-1$ and have $M+L-1$ non-zero elements$^{(1)}$ because of the $(N-M)\textrm{-length}$ zero-padding, which should be greater than $L$ for proper zero-padding designs. Hence, the information is spread within the length of $M+L-1 \leq N$. ... $\endgroup$ – AlexTP Aug 30 at 8:05
  • $\begingroup$ @Fatima_Ali By using $G$ of size $N$, you have the chance of be able to collect all the possible information and, therefore, better results than using $G$ of size $M$ where you have thrown away something. The performance difference depends on many things including but not limited to how padding $N-M$ is greater than delay spread $L$, what is thrown away by using $G$ of size $M$ and how $G$ are designed$^{(2)}$ . $^{(1)}$ element is of general meaning. $^{(2)}$ the "optimal" $G$ for size $M$ processing is not necessarily optimal for the case of size $N$. $\endgroup$ – AlexTP Aug 30 at 8:05
  • $\begingroup$ @AlexTP I added the information you asked about into the question. Could you please reply me as an answer with more details "to accept your answer". $\endgroup$ – Fatima_Ali Aug 30 at 12:27

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