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');

• 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? – Marcus Müller Aug 28 '20 at 20:32
• 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. – Fatima_Ali Aug 29 '20 at 2:19
• @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 $M+L-1 \leq N$. ... – AlexTP Aug 30 '20 at 8:05
• @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 $N$. – AlexTP Aug 30 '20 at 8:05