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I have read a paper that performs Channel Estimation of Wireless Channel as follows.

A training sequence of length $N$, lets call it $a_N(i)$ for $i\in[0:N-1]$ is repeated twice then sent out over the channel.

Assume the transmit sequence formed of this two sequeses is denoted by $s_{CE}(t)$.

The received signal is given by (assuming channel of length $T_{CH}$) is $$r_{CE}(t) = \sum_{t'=0}^{T_{CH}-1}h(t')s_{CE}(t-t')+n(t)$$

The authors claim that if one takes the autocorrelation between a sequence $a_N(i)$ and the received signal then we can mathematically write the autorcorrelation as

$$R(t)= \frac{1}{N}\sum_{d=0}^{N-1}r_{CE}(t+d\underbrace{-N+1}_{????}) a_{N}^*(d) $$

My question is why is there a need for the term I have underbrace $-N+1$. I thought that the autocorrelation is in general expressed as

$$R(t)= \frac{1}{N}\sum_{d=0}^{N-1}r_{CE}(t+d) a_{N}^*(d) $$

Thanks looking forward for your view!

Update: The following is the reference

http://ieeexplore.ieee.org/xpl/login.jsp?tp=&arnumber=4699819&url=http%3A%2F%2Fieeexplore.ieee.org%2Fxpls%2Fabs_all.jsp%3Farnumber%3D4699819

I particular (25) and (26) are my main concerns.

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    $\begingroup$ Can you please give a reference for the paper? $\endgroup$ – Oliver Jun 1 '15 at 4:40
  • $\begingroup$ Are you sure this is for channel estimation and not for timing synchronization? And shouldn't this be called crosscorrelation? $r_\mathrm{CE}$ and $a_\mathrm N$ arre different signals after all. Finally, the formulas seem to mix continous and discrete time signals, since $t$ is continous time (I guess?) and $N$ is an integer number? Or is this just an odd notation? A reference to the original paper would be nice. $\endgroup$ – Deve Jun 2 '15 at 7:33
  • $\begingroup$ Thanks @Deve I will add the reference to paper. Please let me know what you think. $\endgroup$ – Tyrone Jun 2 '15 at 13:53
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I think it's just a trick which makes $R(t)=0, \mathrm{when}\quad t<0$.

First, $-N+1$ is just a time offset and not affect the essence of function.

$r_{CE}(t)$ is bounded to $[0, 2N+T_{CH}-2]$, and $a_N(t)$ to $[0, N-1]$. In the form you gave, $R(t)$ is bounded to $[-N+1, 2N+T_{CH}-2]$. May be the author want the non-zero value of $R(t)$ also starts at 0, so he just move it right with $N-1$ length, which causes the $-N+1$ term in the equation.

--------- Why $R(t)$ is bounded as that -------------

This is just a explanation to give some intuition, not a formal proof. Consider two function $f(t)$ and $g(t)$, which $f(t)=0, t \notin [a, b]$ and $g(t)=0, t\notin [c, d]$. Now we want to calculate the sum of products, i.e., $S=\sum_t f(t)g(t)$, $[a,b]$ and $[c,d]$ must be overlapped, right? Thus we get $a \le d $ and $c\le b$.

Get back to the $R(t)=\sum _d r_{CE}(t+d)a^*(d)$(note here the argument is $d$), $a^*(d)$ is bounded to $[0, N-1]$ and $r_{CE}(t+d)$ to $[-t, L-1-t]$, in which $L=2N+T_{CH}-1$ is the length of $r_{CE}(d)$. Here we get $a=0, b=N-1, c=-t, d=L-1-t$. Using $a \le d $ and $c \le b$, we get $-(N-1) \le t \le L-1$.

As to why $r_{CE}$ is bounded as that, you can make a same process to understand it.

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  • $\begingroup$ thanks, can you please explain why $R (t)$is bounded as you showed? $\endgroup$ – Tyrone Jun 1 '15 at 10:05
  • $\begingroup$ @Tyrone I updated my answer and hope it helps. $\endgroup$ – nanoix9 Jun 2 '15 at 2:03

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