[SOLVED] Perfect Channel Information

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James Crawford

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What exactly does it mean when it says "Channel information is perfectly known at the receiver?".

Let's say that we have a channel with CIR h and additive Gaussian noise n. The transmitted training signal is expressed as a circular matrix S. Thus, the received signal is [Note: Lower-case denotes vectors, upper-case denotes matrices]

y = Sh + n

The estimation of the received signal is

y' = Sh'

Hence, the estimation error is

e = y - y' = y - Sh'

My question is: By appending the received signal with the channel estimation error does this imply channel information is "perfectly known" at the receiver? If not can you please explain to me what is meant by perfect channel information and how exactly I would implement this concept to a simulation.

Any help, comment, suggestion or whatever would be really appreciated .
 

I don't quite get what you mean by " appending the received signal with the channel estimation error", but let me try to answer based on what I understand from the question.

But I think the point you are missing out is, the error that you estimate at each instant (or OFDM symbol) is different. So you can't use error you got at one instant to get a "perfect channel information" at another instant since the data as well as the channel (possibly) is always changing.
 


It simply means no estimation error is assumed. If your received signal is y=hx+n, then the receiver has exact value of h and not an estimate of it h' with error e=h-h', i.e. e=0 for the perfect channel knowledge assumption. Those assumptions are not realistic, but they are frequently used to study other aspects of communication systems.
 
Thanks David, I figured it out. I was confused because I thought that if you could new knew the exact channel information then you may as well consider your received signal as y=x+n.
 

Thanks David, I figured it out. I was confused because I thought that if you could new knew the exact channel information then you may as well consider your received signal as y=x+n.

Knowing the channel doesn't eliminate it, because its equalization will affect the noise statistics as well. Anyway, good you found the answer.
 

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