We know that the use of cyclic prefix (CP) in an OFDM turns the linear
convolution (between the channel and the transmitted signal) into a
circular convolution. This is efficient to do because then we can easily
equalize the channel.
In most books it is referred that CP combats ISI and ICI.
With ISI i am fine.
I know that a circular conv between the channel (L points ) and the
signal (N points) produces a signal of N points.
On the other hand the linear convolution produces a signal of N+L-1
I understand that in this way there is a problem with the DFT but I can't
understand why it causes ICI ;
Reply by mnentwig●February 25, 20082008-02-25
I think it simply means that the cyclic prefix allows you to get a full
period length into the FFT. In other words, the discontinuity between any
two symbols is omitted. One could argue that you'll get ICI as soon as
there is no full number of cycles within the FFT window (for example one
end chopped off).
Maybe there is a better explanation, though, and I just don't see it right
Reply by julius●February 25, 20082008-02-25
On Feb 25, 4:20 am, "mnentwig" <mnent...@elisanet.fi> wrote:
> I think it simply means that the cyclic prefix allows you to get a full
> period length into the FFT. In other words, the discontinuity between any
> two symbols is omitted. One could argue that you'll get ICI as soon as
> there is no full number of cycles within the FFT window (for example one
> end chopped off).
> Maybe there is a better explanation, though, and I just don't see it right
Marcus, one problem is that there are conflicting definitions of ICI.
At the least, it means interference from other tones belonging to
the same OFDM symbol. But what about from other tones belonging
to other OFDM symbols? Is that called ISI?
But at any rate, we know that y[n] = x[n]*h[n] in the usual
The transmitted signal x[n] comes from an N-point inverse DFT, and
y[n] will be the input to an N-point DFT at the receiver. Let's say
we look at only one OFDM symbol and that, x[n] = iDFT( X[k] )
With the correct amount of cyclic prefix,
DFT(y[n]) = DFT(x[n]*h[n])
DFT(y[n]) = DFT(x[n]) . DFT(h[n])
Y[k] = X[k] . H[k]
Without the cyclic prefix, the DFT will not orthogonalize the
To the OP: In other words, try writing the convolution as a matrix
multiplication: y_vec = H_mat x_vec. With cyclic prefix, the DFT
orthogonalize the matrix corresponding to the channel since circular
convolution is orthogonalized by the DFT operation.