Hello group 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 points. I understand that in this way there is a problem with the DFT but I can't understand why it causes ICI ;

# Why linear convolution in OFDM causes ICI;

Started by ●February 21, 2008

Reply by ●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 now...

Reply by ●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 > now...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 notation. 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 that 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 convolution with h[n]. 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 will orthogonalize the matrix corresponding to the channel since circular convolution is orthogonalized by the DFT operation.