# Equalization

Learn how the effect of a wireless channel is removed through the equalization process.

## We'll cover the following

After the insertion of the CP, the signal is ready to be sent over the wireless channel. This is done through a process known as **upconversion** in which the modulated signal is mixed with a sinusoid of a particular frequency intended for the signal.

## Role of channel estimates

We have learned that a linear convolution occurs between the transmitted signal $x[n]$ (that has a cyclic prefix attached) and the wireless channel $h[n]$.

$y[n]=x[n]\ast h[n]=\sum_{m=0}^{N-1}x[m]h[n-m]$

This is true for $n=0,1,\cdots,N-1$. We have also investigated in an earlier lesson why this results in their spectral product. At the FFT output, we have

$Y[k]= X[k]\cdot H[k]$

for $k = 0,1, \cdots,N-1$. Clearly, each symbol $X[k]$ is distorted by a complex number $H[k]$. To recover all $N$ symbols, we either need to know the wireless channel coefficient $h[n]$ or its DFT $H[k]$.

For this purpose, known sequences (*pilots or training*) are sent along with the information symbols. At locations where the symbols $X[k]$ are known, the channel can be estimated as:

$Y[k]=X[k]\cdot H[k] \quad \longrightarrow \quad \hat H[k]=\frac{Y[k]}{X[k]}, ~~\text{for all~}k$

## Equalizer

Since all the remaining locations at which the data symbols are known, these channel estimates can be used as:

$Y[k]=X[k]\cdot H[k] \quad \longrightarrow \quad\hat X[k] = \frac{Y[k]}{H[k]},~~\text{for all~}k$

Let’s see this process in code. To avoid complexity, we assume that channel estimates $\hat H[k]$ are already available at the receiver.

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