Figure 1 (above) shows an example of a PCB channel response. You can see how the signal is 60 dB down at 28 MHz, just where you want to be for 56 Gbit/s NRZ; F igure 1. This ADS simulation of a Wild River Technology channel-modelling platform shows its frequency response. Vertical scale: dB, horizontal scale: frequency in GHz.
We must do a lot of work to help the receiver recognize the resulting waveform as a signal. In addition to careful layout and use of quality components—all in a cost-optimized way, of course—equalization does a lot of work.
Equalization effectively removes the channel response by applying the inverse transfer function of the channel to the signal. That is, if the transfer function of the channel is G(s), where
s = jω + α,
then we're after an equalizer whose transfer function is G -1(s). If we call the transmitter signal Tx(s) and the unequalized received signal Rx(s), then
G(s)Tx(s) = Noise(s).
Apply equalization at the receiver and you get
G-1(s)Rx(s) = G -1(s)G(s)Tx(s) + G -1(s)Noise(s) = Tx(s) + G -1(s)Noise(s)
which, except for the noise, is exactly what we want.
Here's another way to think of it that I find more intuitive: the channel distorts the signal, so why not pre-distort the signal in such a way that the channel itself removes that distortion? In other words, pre-distort the transmitted signal in a way that includes the inverse channel frequency response so that the channel cancels the pre-distortion.
Lets call Symbols(s) the signal that we want. It has the wide open eye that we’d like the receiver to see after the transmitted signal Tx(s) has been through the channel. Then the pre-distorted signal should be as close to G -1(s)Symbols(s) as we can get. Now go back to the simple relationship between