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[BP] Thru channel gain



At the July meeting i suggested that Signal to Crosstalk ratio (ICR)
be measured by the ratios of two integrals, one for signal one for
crosstalk.  Adam showed us, in healey_c1_0505, how to do the
crosstalk integral but i did not have a good signal integral ready
then.  I now proposethe method shown below.

                     charles


-- 
|--------------------------------------------------------------------|
|       Charles Moore
|       Agilent Technologies
|       ASIC Products Division
|       charles_moore@agilent.com
|       (970) 288-4561
|--------------------------------------------------------------------|
DFE response to the signal at the end of a channel.

                                           Charles Moore 2005 July 28

A transmitter transmitting NRZ data, D(j), D(j)=+1or -1, into a channel 
will produce a disturbance at the far end:

    Rx(t)  =  signal at Rx due to Tx.

To this is added noise and interference which i will ignore for this analysis.
To this is also added a feedback term from the DFE receiver this gives

    DFEin(t) = Rx(t) + FB(t)

This sum is then sampled at the baud rate at what i will assume is optimum phase.

    DFEin( i*Tb + tp) = Rx(i) + FB(i)

    Rx(i) =  Sum( k(i-j,tp)*D(j))   ( -infinity < j < i)

where 

    Tb is the baud time
    tp is a small time shift
and k(m,tp) defines the channel. That is:

     k(m,tp) = h(m*Tb+tp) 

where h(t) is the entire channel impulse response.

    If we then select some n so that we expect that DFEin(j+n) can be quantized
to reproduce D(j) delayed by n baud times + tp, then:

    Rx(i) =  Sum( k(n+l,tp)*D(i-n-l) + FB(i)  (-n < l < infinity)

Since this is a DFE,

    FB(i) = Sum( Kfb(l)*sign(Rx(i-l))  ( 1 <= l  < infinity )

and since sign(Rx(i)) = D(i-n)

    FB(i) = Sum(Kfb(l)*D(i-n-l))       ( 1 <= l  < infinity )

    DFEin(i)=    Sum( k(n+l,tp)*D(i-n-l)    (-n < l < infinity)   
              +  Sum(Kfb(l)*D(i-n-l))       (1 <= l  < infinity )

If we adapt so that 

              Kfb(l)  =  - k(n+l,tp)  for (1 <= l  < infinity )

    DFEin(i)= Sum( k(n-l,tp)*D(i+l-n))   ( 0 <=  l  < n)

which is to say that with proper DFE equalization all post ISI is corrected
and only the desired signal (k(-n,tp)*D(i)) and precursor ISI 
(Sum(k(n-l,tp)*D(i-n+l) (1 =< l <n)) is left.  If we ignore precursor ISI beyond 
the first term, which from what i have seen of typical channel pulse responses 
is reasonable:

   DFEin(i) ~=  k(n,tp)*D(i-n) + k(n-1,tp)*D(i+1-n)

based on what i have seen, k(-n) and k(1-n) will have the same sign
so if D(i-n) and D(i+1-n) have opposite signs:

   DEFin(i) ~= D(i-n) * (k(n,tp)-k(n-1,tp))

I will define:

   DFEgain = k(n,tp)-k(n-1,tp)

and the EYE opening due to ISI is 2*DFEgain.  Clearly we want to select n and tp so
as to maximize DFEgain.  We can do this by defining:

    hGain(t)= h(t)-h(t-Tb)

and finding the maximum value of hGain:

   DFEgain = max(hGain(t)) 

   and n is defined as trunc(t(max)/Tb)
       tp is defined as t(max)-n*Tb

   This will occur at the top of the region of fastest rise on the leading 
edge of the pulse response.

In the frequency domain, with H(f) the Fourier transform of h(t)

   HGain(f) = H(f)*(1-exp(-2*j*pi*f*Tb)
            = H(f)*exp(-j*pi*f*Tb)*(exp(j*pi*f*Tb)-exp(-j*pi*f*Tb)) 
            = H(f)*exp(-j*pi*f*Tb)*2*j*sin(pi*f*Tb)

We can compute DFEgain from HGain(f) if we know the value of (n*Tb+tp), using
a single point inverse Fourier transform:

   DFEgain  = integral(exp(j*2*pi*f*(n*Tb+tp))*HGain(f)*df) (-infinity < f < infinity)

Without knowing the value time delay for maximum we note:

              exp(j*2*pi*f*(n*Tb+tp))*HGain(f) <= |HGain(f)|

so a useful upper bound on DFEgain is

   DFEgain <= 2*integral(|HGain(f)| df)  ( 0 < f < infinity)


I computed DFEgain and 2*integral(|HGain(f)| df) for the Tyco Case6 channel with
no package effect and a 35ps rise and fall time, and found the integral of the
absolute value to give a value 20% high.  I think that this is an indication that
the integral will be a useful measure of the effective channel gain.

Note on Tx equalization:

   1.  Any Tx de-emphasis equalization will reduce the integral of the absolute
       magnitude.  This is an indication, but not proof, that if ideal DFEs are
       available, EYE size will be greatest with no Tx equalization.

   2.  Signal to Crosstalk ratio most likely will be more important than EYE 
       size.  If Crosstalk is more weighted toward low frequency than HGain
       Tx equalization will help the ratio.  We can look at the ratio

                RXT  =  |SDD21XT| / (|SDD21Thru| * sin(pi*f*Tb))
       and see if it generally rises or falls going from DC to Nyquist and
       if it rises, minimize Tx equalization, if it falls more Tx 
       equalization may help by minimizing crosstalk.