[gmx-users] H-bonding autocorrelation function
fyyan at andrew.cmu.edu
fyyan at andrew.cmu.edu
Sun Jul 12 05:13:10 CEST 2015
Hi,
I have a question about the H-bonding autocorrelation function, if I have
a trajectory, say
1-1-1-1-1-1-0-1 (1 is H-bonding, 0 is no H-bonding), the time step is
0.006 ps, the correlation is equal to <h(0)h(t)> / <h(0)*h(0)>, my answer
is
at t = 0.006, correlation = 7/7 = 1;
at t = 0.012, correlation = 5/7 = 0.714;
at t = 0.018, correlation = 5/7 = 0.714;
at t = 0.024, correlation = 4/7 = 0.571;
at t = 0.03, correlation = 3/7 = 0.429;
at t = 0.036, correlation = 2/7. = 0.286,
at t = 0.042, correlation = 1./7. = 0.143,
at t = 0.048, correlation = 1./7. = 0.143,
at t = 0.054, correlation = 0.
However, when i check gromacs, I found it gives a different answer, here
is the command I use "g_hbond -s md.tpr -f new.gro -ac -dist -temp 400 -b
72.444 -e 72.486 -dt 0.006", it uses only the time between 72.444 and
72.486 ps with a timestep of 0.006 ps, and the following is the h-bond
number file obtained from gromacs,
time (ps) H-bond ( H-bond without angle requirement)
72.444 1 1
72.45 1 1
72.456 1 1
72.462 1 1
72.468 1 1
72.474 1 1
72.48 0 2
72.486 1 1
and the hbac gives ,
time (ps) C(t)
0 1
0.0059967 0.972222
0.012001 0.933333
the following is from hbac.xvg,
@ s0 legend "Ac\sfin sys\v{}\z{}(t)"
@ s1 legend "Ac(t)"
@ s2 legend "Cc\scontact,hb\v{}\z{}(t)"
@ s3 legend "-dAc\sfs\v{}\z{}/dt"
0 1 1 0 235.274
0.0059967 0.411764 0.972222 -9.93411e-09 117.637
0.012001 -0.411764 0.933333 0.166667 -0
Am my calculation wrong? Can anyone help me to resolve this? Thanks!
Thanks!
Fangyong
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