# [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

```