1
Biochar Water Retention Random Variables
From P. Sherman 4/20/15 in relation to STAT447 Project Ideas
Bioc har (blue) & Control (green) Moisture Content
0.4
0.35
Moisture Content
0.3
0.25
0.2
0.15
0.1
0.05
0
1000
2000
3000
4000
Time (30 min. intervals)
5000
6000
Biochar (blue) & Control (green) Moisture Content
0.36
Moisture Content
0.34
0.32
0.3
0.28
0.26
0.24
2550
2600
2650
2700
2750
2800
Time (30 min. intervals)
2850
2900
Figure 1. Moisture content for biochar (blue) and the control (green).
For any given episode, for either X 1 (t ) (biochar) or X 2 (t ) (control) we have behavior of the general form:
X (t )  X (t 0  ) 
X (t0 ) e  t /
 Y (t )
;
t  t0 .
(1)
2
The random variables X (t ) and Y (t ) are indexed in time. Hence, the collections {X (t ); t  t0 } and {Y (t ); t  t0 } are called
random processes. Equation (1) also includes other random variables.
One is t0  , which is the time that a ‘significant’ rainfall begins. To be consistent with random variable (upper case)
notation, we would describe this as T0 . If we define this random variable via the equivalent events
[T (t )  T0 (t )]  [ B(t )  1] and [T (t )  T0 (t )]  [ B(t )  0] , then {B(t )} is also a random process. In fact, it is a Bernoulli
random process that describes the dynamics of when significant rainfall occurs. With this, it follows that the random
variable (1) is conditioned on the event [T (t )  T0 (t )]  [B(t )  1] .
The random process {Y (t ); t  t0 } , which might be loosely termed the ‘noise’ process, is also conditioned on this event. It
is needed to capture the uncertainty of the model random process {X (t0 ) e t / ; t  t0 } . This process is called a
deterministic random process. This is because, given the event [ X (t0 )  x(t0 )] , the process is deterministic. This assumes
that the time constant, τ, is the same for any episode. Under this assumption, then τ is is an unknow parameter that one
would certainly want to estimate.
The random variable X (t0 ) represents the amount of immediate moisture above the level X (t0 ) that the rainfall imparts
into the ground. The random variable X (t0 ) is the moisture level just prior to the rainfall.
It needs to be emphasized that the above random variable setting applies to each of X 1 (t ) (biochar) and X 2 (t ) (control).
Hence, this is a rich setting for a STAT447 project. Perhaps the most interesting focus of such an investigation (given that
this is a course project, and not a dissertation :) might be one that focusses on the unknown parameters  1 and  2 .
From http://en.wikipedia.org/wiki/Exponential_distribution :
Not to be confused with the exponential family of probability distributions.
In probability theory and statistics, the exponential distribution (a.k.a. negative exponential distribution) is the
probability distribution that describes the time between events in a Poisson process, i.e. a process in which events occur
continuously and independently at a constant average rate. It is a particular case of gamma distribution. It is the
continuous analogue of the geometric distribution, and it has the key property of being memoryless. In addition to being
used for the analysis of Poisson processes, it is found in various other contexts.
2. A Closer Look at the Process X (t )  X (t0 ) e t /
The model X (t )  X (t0 ) e t / is the moisture content associated with an initial condition that is caused by rainfall. While
this is in some ways a reasonable model, the fact is that the rainfall extends over a period of time. Hence, it is actuall an
input that results in the output X (t ) . Consider the input/output model described by
x (t )  ax(t )  br(t ) ; x(t0 )  x0
(2.1)
3
where r(t ) is the input rainfall level and x(t ) is the resulting moisture content. To solve (2.1) we first take its Laplace
transform, giving
[sX ( x)  x0 ]  aX ( s)  bR( s) .
(2.2)
Solving (2.2) for X (s) gives
 1 
 1 
X ( s)  
b R( s)  
 x0 .
s

a


sa
(2.3)
In any table of Laplace transform pairs, you will find the pair:
e a t
1
.
sa

(2.4)
From (2.4) it follows that the righmost term in (2.3) has the equivalent time domain form:
x0 e a t
 1 
 
 x0 .
sa
(2.5)
The time domain equivalent of the left term on the right side of (2.3) is given be the convolution integral
t

b e  a  r (t   ) d
0

 b 
 
 R( s )  G ( s ) R( s ) .
sa
(2.6)
To those with little or no experience with Laplace transforms, (2.6) may be a tad intimidating. However, if we assume
that r(t ) has a suitably simple form, then it will soon hopefully become less so. The advantage of using Laplace
transforms is highlighted in (2.6). Specifically, the intimidating convolution integral is equivalent to simple multiplication
of Laplace transforms. The Laplace term G(s) is called the transfer function associated with the model (2.1).
Suppose that r(t ) is simply a short duration of constant rainfall of a given rate r . Then we have the following Laplace
transform pair:
 1  es  
  R( s ) .
r(t )  r us (t )  us (t  )  r 

s


(2.7)
The function u s (t ) in (2.7) is called the unit step function. The Laplace transform of u s (t ) is 1/s. We have also used the
following time shift property: f (t )  F ( s)  f (t  )  e s F ( s)
Example 1 Assume that the moisture time constant is   50 min. (so that a  .02 ), that the moisture/rainfall scale
parameter b  0.1 , that the rainfall rate is r  10 cm3 / min , and that the duration  Also, assume that the initial moisture
condition is x0  .5 The model transfer function is:

G( s) 
0.1
.
s  .02
(2.8)
4
1  e  s  
R( s )  r 
.
 s 
The rainfall input is:
Moisture Response to i.c. + Rain w duration 5min.
Moisture Response to i.c. + Rain w duration 20min.
6
18
rain
forced
initial condition
total
5
14
4
3
2
rain
forced
initial condition
total
16
Moisture (units)
Moisture (units)
(2.9)
12
10
8
6
4
1
2
0
0
50
100
150
Time (min)
200
250
300
0
0
50
100
150
Time (min)
200
250
300
Figure 2. Moisture response to uniform rainfall over   5 min . (left) and   20 min. (right).
From this figure, we see that the model (2.1) is reasonably good at capturing the dynamics in Figure 1. The influence of
longer rainfall duration is evident, and is quantifiable. Such a model could serve as the basis of a ‘truth model’ to achieve
simulations that are similar to the data shown in Figure 1.
5
Matlab Code
%PROGRAM NAME: moisture.m
tau=50; a=1/tau; b=0.1;
%Model parameters
rbar=10; delt=20; %Rainfall parameters
%----------------s=tf('s');
G=b/(s+a); %Model transfer function
dt=.01; tmax=300;
tvec=0:dt:tmax-dt;tvec=tvec'; %Time array
ntmax=tmax/dt;
ndelt=delt/dt;
% FORCED RESPONSE:
r=zeros(1,ntmax);
r(1:ndelt)=rbar*ones(1,ndelt);
figure(1)
plot(tvec,r/rbar,'k','LineWidth',2)
hold on
xr=lsim(G,r,tvec);
plot(tvec,xr,'LineWidth',2)
%INITIAL CONDITION RESPONSE:
x0=1.0;
xic=x0*exp(-a*tvec);
hold on
plot(tvec,xic,'g','LineWidth',2)
x=xr+xic;
plot(tvec,x,'r','LineWidth',2)
legend('rain','forced','initial condition','total')
xlabel('Time (min)')
ylabel('Moisture (units)')
title(['Moisture Response to i.c. + Rain w duration ',num2str(delt),'min.'])
grid
X (t )  X (t 0 ) e  t /   Y ( t )
(2)
is the solution to the differential equation model
Let Δ denote the time-domain sampling interval. [In the case of the above data, Δ=30 minutes.] Then
X (k)  X (t0 ) e k /   Y (k) .
(2)
6
For notational convenience, write (2) as:
X k  X 0 k  Yk