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We have been discussing the development of certain descriptors of random processes, which
help us to model failures of randomly vibrating systems. So, in this
lecture, we will be discussing more on envelope and phase processes associated with a given
random process. Before that, we quickly recall what we have been
doing; we have solved this problem of characterizing, the number of times a level alpha is crossed
in 0 to T, by a random process X of t and this is a counter that
we setup and this lower case n (alpha, t) gives a rate or crossing of level alpha. And
when X of t is a Gaussian random process, we have been able to characterize,
the some of the lower order moments of these rates.
We also ask the question on number of peaks above a given level alpha and again we setup
a counter and were able to characterize its properties for Gaussian
random processes. Based on certain heuristics assumptions, we also derived the probability
density functions of peaks for both narrow banded and broad band
processes and this was the expression that we obtained. Here, epsilon is a bandwidth
parameter that helps us to characterize with other processes, narrow banded,
broad banded or somewhere in between, we also characterize the so-called fractional occupation
time, that is the fraction of time, that a random process spends
above a level alpha in a given duration 0 to T and we were able to derive its expected
value, for a Gaussian random process.
I also talk briefly about the notion of envelope and phase processes, during the last lecture.
So, we consider, for example, an un-damped free vibration of a single
degree freedom system and the equation of motion is x double dot plus omega square x
equal to 0 and if system start from initial conditions x naught and x naught
dot, we can write this solution as x of t is R cos omega t minus theta, where R is the
amplitude of x of t and which is the function of the initial conditions and the
natural frequency of the system. Similarly, the phase angel theta is the function of initial
conditions and the natural frequency and this R has a property, that it is
greater than or equal to modulus of x of t for all t and this is called the envelop of
x of t, in this case. Similarly, for a damped free vibration problem,
we could show, that the response can be written as e raise to minus eta omega t R cos omega
d t minus theta, where
R is again described in terms of system natural frequency damping and initial conditions and
the quantity R into e raise to minus eta omega t, can be thought of as
the envelop for this response. Now, if the same system is driven harmonically, again
we can show, that the response in steady state can be written as X s t, which is
a static response into a dynamic magnification factor into cos omega d t minus theta and
this quantity X of s t into DMF can be thought of as the envelop of
response for this system. And similarly, theta is the phase angle for this system and it
is dependent on system natural frequency damping and the driving frequency.
Now, what happens is the system is now driven by an arbitrary force f of t, can we get an
envelope representation for the response, in this case. So, we start with
the case, where the system starts from rest, that initial displacement is 0, initial velocity
is 0 and we assume that system is under damped. So, the complete solution
of this equation is given by the Duhamel integral 0 to t h of t minus tau into f of tau d tau
and the h of t minus tau is given by the first two terms here and f of tau is
an excitation. Now, what we could do is, we can expand this sin omega d t, this term and
write it as sin omega d t cos omega d tau minus cos omega d t sin omega
d tau f of tau d tau, now the integration with respect to tau; therefore, terms involve
in time can be pulled out of this and I can write this integral as A of t into sin
omega d t plus B of t into cos omega d t, where A of t and B of t are these integrals,
A of t is the 0 to t 1 by omega d e raise to minus eta omega t into cos omega d
tau d tau f of tau d tau and similarly B of t is given by this.
So, from this expression, we can proceed further and write x of t as R of t cos omega d t minus
theta of t. Here, A of t, that is, that integral just now I showed, is
written as R of t into cos theta of t and B of t is written as R of t sin theta of t.
So, R is square route of A square plus B square and theta is tan inverse B by A; so,
that would mean, even in this case, we can write the response in terms of an envelope
R of t and a phase theta of t. So, this kind of representation is quite useful in
characterizing the dynamic response and question would naturally arise, how to use such descriptions
in charactering random processes. In alternative
interpretation for the envelope can be obtain by considering R square of t as x square of
t plus x dot square of t divided by omega d square.
So, this damped naturally frequency can be approximated by the un-damped natural frequency
and we can write for omega n square k by m and you can rewrite
this as 2 by k into k x square by 2 plus m x dot square by 2. Therefore, R square of
t can be taken to be proportional to the total energy, which is sum of kinetic
energy plus potential energy. Now, whenever x dot is 0 or whenever x is maximum, R of
t passes through the maximum values of x of t, that is, R of t is x of t,
whenever x dot of t is 0; the condition x dot of t equal to 0 is the condition for x
of t to reach its extreme values, so that would mean, R of t passes through extreme
of x of t. If x of t is a sample of a narrow band process,
R of t passes through all the peaks; so, we can expect that since we have already studied
peaks, you could expect
that properties of an envelope and properties of peaks, in some sense would be similar,
but that has to be actually verified; in fact, when we characterize the
probability density function of peaks, we had used a heuristics argument which was not
mathematically rigorous, but we could obtain an expression for probability
density function of peaks which could prove useful, if it is acceptable. But doing the
course of the following discussion, we will show that by following a more
rigorous approach, we can show that R of t indeed shares some of the properties of the
PDF of peaks that we obtained heuristically.
To clarify, the notion of an envelope further, we can consider a signal x of t is A into
1 plus epsilon cos omega m of t into cos omega t; the blue line that you see
here is actually this function x of t. Now, if you look at the multiplier A into 1 plus
epsilon cos omega m t, that is shown in the red line here. This is the actually the
envelope, this is e naught of t is A into 1 plus epsilon cos omega m t which multiplies
cos omega t and this line is minus of that. So, there are pairs a pair of curves,
which actually bound the function x of t; this green line shows only this component
E of t which is A epsilon cos omega m t, this part, a into epsilon that is a green
line. So, you can see that, this is much slowly varying than the x of t itself and it bounds
the x of t; therefore, if you are interested in highest values of x of t and so
on and so forth, it may be much easier to study an envelope than the blue line.
So, now, we will pose this question, how do we generalize the notion of the envelope and
phase to describe random processes. So, again you see, this is the sample
of a narrow band process, so the envelope should pass through, you know something call,
it will release to should pass through all this peaks, that is what
intuitively we expect, but now we need to formalize this notion.
If you look at the plot of x dot of t verses x of t, a narrow band process has this type
of character, it does not fill up the entire space, it occupies, you know, certain
space which is not, if x of t is a broad band process, it will simply fill up this space;
so, this another feature that we need to bear in mind.
Now, to obtain an envelope representation for a random process, we begin with X of t,
let it be a zero mean stationary Gaussian random process and we will
represent this random process, in terms of a Fourier series as shown here. This we have
discuss in one of the earlier lecture, I am recalling what we discussed; here
a n and b n are random variables and we can assume them to be normal distributed zero
mean and say standard duration sigma n; a n and b n are mutually
independent and identically distributed; so, that properties clarified here and using these
properties, if you want, say, mean of X of t, you take expected value of
this; we know expected value of a n is 0, expected value of b n is 0; therefore, expected
value of X of t is 0.
Similarly, we can find the auto covariance expected value of X of t into X of t plus
tau and we can show that, that auto covariance is given by a function, which is
function of only tau, that would mean, the process is a wide sense stationary random
process, but since X of t is Gaussian, because a n and b n are Gaussian and
we are adding Gaussian random variables X of t also would be Gaussian and therefore,
X of t is the strong sense stationary process.
Now, I also shown in the previous lecture, that if we now start with a power spectral
density function made up of a set of Dirac delta functions and if we compute
the auto covariance of this, it has this form by using the Fourier transforms and if we
compare this form with the auto covariance of the signal, that we just now
described. We can see that, these two definitions will agree, if sigma m square is chosen to
be this; that means, for the process that we described here, this process
the power spectral density function will be of this form. So, if we are given a continuous
power spectral density function like this and if we discretize this into... If
you discrete frequencies, we can represent a Gaussian random process, in terms of a Fourier
series with random amplitudes; so, that is, the, you know result that I
will be using shortly.
Now, before I proceed, we can also notice that there is an alternative representation
slightly different from, the one that I described just now. So, to clarify that, let
us consider X of t to be a zero mean stationary random process, defined as X of t n equal
to 1 to infinite A n cos omega n t minus theta n; here, this A n are
deterministic constants, they are not random variables, the only quantity, that is random
on the right hand side are this theta 1, theta 2, theta 3, etcetera we assume
that these theta n are form an iid sequence of random variables with a common probability
distribution function, which is uniformly distributed in 0 to 2 pi.
Now, let us study the property of this random process; suppose, you are interested in mean
of X of t, you have to take expectation of X of t and to do that, if you
expand this cos omega n t minus theta n using this identity, we can show that, this expected
value is given by this expression, where the expectations of cos theta n
and sin theta n need to be evaluated and since theta n are uniformly distributed in 0 to
2 pi, these two integrals are 0, that would mean, mean of X of t is 0.
Now, following the definition of auto covariance of X of t, we find now the expected value
of X of t into X of t plus tau. So, since A n are deterministic and theta n
are iid sequence, we can manipulate this expression and show that, the auto covariance is indeed
given by A n square cos omega n tau summed from n equal to 1
to infinite; so, this process also has a similar structure of auto covariance as we studied
just, where there was summation of A n cos omega n t plus b n sin omega n
t, where A n and B n were random. Similarly, the Fourier transform of this, which will
give the power spectral density, will also be a sequence of Dirac delta
functions centered at omega n; so, the two representations at the level of auto covariance
and psd yield the same results.
So, X of t in this case, again is a wide sense stationary random process and we can show
that X of t is Gaussian, because we are adding random variables which are
identically distributed and which are independent and we can expect that X of t would be Gaussian
indeed, that would be the case and therefore, even this process,
would be a strong sense stationary process. Now, based on these definitions of X of t,
we can now introduce the notion of envelope and phase process for a
random process; this definition follows the one that is, proposed by Rice's in nineteen
forties. So, we begin by using the Fourier representation X of t is n equal to
1 to infinite a n cos omega n t plus b n sin omega n t and A n and B n are Gaussian random
variables mutually independent and omega naught is one of the basic
frequencies, that is the parameter in this model.
We rewrite this terms cos omega n t sin omega n t as cos of omega n minus omega r t plus
omega r t; similarly, b n sin omega n minus omega r t plus omega r t,
where omega r t, omega r is a central frequency, we will clarify the meaning of this in due
course. Now, I can now manipulate this expression, I can expand this
cos of omega n minus omega r t plus omega r t using cos of a plus b identity; so, I
rewrite this expression in this form. So, the first term correspond to the first terms
correspond to this and the second term corresponds to this; now, the summation is on n, therefore
terms involving omega r can be pulled outside.
So, I can rewrite this as coefficient of cos omega r t is collected in one place, that is what is contended in
this braces and coefficient of the sin omega r t is
collected in one place. so the first term inside the brace, I call it as I c of t and
the second term, I call it as I s of t right, where I c of t and I s of t are indeed this
summations as depicted here. I c of t is again a Gaussian random process, because a n and
b n are Gaussian and I s of t by the same argument is also a Gaussian
random process, having zero mean and you can show that stationary also.
Now, I c square of t, if you mean is 0 so for variance if you want to find it is expected
value of I c square of t, you can show that this is same as the variance of x
square of t, similarly I s of t is given by this and based on this we can show that variance
of I s of t is again equal to x square of t.
Now, I now introduce this substitution I c of t is a of t into cos theta of t; I s of
t a of t into sin theta of t, where a square is I c square plus I s square and theta of
t is
tan inverse I s by I c; using these notation, now I am able to write as X of t as a of t
cos omega r t plus theta of t; a of t is the envelope process associated with X of
t theta of t is the phase process associated with X of t; a of t is square route of plus
I c square plus I s square, that would mean, it is a non-linear transformation on
to Gaussian random processes, so a of t would be non-Gaussian. Similarly, theta of is a
non-linear transformation on ratio of two Gaussian random processes;
therefore, theta of t also would be non-Gaussian. Omega r is a central frequency associated
with X of t; so, this is the envelope representation for a random
process.
We will quickly consider the alternative representation that we used, where X of t a was written as
n equal to 1 to infinite A n cos omega n t minus theta n, where
A n were deterministic. Here again what I will do is, I will rewrite this as cos of
omega n minus omega r t minus theta n plus omega r t; again expand, collect terms
which multiply cos omega r t and sin omega r t and I will be able to write this as E
c of t into cos omega r t plus minus E s of t into sin omega r t; these are again
two summations E c of t is this summation, first term and E s of t is a second summation,
these are again stationary random processes, having properties quite
similar to that of X of t.
Now, if I now introduce the notation E c of t is a of t cos theta of t and E s of t is
a of t sin theta of t, I can write X of t as a of t into cos omega r t minus theta of
t,
where a of t is square root of E c square plus E s square, which is the envelope process;
theta of t is tan inverse E s of t divided by E c of t, this is the phase
process. So, using the two alternative representations, we get similar representation for the envelope;
they may differ in some details, but in essential in essential
they are quite similar.
The question now is we have defined envelope and phase processes, they are non-Gaussian,
even when X of t is Gaussian. So, we are we seem to be making the
problem complicated, the notion of envelope and phase processes would be useful, if we
can determine their probability distributions. The basic idea is that,
envelope varies lot more slowly than the parent process, therefore, describing a slowly varying
function is easier than describing a rapidly varying function; so, that
is a basic expectation, but that expectation would be met, only if we are able to determine
the requisite probability distribution functions of these two random
processes. A random processes is completely described
in terms of its joint probability distribution and density functions and unless, we are able
to say something useful
about damp. This notion of envelope and phase which essentially introduces non-linear transformation
on the parent process, likely to be not helpful, but
fortunately the problem of finding probability distribution of envelope and phase processes
is solvable, especially when X of t is a Gaussian random process and
we will see later, that this could be done even for a few non-Gaussian random processes.
So, to see that, we will start with the following definition; we introduce
two quantities A of t and B of t, which are random processes and we define X of t as A
of t cos omega t plus B of t sin omega t and further more we put A of t is R
of t cos phi of t and B of t is R of t sine phi of t, so X of t itself now can be written
in the form of R of t cos omega t plus phi of t, where R of t is square root of A
square plus B square and phi is tan inverse B by A.
By definition, we say that R of t is amplitude process envelop or amplitude modulation of
X of t, they are all synonyms; phi of t is also is called a phase process or
phase modulation of X of t; omega is known as a carrier frequency or carrier frequency
or central frequency. Now, what is the problem, now the problem is we
started with definition of A and B, suppose, we are given the joint probability distribution
function of the process A of t and B of t, can we find the joint probability
distribution function of R and phi, which is envelope and phase.
So, we know A is R cos phi and B is R sin phi; this transformation of random variables
can be handle using the rules of transformation of random variables, this is
not very complicated, so we find the Jacobean or its inverse and we show that this is J
2 power of minus 1 is 1 by J is R and consequently, we get P of R phi as r P
A B (a, b) with a and b evaluated r cos phi n r sin phi n. If A and B are jointly Gaussian,
I can right the two- dimensional probability density function, in terms of
standard deviation of A standard duration of B and correlation coefficient between A
and B and in that, is of this form ,we are taking that A and B have zero
mean.
So, now we can substitute this expression into this identity and try to get the probability
density function between joint probability density function of r and phi and
if we do that, we get this expression, which is a joint probability density function between
r and phi. If we want the marginal probability density function of r, you
have to integrate the joint density function between r and phi with respect to phi; r takes
values from 0 to infinity, phi takes values from 0 to 2 pi. Similarly, you
want marginal density function of phase, this is 0 to infinity P R phi r , phi dr and phi
varies from 0 to 2 pi; so, the problem is in, in some sense, the, at this level is
solved.
Indeed for the this particular joint probability density function, we can evaluate these integrals
and we can show that, the envelope process, the first order
probability density function has this form and the phase process has this form; it is
not uniformly distributed between 0 and to 2 pi, in that sense, is not there it is
always characterless, it has some properties. Now, this I naught is a Bessel's function
of the first kind and this distribution we called it as a generalize Rayleigh
distribution; so, r is a generalize Rayleigh random variable and phi is a generalize beta
random variable; this is a beta distribution. This I naught incidentally is the
definition of I naught is displayed here, it is integral 0 to 2 pi exponential of b
cos theta d theta is 2 pi I naught of b, where I naught of b is a modified Bessel's
function of argument b and order 0. This is a tabulated function, so you can obtain the
value of I naught of b with reasonable effort.
We can now consider a special case, where we can assume that a and b are uncorrelated
and they are identically distributed with same stand deviation. In this
case, the joint density function is given by this, because here moment r a b becomes
0, some of these terms drop of and it is possible to simplify that and we get this
expression. Now, cos square plus cos square phi plus sine square phi is 1, therefore we
really get an expression which is lot simpler than the case when r a b is not
0. And if we now find the marginally probability distribution function of r, we get Rayleigh
random variable and if we find the marginal distribution of the phase
angel, we find that the phase angle is uniformly distributed and also, we can show that, the
envelope and phase are statistically independent.
So, mind you this is for this special case, where a and b are uncorrelated and sigma a
is equal to sigma; the more general result has been obtained previously. As I
said at the beginning of the lecture, envelope and peaks share common properties, because
envelope passes through all the maximum values of X of t and
therefore, it is not surprising, that for the envelope, we obtained a Rayleigh probability
distribution function, because the same result was obtained earlier by
studying peaks and by using heuristic argument, a valid for narrow band random processes.
The heuristic approach based on which we derived the Rayleigh model
for the peak distribution, thus in a way stand justified since using a more rigorous argument;
we have arrived at Rayleigh model for the envelope, in a way, that the
ad hoc assumption that we made seen to be justified.
Another important thing that we should notice is the first order probability distribution
of properties of amplitude and phases are independent of the choice of
central frequency. We can ask slightly more involved questions, for example, can we find
the joint probability distribution function of R of t 1, R of t 2, phi of t 1
and phi of t 2, that means, R of t is a random process, can be determined the second order
probability, characteristic second order moments or second order
probability density function; so, to do that, we consider X of t, at t 1 and t 2, so I get
A of t 1 cos omega t 1 plus B of t 1 sin omega t 1 and so on and so forth. So,
the quantity A of t 1, A of t 2, B of t 1 and B of t 2 are related to R of t 1, R of
t 2, phi of t 1 and phi of t 2, through these four equations. So, we will consider this
as A 1, B 1, A 2, B 2 and R 1 phi 1, R 2 phi 2 and we rewrite this, in this from and the
here, again this is the problem of transformation of random variables, we are
considering this transformations at two time instance, there are four random variables
transform to produce four more random variables; so, we can apply the rules
of transformation, we need to evaluate the Jacobean, which is determinant of a 4 by 4
matrix and in this case, it terms out, that the some of the intermediate steps
are displayed here, 1 by j turns out to be R 1 into R 2.
Therefore, now, the joint density function between r 1, r 2, phi 1 phi 2 is obtained
in terms of joint density of a 1, b 1, a 2, b 2 using this identity r 1 r 2 is 1 by j
and
we for a 1, a 2, b 1, b 2 we have to use these transformations; so, from this, if I want
now the joint density second order probability density function of the
envelope process. This can be obtained by finding the marginal density of this joint
density, we need to integrate with respect to phi 1 and phi 2 over 2 to 2 pi;
actually, this strictly speaking, this has to be written as t 1, t 2, because r of t
is a random process and we are considering two time instants. Now, if x of t is a
Gaussian random process with zero mean and a and b become Gaussian; in fact, this integration
can be done and one can show, that the second order probability
density function is indeed given by this joint density function. These details can be worked
out, that I leave it as a matter of an exercise for you to verify. So, any
case, the basic result, is that, we are able to find out first order and second order probability
density functions of envelope and phase process, although I have
displayed here the result for the amplitude process, you can also get the second order
probability density function of phase process also by a similar exercise,
where I integrate from 0 to infinity, this quantity with respect to r 1 and r 2.
Now, in our studies on level crossing and peak etcetera, we found that, if X of t is
a random process, the number of times the level alpha is crossed depends, if you
want characterize, that we need to get the joint probability density function between
the process and its derivative; so, that leads us to the question, if I am now
interested in finding, for example, the number of times the envelope process crosses a level
alpha, then I will need the joint probability density function between
the envelope and its derivative at the same time instant; so, it is a fairly complicated
question, mind you A of t is a non-Gaussian random process. Now, however
the nature of transformations involved are not very complicated, therefore a solution
to this could be obtained and that is what I will briefly out line. So, I have A
of t is R cos phi of t, B of t is R of t sin phi of t, from this, if I now evaluate A dot
of t it will be R dot cos phi minus R phi dot sine phi of t. And similarly, B dot will
be R dot sin phi of t plus R of t phi dot cos phi of t, that mean, I am differentiating
sin phi of t and I get these terms. So, there are now four random variables which
are transformed through a set four non-linear equations leading to four new random variables
and we can now do
the address, the problem of finding the joint probability density function of p R R dot
phi phi dot, all of that evaluated at the same time instant t, this is doable, you
have to now evaluate the Jacobean, which is 1 by j, in this case, which is determinant
of a 4 by 4 matrix and we can go through this calculation; initially, it may
look quit complicated, but you should notice that, there are several zeros in this and
the expansion of determinant is lot more simple, then what it appears at the
first side and you can show that, 1 by j is indeed r square. So, I have shown some intermediate
steps to assist you in verifying this.
So, we have this now the formal solution, joint density of r r dot phi phi dot evaluated
at t is given, in terms of r square which is 1 by j p of A A dot B B dot, where
a b a dot b dot are related through these relations. So, in principle I have obtained
the four-dimensional joint probability density function between r r dot phi phi
dot; now, if I want only r joint density function of r r dot, I have to carry out a twofold
integration with respect phi and phi dot these are tedious, but do especially
if certain simplifications are made on properties of A of t and B of t. Now, if A of t and B
of t are zero mean Gaussian stationary random processes, such that A
square expected value of A square expected value of B square are equal and that we have
seen a while before, that there indeed equal for A X random process X
of t and if we further assume that, A of t, A dot of t, B of t and B dot, B dot of t are
independent and we impose a condition expected value of A dot square of t
and B dot square of t is sigma 1 square, which is not the variance of the velocity derivative
process, it is something different. We can now consider, for example, if we take
now X of t A of t cos omega t plus B of t sin omega t, X dot of t, I can write in this
form and we can actually
evaluate X dot square of t, which I need here as sigma X square is equal to sigma m square
plus omega square sigma X square, where sigma 1 square is related to,
you can show this, this is sigma 1 square, where in evaluating this, you will see that
A dot and B dot are sitting here and that is why we get sigma 1 square here.
Now, I leave it as an exercise fairly, Langley exercise, for you to verify that this fourth
order join density function is given by this; you have to verify whether
these statements are true. The range of r is 0 to infinity r dot is minus infinity to
plus infinity phi is 0 to 2 pi phi dot is minus infinity to plus infinity. You can,
after
determining this, you could find the marginal density of r r dot by integrating from 0 to
2 pi for phi, minus infinity to plus infinity for phi dot and if you do this, you
get this non-Gaussian two-dimensional probability density function; you could also determine
the second order probability density function of phi phi dot
evaluated at same time t. There is one research paper by Langley in
1996 which appeared in journal of sound and vibration, where some of these issues are
discussed in greater detail; so, if
you would like to solve this address, this exercise, attempt this exercise, I would encourage
you to go through this paper.
Now, as I said at the beginning, if we know the joint density function between process
and its time derivative and the same time instant, we can characterize the
number of crossing of level alpha by the envelope process R of t, I mean, we can characterize,
in fact, the average rate of crossing of level x i by the random
process R of t, where the crossings are taken to be with positive slopes is given by this
expression and we could use the result, that we derived just now and show
that, the this rate is indeed given by this expression; please notice that, R of t is
non-Gaussian, so this expression is not similar to what we got for a Gaussian
random process slightly different.
Now, one more thing that we should notice is when we are talking about crossing of level
alpha by a random process X of t, if a process is narrow banded, the
crossings tend to occurring clumps, what does it mean. Now, you consider this blue line,
which is you can say that is a sample of narrow band process and suppose,
you have interested in crossing of this level alpha, now you follow the crossing of this
red line by the blue curve, here there is one crossing and you see that
moment one crossing occurs, there are four crossings, after that, there is a line, you
weight here for this time and again there are crossings; similarly, here crossing
is in a clump, so this is known as clumping. Whereas, you look at the envelope here, this
crossing, the next crossing with positive slope occurs here, that means, the
time between two crossings is well separated for the envelope process than for a narrow
band process. Now, what is the signification of this result;
when we were modeling the number of times the level alpha is crossed by the propose
X of t, we proposed the use of a
Poisson random variable, that counting process be modeled as a Poisson random process was
our preposition. In Poisson model, we, the events are taken to be
independent. Now, the assumption of independents is more likely to be valid, for an envelope
than for the parent process, because for a parent process, moment
this is cross, that there, it going to be several crossings, that would mean, this crossing
and this crossing are unlikely to be stochastically independent, there is a
element of dependence, because they occur in clumps; whereas, here for the envelope,
there is no such restriction, because a time spent between two successive
crossings is longer and therefore, the assumption of independence is likely to be more acceptable
here.
Later on we will see that their implications of these features; to characterize is clumping
effect, we define what is known as clump size and that is the average
clump size is defined as the ratio of rate of crossing of level x i with positive slopes
by the parent process to the rate of crossing of the same level, by the analog
process with positive slopes and for the process, that we have been studying Gaussian random
process, the expression for this is obtained as shown here. So, as I
said, the Poisson model for number of level crossings is more appropriate for R of t than
for X of t.
Now, we now move to the description of random process by at another criteria, we consider
now the time required by X of t to reach a level alpha for the first
time. So, quickly let us recall, suppose this is X of t, the blue line is X of t, this is
X of t and the red line is the level alpha. For this trajectory, the time required for
X of t to cross alpha for the first time, we shown by this pink line; for the next realizations,
this crossing occurs fairly early and the time required for first
occurrence of crossing is much less, whereas here it takes quite a long time or in another
words, for every sample realizations, if you observe the time required for
first crossing of level alpha, you will see that those observations can be interpreted
as outcome of a random experiment and therefore, that itself is a random
variable.
So, we probably is, therefore, we introduce that as T f of alpha, that is time required
for crossing of level alpha for the first time, is a real valued random variable
taking values in 0 to infinity. Now, the question is, given the complete description of X of
t can we characterize this random variable, can we obtain its probability
distribution function or its movement or what we can do about it; this problem is known
as the problem of first passage problem, barrier crossing problem or out
crossing problem, they are all synonyms and it is a very important problem, in the study
of reliability of dynamical systems, because this time for first passage can
be interpreted as a life time of the system. So, the level alpha could be the crossing
of some prescribed stress metric at a given point and if that stress test is level
crossed, we define that as failure, so how much time the structure takes to cross that
level for the first time; so, that tells us what the life time of the structure is.
Now, we use now Poisson model for this number of crossing of level alpha by 0 to T, this
we have seen earlier. So, X of t is a random process and alpha is a level
and we assume that, the threshold level alpha is high, so that the crossing is a rare event
and crossing times are mutually independent. And under these
assumptions, we show that, we can use the model that N (alpha, 0, T) is a Poisson random
variable with this probability distribution function. The parameter
lambda here is a rate of crossing of level alpha and for a random process, this we have
already determined to be N of expected value of N (alpha, T) and we have
derived this for Gaussian random process and also for the envelop process. If X of t is
a stationary Gaussian random processes with zero mean, we have shown
that, this is the expression for this rate parameter lambda.
So, the probability distribution function for the number of times the level alpha is
crossed, that is probability that N equal to k, is given by this expression
essentially the Poisson module with alpha lambda the parameter lambda given in terms
of expected value of N (alpha, T).
Now, what is its relation to the problem of first passage times; now, if you look at probability
of first passage time being greater than or equal to t, this probability
is same as the probability, that there are no crossings of level alpha in 0 to t or in
other words, probability that crossings with positive slopes of level alpha is
actually equal to 0; if the first passage time is greater than t, this should be equal
to 0. Now, we have no postulated a model for it and this is what we get, in terms
of rate of crossing of level average rate of crossing of level alpha; from this, now
we can get the probability distribution function which is 1 minus P of t of f alpha
greater than or equal to t; mind you, that this lower case t, which appears here is actually
the state variable now; t f is a random variable, t is a state variable and
this is a distribution function which is given here. If you want the probability density
function you have to differentiate that with respect to t, this lower case t and
we get this as the model for the probability density function for the first passage time.
So, what are the parameters involved here, the level alpha the variance of
the parent process, the variance of the derivative process and of course, the time t with the
state.
This is actually nothing; this density function actually corresponds to the probability density
function of an exponential random variable. So, probability
distribution function is 1 minus e exponential minus lambda t and probability density function
is lambda into exponential minus lambda t, where lambda is this rate.
So, the expression just now I showed, this expression is essentially this written with
lambda in these places. So, T f under these hypothesis, that is a using Poisson
model for level crossings, we get exponential model for the first passage time; we could,
of course evaluate moments of this first passage time its variance and so
on and so forth, for example, the expected value of first passage time can be shown to
be given by 1 by lambda.
Now, what happens if X of t is a non-stationary zero mean Gaussian random process; so, let
us consider that, let X of t be a non-stationary zero mean Gaussian
random process with auto covariance R x x (t 1,t 2). Now, we have solved this problem,
how to find the average rate of crossing of level alpha by the process X of
t, this is the problem that we have considered before and we got this is a expression, that
we need to solve and we have shown that, this rate is given by this fairly
complicated formal. The parameter sigma x, sigma x dot and r which is the correlation
coefficient are all now time varying; so, this rate itself is time varying
because process is non-stationary.
Now, how do you, I, find the first probability distribution function of the first passage
time here; we had one minus exponential of minus lambda t, where lambda
was a constant, now, we have to replace it by an integral it is one minus exponential
0 to t, this rate into d tau. So, this is fairly involved, because parameters inside,
that are, that are present in the expression for a n X plus are function of time which
characterize the non-stationary trend of the random process X of t and that
need to be evaluated, I mean, that integration has to be performed over time, to evaluate
this probability distribution function. So, obviously this is doable, but it is
more complicated than the case of a stationary random process.
So, the problem of first passage time therefore can be tackled, if you can correct as level
crossing problem. So, when we started talking about level crossing
problem, this connection to first passage time was not very obvious, but if you now
trace back the argument, that we have used we started with level crossing
problem and for high levels of crossings, we use Poisson model and based on Poisson
model, now we are able to solve the problem of first passage times.
Now, one of the assumption that we made as I mention, this is the important assumptions
that is crossing times are mutually independent; now, we noticed, when I
discussed about clumping facts in narrow band random processes, the assumptions that crossing
times are mutually independent is unlikely to be valid for a narrow
band process, because of this occurrence of clumps, but on the other hand, for the same
process, the envelope process can be thought of as, I mean, this
assumption of crossing times being mutually independent is likely to be more valid for
envelope, because the time difference between crossings are longer and a
more random for envelope process.
So, based on that, we could look at first passage time for envelope process R of t;
so, we have actually derived this rate for a actually for a specific case; in
general, since we know the expression for the join density between the envelope and
its time derivative, in principle, this rate can be evaluated, but for under
certain simplified assumption, we have shown that, this rate is given by this; therefore,
the first passage time of this non-Gaussian random process right, this a
fairly complicated question, has been tackled and we get this model for first passage times.
This is much likely to be, much more realistic than the previous model
that we obtained here.
Now, in the next lecture, I will be considering the most important descriptor, namely the
maximum value of a random process in a given time interval 0 to T, this is
what primarily we are interested in engineers; suppose, we are considering time duration
of say 0 to 1 second and we are interested in the highest value of X of t,
this black dot shows, that number for this realization; this is for the second realization;
this is for a third realization. Clearly, for different realization of X of t, this
highest value can be thought of as an outcome of a random experiment; therefore, it is a
random variable itself and it is a continuous random variable. So, the
problem is, given the complete description of X of t, what is the probability distribution
function of this extreme value. I will show in the next lecture, that the
solution to this problem is again intimately connected to the problem of level crossings
Poisson model for level crossings for
high levels and the solution of first passage problem; all these are inter linked and this
is what we will consider in the next lecture and we will conclude this lecture
at this stage.