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So, in the previous lecture, we solved some problems covering topics in theory of probability;
we will continue with that; we will discuss some problems related to theory of random
processes. So, again there are set of problems, that I will be discussing in this lecture,
they are not particularly ordered in any sequence of increasing complexity or any such scheme,
they just go by topics. So, we will now start with the first problem, the problem is as
follows. Consider the vector random variable, Y given
by Y 1, Y 2, Y 3; it is 3 cross 1 random vector; it is given that Y is normal with mean vector
mu and correlation matrix r given as shown here; mean is 1 2 3 and correlation is matrix
is this; we now found the random process X of t is Y 1 plus Y 2 t plus Y 3 t square.
So, the problem is to find the mean autocorrelations and cross correlations of X of t and X dot
of t. Y is a 3 cross 1 vector; so mean is 3 cross 1 and correlation matrix is 3 by 3;
it is symmetric.
Now, we define a vector N of t as 1 t t square, so that X of t can be return as n transpose
Y. Thus, now expected value of X of t will be n transpose into expected value of Y, which
is 1 t t square and this is 1 2 3; this will be 1 2 t 3 t square. Now, we can differentiate
the mean and get the mean of the derivative; in this particular case, it will be B plus
2 C t, expected value of B plus 2 C t, which is 2 plus 6 t.
Now, this matches with the derivative here, 2 plus 6 t so that is find as we expect. Now,
the expected value of X of t 1 into X transpose t 2 will be N transpose t 1 Y Y transpose
N transpose t 2. So, with that, if we write now N transpose, this is 1 t 1 t 1 square
transpose and this matrix Y Y transpose and N t 2 is 1 t 2 t 2 square. So, if we carry
out this multiplication, we get this expression for R X X of (t 1, t 2) and for t 1 equal
to t 2 equal to t, we get 4 minus 2 t plus 21 t square plus 19 t 2 the power of 4.
So, now, we have expected value as given here, 1 plus 2 t plus 3 t square and correlation
function autocorrelation function is as shown here and from that, I got the mean square
value and I can now get the variance as mean square value minus square of the mean. If
I do that, I get this expression and that I have shown here just to make sure that calculations
are right one check is that, variance is positive and it is to that extend, the answers are
reasonable.
Now, this is the autocorrelation function for X of t. Now, if we want now cross correlation
between X of t 1 and X dot of t 2, we can begin by differentiating this autocorrelation
function and we get this expression minus 1 plus 9 t 1 plus 12 t 2 plus 38 t 1 square
t 2. We could check this by directly evaluating expected value of X of t 1 into X dot of t
2; X of t 1 is Y 1 plus Y 2 t 1 plus Y 3 t 1 square; X dot of t 2 is Y 2 plus 2 Y 3 t
2 and if we carry out this expectation operation, we get what we got earlier by differentiating
the autocorrelation function of the parent process.
Similarly, we can find out now the autocorrelation of the derivative process by differentiating
the autocorrelation function of X of t with respect to t 1 and t 2 and we get this function.
So, this completes the solution to the problem; this helps you to manipulate simple properties
of a random processes first and second order properties.
In the next problem, we consider a random process X of t, which is a into exponential
j omega t minus theta; a is a deterministic constant; j is square of root minus 1 complex
number; capital omega is a random variable with probability density function p omega
of omega and characteristic function phi omega of lambda; theta is a random variable that
is independent of omega and distributed uniformly in minus pi to plus pi. So, we are asked to
determine autocorrelation and power spectral density function of X of t and show that they
have certain properties.
Now, X of t is a into exponential j omega t minus theta; therefore, expected value of
this is a, is a constant. So, we get expected value of j omega t into expected value of
minus j theta, because omega and theta are independent, I can multiply these expectations
since theta is uniformly distributed between minus pi and or 0 to 2 pi, e raise to j minus
theta is cos theta minus j sin theta and average over 0 to 2 pi that would be 0 therefore,
the mean would be 0. Now, the expected values of X of t into X
conjugate t minus lambda; X of t is a complex valued random process. So, when we find autocorrelation,
we have to take the conjugation; if we do that, we get this to be a square into expected
value of j omega lambda; this is nothing but if you look at the expectation here, it is
nothing but the characteristic function of random variable capital omega. So, that would
mean R x x of lambda is a square into characteristic function of lambda omega.
Now, the Fourier transform of this if we take, that is S x omega will be the Fourier transform
of this function since this is the characteristic function of or proportional to the characteristic
function of capital omega the, Fourier transform of this will be the probability density function.
So that is what we are asked to show that, autocorrelation function is proportional to
the characteristic function of omega and power spectral density function is proportional
to probability density function of omega. We will return to a similar problem slightly
later and make some more observations on this solution.
Another exercise in manipulating simple random processes, we define a random process Y of
t as X of t plus 2 X t minus tau plus X of t minus 2 tau, where X of t is a 0 mean stationary
random process with power spectral density function; S x x omega is C divided by omega
square plus lambda square alpha square. The problem on hand is to determine the power
spectral density function of Y of t.
So, we need to evaluate the mean and auto covariance of Y of t and take the Fourier
transform. We could do it in alternate way, but this is what we are trying to do now.
So, if Y of t is X of t into 2 X t minus lambda plus X of t minus 2 lambda, the expected value
of Y would be 0, because X of t has 0 mean. Now, I can write an expression for Y of t
plus tau by replacing t by t plus tau and I get this expression. Now, if I multiply
this with this and take an expectation, we can show that we will get somewhat long looking
expression, which involves for instance, X of t into X of t plus tau is R x x of tau;
X of t into 2 X of t into 2 X t minus lambda plus tau is this so on and so forth. So, we
have to simply this, to get the required expression.
Now, we have now determined the auto correlation of Y of t, which is the auto covariance also
since mean is 0. Now, we need to find the power spectral density function now, before
we do that, we can quickly do a small calculation. If S of omega is the Fourier transform R of
tau, if you now consider Fourier transform of R of tau plus a, it will be R of tau plus
a exponential E raise to minus omega tau d tau. And if I now substitute tau plus a as
u, I will get this expression from which it follows that the Fourier transform of R of
tau plus a will be exponential i omega a into S u u of omega; S u u of omega is a Fourier
transform of R of tau.
So, the time delay or a time shift in R of tau leads to a multiplier of the kind E raise
to i omega a in the frequency domain. Now, using that, we can now write the expression
for quantities like R x x of tau minus lambda, tau plus lambda, etcetera we can now take
the Fourier transform and use the result that we just now obtained. We get the power spectral
density function of Y in terms of power spectral density of X along with these exponents and
if we rearrange that and use the definition of exponential function E raise to i theta
is cos theta plus i sin theta, we can show that the power spectral density function is
given by this expression.
A plot of this is shown here; the blue line is the power spectral density function of
X of t and this red line is the power spectral density of Y of t. A similar exercise, here
we considered two independent random processes X of t and Y of t which have 0 mean and are
jointly stationary; they are independent therefore, individually individual stationarity implies
joint stationary. Define a new process Z of t which is X of t into Y of t minus lambda,
where lambda is the deterministic constant. So, the problem is to find out the power spectral
density function of Z of t. Now, Z of t is X of t into Y of t minus lambda; so expected
value of Z of t is expected value of this product, but since X of t and Y of t are independent,
I can multiply the expectations and since X and Y have 0 mean, the mean of Z of t becomes
0. Now, we will construct the product Z of t
into Z of t plus tau so that will be X of t into Y of t minus lambda X of t plus tau
Y of t minus lambda plus tau. So, if we complete this calculation, we can show that the auto
covariance of Z of t is obtained as product of auto covariance of X of t and Y of t.
Now, we are asked to find the power spectral density function. So, multiplication in time
domain is the convolution in frequency domain. So, to obtain the power spectral density of
Z of t, we need to convolve the power spectrum power spectral density function X of t with
power spectral density of Y of t. So, this is a required answer; a slightly different
kind of problem.
In next couple of problem, we will consider statically loaded beam structures, where load
is a random field evolving in space that is a kind of problem that we will be considering.
So, to start with, we will consider a simply supported beam of span L, which carries a
distributed load f of x. The load is modeled as a segment of a stationary random process
as f of x is equal to F not into 1 plus epsilon xi of X such that this X i of X has 0 mean
and it is a white noise with unit strength. So, we are asked to find the bending moment
at the mid span and joint probability density function of reactions at the two supports;
this is the problem on hand.
So, we can begin by considering the reaction; this is the beam and this is the load; this
is notionally shown f of x is given to be a white noise, so we cannot really the white
noise is not physically realizable, but this is schematically shown here. So, we take moments
of forces about point a, I get reaction B into L is f of x d x into x, that is the moment
due to... So, we consider an elementary strip this and
this load is f of x d x. So, this is the moment; from this, I get R B as 1 by L x into f not
1 plus epsilon psi of x d x or in a slightly simplified form we get this. Now, we take
the expectation, first term is deterministic; it stays as it is; the next one is F not epsilon
by L x into expected value of X i of X; X i of X is the 0 mean therefore, the mean of
a reaction is F not L by 2.
Now, I will consider the variable F R B minus F not of L; this is another random variable,
but now it will have 0 mean and it is given by F not epsilon divided by L into 0 to L
x psi of x d x, the mean of this is 0. Now, if you find the mean square value, we have
to square this; so a single integral become double integral and if you take expectation,
we have inside these integral expected value of X i of X 1 into X i of X 2, which is given
to be white noise since psi of x is the white noise, this will be a direct delta function
and one of the integration can be done easily and this is followed by the next integration,
because this is the simple function, I get the variance as this.
So, what we have shown is reaction B is normally distributed with mean this and standard deviation
this. Now, similarly, we can show the expression for reaction A; so here we take moment about
B and the expression will be this and if you simplify and follow the steps that we are
just now outlined for finding reaction B, we will find that mean of reaction A is again
this number and standard division will be this. So, as you can expect, since structure
is symmetric, loading is symmetric, the two reactions are identically distributed.
Now, how about the correlation covariance of reaction A and reaction B? If you do that
exercise, to do that exercise, you have to take the expected value of R A minus mean
of R A into R B into minus mean of R B. So, if you multiply this, each one is single integrals
so the product will be the double integral and we get this and we can now carry out integration
with respect to X 1 and X 2. We again note that, expected value of X i of X 1 into X
i of X 2 is direct delta function. So, one integration can be done quickly; the other
integration is also straight forward, we get this as the covariance.
Since R A and R B are linear functions of X i of X, we are looking at quantities like
integrals of Gaussian random processes. So, linear operation on Gaussian random processes
preserve the Gaussian property therefore, R A and R B will be Gaussian and in fact,
they are jointly Gaussian and this is the mean and this is the covariance function.
Now, similarly, we can find out the bending moment at the mid span that is one of the
part of our question so that is given by this expression and we can manipulate this and
I have not filled up these details; I will leave it as an exercise to complete the calculations.
So, first you find the mean and then find the subtract the mean from M and find the
expected value of this square of the difference, you get the variance and that is what you
need to do by following the steps, which we have done for finding reactions at A and B.
Now, a similar problem, now we have a cantilever beam, it carries a randomly distributed load
as shown here, q of x is the randomly distributed load; the load q of x is modeled as q of x
into q not 1 plus epsilon f of x; f of x is the 0 mean that expected value of f of x 1
into f of x 2 is not a function of X 1 minus X 2, therefore, the process is not stationary;
this random field is not stationary. The question that is being asked is determine the bending
moment at a section x measured from the free end, that means, what is the bending moment
at this section.
So, bending moment at that section can be easily found out. You find out the bending
moment due to an incremental load q of psi into d psi, take moment about that section
we get this and then, integrate from 0 to x. Now, you manipulate this expression; we
get the bending expression for bending moment to be this. Now, mean of M of x is given by
this; we have we have told that the expected value of F of psi is 0 therefore, the mean
value will be q not x square by 2. Now, I will detect from M of x, q not X square
by 2 as here and square it and take expected value, I get the variance of bending moment
at x and that this is the expression that we have to deal with. Now, the auto covariance
of F of psi is given to be of this form; now you can recognize that this is as the form
of the Gaussian probability density function; so you can quickly see that we are writing
expression for expected value of Gaussian quantities therefore, evaluation can follow
simple rules.
So, following that, but limits are from 0 to X that has to be bound in mind. So, if
we do this, there are various terms; we will get these four different terms; this is straight
forward, you can check if these are right.
Then we get the expression, if you simplify, variance of bending moment is given through
this expression, where this capital psi of x is the Gaussian integral, 1 by square root
of 2 pi 0 to x e raise to minus x square by 2 d x. So, the applied load is a non-stationary
random process, we are able to get the variance of the bending moment at any point x.
So, the next problem is again on a cantilever beam. Here first go through the problem; a
cantilever beam of span L carries a series of concentrated loads. The point of application
of these loads is distributed as Poisson points on 0 to L. The magnitude of the loads is modeled
as a sequence of ii-ds with a common Rayleigh distribution with parameter sigma. So, we
are asked to determine the characteristic function of the reaction at the support.
So, what is happening is w 1, w 2, w 3, w n are Poisson points and reaction R and reaction
M are to be determined. We are focusing on reaction R; reaction R is nothing but summation
of N equal to 1 to N N of L w n; this N of L is a Poisson random variable and w n is
a ii- d sequence of Rayleigh random variables. So, we are required to find the probability
distribution of N. Now, what we are given is N of L is Poisson therefore, probability
of N of L equal to n is E raise to minus a L a L to the power of n divided by N factorial
and w that is w 1, w 2 3 w 2, w 3, etcetera form a ii- d sequence with a common density
function p w of w and that is Rayleigh. We are asked to find characteristic function
of R.
So, what is characteristic function of R? It is expected value of i omega R that is
exponential of i omega N to 1 to capital N of L w i. So, what we do is, we condition
on N of L, find a conditional expectation and then, take expectation of N of L with
respect to distribution of N of L. So, we begin by considering the situation when there
are no loads on the structure. So that is probability of N of L equal to 0 plus summation
from k to 1 to infinity exponential i omega n to n from 1 to k w i condition on N of L
equal to k and probability of N of L equal to k.
So, probability of N of L equal to k is given by this E raise to minus a L a L to the power
of k by k factorial and this is nothing but the characteristic function of the ii-d sequence
w, we need to multiply them, because they are all independent; so this become phi w
of i omega to the power of k. So, we can rearrange these terms; there is
a L to the power of k here and phi w i omega to the power of k; if we arrange these terms
together we can show that this characteristic function is nothing but exponential of a L
phi w of i omega minus 1. This phi w of i omega is the characteristic function of Rayleigh
random variable, that we can show I leave it as an exercise; it involves slightly tedious
integration. You can show that this is given by the characteristic function of Rayleigh
random variable phi is given by this; so you substitute that here and you got the solution
to the problem that is post.
So, another simple example, the problem here is, we are given auto correlation function
beta exponential minus alpha modulus t 2 minus t 1 sin gamma t 2 minus t 1. Question is,
can this be a valid auto covariance function of a 0 mean random process. It is given that
alpha, beta and gamma are positive. If you recall, the required characteristics of an
auto covariance functions is should be symmetric. So, this function is symmetric, because if
you exchange t 1 and t 2, the function remains unaltered therefore, function is symmetric
so it passes this test.
Now, R of (t, t) should be greater than 0 so that if you see here, there will be a problem,
because sin gamma of 0 is 0; so the variance becomes 0. So, since the function is not positive
definite, we can we can answer to this question is, this cannot be an auto covariance function
of a random process. Another example, where you need to manipulate a random process and
its derivative; so consider X of t to be a stationary random process with 0 mean, we
define Y of t as X of t plus a of t into X dot t minus lambda, where lambda and a of
t are deterministic. Determine the auto covariance of Y of t. The auto covariance of X of t is
given to be sigma square exponential minus alpha mod tau into 1 plus beta mod tau.
So, this a reasonably straight forward exercise, which we need to do carefully. So, Y of t
is given as X of t plus a of t into X dot t minus lambda. The expected value of Y of
t is 0, because X of t has 0 mean therefore, derivative process will also have 0 mean;
so this is 0. So, the auto covariance is given by expected value of Y of t into Y of t plus
tau. So, you need to write expression for Y of t and for Y of t plus tau and multiply
all those terms and if you carefully do that, the answer that we are looking for is R x
x of tau plus a into t plus tau R x x dot tau minus lambda and so on and so forth.
So, here the question is, we need to evaluate R x x dot, R x dot X and R x dot X dot. We
are given R x of tau so we need to use this formula to evaluate that. So, R x x of tau
is sigma square exponential of this this into this multiplication factor; if you differentiate
this, differential of modulus of tau leads to term called signum of tau, that needs to
be handle and then, if you simplify, we will get this as a first derivative.
Now, this is nothing but R x x dot, we can get that quickly. Now, we can write signum
of tau was u of tau minus u of minus tau and d u by d t of as direct delta function and
thus get the higher derivatives also and go back and substitute into this expression,
we will get the required auto covariance of Y of t.
Now, you please notice that, R y y of (t, t) plus tau, Y of t is not stationary, because
this R y y (t, t) plus tau is not a function of t alone, simply because there is a deterministic
function a of t; because of that, this Y of t is non-stationary. But auto covariance is
obtained as some kind of super position of R x x of tau, each one of is a function of
tau only, because X of t is stationary.
Another simple problem, an undamped single degree of freedom system is set into free
vibration by imparting random initial displacement and velocity. Characterize the system response.
Determine the conditions under which the response can become stationary. So, these are reasonably
simple exercise. So, the problem is x double dot plus omega square x is equal to 0; x of
0 is u; x dot of 0 is v, where u and v are random variables.
So, x of t is a cos omega t plus B sin omega t. If you use the required initial specified
initial conditions, I get x of t as u cos omega t plus v by omega sin omega t. Expected
value of x of t is expected value of u into cos omega t plus expected value of v divided
by omega into sin omega t. Since, we are not yet given the information on expected value
of u and v, this is the answer at this stage.
Now, you find auto correlation; you have to multiply u cos omega t plus v by omega sin
omega t with x of t plus tau and carry out the calculations and this is the answer that
we get. But the question that we are asked is under what conditions on u and v, the process
can become stationary? Now, if you simplify this expression, if we take if you look at
auto correlation of x of t, this is the expression here and for x of t to be stationary, this
R x x of (t, tau) should be function of tau alone.
So, for that to happen, in this expression if expected value of u v is 0 and expected
value of u square and v square by omega square are equal, we get auto correlation to be function
of time difference alone, but mean is still a function of time. So, we need to make mean
equal to 0; see mean is expected value of u into cos omega t plus this into sin omega
t. So, this the way that this can become in time invariant is that, the expected value
of u and the expected value of v should be equal to 0. So, under this condition, the
process becomes wide sense stationary.
Here, we consider an input F of t which is 0 for t less than 0 and equal to e raise to
minus 2 t for t greater than or equal to 0 to a linear system, we observe this output.
y of t is the output half e raise to minus 2 t minus e raise to minus 4 t, given this
information. The system is now excited by a Gaussian white noise excitation with unit
strength. Determine the power spectral density function of the steady state response.
So, notice that we are not given the governing differential equation here, but input output
relation in time domain is given. Now, f of t is exponential minus 2 t U of t. So, the
Fourier transform of this is, you can quickly verify, it is 1 one by 2 plus i omega. Y of
t is given as half of E raise to minus 2 t minus E raise to 4 t into the step function.
The Fourier transform Y of t can again be derived and that is shown here and based on
this, we can determine the complex frequency response function, which is ratio of Y of
omega to f of omega.
So, if we do that, I get the H of omega is 1 plus 4 1 by 4 plus i omega. Now, if the
system is now driven by white noise, the output power spectral density function will be square
of this transfer function into unit. So, the output power spectral density function is
therefore, this. Now, you can to gain a bit of inside in to this, if you consider the
dynamical system x dot plus beta x equal to E raise to minus alpha t, the starting from
rest, we can see that the solution to this problem will be in this form and if we take
now the 0 initial condition, we can get x of t is 1 minus 1 by beta minus alpha this
that would mean that the given problem the underlined dynamical system is simply x dot
plus beta x equal to some F of t. So, now, on this system, if you apply white
noise, you can show that that you will get the same transfer function that you got for
this system is 1 plus beta plus i omega and you can show that power spectral density function
would be similar to what we got by using other argument.
In this example, we consider a non-Gaussian random process. So, here we are considering
X of t to be F sin t plus phi plus Y of t; Y is P is deterministic; this is P, is deterministic,
capital phi is a random variable distributed uniformly in 0 to 2 pi and Y of t is a 0 mean
stationary Gaussian random process. You can also assume that Y of t and phi are independent.
Now, we are asked to determine the joint p d F of X of t and X dot of t. The question
is are X of t and X dot of t uncorrelated or independent? So, X of t is F sin t plus
phi plus Y of t. You take expected value phi is uniformly distributed between 0 to 2 pi;
so expected value of this would be 0; the expected value of this also would be 0; therefore,
expected value of X of t would be 0. Now, if you form the product X of t into X of t
plus tau into expectation, we get this expression, where we use the fact that phi and Y are independent
and if you simplify this, we get thus F square by 2 cos tau plus R y y of tau.
So, the process is the auto covariance function is the function of time difference; so we
can conclude that X of t is wide sense stationary. Now, how about the joint probability density
function? So, we consider X of t to be F sin t plus phi plus Y of t; X dot is F cos t plus
phi plus Y dot of t. First what we will do is, we will find out the joint density function
condition on phi. So, this will be P y y dot Y comma Y dot, where Y is X minus F sin t
plus phi Y dot is X dot minus F cos t plus phi; we are finding conditional probability
density function. Now, conditioned on phi, X of t is a Gaussian
random process, because Y of t is Gaussian and it will have now the mean which is X minus
F sin t plus phi; so if you mean will be sin t plus phi. So, if you write this Gaussian
density function, this will be the conditional density function. Now, the unconditional density
function, we have to do carry out integration with respect to the probability distribution
of phi that is as shown here.
Now, if you make this substitutions, you can carry out this; we can show that P x of x
dot, we can square this and take out terms, which are free from psi and look at the terms
which contain only this sin sin cos psi; you can see that this integral is nothing but
Bessel's function i naught and this has this form, this is first term and this is second
term. So, it is clear that x, x of t is a non-Gaussian random process and if you look
at the marginal density again, the same logic can be used; probability density function
of x condition on phi is this and P x of x is the unconditional density function will
be will involve another quadrature with respect to phi and if you carry out this again, you
will get answers in terms of Bessel's function. If you multiply the marginal density, you
can verify that you would not get the joint density function.
So, from this, what is the conclusion that we can draw? X of t and X dot of t are uncorrelated,
because they are stationary random process processes and X of t and X dot of t are not
independent. So, it is an example of non-Gaussian random processes, where process and it is
derivative are uncorrelated, but they are not independent.
Now, while discussing random variable, we talked about what is known as Cheybychev inequality.
Now, what happens to that inequality when we extend the logic to random processes? So,
let X of t be a random process with mean mu X of t and variance sigma X square of t; now
the inequality that we are asked to show is probability of modulus of X of t minus mu
X greater than or equal to epsilon for some t in a to b is less than or equal to a quantity
which involves the variance of X evaluated at a and b and an integral over a to b.
So, this is the generalization of cheybychev inequality and this can be proven. I have
indicated the steps; I would not like to go through this; you can take a look reasonably
straight forward. Now, let X of t be a stationary random process with 0 mean and auto covariance,
which has the form of a Gaussian probability density function. The question is how many
times can we differentiate this process? And we are asked to determine probability of X
dot of t is less than or equal to 0.75, if it is given that the process is Gaussian and
sigma is 1.
Now, R x x of tau is given to be this; from this, if I take the Fourier transform, I get
the power spectral density function to be this. Now, if you consider the spectral moments,
we can show that the since this exponent is decaying as minus omega square, all these
integrals will exist for n equal to 1, 2, n, etcetera. In spectral moments, that 0 to
4, 6, etcetera are nothing but auto covariance of the parent process and there derivatives
evaluated at tau equal to 0 and since all these are finite, it follows that X of t is
differentiable in the mean square sense to any order n; so it is a fairly smooth process.
Now, we can differentiate the auto covariance function and we can evaluate the various quantities
at X equal to 0. So, R x x of 0 is 1; from this, it follows. So, if you differentiate
these twice, you get R x x dot and if you put tau equal to 0, we get 1 by square root
of 2 pi. And we can now write therefore, p X dot of X dot is given by this and from this,
we are asked to find out probability density probability of X dot of t is less than or
equal to 0.75, which is an integral minus infinity to 0.75, this quantity this is given;
so the answer that we are looking for is this number 0.97.
This is an interesting problem; here, we consider a random process N of t to be a Poisson random
process with arrival rate lambda. Now, I define a new random process X of t which is minus
1 to the power of N of t. Now, we are asked to find a mean and covariance of X of t; X
of t in the literature is known as semi random telegraph signal. So, X of t obviously, N
of t is an integer value random process. So, whenever N of t is even, X of t will be 1
and N of t is odd, X of t will be minus 1. So, sample will look like this, depending
on even odd combinations of n of t.
So, X of t is minus 1 to the power of N of t. So, X of t is 1, if N of t equal to 0 or
N of t is even; it is minus 1, if N of t is odd. So, what is probability that X of t equal
to 1? It is nothing but probability of N of t is 0 or N of t is even and that probability
since it is Poisson distributed, I can evaluate and this I get to be that expression and if
you carefully look at this series inside the bracket, we recognize that this is cosine
hyperbolic term. Now, how about the case, where X of t is minus
1? We have to sum over all integers- odd values of n- and if you do that, I get E raise to
minus lambda t sin h lambda t. Therefore, the expected value of X of t would be X of
t takes only two values, with one probability is this; other probability is this. So, probability
of X of t equal to 1 into 1 plus probability of X of t equal to minus 1 into minus 1. So,
if you do that, we get the answer as 2 into E raise to minus 2 lambda t; so this is the
expected value of X of t.
Now, what is correlation function auto correlation? Again X of t takes two values- plus 1, minus
1; therefore, the product X of t into X of t plus tau is 1, if there are even no of occurrences
in t to t plus tau; otherwise, it is minus 1. So, based on this argument, we can write
the expectation of X of t into X of t plus tau and we can show that the auto correlation
function is the function of modulus of tau. So, this process is wide sense stationary,
that is, if I remove the mean, that is, if I define another process, where X of t minus
2 E raise to minus lambda t, if I consider that process, that will have 0 mean and it
is wide sense stationary. I will return to some comments on this example shortly.
We will consider another problem; here, we consider a random walk that is performed on
a two-dimensional plane with a uniform step size of delta. At every step, the direction
alpha i is a random variable and alpha i alpha i's taken to be iid sequence with a common
p d F that is uniformly distributed in 0 to 2 pi. Find the distribution of the x- coordinate
after n steps. That means, we start here, we draw a random variable to be distributed
uniformly between 0 to 2 pi and this is our realization. So, we take one step, then next,
this step, this step, etcetera so after n steps, I am here and I am asking what is this
X?
So, X is nothing but i equal to 1 to n delta cos alpha i, where alpha i are the sequence
of random variable that we have generated. So, what is the characteristic function of
this? It is exponential expected value of exponential i omega X and it is this and since
alpha i's are all independent, I can get this. And we can evaluate this expected value of
i omega delta cos alpha i and given that alpha i alpha i is uniformly distributed between
0 to 2 pi, I get this and this integral is nothing but Bessel's function j naught.
So, phi X of omega which is a characteristic function of this random variable X is actually
the j naught of omega delta to the power of n. Now, we have to invert it to get the probability
density function and if you do this, we get this integral; this is the integral. And now
if you do a series expansion for j naught to the power of n and consider the limit of
n tend into infinity, number of steps going to infinity such that delta of square root
of n goes to C that is a kind of limit that we used for taking a one-dimensional random
walk to a Brownian motion process in the same sense, this is done; if you do that, we can
show that this process is this resulting random variable is Gaussian.
So, this is an illustration of application of central limit theorem. So, you get the
limiting density function to be Gaussian. If that n tend to infinity is not reached,
the answer that we are looking for is given by this integral.
So, another example which will help you to manipulate simple random processes.; let the
time interval 0 to t be divided into a sequence of equal intervals of length T, that means,
I am considering some n T. Consider a sequence of n Bernoulli trails, that means, at every
step probability of success is half. Now, I define X of t equal to 1, if success if
we observe success on nth trails; if it is minus 1 it is minus 1, if we observe a failure
so, where t is between n minus 1 capital T into N t.
So, the problem is find mean and auto correlation of X of t. This is the second bit to this
problem, where we define another random variable e, which is distributed uniformly in 0 to
capital T and independent of X of t. We define Y of t as X of t minus e. So, the problem
is again determine the mean and auto correlation of Y of t.
This is the parallely simple exercise that needs to be carefully done. So, X of t is
1 if success on n th; trail X of t is minus 1 if failure on n th trail. So, X of expected
value of X of t is therefore, 1 into half minus 1 into half which is 0; the mean square
value is again we can easily find out, it is this.
Now, the product X of t 1 into X of t 2 is 1, if both t 1 and t 2 are contained in this
interval n minus 1 capital T into 2 n t otherwise, it is 0. So, that completes the solution to
this problem. Now, we introduce the this epsilon that is the random variable epsilon; now again
we can find expected value on auto correlation of Y of t. We can first find the expected
value condition on epsilon and then, integrate with respect to distribution of epsilon. If
we do that, we can show I will leave this as an exercise that the auto covariance of
correlation of X of t in this case that is Y of t in this case will be given by this
function.
So that is how it looks like here; it is a triangle over the step minus t 2 two steps
minus t 0 to t and it is 0 elsewhere. Now, this is a slightly question related to existence
of random processes. So, we are given a positive function S of omega and we are asked to find
a stochastic process whose power spectral density S of omega.
Now, there are two solutions; I will start with the second solution what I will do is
I will determine a square, which is area under S of omega and define a function F of omega,
which is S of omega by a square. So, clearly area under F of omega is 1 and it is positive.
We assume that S of omega is equal to S of minus omega; so consequently what happen F
of omega is F of minus omega. So that would mean based on these two properties, we can
conclude that F of omega has the properties of a probability density function of a random
variable. Now, I define a random process X of t as a
cos omega t plus 5, where omega and phi are random variables with omega having probability
density function F of omega and phi is uniformly distributed between 0 to 2 pi and omega is
independent of phi. So, you consider now expected value of X of t, we can show that it is 0
you can start by you can find the mean condition on omega and then, integrate with respect
to omega condition on omega this mean is 0 therefore the answer is 0.
The auto correlation function can be evaluated; this is reasonably straight forward and you
can show that the auto correlation function is a square into F of omega, which is what
we are expecting. that is Auto correlation is this the power spectral density function
is this which is what we are expecting.
Now, interestingly how does a sample of this function look like? They are If we assume
say this as 1 they all are harmonics, where frequency and phase are randomly distributed,
but the power spectral density function can be specified by the is a part of the specification
of the problem.
By that what I mean is if power spectral density function is of the form say alpha square plus
omega square. If i now assume that this is the output of alpha X equal to psi of t, the
power spectral density can be shown to be similar to this and samples of this will look
like this. But following the logic that we outlined here, X of t is a cos omega t plus
phi, this also has the same power spectral density. So, these two samples come from two
different random processes, whose power spectral density functions are the same.
So, this would be question us to the fact that power spectral density function is an
ensemble property. The two time histories that we see here represent samples from two
different processes having the same mean and power spectral density function.
Just to summarize, we saw that one example earlier X of t is minus 1 to the power of
N of t, where samples were like square waves. So, in this if we detect that mean, this quantity
had power spectral density function or auto covariance function, which is same as the
power spectral density of this; the power spectral density of response of this system.
But in this case, the sample looks something like this; in the first case, it is a square
wave pulse; in this case, it is harmonic.
So, while interpreting power spectral density function in modeling, you have to take cognisense
of this fact. At the end of the day, it is a not a sample property; it is an ensemble
property. So, this is what I am showing here; this is harmonic function; this is an erratic
function and this is a square wave function. It is intuitively not clear; if i say that
these three samples come from three different processes having the same power spectral density
function.
Now, there are few properties of random processes with independent increments, that I will take
up in the following lecture and that discussion will be followed by some which problems on
first persist time and extreme value distribution and so on and so forth and then, we will consider
some discussions on the notion of say factor of safety and probability of failure.
The notion of factor of safety essentially originates from deterministic outlook, whereas
probability of failure when we talk about how we are modeling uncertainty using theory
of probability random variables and random processes so how they are related. So that
is one of the questions, will briefly consider the philosophical issue there and then, we
will consider some more problems in response of dynamical systems to random excitations;
specifically, we will consider excitations which are non-Gaussian in nature and systems
which are non-linear and we will apply in some problems the theory of Markova processes
and derive the response moments and all these will consider in the following lecture.
So, we will close the present lecture at this junction.