. So, towards the end of wavelet us, we have
to understand uh the following that there
is a resolution that is happening at different
scales right and there is a resolution in
time. And what about in the frequency is a
resolution the same in frequency right, I
mean imagine I have have a short burst of
a pulse right. And I can measure the the burst
or I can measure some event in the a of the
signal in and the variance of the signal in
in in in time and some statistical properties
of the signal in time, and can I have the
same sort of resolution that I can get in
the frequency domain. So, this is one of the
questions that one would naturally think about.
So, fortunately for us somebody did investigation
on this time frequency uncertainty principle.
This follows a kin to the the position momentum
uncertainty Heisenberg's uncertainty principle
that we are aware in quantum mechanics right.
And the same sort of idea is exists actually
in in in when we look at time frequency uncertainty
in signal processing. And there is a more
deeper theory using operators in mathematics
and will not get there ok.
So, let us begin consider 
a finite 
energy signal right. If it is a finite energy
signal which means integral minus infinity
to plus infinity, I mean you can put any range
for this, I mean this is very general, I mean
you really will not ideally reach minus infinity
and plus infinity, but this is just for pedagogical
purposes and if you see some range or you
just have to put the appropriate range here
in your definite integral. So, this energy
is strictly less than infinity .
Now, let us assume that the signal is centered
around 0 I would say precisely at 0 both in
time and frequency, that is it is 0 mean if
it is not 0 mean you can always subtract the
bias and get it to 0 mean ok. Now, when we
think about signals we discussed in the very
beginning of module 1, when we looked into
signal geometry, that we can think about different
kinds of averaging statistics 1 is just a
normal time average youre given the samples
you compute all these statistics, whether
it is mean or the variance or second moment
third moment so on and so forth right. You
can compute all the moments across the signal
or you could do the statistical average right
statistical average. So, let us look at the
time average time average ok.
Now, let us compute 
the variance in time and frequency 
by usual 
time averaging ok.
Now since we said we are dealing with 0 mean
we can compute the variance in time sigma
t square, as integral minus infinity to plus
infinity t square mod s of t square dt and
there is a normalization factor here, which
is the square of the l 2 norm of the signal
right. And why should we do this is a question
that you might want to ask why should I normalize?
And, if we treat the signal as a random signal
there should be some distribution over which
we have to weigh it right. So, if you take
if you fold this norm inside right. So, if
you can you can you can basically rewrite
this as essentially integral minus infinity
to plus infinity t square mod of s of t square
dt divided by the l 2 norm . Now can you appreciate
that this is like 
like your density right this is like your
like your PDF probability density function,
because when you integrate this this this
is basically this integrates to 1 right. So,
therefore, you can compute a moment a second
moment like this ok.
Now, we will do this in frequency computing
the variance in the frequency domain, we get
sigma omega square is integral minus infinity
to plus infinity omega square modulus of s
of omega square d omega divided by the norm
l 2 norm over the spectrum right. Again you
can interpret this as a density like what
we did earlier right there is no big deal.
So, this is also like you can interpret this
as ok.
So, now, I think we are sort of set with our
with our with our problem. Now, you can also
think about in the notion of quantum mechanics,
you can think about this as position this
is momentum and you can think about these
operate or these quantities and you are also
looking at the variance here right or some
other 2 set of quantities.
Now, by Parseval's theorem from your undergraduate
norm s of t squared l 2 norm of s of t is
the same as s of omega square right. Because,
if you apply the Fourier transform you are
not going to alter the energy in the signal,
otherwise it would be the transformation is
useless if if you lose the energy right, because
of energy conservation property . So, notation
wise this is easy for me I can just write
it like this, some norm of s square I will
get rid of t and omega I am saying norm of
s square because it is the same whether it
is time or frequency ok.
Now, let us consider 
the following product, which is the variance
in time and the variance in frequency I consider
this product . This is essentially integral
minus infinity plus infinity t square mod
s of t square dt times. Now, we can interpret
the omega times modulus of s of omega as 
this quantity, because from duality if you
take d by dt of s of t in the frequency domain,
it is you are multiplying by g omega right
and the modulus will pull the magnitude of
j to be one right and you have omega square
s of omega square modulus of s of omega square
right.
So, this is an important step divided by power
4, because you have a norm s square for this
term and another norm s square for this term.
So, you have a power 4 in the denominator
and in this quantity it is basically due to
Fourier transform property right take the
Fourier, it is j omega right. So, you have
this additional jmo this j omega quantity
which justifies what we have.
So, now, consider integral minus infinity
plus infinity modulus of I will say t square
times s of t square dt consider this this
can be written I can fold the t inside the
modulus there is no problem with it. So, I
can write this in this form modulus t times
s of t square dt. And you can interpret this
quantity as the l 2 norm square of t time's
s of t ok. See how we are kind of building
the the logic right from starting from the
variance to interpreting this into the norm.
And, similarly 
integral minus infinity to plus infinity modulus
of d by dt s of t 
square dt can be interpreted as the l 2 norm
of the derivative of the signal right, l 2
norm square of the derivative of the signal
. Now, we have a product of 2 norms. So, your
intuition should tell that there is Cauchy
Schwartz somewhere kicking in ok. So, let
us apply Cauchy Schwartz inequality .
So, the reason states that if you look at
the norm of 2 functions, that are possibly
complex then the modulus of the in the product
square is basically less than or equal to
the norm square of the product of the individuals
ok.
Now, using this we can say and say this is
relationship a using A, we can write sigma
t square times sigma omega square as a quantity,
which is greater than or equal to because
now it is existing in product form exactly.
So, therefore, the inequality is this way
right one over norm of s l 2 norm of s power
4 times I bring the modulus integral minus
infinity to plus infinity t times s of t times
there is one careful observation, that you
have to make where I am bringing in the conjugate
of this signal 
under the hermitian inner product form .
Now, this quantity that we have modulus of
some integral square so, this is this can
be a complex quantity. So, we can simplify
this little carefully. So, this is one over
the norm of s power 4 times I can write the
absolute value of the real part of integral
t times s of t times d by dt of s of t bar
dt square why? Because, you know for complex
numbers modulus of real of z is certainly
less than or equal 2 modulus of z right. This
is a straight forward result.
So, you take just for a cross check 3 plus
4 j is that vector the length is 5, then the
real is 3 and when the complex part is 0 then
it coincides with equality. So, you you have
this relationship .
Now, real part of some complex quantity z
can be written as one half z plus z conjugate,
this is a straight forward result . Now consider
the real part of this integral minus infinity
to plus infinity t times s of t times derivative
of the signal conjugate dt right . I can write
this as 1 half I will pull the t factor outside
because it is just a scalar it is time variable
it is a scalar right .
So, we can write it as s of t times derivative
of s bar of t plus s bar of t derivative of
s of t right using this this room using this
step one half of of z plus z conjugate I just
did that yeah right. So, I just take this
as my z here I just conjugate it 
I will know I will come to the point I am
just yeah I need the integration there yes
a real part . So, I just have t times I just
I am looking at this term here I am just rewriting
this thing here and I just have to put if
you are ok with I am just focusing on this
term just the argument of the of the integral
ok .
Now, let us focus on the term within the integral
let us focus on the term within the integral
. So, this is basically I would say term 
within the integral is basically half t times
d by dt of modulus of s of t square why? Because,
you can write modulus of s of t is s of t
is power of t now you take the derivative
using chain rule and it automatically comes
ok.
So, therefore, sigma t square times sigma
omega square is greater than or equal to now
I have to square this quantity right if it
is one-fourth, I have a norm s power 4 and
then have an absolute value here then this
integral minus infinity to plus infinity t
times d by dt of modulus of s of t square
dt .
Now, still it looks complicated let us consider
integral minus infinity to plus infinity t
times derivative of modulus of s of t square
dt and perform integration by parts right.
We have the familiar this is what I recall
from my high school I late this is a inverse
function logarithmic. So, on and. So, forth
right you you use that last is exponential.
So, you apply this rule integration by parts
.
So, what we get is t times s of t square minus
infinity to plus infinity minus integral,
because I take the derivative I integrate
that I get this and then I take a differentiate
time, then I get one there right first function
integral or second function minus integral
of second function derivative of the first
function ok.
Now, this quantity goes to 0 and this is basically
interpretable as negative norm of s square
. Therefore, sigma t square times sigma omega
square is greater than or equal to 1 upon
4 times minus norm s square and if you square
this quantity because I have that modulus
of that integral square is what I have. So,
this is basically just 1 by 4, because norm
s power 4 and norm s power 4 cancel in the
numerator and denominator right. It is a very
interesting result, I mean it is a very interesting
result that you look at the variance in time
and the variance in frequency and and that
is lower bounded by 1 upon 4 take any signal
take any signal any s of t.
It is at least this much that you will live
with your uncertainty in your measurement
in time and your measurement in in frequency,
it is a very very profound ah profound result
and this result was proved by no less than
the Nobel prize winner Dennis Gabor who is
the father of holography. So, let us say the
roots this this is originally proved by Dennis
Gabor in 1946 and of course, Dennis Gabor
was great in many areas including in signal
processing the signal processing result, but
of course, he is a Father of Holography .
So, now we have a very important principle
uncertainty inequality or uncertainty principle
in signal processing, that we cannot simultaneously
localize in time and frequency no matter what?
So, a follow up question arises when does
this equation when when is the inequality
is satisfied with equality or when is the
equality satisfied, when can we get this actual
limit of 1 by 4.
So, for this we asked the next question when
can we achieve equality that is sigma t square,
sigma w square, sigma omega square, equals
one-fourth we need to I need to get to this
lower bound . Now fortunately we have a solution
from Cauchy Schwartz again that we are comfortable,
we can say it happens when t times s of t
is some constant k times derivative of s of
t and I assume at the moment I just will remove
the conjugation here and I will just assume
s of t assuming real signals.
Now the k is this factor right I mean when
this k times the derivative of s of t, then
this is equal the inequality becomes a equality
right. Now, we have a differential equation
here. So, let us 
group the terms and integrate both sides . So,
how we do this we say derivative of s of t
by s of t is t upon k times dt and then I
need to do an integration on both the sides
ok
So, if I integrate 
I get log s of t is t square upon 2 times
k plus some constant of integration .
Now, s of t is of the form e power t square
upon 2 k plus some constant, which is if you
write it carefully it is e power times c times
e power t square upon 2 k let me call this
as some constant a times e power t square
upon 2 k well we have done the integration,
but let us see if things are making physical
sense for us right.
So, if you recall in the very beginning I
said let us consider finite energy signals.
Now I have a solution s of t is of this one
a times e power t square by 2 k and if you
if you let k to be a positive constant the
energy in s of t would be infinite potentially
it could be infinite right. So, therefore,
if s of t is to be a finite energy signal,
then k must be a negative real number ok.
So, therefore, let b equals minus k right
let b is some negative k. So, if I did that
then s of t is some a times e power minus
t square by 2 b .
Now, this is easy for us to think about because
this is like has the form 
of a Gaussian pulse, you are not making it
a distribution I mean if you if you want if
you can choose a to be one upon root of 2
pi b, then it becomes like a distribution
t degrades to one otherwise this is just a
Gaussian pulse .
Now, this basically led to Gabor transformations.
So, this is form this form led to Gabor transforms
I mean I am not going to deal with Gabor Gabor
transforms, I am not going to deal with Gabor's
Gabor's transforms, if we were to go people
to go into into more details of time frequency
analysis we would start possibly with this
as our first step and then go into the details
ok.
So, but and you can also appreciate ah why
you think about the Fourier transform of a
Gaussian pulse why this has the same, shape?
And Gaussian has very interesting consequences
one of this is this uncertainty relationship,
the Gaussian PDF has the maximum uncertainty.
If you if you look at the differential entropy
right and evaluate this for the Gaussian PDF
it gives you the maximum uncertainty right.
And that is the reasons why you look people
look into Gaussian PDFs you know when they
are in their study of signals with noise and
particularly noise having Gaussian PDF. And
Gaussian PDF again comes from your central
limit theorem results.
So, somehow whether it is consequential or
it just nature is providing you that way it
is just there there are a lot of very interesting
properties with Gaussian, whether it is distribution
or a pulse or a transform or whatever it is
and you see this again and again in signal
processing and information theory ok.
So, we will so, we will we will stop at this
stage I think this is something which you
have to really understand. And, before we
conclude I would sort of give you a homework
problem to ponder ponder on how 
the time frequency 
uncertainty principle applies, in the context
of wavelet decomposition 
at different scales .
So, we have studied the wavelet decomposition
and reconstruction. So, at every stage of
wavelet decomposition we have a certain resolution
right, it is a time resolution. Now think
about if this transformation can meet this
this bound or how close or how far away is
it from the uncertainty principle, if you
were to do localization through wavelet us
ok this is something to think about this may
or may not appear on the exam, but this is
a hint for you to think through this homework
exercise. So, we will stop here and we can
go to questions ok.
