 Welcome to another module in this massive
open online course So we are looking at the Hilbert transform for the generation of (oyo)
single sideband modulated signals and we also looked at a frequency domain description of
the Hilbert transform alright So now let us look at the time domain description
that is the impulse response of the Hilbert transform So we are looking at the impulse
response of the (Hemb) Hilbert transform the impulse response of the Hilbert transform
and we have denoted this by hHT of t this is the impulse response of the Hilbert transform
alright We are denoting it by hHT of t to start with
let us recall that the frequency response of the Hilbert transform is minus J sgn F
which looks something like this remember for convenience I can always denote it by this
it’s j for f less than 0 minus j for F greater than 0 and F equal to 0 this is 0 So this
is your minus j sgn F to derive the impulse response of the Hilbert transform we already
said that we are going to use that derivative property of the Fourier transform alright
And I am going to talk about that shortly But first let us look at the derivative of
the sgn function let us go back looking at sgn of t okay So this is your sgn of t it
is one for the greater than 0 minus one for t less than 0 and t equal to 0 it is 0So sgn
of t we already know sgn of t equals one for t greater than 0 0 t equal to 0 minus t for
t less than 0 And now you can see the derivative of sgn
of t now here in this region sgn of t that is for t less than 0 sgn of t is constant
So in this region d by dt of sgn of t equal to 0 for t less than 0 right You can see it
is flat For t less than 0 sgn of t is minus 1 So d by dt the derivative is 0 because it
is a constant for t less than zero Similarly for t greater than 0 sgn of t is
again a constant it is equal to one so therefore the derivative is again 0 so in this region
also d by dt sgn of t equal to 0 for t strictly greater than 0 alright So for t both t strictly
less than 0 and t strictly greater than 0 the derivative or sgn of t is 0 because sgn
of t is constant in both these regions The only place where the derivative of the sgn
sgn function that is sgn of t is nonzero is at the point t equal to 0 where it is transitioning
from minus 1 to minus it takes a step jump from minus 1 to 1 and there the derivative
you can see since the magnitude is changing from minus 1 to 1 that is magnitude is changing
by 2 the derivative is 2 delta t alright So in this position at t equal to 0 at t equal
to 0 there is a step change of magnitude equal to 2 therefore d by dt of sgn t equal 2delta
t at t equal to at t equal to 0 Therefore the derivative of this function so you can
see it is constant step change at t equal to 0 and again constant for t greater than
or equal to 0 for t greater than 0 therefore the derivative is basically twice delta t
alright Since the change occurs only at so t equal to 0 So therefore d by dt of sgn t equals two delta
t which is nonzero only for t equal to zero okay which implies now let us start with the
Fourier transform of the derivative I will show why this is convenient to find the Fourier
transform the original function Fourier transform of d by dt of Fourier transform of d by dt
of sgn t d by dt of sgn t is 2delta t therefore the
Fourier transform of d by dt sgn t equals the Fourier transform of 2delta t but the
Fourier transform of 2delta t is simply 2 because of Fourier transform or we can write
one more step that is twice the Fourier transform of delta t which is equal to two right Fourier
transform of Delta t is simply one over the entire frequency domain However we also have a result which states
that the derivative of that is the Fourier transform of the derivative of a function
that is if consider this is the derivative property of the Fourier transform states that if a
signal x(t) has Fourier transform X(F) than the derivative of x(t) that is if I look at
the derivative of x(t) the Fourier transform of the derivative of x(t) is j 2pi F times
X(F) The Fourier transform of the derivative of a signal x(t) is j 2pi F X(F) where X(F)
is a Fourier transform of x(t) And using this property the Fourier transform
of the derivative of sgn t therefore the Fourier transform of the derivative that is the FT
of the derivative of sgn t that is j 2pi Fourier transform of sgn t however we have already
seen that the derivative of the Fourier transform of sgn t from here We have seen that the derivative
of the Fourier transform of sgn t is 2 alright and therefore this is also equal to from above
this is also equal to this result we have from above implies that these 2 must be equal So this implies we must have j2piF Fourier
transform of sgn t equals 2 alright We have already shown that the derivative of the Fourier
(trans) that uhh that the derivative that is the derivative of uhh sgn t is 2delta t
therefore the Fourier transform of the derivative of sgn t is 2 and we also it is all from the
derivative property of the Fourier transform we are saying that the derivative of sgn t
has a Fourier transform that is j 2pif times uhh the (fou) the Fourier transform of sgn
t therefore the issue quantities must equal be equal which implies j 2piF Fourier transform
of sgn t equals 2 which implies that Fourier transform of sgn t equals 2 by j 2piF which
is basically equal to 1 by jpiF therefore sgn t Fourier transform sgn t has the Fourier
transform 1 by j 1 over j pi F this is the property that we have been able to sgn t has
the (pou) Fourier transform 1 over jpiF alright And now let us use (pru) duality let us now
use duality and using duality remember duality states
that if x(t) has duality property of Fourier transform which states that if x(t) has Fourier
transform X(F) than x(t) has Fourier transform x of minus F using the duality property we
have sgn t has Fourier transform 1 by j phi So using
the duality property of sgn t has Fourier transform 1 by jpiF 1 by jpit has Fourier
transform x of minus F that is sgn of minus F Now remember sgn function is an odd function
right Because it is 1 for t (grea) sgn t is 1 for t greater than 0 minus 1 for t less
than 00 at equal to 0 this is an odd function So sgn of minus F is minus j sgn F This is
sgn in the frequency domain by the way so this is minus which is exactly what we want
for the Hilbert transform this is minus j sgn F this implies that 1 by j pit has Fourier
transform minus sgn I am sorry not minus sgn of minus F is minus minus sgn F so 1 by jpit
has Fourier transform minus sgn F which implies Now taking the j to the other side 1 by pit
has the Fourier transform minus j sgn F 1 by pit has the Fourier transform of minus
j sgn F now minus j sgn F is nothing but recall that minus j sgn F is nothing but the impulse
response of the Hilbert transform HHt of F which implies that 1 by pit which is the inverse
Fourier transform of this must be nothing but the impulse response of the Hilbert transform
therefore the impulse response of the Hilbert transformer is 1 by pi And that is what we have derived H of HT of
t equals 1 by pi 1 by equals 1 by pi t okay So that is a property
that we have been able to show that the impulse response of the that is if remember the Hilbert
transform is a linear time invariant system whose Fourier transform is given by minus
j sgn F In the time domain its impulse response is 1 by pi t therefore now if you look at
the Hilbert transform remember the Hilbert transform is an LTI system correct This is your Hilbert transform
is an LTi system if you pass a signal m(t) outcomes the Hilbert transform m hat of t
this is your signal this is its Hilbert transform of m(t) of m(t)
and now this system has uhh (impu) Fourier transform we have already seen this equals
minus j sgn F in fact this is a phase shifter correct Remember this is a phase shifter that
is it shifts the phase of all the positive frequency components by minus pi by two all
the negative frequency components by pi by two This is a phase shifter it has (impu)
frequency spectrum the Fourier transform is minus j sgn F and the frequency And now the impulse response we have derived
the impulse response hHT of t equals one by pi t which implies in the time domain the
Hilbert transform can be represented equivalently as m hat of t equals the input signal remember
when you pass input signal through LTi system signal gets convolved with the impulse response
so m(t) convolved with 1 by pit This is the time domain representation of
the Hilbert transform this is the time domain
representation of the Hilbert transform and this is remember convolution this is a time
domain representation therefore it is a convolution in time domain
That is convolution of the input signal with the impulse response Now frequency domain representation that is
the output Fourier transform remember convolution in time domain is multiplication in the frequency
domain so this is MF into H HT of F which is basically your MF times minus j a product
correct This is a product or you can write this as minus j minus j sgn F times okay So this is the product in the frequency
domain So this is the frequency domain representation of HT Hilbert transform and this is a multiplication in the frequency domain alright So in this module we have developed the time
domain representation of the impulse of the Hilbert transform that is we have calculated
the impulse response of the Hilbert transformer using 2 properties one is the derivative property
of the Fourier transform and the duality property of the Fourier transform to demonstrate that
the impulse response of the Hilbert transform is 1 over pi t which corresponds to the spectrum
of the Fourier transform of the Hilbert (tress) transform which is given by or the spectrum
of the Hilbert transformer which is given by minus j sgn F alright And therefore Hilbert transform can be represented
in the time domain also equivalently by the convolution m hat of t which is a Hilbert
transform of m(t) is given by the convolution of m(t) with 1 over pi t However as you can
see since 1 over pi t is a messy signal correct 1 over pi t Uhh is a quite a messy signal
therefore it is much more easily and readily expressed in the it is much more this signal
is much more readily expressed in the Uhh in the Uhh the Hilbert transform the operation
of the Hilbert transform is much more readily expressed in the frequency domain rather than
the time domain So the frequency domain specification of the
Hilbert transform which is in the phase shifting property that is minus j sgn F alright where
it’s Uhh which shifts the phase of all the positive frequency components by minus pi
by 2 and the negative frequency components by pi by 2 is much more simpler compared to
the times domain description of the Hilbert transform alright So we will stop this point
over here and continue with the other aspects in the subsequent modules thank you

• ### Landon Barnett

An incredible, clear, well thought out series of videos. Thank you very much.

• ### UTkARSH

you are great sir >>thank you very much sir

• ### gurmeet singh

thanks sir

• ### Santosh Patil

Hats of you sir.

• ### jay ghiya

Most useful and productive 21 mins of my life . Thank u sir🙇🙇🙇