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Today, we shall be discussing about frequency domain representation of discrete signals.
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Before we go to the frequency domain representation,
we just briefly discuss about how to represent
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a continuous domain, the dynamic system in
a difference equation mode, okay then we will
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take up frequency domain representation.
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Suppose, you have a differential equation
in the continuous domain; 2 into d square
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x by d t squared plus 3 into d x by d t plus
2 into x, is some forcing function f t. Now,
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we have approximated d x by d t as, some first
order approximation of this kind x n minus
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x n minus 1 by T, where T is the sampling
time, all right.
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That means, if there is a continuous function;
if you want to evaluate at some interval,
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at some interval this function is given then
the derivative is approximated as the difference
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between the two successive points and the
slope, okay. d square x by d t squared; similarly,
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will be x n minus twice x n minus 1 plus x
n minus 2 by T squared, this we observed last
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time. So, this differential equation can be
written as, 2 into x n minus 2 x n minus 1
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plus x n minus 2 by T squared plus 3 into
x n minus x n minus 1 by T; 3 into d x by
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d t plus 2 into x n is equal to f n, okay,
all right.
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This you can simplify, this you can simplify
multiplied by T squared and it will appear
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finally in this form; some a into x n plus
b into x n minus 1, plus c into x n minus
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2 is equal to some f n.
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The right hand side could have been also,
given in a differential form f t plus some
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constant time, of dash T and so on. So, that
would have generated f n, f n minus one and
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so on okay. So, this is a general form of
differential equation, we can write in the
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general form, some f n plus a d times f n
minus 1 or f n minus two and so on; so this
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is the difference equation form. That is in
the discrete domain, we write a dynamic system
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equation in this form.
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Now, we come to frequency domain representation
before we go to the discrete domain, let us
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concentrate on the frequency response of system
in the continuous domain. You are given a
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transfer function G s, you are having an input
f t corresponding output is y t. You convert
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f t into F s, y t in the transform domain
it becomes Y s.
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So, F s if you remember is G sorry F s times
G s is Y s in the Laplace domain or in the
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frequency domain, you have also studied the
frequency response; if f t is say A sin omega
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t, a sinusoidal input. If you give a sinusoidal
input, what will be the corresponding output?
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The output can be evaluated very easily and
at the steady state condition, it will be
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A this is multiplied by a gain function which
can be obtained by putting j omega, all right.
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And then sin omega t, there will be we an
additional phase which will be the angle associated
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with G j omega; if i substitute for s equal
to j omega then this will be G j omega magnitude
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and angle phi. So, that phi gets added with
this signal. So, this will be the response
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of the system y t okay. So in the discrete
domain, in the discrete domain we would like
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to study; if you are giving a discrete signal
a discrete signal, discrete sinusoidal function,
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what will be the corresponding output, let
us see, what it means.
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Suppose this is a sinusoidal function, you
are sampling it at, regular intervals of time
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T, okay. So, t is equal to n T, all right.
We can say; we are exciting exciting it by
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a complex sinusoid, I can always write this
e to the power j omega t. What will be the
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corresponding discrete domain representation;
e to the power j omega n T, okay. So, that
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gives me e to the power j n. This capital
omega henceforth shall be representing the
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analogue domain frequency as capital omega
that is radian per second.
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And omega into T is also a frequency, we call
it radian frequency will represent by small
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omega, okay. Henceforth, this small omega
will be used to represent discrete domain
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frequency which is only radians, all right.
And this is in radians per second which is
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two p i into the normal frequency in hours,
okay.
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So, this will be e to the power j omega n
all right, e to the power j omega n. Now,
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we are exciting the signal by e to the power
j omega n; that means a sinusoid which is
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discretized. So, what we are trying to see
is, if I excited by a signal of this kind;
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this is the kind of pulse that you are sending
all right, that is it is envelope is a sinusoid
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but is discrete version is taken as a sequence,
all right.
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You are exciting it by this kind of a signal.
And this will give you an output which is
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also measured at regular intervals. Whatever
is output, you are measuring that output also
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at regular intervals, what will be the ratio
of the magnitudes and what will be the changing
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phase; okay that will be depending on the
system’s behaviour. So, the ratio of the
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magnitudes will give you G j omega something
corresponding to similar to this function,
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all right. And the phase associated, that
is in find there is a phase shift, so phase
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will be associated with, again that function
G j omega in the continuous domain.
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So, we shall be writing something like e to
the power j omega; will come to this very
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soon and angle G e to the power j omega. In
the discrete domain, we do not write j omega,
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we write e to the power j omega because it
appears in this form, okay it is a matter
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of convention.
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Now, you are observed; if I excite a system
whose impulse response is h n, if I excited
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by x n the corresponding output is x n convolution
h n, this we observed last time. Now, suppose
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x n is e to the power j omega n, so what will
be the output? Convolved with h n; okay and
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what was the convolution expression, if remembered
summation e to the power j.
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Okay, let us excited by e to the minus j omega
n, it will be just e to minus j omega n. So,
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minus j omega n minus K, h K, do you all agree?
I can write e to the power summated over K,
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e to the power minus j omega n can be taken
out; and I can write h K
e to the power all right, let us put it as
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plus itself, okay h K e to the power j minus
omega K summated over K, let us write plus
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j omega.
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So now you see, if I excite the system by
an exponential complex function like this,
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the response is; the same function e to the
power j omega n, multiplied by this h K e
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to the power minus j omega K. For a given
h K, this is a unique function, a frequency
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domain function okay. I call this as capital
H e to the power j omega, all right.
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This represents the quantity here h K e to
the power minus j omega K, this is known as
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frequency response of the system or discrete
time Fourier transform of the system, in the
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discrete domain; in the time domain it is
discrete, so discrete time Fourier transform.
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It is a Fourier transform of the system, h
n is converted to H e to the power j omega,
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all right.
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Now, what is the nature of this function,
let us see. It is very similar to e to the
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power j omega H; e to the power j omega is
very similar to, G j omega in the continuous
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domain okay. So, this complex quantity, this
is also complex quantity.
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Suppose, I have for a system h n is given
as h 0, h 1, h 2, h 3. Like this, 2 minus
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1 plus 1 and 0.5, these are the values. What
would be the frequency response? H e to the
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power j omega corresponding to this, it will
be 2 plus e to the power minus j omega into
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minus 1 plus e to the power minus j 2 omega
into plus 1 plus e to the power minus j 3
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omega into 0.5, okay. It is see; h 0 plus
h 1 e to the power minus j omega plus h 2
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e to the power minus j 2 omega plus h 3 e
to the power minus j 3 omega, is it not? So,
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I have just use that okay.
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You can evaluate this for different values
of omega. So, one we take omega equal to 0,
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0.1 radian, 0.2 radian, 0.3 radian and so
on, you can compute it. You can take instead
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of such discrete values of 0.1, 0.2; you might
as well take 0.1 p I, 0.2 p I, 0.3 p I, that
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will be very convenient.
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Then you can go to 2 p I, then 2.1 p I, 2.2
p i and so on, okay. Substitute these values
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and then see how it changes. Now, this kind
of a function is periodic; can you see this,
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if I take a step of two p i that is say, omega
equal to point one p i, okay what will be
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the value?
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H e to the power j 0.1 p i, it will be 2 plus;
okay minus 1 into e to the power minus j 0.1
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p i plus 1 into e to the power minus j into
2 into 0.1 p i plus 0.5 into e to the power
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minus j, 3 into 0.1 p i, okay, whatever be
that value. And if I take 2.1 p I, that means
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after a jump of two p i, what will be the
value; 2 plus minus 1 into e to the power
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minus j 2.1 p i plus all other terms. What
is 2.1 p i, e to the power minus j 2 p i and
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minus 1 p I; so it will be 0.1 p i e to the
power minus j 2 p i minus j 4 p I, all will
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be equal to 1.
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So, from the second term and third term will
get, it e to the power minus j 4 p i to the
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minus j 6 p i. So, finally the same terms
will be reappearing. So, it is having a periodicity
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of two p i, okay.
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If we make a plot; therefore suppose at 0,
what will be the value of this function at
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omega equal to zero?
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It will be just two; these all be reducing
to one, sorry e to the power 0 p i. So, e
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to the power zero p i, how much is it?...
All it will be reducing to 1, so it will be,
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is it all right?
So, this this will be 2 plus minus 1 plus
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1 plus 0.5; whatever be that value it will
be coming here. At 0.1, we have measured the
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magnitude and phase, so if I take the magnitude;
now if you take minus omega, if I take minus
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omega these signs will be just opposite, okay.
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Your series was h 0 plus h 1 e to the power
minus j omega plus h 2 e to the power minus
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j 2 omega and so on. So for any value of omega,
whatever imaginary part you get for minus
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omega, you get just the conjugate, okay. So,
the magnitude remains same, it is the phase
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which will be opposite, all right.
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So, magnitude will be for both positive and
negative direction of omega, you will get
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the same magnitudes. So, it may be
function like this, okay but then
it is symmetric; it will have a repetition
after every two p i. On this side also, if
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you produce it backward then it will be after
minus two p i, minus four p i and so on, okay.
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So, there is a period of 2 p i, the values
will be lying here. So, 0.1 p i and 2.1 p
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i will be identical, is it all right? The
response of the system corresponding to 0.1
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p i and 2.1 p i will be same.
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What is 2 p I, what is it in terms of absolute
frequency?
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Suppose, you are having a system where you
are having a sampling time corresponding to
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a frequency of 2 kilohertz, is 0.5 second
or 0.005, 0.005, T is equal to 0.005 second,
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okay that means 2 kilohertz.
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So, one kilohertz will correspond to half
that sampling frequency, all right. So, omega
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any omega into T which will give you the corresponding
frequency; say hundred hertz, if I consider
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100Hz, what would be its a presentation in
the discrete frequency domain? It will be
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100 multiplied by 0.005 radian, all right
so, 0.5 radian all right.
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2 KHz will correspond to…. is it all right?
Point, so hundred hertz, actually 100 Hz is
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100 into 2 p I, okay so many radians per second,
all right? So, its omega is corresponding
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to 628 radians per second. So, omega into
T will correspond to 628 into 0.005, so it
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is not 0.5 radian into 2 p I, so 6.28 so that
will be the radian, okay.
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Similarly here, so this is 3. Or 0.314, 0.314
radian, is it all right? No, 3.14 radian okay,
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hundred hertz will correspond to three point
one four; it is hundred, one tenth it will
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be, you see there is a 0 here… One by two
thousand is three 000, is it all right, okay
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there was a slip. So, this will be equal to
0.314 radian, all right.
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So what I meant is, in terms of the sampling
frequency, sampling frequency corresponds
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to 2 p i.
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In the radian domain, sampling frequency will
correspond to 2 p i, is it not, because; omega
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s into T is 2 p i into f s into T and f s
into T is 1, so that gives me 2 p i radians,
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okay. So, any fraction of that will be, that
fraction of two p i. So, hundred hertz was
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one twentieth of two kilo hertz, so one twentieth
of two p i.
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So, 100 hertz therefore and two kilo hertz
plus hundred hertz, that is 2100 hertz will
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appear identical; the response of the system
correspond to corresponding to hundred hertz
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and twenty one hundred hertz will be identical,
is it not, that is what you have observed.
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1, 0.1 p i and 2.1 p i, 01 p i corresponds
to say; one tenth sorry, 0.1 p i may be corresponding
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to some frequency like hundred hertz say,
and this would be 2.1. That means at a gap
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of that sampling frequency, two p i correspond
to sampling frequency you get the identical
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response.
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So, 100 Hz and 2100Hz will give me the same
response when the sampling frequency is 2
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KHz. So, if you change the sampling frequency
then only you can distinguish the response
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corresponding to these these two different
frequencies, not otherwise, number one.
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Number two; if I change the sampling frequency
then the response due to these two will not
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be same, okay. So, sampling frequency is very
important, important. So any omega and 2 p
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i plus omega, 4 p i plus omega will have identical
response; physically what it means, let us
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see.
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Let us have a sinusoid, I am taking just a
cosine function; see varying slowly…….
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these are the sampling points, okay. I am
also having….. another sinusoid like this,
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okay. Now, I am sampling at this rate; see
the sampling rate is relatively slow, means
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the time gap is sufficiently high sampling
time is quite high. You have scanned these
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values from this signal, somebody else has
got another signal he has scanned at the same
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rate but they also fall on this line. Now
which particular signal from these values,
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these discrete values which particular signals
can you identify, there is an ambiguity.
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I could have had many more such signals, many
more such signals which will have these as
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the sample points. I am not drawn it for many
other frequencies, it is a little difficult
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to draw it here. Anyway I could have shown
you, many more signals can be represented
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whether same discrete values, okay.
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So, there is an ambiguity. So what should
be the; suppose the signal is having is like
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this, is having lot of kings here and there,
that means number of frequencies are involved.
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And what should be proper sampling rate; so
that we can trap, all possible frequencies.
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So, the highest frequency that is present
here will have, like this say you take the
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maximum number of variations in that signal;
that means where that high frequency component
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appears and then sample it there, okay let
us see what it means.
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Any function, I sample two points, these are
the two values; if I just all okay, let us
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take these two values from this point to this
point the function has gone this way, it could
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have gone this way also. So, I could have
represented with the help of these two sampled
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points, I could have represented an analogue
function; the continuous domain function which
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has a part like this, that is also possible.
So, there is a path like this either a convex
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or a concave form.
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If you want to determine, if you want to determine
if between the two successive points; there
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00:34:41,899 --> 00:34:48,899
should not be any ambiguity then it should
not go through a maxima, without having another
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00:34:49,339 --> 00:34:55,139
sampling. If I would have had another sampling
here then I could have determined, it has
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gone this way and not this way, another sample
point should have been here; that means between
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00:35:01,539 --> 00:35:08,539
say on the two sides of a maxima or a minima,
I should have sampling okay. That means, if
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00:35:12,970 --> 00:35:19,970
between two points there is a maxima or a
minima then I must have another sampling there,
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00:35:20,239 --> 00:35:21,670
okay.
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00:35:21,670 --> 00:35:28,670
Let us consider this, if I sample it here
and here both may may mean zero; it could
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00:35:31,430 --> 00:35:38,430
have been this way or this way, this is half
the frequency present. So, I must have at
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00:35:39,599 --> 00:35:46,599
least a sampler which will sample values in-between.
So, if I can sample somewhere here then I
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00:35:46,960 --> 00:35:53,960
know it is going this way, it cannot be this
way, okay. This is half the frequency, so
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00:35:54,049 --> 00:36:01,049
I must have a rate faster than half that frequency,
sorry this is corresponding to this is twice
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00:36:02,359 --> 00:36:04,099
the sampling frequency.
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00:36:04,099 --> 00:36:11,099
See, if I sample at this rate, sampling frequencies
twice this frequency, okay. So, it must be
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00:36:11,880 --> 00:36:18,880
more than twice that frequency, okay, physical
it means this. If I can have a sampler whose
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00:36:20,170 --> 00:36:27,170
sampling rate, your sampling at this rate
which is twice the frequency of the signal.
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00:36:29,569 --> 00:36:36,099
This a highest possible frequency presents
here, so if I want to trap that, then I must
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00:36:36,099 --> 00:36:41,400
have a sampling frequency which is higher
than twice that frequency; that is sampling
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00:36:41,400 --> 00:36:47,259
time should be half of this less than half
of this.
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00:36:47,259 --> 00:36:54,259
This will derive sometime later also, mathematically….
Let us consider for the time being, let us
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consider Laplace transform of a continuous
domain.
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00:37:29,009 --> 00:37:36,009
Hurriedly, we shall go through the standard
Laplace transforms then we will go for disc
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00:37:38,259 --> 00:37:45,259
domain transform. What is a Laplace transform
of function f t; which is say e to the power
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00:37:49,609 --> 00:37:56,609
minus a t? So, we know this is 1 by S plus
a, okay. So, if I put a equal to alpha plus
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00:38:02,549 --> 00:38:09,549
j beta then Laplace transform of e to the
power minus alpha t, e to the power j beta
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00:38:15,200 --> 00:38:22,200
t; would be 1 by s plus alpha plus j beta,
okay which is S plus alpha minus j beta by
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00:38:34,109 --> 00:38:38,660
S plus alpha whole squared plus beta squared.
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00:38:38,660 --> 00:38:45,660
If you segregate the real and imaginary part
then the real part of this will be e to the
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00:38:52,579 --> 00:38:59,579
power minus alpha t cosine beta t; so Laplace
transform of this will be real part of this
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00:39:01,440 --> 00:39:08,440
which is S plus alpha by S plus alpha whole
squared plus beta squared. Why, I am doing
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00:39:12,440 --> 00:39:19,440
this, is this that by the same logic; we should
be deriving the Z transform of a different
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00:39:22,839 --> 00:39:29,839
time domain function, discrete time domain
functions which need not go through the entire
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00:39:29,950 --> 00:39:30,460
derivation.
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00:39:30,460 --> 00:39:37,460
Similarly if I take the imaginary part, e
to the power minus alpha t sin beta t will
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00:39:41,789 --> 00:39:48,789
be beta divided by from both sides minus beta
t minus j will get cancelled, so it will be
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00:39:51,900 --> 00:39:58,900
beta by S plus alpha whole squared plus beta
squared.
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00:40:03,479 --> 00:40:09,779
If you differentiate, once again e to the
power minus a t, if it differentiate with
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00:40:09,779 --> 00:40:16,779
respect to a, a is the common quantity, common
variable. So, if I differentiate with respect
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00:40:17,609 --> 00:40:24,609
to a, I will get t into e to the power minus
a t; and that gives me on this side, 1 by
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00:40:31,170 --> 00:40:38,170
S plus a whole squared, we can the negative
sign will get cancelled.
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00:40:41,619 --> 00:40:48,619
You differentiate once again, t square into
e to the power minus a t, will give you 2
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00:40:48,799 --> 00:40:55,799
by S plus a to the power 3. And successively,
if you keep on differentiating with respect
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00:40:58,239 --> 00:41:05,239
to a; we will get factorial n by S plus a
to the power n plus 1, okay. If you put a
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00:41:14,799 --> 00:41:21,799
equal to alpha equal to zero, you get the
Laplace transform for a pure sinusoid sin
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00:41:21,969 --> 00:41:28,969
beta and for sin beta 2, will be makings of
this relation very soon.
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00:41:32,369 --> 00:41:39,369
Now, you come to discrete domain transform,
known as Z transform. We defined Z transform
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00:41:53,359 --> 00:42:00,359
of a sequence x n, as say transform will be
shown like this; capital Z of the function
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00:42:05,559 --> 00:42:12,559
x n or the sequence x n, is equal to x 0 where
I write x 0 as nothing but x at 0, in many
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00:42:16,729 --> 00:42:23,729
books they write x at 0 also it is one and
the same thing…wherever retains okay. K
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00:42:35,920 --> 00:42:41,160
may tend to infinity; if it is an infinite
sequence then it may tend to infinity, it
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00:42:41,160 --> 00:42:48,160
might be finite sequence also, okay.
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00:42:48,160 --> 00:42:55,160
So, let us see what would be the Z transform
of a delta function; it is 1 then 0, 0 the
241
00:43:05,309 --> 00:43:12,309
sequence is like this, is it not? A delta
function is existing only at n is equal to
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00:43:12,539 --> 00:43:16,930
zero, after that it is all zero. So, what
will be the Z transform of this sequence?
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00:43:16,930 --> 00:43:23,049
It will be 1 plus 0 into z to the power minus
1 plus 0 into z to the power minus 2 and so
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00:43:23,049 --> 00:43:27,859
on, which is 1.
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00:43:27,859 --> 00:43:34,710
What is the Laplace transform of a delta function?
That is also one. For a delta function, both
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00:43:34,710 --> 00:43:41,710
in Laplace domain and Z domain the transform
function is one. What is the Z transform of
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00:43:43,809 --> 00:43:50,809
a delayed impulse, say delayed by 2 steps,
2 intervals? So it will be 0, 0, 1, 0, 0 this
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00:43:58,099 --> 00:44:01,549
kind of a sequence, okay.
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00:44:01,549 --> 00:44:08,549
So, what will be the Z transform of this?
0 plus 0 plus 1 into z to the power, minus
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00:44:09,029 --> 00:44:16,029
2, is it not? So, it is z to the power minus
2. So, if there is a delay of two steps; one
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00:44:25,460 --> 00:44:32,460
gets multiplied by z to the power minus two,
okay any x n okay, any x n.
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00:44:40,119 --> 00:44:47,119
If, I shift by some n steps it will look like;
0, 0, 0 after n steps only, it will appear,
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00:45:03,289 --> 00:45:10,289
is it not? So, what will be the corresponding
Z transform? So, 0 plus 0 plus 0 then starts
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00:45:17,079 --> 00:45:24,079
with x 0 into z to the power minus N plus
x 1 into z to the power minus N plus 1 bracket
255
00:45:27,680 --> 00:45:34,680
and so on. So, z to the power minus n can
be taken common, it will be x 0 plus x 1 z
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00:45:38,849 --> 00:45:45,849
to the power minus 1 and so on, which is nothing
but x z, okay.
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00:45:48,369 --> 00:45:55,369
So, x z gets multiplied by z to the power
minus n, if there is a shift of n steps. In
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00:45:57,769 --> 00:46:04,769
the Laplace domain, if we remember; if f t
gives you F s then f t minus tau was represented
259
00:46:10,589 --> 00:46:17,589
by F s, it was getting multiplied by S tau,
e to the power minus s tau. So similar to
260
00:46:18,150 --> 00:46:22,759
this, here it gets multiplied by z to the
power minus N, if the function gets shifted
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00:46:22,759 --> 00:46:29,759
by n steps.
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00:46:31,079 --> 00:46:38,079
Let us now consider a unit step, what be the
Z transform of a unit step?
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00:46:41,640 --> 00:46:48,640
Unit step is given by 1, 1, 1 okay, so what
will be Z transform? So, U z, I write U z
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00:46:59,690 --> 00:47:06,690
is 1 plus z to the power minus 1, z to the
power minus 2 and so on up to infinity and
265
00:47:08,319 --> 00:47:15,319
this is a G p series;1 by 1 minus z to the
power minus 1, okay. What will be an exponential
266
00:47:26,039 --> 00:47:33,039
sequence, c to the power n? c to the power
n is basically, c to the power zero is 1,
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00:47:36,539 --> 00:47:43,539
c, c squared, c cube and so on, all right.
268
00:47:45,119 --> 00:47:52,119
Suppose, c is less than one and so on, then
what will be the Z transform? 1 plus c z inverse
269
00:48:01,749 --> 00:48:08,749
plus c square z to the power minus 2 and so
on. And this is again, c z inverse is a common
270
00:48:11,089 --> 00:48:18,089
ratio; so this is again a G p series which
can be written, as where this quantity must
271
00:48:23,380 --> 00:48:30,380
be less than 1, c z inverse must be less than
1 then only you can represent this, this means
272
00:48:37,799 --> 00:48:44,799
z should be greater than, okay.
273
00:49:00,729 --> 00:49:07,729
Suppose; you are having a sequence, a function
1 minus 0.8 z to the power minus 1, so c is
274
00:49:14,809 --> 00:49:20,119
corresponding to 0.8, okay. What will be the
corresponding inverse transform? Inverse transform,
275
00:49:20,119 --> 00:49:24,469
inverse transform like Laplace transform;
we just take the table and then see the values
276
00:49:24,469 --> 00:49:26,589
of the corresponding inverses.
277
00:49:26,589 --> 00:49:33,589
So, corresponding to 1 minus c z inverse,
1 by 1 minus c z inverse I know, the sequence
278
00:49:34,479 --> 00:49:41,479
is c to the power n. So, c is 0.8, so it will
be 1, 0.8, 0.64 and so on, okay, this is the
279
00:49:54,079 --> 00:50:01,079
sequence. If, I have 1 by 1 plus 0.8 z inverse
then c is minus 0.8, so it will be 1 minus
280
00:50:08,630 --> 00:50:15,630
0.8 plus 0.64 and so on. So, it will be like
this, 1 minus 0.8 plus 0.64 is progressively
281
00:50:23,979 --> 00:50:30,979
decreasing, but it is alternative in sign,
okay.
282
00:50:35,400 --> 00:50:42,400
What would be z to the power minus 2 by 1
plus 0.8 z to the power minus 1; what would
283
00:50:44,609 --> 00:50:51,609
be the corresponding time domain representation?
So, z to the power minus 2, we have observed
284
00:50:52,660 --> 00:50:58,359
if a function is delayed by n steps then the
original function gets multiplied by z to
285
00:50:58,359 --> 00:51:04,329
the power minus n. So, we know the inverse
corresponding to 1 by 1 plus 0.8 centi- inverses,
286
00:51:04,329 --> 00:51:07,190
this one, so the whole thing gets shifted
by two steps.
287
00:51:07,190 --> 00:51:14,190
So, it will be zero, zero then one then minus
0.8 then 0.64 and so on, it is a sequence
288
00:51:20,499 --> 00:51:27,499
of this type, okay. See, if I multiplied by
z to the power minus 2, the sequence of this
289
00:51:30,369 --> 00:51:31,200
type.
290
00:51:31,200 --> 00:51:38,200
Now, let us come to another interesting function;
what will be okay, what will be the inverse
291
00:51:45,829 --> 00:51:52,829
of 1 plus z to the power minus 1? It is minus
1 to the power n, so it will be 1, minus 1,
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00:52:01,979 --> 00:52:08,979
1, minus 1, 1, minus 1 and so on, okay. What
will be 1 plus z to the power minus 2, 1 plus
293
00:52:12,979 --> 00:52:19,979
z to the minus 2? Now, look at it, look at
it this way; z to the power minus 1 corresponds
294
00:52:27,109 --> 00:52:34,109
to one delay, if I had two delays then that
correspond to z to the minus two, all right.
295
00:52:34,180 --> 00:52:41,180
So, I will get the same sequence with two
delays. So, this is 1, minus 1, 1, minus 1
296
00:52:51,269 --> 00:52:56,209
and so on, in-between there will be only zeros,
all right. I do not have anything like z to
297
00:52:56,209 --> 00:53:03,209
the power minus one, see in-between I will
just put some zeros, okay. Now, you tell me;
298
00:53:08,799 --> 00:53:15,799
z to the power minus 1 by 1 plus z to the
power 3 minus three, what will be the z inverse
299
00:53:17,519 --> 00:53:24,519
of this? Shifted sequence, so its starts from
here; so it will be one then it will be two
300
00:53:28,539 --> 00:53:42,639
zeros minus 1, two zeros plus 1, two zeros minus 1 and so on, okay. So in the next class,
301
00:53:42,650 --> 00:53:44,219
we will stop here today.