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We just give you the definitions of Z transform, we shall continue with that. We will see some
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other important properties of Z transform.
We defined, x z as sigma x n z to the power
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minus n, n varying from minus infinity to
plus infinity, z is a complex variable. Now,
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if x n is causal, if x n is causal; causal
means it exists only for n greater than or
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equal to 0, all right for n greater than or
equal to 0, x n exists, okay. This is non-zero,
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may be zero but x n is 0 for all negative
n, this a causal sequence.
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Now, x z is a power series. Let you can see,
x z it may not converge at all, it may or
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may not converge. So, let us investigate under
what situation it may not converge, the region
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in which it converges we call region of convergence
or in short R O C, okay. So will put the boundaries
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where it will be converging or it will not
be converging.
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Let us take the following series, x z equal
to 1 plus 2 z 2 plus 3 z 3 plus 4 z 4 plus
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3 z 5 plus 2 z 6 plus z to the power 7. What
is this sequence; in terms of z to the power
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minus one, if we try to see. It is a sequence
which exists in the negative domain of time
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okay, zero, one, two, three, four, three,
two, one and then zero. So minus 1, minus
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2, minus 3, minus 4, minus 5, minus 6, minus
7 this is n. Now, for such a sequence where
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for negative values of n only x n exists and
this is all zero. The value of x z will be
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this or does it exist for all values of z,
at z is equal to infinity, x z becomes infinity;
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we would like to see when the function becomes
infinity, that means it is not converging.
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When z is very very large as it is tending
to infinity, the function exist also tends
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to infinity. So, the function exists except
at z is equal to infinity. Let us take, the
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similar sequence but with a shift.
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Suppose, we have the same sequence with a
shift four, three, two, one, three, sorry
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sorry this is two, one, zero; so minus 1,
minus 2, minus 3 and then this is 1, 2, 3,
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0. What will be the value of x z? z to the
power minus 3 starts from here plus 2 into
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z to the power minus 2 plus z to the power
minus 3 plus z to the power minus, is it z
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to the power minus 4? It should be minus 1,
then 4 into z to the power 0 plus 3 into z
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to the power it is sorry, plus 2 plus 1, is
it all right, now 2 into z to the power minus
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2 plus 3 into z to the power minus 3, okay.
Now, what are the values for which x z becomes
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infinity? z is equal to 0 it becomes infinity,
z is equal to infinity also it becomes infinity.
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So, its region of convergence will will exclude
these two points; at these two points the
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function becomes infinity, is that all right?
If we shifted further by four more steps then
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we will have a function like this, zero, one,
two, three, four, three, two, one okay.
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What will be the sequence, corresponding sequence
z to the power minus 1? It is x z, z to the
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power minus 1 plus z to the power minus 2
into 2 plus 3 into z to the power minus 3
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and so on, at the end z to the power minus
7, 1, 2, 3, 4, 5, 6, 7 all right. Now, when
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this function becomes infinity that you have
to identify, that you can see from here; when
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z is equal to 0, anybody? z is equal to zero,
okay when z is equal to zero x z is infinity.
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So, a sequence that we have considered in
the three cases is having the same nature
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of variation, but it is shifted all right
from left to right and the region of convergence
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also changes; earlier it was not converging
for z is equal to minus z is equal to infinity
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then z is equal to 0 and infinity and now
z is equal to 0. So, region of convergence
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will be depending on the nature of the polynomial,
nature of the function x z that you get.
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Let us consider this sequence, well known
step function which is one, one, one and continues.
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So, what will be corresponding Z transform?
x z, 1 plus z inverse plus z to the power
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minus 2 and so on; it will be 1 by z inverse,
it is g p series provided, z inverse is less
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than 1 or z is greater than 1, okay.
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So, the region of convergence, z is a complex
quantity. So, in the complex z plane it will
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define a unit circle, okay. So, this is a
forbidden range and this is R O C, the space
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outside is a region of convergence; region
of convergence is the space outside, this
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is real and imaginary part, all right. Now,
for this sequence you can see for yourself,
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if I put z is equal to 2 then it is 1 plus
half plus x at 2 becomes 1 plus half plus
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1 fourth and so on.
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And this a g p series with reduced magnitude
of these terms, so this will be tending to
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1 by 1 minus half which is 2. If I put z is
equal to half then it becomes 1 plus 2 plus
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4 plus and so on, tends to infinity, okay.
So, any point inside if I choose, I have just
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taken half you can say choose any good, any
value, any complex value it will be diverging.
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So for a causal system, for a causal system
of finite sequence length of finite length
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of sequence; x z converges because you have
got finite number of terms, okay except at
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zero, except at zero your causal sequence
means, the terms will be x zero plus x one
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into z to the power minus one and so on, all
right.
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It will be causal sequence means; it will
have some value here, some value here and
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so on. If I put z is equal to zero, z to the
power minus one will tend to infinity. So
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all the terms will be giving you finally,
x z will tend to infinity, at z is equal to
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zero. For infinite sequence, that is sequence
of infinite duration x z converges outside
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a circle. The circle is bounded by the poles
okay; poles means if it is an infinite sequence,
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all right if you can resolve it in terms of
some A z by B z, then the roots of B z will
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be called poles.
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So, at those points at those values the function
will try to blow up. This becomes zero means,
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the function will be having an infinite values.
So at poles, the value of the function will
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be infinity all right so that will be excluded,
the region of convergence will have to exclude
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those poles, all right. You must have studied
in complex algebra, analytic functions; so
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wherever there are poles, there will be the
function will try to go to infinity, so for
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an analytic function it excludes those poles,
is it not?
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For stable sequences, we have not yet defined
stability, what stability? So, next we will
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take up stability, for stable sequences the
R O C, the region of convergence always encloses
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the circle of unit radius; this point we will
discuss after we define what stability is,
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okay. Stability of a sequence of a system;
when we define the impulse response of a system,
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impulse response of a system h n, this dictates
the barrier of the system all right.
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Now, what are the conditions on h n, what
are the conditions on h n for stability? What
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do you mean by a stable sequence or a stable
system? If this is driven by an input which
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is bounded then output must be bounded; that
means if x n is bounded
for a bounded input sequence, if x n is bounded
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then y n also should be bounded for a stable
sequence h n, all right. If that is violated,
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then the system is unstable all right.
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What is it actually telling us, it says that;
if I give a input sequence whose magnitude
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does not grow beyond a certain value, whose
maximum value is limited then the output will
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also have a maximum value which will be limited,
which will be less than infinity, output cannot
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tend to infinity if it does, then it is an
obstacle system, all right.
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So, let us see the proof of this. Let the
magnitude of x n the maximum value, be restricted
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to M x, so x n can be at the most equal to
less than M x. Now y n is sigma x K okay,
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x n minus K into h K, is it not? So, y n magnitude
will be less than or equal to magnitude of
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x n minus K into magnitude of h K, do you
all agree? It has to be less than the product
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of the two magnitudes and then submitted.
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So that means, it must be less than if I take
the maximum value of this which is M x into
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h K; it is definitely less than this product,
I taken the maximum value of this summation.
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I can put the summation, since this is a constant.
Summation sign here, so y n if this summation
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is denoted as sum sum h, I call it sum h to
me; it must be less than this product all
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right.
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Therefore, sum h must be finite for magnitude
of y n to be finite, to be less than a quantity
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finite quantity. This to be bounded, that
is if this is to be less than or equal to
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sum M y then M y must be less than M x into
sum h therefore sum h must be less than infinity,
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it must be a bounded quantity, it must be
a finite quantity, okay.
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So this is the condition for stability of
a system. You add up the magnitude of all
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the values of h n, it must be converging to
some finite value. Now, let us take some standard
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functions and evaluate the Z transform, standard
functions for sequences. We have seen A into
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C to the power n, if you take the summation;
it will be A by 1 minus C z inverse, provided
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this magnitude is less than 1, is it not?
This we saw last time, C to the power n if
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you take the Z transform the series that you
get, this is a common ratio, so that magnitude
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must be less than 1.
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If I differentiate with respect to the common
quantity C, as you have done in case of Laplace
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transform, if you differentiate with respect
to C differentiate with respect to C then
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I get, A n C to the power n, mind you; I am
considering sequence for which exist for n
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greater than equal to 0, that is we are taking
only causal sequences, okay. And I list out
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the corresponding region of convergence.
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So, A n into C n will give me n minus 1, thank
you. If I differentiate, I will get A by 1
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minus C z inverse whole squared into z to
the power minus 1 okay, minus 1 and minus
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1 make final 2 minus 1 will give me plus,
what will be the region of convergence? The
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same C z inverse must be less than 1. If I
put C is equal to 1, what do you get; A into
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n will give me A into z inverse by 1 minus
z inverse whole square, what is A into n,
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if you remember what is A into n?
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A into n
is basically ram function, moment I have defined
n greater than equal to 0, so all these functions
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should have their implied u n as a multiplier,
all right so, it is basically a ram function,
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it is implied though it is not mentioned because
I am taking only n greater than equal to 0.
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So, A z inverse by 1 minus z inverse squared,
now what are the regions of, what is the region
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of convergence, I leave it to you to find
out the condition where it becomes infinity,
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okay.
Similarly, A n squared u n
try to check whether you get this; I can always
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multiply by z to the power minus 2, z to the
power plus 2. So, this will become A z divided
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by z minus 1 whole square, okay. In many books
we will find, they write in terms of z all
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right, Z transform and this becomes z minus
1 whole cube; derivative it yourself and again
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z greater than 1, z inverse less than 1 means
z greater than 1, okay.
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If C is, once again we come back to the first
relation involving C. Suppose, C is equal
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to e to the power minus alpha plus j beta
then C to the power n means e to the power
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minus alpha n e to the power j beta n, do
you all agree? Now, what will be the Z transform
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of this sequence? So, A into e to the power
minus alpha n into e to the power j beta n,
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you just put in place of C this quantity.
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So, it will be 1 by 1 minus e to the power
minus alpha, e to the power j beta into z
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to the power minus 1, okay. I just substituted
for C e to the power minus alpha plus j beta,
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if you segregate the real and imaginary part;
what do you get? Let us see.
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Separating the as you are done in the case
of Laplace transform, separating the real
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and imaginary parts and equating them on both
sides; we get Z transform of e to the power
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minus alpha n cosine beta n, as z squared
minus z cosine beta by z squared minus twice
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zeta twice z e to the power minus alpha cos
beta plus e to the power minus twice alpha,
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okay. Some books write e to the power minus
alpha is just r naught, okay.
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So you can do that, because e to the power
minus alpha appears as it is. One may write
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also this as, 1 minus z inverse cos beta,
this is also written in many books in this
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form; e to the power minus alpha cos beta
plus e to the power minus twice alpha z to
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the power minus 2. Similarly, Z transform
of e to the power minus alpha n sin beta n,
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will be z to the power minus z to the power
minus 1 sin beta divided by the same denominator,
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okay. We shall be using it later on in partial
fractions in certain situations that we will
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see very soon, I will keep it aside.
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I would like you to try, try to find out the
R O C’s determined X z and R O C’ s for
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the two functions; x n equal to cos n and
sin n, okay, alpha n, cos alpha n, sin alpha
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n. And second, prove that Z transform you
try this yourself e to the power n, I am writing
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this C is basically A, okay; A cos n angle
of p plus beta is C z by z minus p plus C
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star z by z minus p star, A sorry I have change
it to A, okay to avoid confusion with a earlier
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C, okay.
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So, A is a complex quantity all right. Now
let us see, how to in you can evaluate the
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Z transform of many such functions with the
same simple principle of addition.
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Now, let us see how you get the inverse of
these functions? Given a function X z, how
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do you find out the sequence x n that will
be our next task, inverse Z transform okay.
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So, there are three standard methods very
simple, some of you must have studied in control
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systems also. Power series expansion, second
one is partial fraction
and third one is residue method, residue calculation,
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okay.
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So the first one, X z suppose; we give you
X z as a 0 plus a 1 z inverse like this a
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polynomial a m z to the power minus m divided
by b 0 plus b 1 z inverse plus b N z to the
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power minus N. By long division, you generate
this as x 0 plus x 1 z to the power minus
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1 plus x 2 z to the power minus 2 and so on;
all right you can divide it and then get these.
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For example, if I have 1 plus 2 z inverse
plus z to the power minus 2 divided by 1 minus
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z inverse; say I have taken a simple quantity,
z to the power minus 2, 0.3561, z to the power
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minus 2, what will be the sequence if X z
is given by this ratio of two polynomials,
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what will be x n? It will be an infinity sequence.
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So, you divide it, take the denominator of
coefficients 1, minus 1m, 0.3561; I am not
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writing z to the power zero, z to the power
minus one, z to the power minus two, just
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put little gap in between the coefficients;
similarly one two one, 1, 2, 1, okay. So,
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the division gives me 1, 1, minus 1, 0.3561.
So, after subtraction I get 3.6439 then next
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division gives me, 3, 3 then minus 3, 1.068l;
three into this much so that gives me, 3.6436
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and so on, okay.
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So, next value will be 3.6436. Now this is
x zero, this is x one, this is x two and so
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on because this is coming as a constant. What
you are getting is 1 plus 3 z inverse plus
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00:33:27,100 --> 00:33:34,100
into z to the power minus 2 and so on, so
this will be endless, so will be the sequence.
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So, the coefficients of z inverse, z to the
power minus 2 and so on, they will represent
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x zero, x one, x two, okay it is as simple
as that.
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Recursively, we can write this as the same
result we can write x 0, x zero is basically
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a 0 by b 0, okay sorry it should be here,
okay. x 1 will be a 1, sorry b 1 into x 0
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00:34:33,109 --> 00:34:40,109
whole thing, divided by b 0. x 2, I write
one or two more terms and then a 2 minus b
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00:34:49,540 --> 00:34:56,540
1 into x 1 minus b 2 into x 0, divided by
b 0.
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So generalizing this, it will be
X n equal to a n minus i equal to 1 to n,
x n minus i b i by b 0. So, you can use, I
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00:35:36,670 --> 00:35:43,670
would request you to write as short program.
Take any polynomial a and b, a z and b z and
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00:35:45,569 --> 00:35:52,569
then get the values of x n from this recursive
relation, okay may be up to n is equal to
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00:35:53,789 --> 00:35:59,910
ten or twelve. Write a short program and then
see for yourself, give the coefficients and
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00:35:59,910 --> 00:36:05,160
then find out the values, okay.
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00:36:05,160 --> 00:36:12,160
The next method is partial fraction, X z equal
to a 0 plus a 1 z inverse etcetera plus a
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00:36:29,210 --> 00:36:36,210
n z to the power minus m divided by b 0 up
to b N, z to the power minus N. If M is equal
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00:36:39,789 --> 00:36:46,789
to N then we can write; if M is equal to N
then we can write this as, some C 0 plus C
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00:36:50,759 --> 00:36:57,759
1 by 1 minus p 1 z inverse plus C 2 by 1 minus
p 2 z to the power minus 1 and so on.
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00:37:03,089 --> 00:37:10,089
C N by 1 minus p N z to the power minus 1
equal to, I can write this as a very general
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00:37:19,690 --> 00:37:26,690
term C K by 1 minus p K, z to the power minus
1 okay; K varying from 1 to N. Now, from here
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00:37:38,579 --> 00:37:44,759
see from the polynomial in the denominator,
we have evaluated the routes so the routes
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00:37:44,759 --> 00:37:50,319
have to be evaluated here and then you can
write this one, for each one of them you know
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00:37:50,319 --> 00:37:52,619
the solution.
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00:37:52,619 --> 00:37:59,619
So, X n will consist of this sequence C K
to the power n summation C K, okay. C K, C
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00:38:14,569 --> 00:38:21,569
1, C 2 etcetera all to the power n, u n all
right. And all we evaluate C K, it will be
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00:38:34,109 --> 00:38:41,109
X z by z into z minus p K evaluated at z is
equal to p K, as you have calculated residuals
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00:38:45,989 --> 00:38:52,989
in case of Laplace transform. If, X z has
multiple roots, multiple roots say; m number
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00:39:04,339 --> 00:39:11,339
of multiple roots m number of multiple roots
then that is say, x z contains a term like
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00:39:17,539 --> 00:39:24,539
this, D i by z minus p K to the power i, i
is varying to m varying from 1 to m, okay.
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00:39:32,089 --> 00:39:39,089
There is m number of routes are there, at
the same place p K then the corresponding
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00:39:41,190 --> 00:39:48,190
residues; so if it is 1 by z minus say 0.5
to the power 3, then I will have a term say
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00:39:49,640 --> 00:39:56,640
D 1 by z minus 0.5 plus D 2 by z minus, sorry
z minus 0.5 squared plus D 3 by z minus 0.5
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00:40:02,920 --> 00:40:09,920
cube. So, these D i’s, the general term
D i can be evaluated
m minus i factorial, sorry excuse me m minus
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00:40:22,210 --> 00:40:29,210
i d z, m minus i, z minus p K, exactly identical
method as you have done in case of Laplace
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00:40:36,309 --> 00:40:43,309
transform, okay.
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00:40:47,229 --> 00:40:54,229
Let us take one or two small examples. Suppose,
X z is given whereas z inverse by 1 minus
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00:40:58,880 --> 00:41:05,880
0.25, z inverse minus 0.375 z to the power
minus 2, I can evaluate the residues; this
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00:41:13,339 --> 00:41:18,670
is factorable you can this is quadratic, you
can find out the factors, okay.
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00:41:18,670 --> 00:41:25,670
So, 1 minus 0.75 z inverse; okay let me write
in the factor factored form first, 1 plus
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00:41:32,109 --> 00:41:39,109
zero 0.5 z to the power minus 1, okay, the
factors are actually like this. So, this I
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00:41:41,719 --> 00:41:48,719
write as A 1 by 1 minus 0.75 z inverse plus
A 2 by 1 plus 0.5 z to the power minus 2,
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00:41:54,319 --> 00:42:01,319
okay. So A 1 and A 2, one may try very simple
combinations of 0.75 and 0.5.
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00:42:03,479 --> 00:42:10,479
You can see A 1 is equal to 0.8; A 2 is minus
0.8. So, X z becomes 0.8 into 1 by 1 minus
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00:42:22,109 --> 00:42:29,109
0.75 z inverse minus 1 by 1 minus 0.5 z inverse.
So, what will be x n, very simple 0.8 corresponding
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00:42:36,660 --> 00:42:43,660
to this it is 0.75 to the power n minus, minus
0.5 to the power n, whole thing multiplied
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00:42:51,029 --> 00:42:53,729
by u n.
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00:42:53,729 --> 00:43:00,729
So, these are closed. This is minus 0.5 this
was 1 plus 0.5 thank you, thank you 1 plus
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00:43:08,079 --> 00:43:15,079
0.5 z inverse. Now, if B z has the quadratic
with complex routes what will you do; if you
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00:43:19,589 --> 00:43:26,589
remember in your early classes, in Laplace
transform we studied the complex routes. If
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00:43:30,619 --> 00:43:37,170
the denominator contains complex routes when
you try to resolve in say, s plus alpha Whole
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00:43:37,170 --> 00:43:41,219
Square plus beta square form, remember?
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00:43:41,219 --> 00:43:48,219
And we tried to say, if you are given some
function A s by B s, all right and B s contains
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00:43:54,369 --> 00:44:01,369
a factor like this then you try to dissolve
it into; some A into some K 1 into s plus
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00:44:04,219 --> 00:44:11,219
alpha by s plus alpha whole squared plus beta
square plus K 2 times beta by s plus alpha
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00:44:14,229 --> 00:44:21,229
whole plus beta square, all right. Corresponding
to this, you get a cosine function as inverse
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00:44:22,369 --> 00:44:29,369
and for this you get a sin function, exactly
similar procedure we shall follow here.
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00:44:30,319 --> 00:44:37,319
So, X z, X z if it is given as A z by B z
and B z is having a quadratic then we can
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00:44:44,349 --> 00:44:51,349
write B z as 1 minus 2 into some e to the
power minus alpha, cos beta z inverse plus
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00:44:56,150 --> 00:45:02,249
e to the power minus twice alpha z to the
power minus 2, okay.
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00:45:02,249 --> 00:45:09,249
So, the quadratic that will be given to you,
quadratic that will be given to you will tell
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00:45:10,839 --> 00:45:16,009
you the last term; z to the power minus 2
whatever is a coefficient that will be corresponding
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00:45:16,009 --> 00:45:22,309
to e to the power minus 2 alphas. Take that
square root e to the power of minus 2 alpha
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00:45:22,309 --> 00:45:26,769
then the middle term should be equally 2 into
e to the power minus alpha into cos beta,
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00:45:26,769 --> 00:45:31,769
so you know cos beta, okay you can calculate
sin beta also.
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00:45:31,769 --> 00:45:38,769
So, this finally you reduce into the same
form, some K 1 times 1 minus z inverse cos
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00:45:45,910 --> 00:45:52,910
beta e to the power minus alpha divided by
listed denominator plus K 2 times, z inverse
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00:46:06,670 --> 00:46:13,670
sin beta e to the power minus alpha by 1 minus
2 e to the power minus alpha cos beta z to
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00:46:20,849 --> 00:46:27,380
the power minus 1 plus e to the power minus
2 alpha z to the power minus 2.
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00:46:27,380 --> 00:46:34,380
So, corresponding x n will be after evaluating
K 1 and K 2, it is just a matching that you
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00:46:35,099 --> 00:46:42,099
have to do for the numerator function. So,
you get K 1 times, what is the inverse of
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00:46:42,789 --> 00:46:49,789
this? e to the power minus alpha, cos beta,
okay. So cos n beta and this is minus alpha
248
00:46:58,489 --> 00:47:05,489
n plus K 2 times e to the power minus alpha
n, sin n beta, okay. So that, gives me K 1
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00:47:10,130 --> 00:47:16,229
into e to the power minus alpha, sorry e to
the power minus alpha n I can take out; K
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00:47:16,229 --> 00:47:23,229
1 into cos n beta plus K 2 into sin n beta.
It is better to write in separate forms, sin
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00:47:29,609 --> 00:47:36,609
n, beta n, cosine n beta. Somebody may write
cosine n beta plus theta, whatever you want
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00:47:42,710 --> 00:47:44,859
okay.
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00:47:44,859 --> 00:47:51,859
Now, sometimes we try to link up the discrete
time Fourier transforms; that is if you remember
254
00:47:56,279 --> 00:48:03,279
earlier we are talking about, frequency response
of the system or frequency domain representation
255
00:48:04,489 --> 00:48:11,489
of a signal, that was x 0 plus x 1 e to the
power minus j omega plus x 2 e to the power
256
00:48:18,529 --> 00:48:22,829
minus j 2 omega and so on, is it not?
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00:48:22,829 --> 00:48:29,039
If you remember, we are writing H e to the
power j omega frequency response of the system
258
00:48:29,039 --> 00:48:36,039
whose time domain response, impulse response
was H n. So, this was h 0 plus h 1 e to the
259
00:48:39,309 --> 00:48:46,309
power minus j omega and so on, okay. You see,
in X z the expression is x 0 plus x 1 z to
260
00:48:54,130 --> 00:49:01,130
the power minus 1 and so on; so from the z
transform you can directly get the frequency
261
00:49:03,499 --> 00:49:09,869
domain representation of the sequence, just
put z is equal to e to the power j omega,
262
00:49:09,869 --> 00:49:15,339
you get the same same function.
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00:49:15,339 --> 00:49:22,339
So, in the z domain what is e to the power
of j omega, it is a unit circle as omega increases
264
00:49:33,709 --> 00:49:40,709
sorry, omega increases along this; e to the
power j omega, omega is equal to say two radians,
265
00:49:44,809 --> 00:49:50,479
means I will go here by two radians.
266
00:49:50,479 --> 00:49:57,479
So these are the points on the units circle
in the z plane; this is a z plane, z plane
267
00:49:58,900 --> 00:50:05,900
is a complex plane okay. That means some specific
values of z which will be defined on this
268
00:50:06,140 --> 00:50:13,140
unit circle, if I take then X z represents
nothing but the discrete time Fourier transform
269
00:50:13,880 --> 00:50:20,880
of the sequence, we collect DTFT, this is
known as DTFT; discrete time Fourier transform
270
00:50:21,539 --> 00:50:25,880
of a sequence X n, is it all right?
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00:50:25,880 --> 00:50:32,880
So just take the unit circle, take the points
on this so that presents x e to the power
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00:50:37,039 --> 00:50:44,039
j omega. Now, when I comeback that is when
is complete two pi, omega 2 p i, again I comeback
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00:50:49,009 --> 00:50:56,009
to the same point, same valuation, all right.
So, I keep on making circles as omega tends
274
00:50:58,499 --> 00:51:05,499
to infinity from zero; if i keep on increasing
the value of omega on the in the z plane on
275
00:51:06,299 --> 00:51:13,299
this circle, on this perimeter, I am just
describing the same points, all right which
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00:51:17,630 --> 00:51:24,390
is one another same thing. This is the point
that you shall take up later on, when I talk
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00:51:24,390 --> 00:51:31,390
about stability of systems and well in the
filter design.
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00:51:32,029 --> 00:51:39,029
As in the case of continuous domain, you have
varied along the imaginary axis of s plane;
279
00:51:41,920 --> 00:51:47,219
along the imaginary axis of s plane you have
varied the frequency from minus infinity to
280
00:51:47,219 --> 00:51:54,219
plus infinity or from 0 to infinity, if you
take 0 to infinity. Basically, points on this
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00:51:56,180 --> 00:52:03,180
represent different sinusoids or exponential
periodic functions of this nature, is it not?
282
00:52:07,239 --> 00:52:14,239
You consider the frequencies of the imaginary
axis. We will try to establish a relation
283
00:52:15,479 --> 00:52:22,479
between s plane and z plane parameters, that
is if I shift a point in the continuous domain
284
00:52:24,009 --> 00:52:31,009
along this, what does it mean in the z plane?
Now that mapping of a particular zone in the
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00:52:34,269 --> 00:52:41,269
x plane onto z plane that means from the continuous
domain, how to translate it into discrete
286
00:52:41,569 --> 00:52:47,390
domain, is that all right. Thank you very much, we will continue with
287
00:52:47,390 --> 00:52:48,349
this in the next class.