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Good morning friends. In the last class, we are observing for exponential functions of
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this current. The sampled version was f n
e to the power alpha T to power n, so this
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we took as constant C. And if alpha is positive,
we observed e to the power alpha T is greater
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than 1; that is C is greater than 1. So, the
response was, well the function f n was something
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like this, okay. Thank you this is for C negative,
C less than 1; C less than 1, that is alpha
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negative okay, it was like this. And for C
greater than 1, it was increasing like this,
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okay.
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If C is less than 0, C is negative; then C
to the power n is some negative quantity,
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all right, to the power n, depending on the
magnitude of that quantity. It may be decreasing
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in magnitude or increasing in magnitude but
it will be appearing alternately with alternate
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signs. One of our friends had asked when do
you get C negative; for alpha positive values
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or negative values, you do not get C negative,
is it not? C was less than one or greater
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than one, but when do you get C negative,
okay. Thank you very much.
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It is for a complex function. Suppose we have,
that is alpha equal to a plus j b, all right;
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it can be negative or positive depending on
that, it will be exponentially increasing
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or decreasing. Let us select the sampling
time such that, b into T is say pi. We select
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the sampling time T, such that the product
of T, at the frequency term b is equal to
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pi, then what happens? e to the power alpha
t, if I take the sampled values; would be
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e to the power a T to the power n, e to the
power j b T to the power n, okay. So, this
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is some magnitude C to the power n, e to the
power j b T and b T is pi; therefore, this
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is e to the power j pi to the power n. And
what is e to the power j pi, minus 1.
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So minus 1 to the power n, magnitude C to
the power n, is it all right? So, it will
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be C depending upon the magnitude of C, if
it is greater than one then it will be more;
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it will start from one, then C, then C squared
and so on, okay.
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Now, suppose we have a function f n. Say,
this is 1, 0, minus 1 by 4, 0 then 1 by 16,
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these are not shown to the scale; so I am
just writing the values of this, it is gradually
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diminishing in magnitude, 1 by 64 and so on.
And there coming alternately plus minus with
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zeros appearing in-between, okay. So, it is
1, 0, minus 1 by 4, 0, 1 by 16 and so on.
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What will be representation of such a function?
Could someone suggest?
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p i by two, okay.
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So, C to the power n, e to the power j pi
by 2, let us try e to the power j pi by 2
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n, you need this. Okay, let us see, C to the
the power n, will take care of the magnitude
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part. Let us see, e to the power j pi by 2
n, what it gives. e to the power j pi by 2
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n, what are the values, for different values
of n? n is equal to zero, it is 1. n is equal
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to one, e to the power j pi by 2, that will
be j. And then minus 1, then again minus j,
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then plus 1, then j and so on, okay.
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So, that does not satisfy this. I have to
have another term; I will to have another
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term which I can write, I will compliment
this j’s. If I add them together and then
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take the average, I will get so I call this
as, some y 1, y 1 series and y 2 series. So,
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y 1 plus y 2 by 2 will give me, the desired
series, okay. So, it will not be e to the
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power j pi by 2 n alone; it will be
and average of this, is it all right?
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Could it be 1 to the power n, n minus 1 to
the power n? Instead of pi, if I take 1 to
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the power n, n and minus 1 to the power n.
So, basically this is half of C to the power
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n. This is j, so j to the power n plus minus
j to the power n, is it all right? This is
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j to the power n plus minus j to the power
n, half of that.
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Next question is, could it have been 1 to
the power n and minus n to the power n? Let
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us see.
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Half of this, what is it give me? 1 then if
I put n is equal to 0, n is equal to 0, 1.
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n is equal to 1, n is equal to 2; again 0,
1, so this will give me a series; 1, 0, 1,
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0, 1, 0 and so on.
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So, far as this term is concerned equivalent
of this term; that gets multiplied by C to
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the power n, so it will be modulated. So,
it will either exponentially decreasing or
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increasing, but the terms that will be appearing
will be alternately coming it positive signs
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and in-between there will be zeros. So, if
you have signals of this kind, where C is
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less than 1; it will be decreasing and if
there are zeros in-between, it will be like
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this, okay.
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We will get back to this kind of exercise
again later on. X n, the series is normally
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written; if it is a finite series then say,
minus 2, 3, 0, minus 1, 2, 5, 1 and this is
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the finite series, I show by an arrow. Suppose
there is an arrow here, that means this is
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the origin. So this is in the negative region
of time, this is in the positive region of
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time, that is the sequence is defined both
in negative and positive directions.
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So this is, at this point it is minus 1, 2,
5 then up to this scale; 1, 0 if this terminated
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here, that means here after this is all zeros,
0 then 3 minus 2 and so on, rest all zeros.
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So by showing an arrow here, you show the
location of the origin okay, this is origin.
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The sequence X n, may be defined over a range,
is a finite range; where M 1 is greater than
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minus infinite and M 2 less than plus infinity,
that means between some M 1 it may be in the
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negative region of time, and some M 2, it
is defined all right, like the previous one.
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Here, this is 0, minus 1, minus 2, minus 3,
so M 1 was up to minus 3. M 2 was up to plus
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3.
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So, what is the sequence length? What will
be the sequence length? One two three four
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five six seven, all right, so three plus three
minus of minus three plus one, okay. So, M
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2 minus M 1 plus 1, this will be the length
of the sequence, okay. M 1 could have been
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here also then also M 2 minus M 1 would have
been this much.
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Like in the continuous domain, we also defined
here in the discrete systems, even and odd
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functions
or even and odd sequences. So any sequence
X n, if it is an even sequence then it will
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be same as X minus n. So, this is the condition
for even sequence; such sequences we shall
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write as, X e n, x even n. Similarly an odd
sequence, we will follow this relationship
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and we will denote these as X o n, okay.
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Any sequence X n, can be written as a sum
of an even component and then odd component,
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okay. Now, how do you find out the even component?
You take X n plus X of minus n, divided by
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2; that gives you X e n, okay. Similarly,
X n minus of X of minus n, divided by 2 will
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give you X o n.
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Let us take an example. We have given X n,
as say 2, minus 1, 1, 0, 3 what will be the
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even sequence? Would someone tell me? What
are values in the negative region of time,
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all zeros? If I have origin here, okay on
this side of, this arrow left side it is all
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zero, so it will be half of this sequence
plus its negative region values, divided by
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two. So, it will be half of all the values.
Should it be the half? Any disagreement? First
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one? 2, then minus 0.5, then 0.5 then 0 then
1.5, is it okay?
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So, I have a sequence, graphically if we show,
if this is 2, minus 1, 1, 0 and 3, it’s
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corresponding even part, this is X n, this
is n, okay all right, it is a break here.
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Then even part will be 2 then it is minus
0.5 then plus 0.5 then 0 then 1.5. So, except
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the first value; others all half, the first
value is remaining as it is, all right because
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when you add with this X, minus zero is also
same as X plus zero so, that divided by 2,
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means the original value. So all other values
will be reduced to half, what with the odd
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sequence?
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X odd, it will be 2 minus 2, 0 then this one
will be the same, so minus 0.5 plus 0.5, 0,
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1.5. So, the odd sequence
be 0, minus 0.5 plus 0.5, 0 and 1.5, okay.
Say, except for the first value others are
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same. So, you mean to say except the first
value and even function and odd function will
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appear identical. So where is the difference?
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So, even function is though; we are showing
it only in the positive region of time, it
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is also having its counterpart in the negative
region which has not been shown. So, the total
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function is a mirror image of this plus this
itself, okay. So X n, if you take a mirror
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image on this side and then club with this
that will be the even part X e n. Similarly,
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if I take a mirror image of this twice, once
about y axis then again about x axis; so I
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get its replica on this side, so first quadrant
and third quadrant symmetry.
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So it will be, for example odd function; if
you take X o n, the total function will be
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0.5, minus 0.5, 0 and minus 1.5, okay. Sorry.
So, this is the total function. Normally,
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we do not show this part, it is understood;
it will be having these values, in the negative
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region it will have just negative of these
values. Similarly, the even part we do not
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show, the left hand side but it is also having
these values.
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So, even function and odd function will be
met much longer in sequence compared to original
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sequence; because the original sequence was
only up to this, okay. The other values on
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this side were zero. Now, you
take some basic elements used for signal processing
operations. One is an adder, you have a sequence
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X 1 n, X 2 n, the output is X 1 n plus X 2
n okay. Then you will have a multiplier; X
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1 n symbol for a multiplier is like this.
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So, multiplier K, so output y n gets multiplied
by a scalar K, k times, x n okay. So, this
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is a multiplier. Then we have the most important
element that is a delay element. If this is
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x n, if this is a unit delay, one sampling
time delay then y n, the output will be x
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n minus 1, okay, if it is a unit delay. The
symbol for a unit delay used in different
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books are; sometimes it is written as, capital
T that is one sampling time, sometimes capital
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D that is a delay standard one delay.
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Mathematicians, they use backward backward
shift operator B. And in the transform domain,
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we also use in text books Z to the power minus
1 okay, in terms of Z transform operator.
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It is similar to using or in in a derivative
circuit. In many books they write d d t; if
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you have f t here, you get f dash t here in
an analogue domain, okay. In the control books,
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they replace it by s. And mathematicians will
find they write as p, d d t operator. In some
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books, they write heavy side operator, capital
D. They are one and the same thing, only thing
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is in different domains, we use different
symbols.
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If, you are working in the time domain only
then you write d by d t but the moment you
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use this as a Laplace operator s, then you
should not write f t; this is not correct,
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you should write its transform F s and corresponding
output, whatever it is. Suppose, this was
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y t then I call it Y s and not y t. Y s it
happens to be s times F s in this case, okay.
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So, the symbol use of the symbol is depending
on what domain you are using. In the frequency
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domain, similarly this will be; if I write
F j omega and this Y j omega then this will
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be straight away j omega, okay. s replaced
by j omega in the continuous domain.
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However, in the discrete domain in the discrete
domain; this is very commonly used symbol,
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though technically it should be used only
when you are using the variables in the Z
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domain. If I am using X z then this has a
meaning, Y z will be, z inverse of X z. I
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should not write X n is in the time domain
then, I should use either B or capital T,
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okay. So, B means backward shift operation
that will give me X n minus 1, which is y
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n okay. But, anyway most of the text books
you find, they use z inverse even in the time
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domain okay. So, this is a common slip but
conventionally it is acceptable; so we will
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also use z inverse as an element, if required.
All these symbols will be used sometime or
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rather, they are one and the same thing.
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So, if you have two delay elements, if I show
two delay elements; if this is x n then what
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will be the output? x n minus 2, okay. Next,
you have another kind of multiplication that
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is; x 1 n gets multiple multiplied by x 2
n, so y n output is x 1 n multiplied by x
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2 n is a modulator, okay. You must have studied
the amplitude modulation, frequency modulation,
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all right.
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This is also used in windowing windowing operation;
that means a function x 1 n, we passed through
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a window window function x 2 n where the function
values will be multiplied by, it will be scaled
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by different amount, depending on the values
of x 2 n, okay. So, basically what you are
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trying to, you are trying to prove; x 1 n
by different amounts, at different instant,
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okay. That proving function is x 2 n that
is windowing.
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So, what is this backward shift operation?
Therefore, on x n, x n is a sequence; I can
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write x n, sometimes we also use x n all right.
So, B on x n will give me x n minus 1, okay.
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So, what is x n minus x n minus 1? 1 minus
B times x n, so this we call; as differencing
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operation on x n. So, on x n if I apply a
differencing operator, I get the difference
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between x n and x n minus 1; it is very similar
to your derivative operation.
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If I have B x n as x n minus 1; what will
be a forward shift operation? Forward shift
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will be, B inverse x n, which will be x n
plus 1, okay. What is this shift really? Meaning,
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suppose you have a sequence x n; this is say
at 0, x 1, x 2, rest are all 0, this is a
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sequence x 0 may be five, four, three. If,
I give it a shift, B of this represents the
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general x n, B of x n will give me; this whole
thing is shifted to the right, shifted to
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the right means first instant, it will be
0 then only x 0 will appear, the x 1, x 2
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and then 0. So, x 2 starts with a 0, if I
give two shifts then 0 0 then x 0, x 1, x
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2.
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So, B of x n will be shifting the sequence;
such that this will be the values. Now, 0,
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x 1, x 2, x 0, x 1, x 2.
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There is another operator, a summator or an
accumulator, accumulator is a proper name.
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So, what you get is summation of x n, all
right; that is x 0, it is similar to your
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integration, summation of the sequence up
to that point n. What kind of output you get,
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if I excide the system with a delta function?
It is delta n is this one; okay this is delta
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n, rest are all 0. Now, in an accumulator
what will be the output? At this instant,
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it is one, next instant also it is one, one
plus zero, one plus, zero plus zero and so
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on. So, the output will be u n, okay; it is
similar to integration.
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If I now, use it because a differencing operator,
differencing operator what do you get? Delta
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n minus delta n minus 1, is it not? What will
it look like? Okay, this is delta n on and
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minus 1; if I use a delta n and a delta function
and then an accumulator, what will be the
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output?
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So, this one is delta n minus delta n minus
1, and a summator delta n will give me u n.
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And this will give me u n minus 1. So, this
will be u n minus, u n minus 1 which is nothing
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00:36:25,609 --> 00:36:32,609
but again delta function, all right. This
is delta n minus delta n minus 1, so delta
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n is coming back as output.
So, one is the inverse of the other, all right.
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So, accumulator is the inverse of a differentiator,
difference in differencing operator okay.
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So, delta and sigma; if we use these notations
is equal to unit, all right and delta was
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1 minus B and therefore sigma will be, can
you prove it? Can I prove it? Let us see,
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what is sigma?
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It is one plus; see sigma, this operator operating
on x n, sigma operating on x n, nearly x n
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plus, is it not, all the values, p s values?
So, this is 1 plus, this is B plus, B square
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and infinite sequence and this is a g p series;
we can write this as 1 by 1 minus B, okay.
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Now, let us use these elements, let us see;
how these elements can be used, to represent
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a different equation or a general equation.
Suppose, you have a difference equation in
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00:39:01,619 --> 00:39:08,619
this form, will discuss later on how such
equations can be obtained, okay.
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00:39:26,700 --> 00:39:33,700
Suppose, we have an equation y n plus b 1
y n minus 1 plus b 2 y n minus 2 is equal
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00:39:40,819 --> 00:39:47,819
to a 0 x n a 1 x n minus 1 plus a 2 x n minus
2. One may write, where y is the output and
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x is the input, okay. So, you can write y
n is equal to a 0 x n plus a 1 x n minus 1
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00:40:04,849 --> 00:40:11,849
plus a 2 x n minus 2 minus b 1 y n minus 1
minus b 2 y n minus 2, do you agree? So, from
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00:40:27,480 --> 00:40:34,480
there, you can see a 0 x n.
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00:40:41,230 --> 00:40:48,230
Suppose, x n is the input then that gets multiplied
by a constant a 0, added with a 1 times one
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00:40:56,420 --> 00:41:03,420
delayed function. So, I take x n here, I tap
it here, put a delay element, you can write
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d or capital D then multiplied by a 1. Then
again take another delay, multiplied by
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00:41:42,660 --> 00:41:49,660
a 2, okay. So, this is a 2 times x n minus
2, plus a 1 times x n minus 1. So, they are
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added together and then finally added with
a 0 times x n.
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This is suppose, this is output y n then I
am yet to add these two terms. So, that y
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00:42:09,859 --> 00:42:16,859
n from the output side, again I can have a
delay; b 1 then b 2, I can add them together
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00:42:45,970 --> 00:42:52,970
and then they are to be negated, okay. So,
these two from these two, if I stretch channels
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and added signals; will be giving you y n,
okay. So there is a multiplier, adder, delay.
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00:43:13,859 --> 00:43:20,859
There are some more special functions, operators,
we use for changing the sampling rate, all
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right. One is a compressor or down sampler
and the other one is up sampler or expander,
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00:43:53,019 --> 00:44:00,019
okay. When we compress it?
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00:44:06,599 --> 00:44:13,599
Suppose, okay let me take the signal here,
there is a signal here; this is a continuous
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00:44:20,589 --> 00:44:27,589
domain signal x t, your sampling at very close
intervals, if you have very close intervals
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00:44:38,130 --> 00:44:45,130
of sampling then you are almost recovering
the continuous domain function, okay depending
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00:44:47,299 --> 00:44:51,900
on the degree of that sampling rate.
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00:44:51,900 --> 00:44:58,900
Now, if you are very very close and if you
are using sample and hold elements that means;
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00:45:00,329 --> 00:45:07,329
your approximate approximating this function
by this type of function, okay, these size
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00:45:12,960 --> 00:45:19,960
constant functions. And if they are very very
close then that error will be minimise. Now,
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00:45:20,460 --> 00:45:27,460
suppose the function is changing very slowly,
all right then you can relax the sampling
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00:45:27,950 --> 00:45:31,329
rate, you can reduce the sampling rate.
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00:45:31,329 --> 00:45:38,329
So, you drop, say we take only alternate values
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00:45:47,230 --> 00:45:54,230
whatever it be this, in-between these values;
I made zeros, will it be same? If the function
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00:46:02,009 --> 00:46:09,009
was changing very slowly and if I sample it
instead of at this high rate, I have got these
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00:46:09,920 --> 00:46:16,920
sample values. Now, I drop in-between, these
values we drop that means, you make them zeros,
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00:46:22,400 --> 00:46:29,400
okay. Then we get samples like this; that
may represent the original signal to the satisfactory
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00:46:35,349 --> 00:46:36,359
level.
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00:46:36,359 --> 00:46:43,359
We could have dropped every third or sorry,
we could have dropped two consecutive values,
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00:46:45,450 --> 00:46:52,450
that is; we could have taken only the, only
every third value x zero then x three then
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00:46:54,150 --> 00:47:01,150
x six and so on. And make x one, x two as
zeros, we could have done that. And then you
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00:47:02,519 --> 00:47:09,519
compress it, then you compress it back to
its original sampling time; then you get the
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00:47:12,150 --> 00:47:19,150
same wave, seen in a very compressed form,
okay.
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00:47:19,400 --> 00:47:24,069
You may not lose much of information, if the
sampling rate original sampling rate was very
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00:47:24,069 --> 00:47:30,349
high or compared to the sampling rate, their
signal is changing very slowly, okay. So,
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00:47:30,349 --> 00:47:37,349
you can keep on trying to decimate to down
sample, keep on doing it till you find that
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00:47:38,829 --> 00:47:45,240
important information are getting lost, till
that time you can do these. Will get back
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00:47:45,240 --> 00:47:52,240
to this problem latter on, what are the conditions
for fixing the sampling rate, for a given
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00:47:52,700 --> 00:47:57,799
signal that we shall discuss in subsequent
discussions.
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00:47:57,799 --> 00:48:04,799
So the basically, in down sampling we drop
the values. These are dropped and then again
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00:48:09,400 --> 00:48:16,400
these are compressed. In up sampling, we insert
some artificial values, zeros; so, you are
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00:48:23,180 --> 00:48:30,180
given a sequence say for example, you may
be given the power demand for campus at regular
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00:48:31,769 --> 00:48:38,769
intervals for one hour, we have the records,
at say 6 o' clock, 7 o' clock, 8 o' clock,
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00:48:42,430 --> 00:48:47,940
in the morning 9 and so on, I have these values.
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00:48:47,940 --> 00:48:54,940
The demand in megavolt may be like this. One
may insert some zero values, because they
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00:48:59,200 --> 00:49:06,200
were not defined, because they were not defined;
you can insert zeros also, will it be same?
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00:49:11,059 --> 00:49:18,059
Basically, this denotes the profile of the
actual variable to a certain extent all right,
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00:49:23,200 --> 00:49:29,319
because the demand is not very varying very
fast; so, we we can sample it at a at an interval
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00:49:29,319 --> 00:49:35,690
of one hour, now if somebody puts zeros here,
though initially, we said that here it is
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00:49:35,690 --> 00:49:40,109
not defined but it cannot be zero, is it not?
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00:49:40,109 --> 00:49:47,109
Now, if I try to make a zero here and then
the profile will be like this, is it not?
250
00:49:52,499 --> 00:49:58,960
It is not same as this. You are bringing it
down to zero, so there is a sharp change taking
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00:49:58,960 --> 00:50:05,960
place which is not representing the original
sequence. A sharp change means high frequency
252
00:50:06,460 --> 00:50:09,660
components, so you have to eliminate this.
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00:50:09,660 --> 00:50:16,660
So, if this sequence is pass through a low
pass filter then you can get back more or
254
00:50:17,420 --> 00:50:24,420
less the original sequence. One may insert
some values by linearly interpolating these.
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00:50:28,869 --> 00:50:35,670
That is take the values in-between, between
6 o'clock and 7 o' clock whatever values you
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00:50:35,670 --> 00:50:41,900
want, at six thirty, it will be the average
of the data that you get at these hours. I
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00:50:41,900 --> 00:50:47,029
add these plus these by two, these plus these
by two; this is a simple linear interpolation
258
00:50:47,029 --> 00:50:54,029
of second order, okay. Then probably, it will
represent to a certain extent, the original
259
00:50:56,710 --> 00:51:02,829
profile, okay.
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00:51:02,829 --> 00:51:09,829
So, for a compressor y n, if this is the output
of a compressor; this is x n and we are down
261
00:51:20,119 --> 00:51:27,119
sampling by M, it is x n M, okay. And rest
of the values are all zero, only the Mth value
262
00:51:38,809 --> 00:51:45,809
is counted, okay. Suppose, M is equal to five
then n you put as one, it will be one into
263
00:51:50,309 --> 00:51:57,309
five, so x five will be treated as y one,
x zero as y zero, x five as y one, x ten as
264
00:51:59,529 --> 00:52:06,529
y two and so on, so this is for a down sampler.
265
00:52:10,359 --> 00:52:17,359
In case of an up sampler, y n will be x n
by L where symbol is, x n is the input, this
266
00:52:37,910 --> 00:52:44,910
is the symbol for the up sampler; it is your
increasing the sampling rate. And y n, x is
267
00:52:45,059 --> 00:52:52,059
equal to n by L where n is equal to 0 plus
minus L plus minus 2L and so on and is equal
268
00:52:57,130 --> 00:53:04,130
to 0 otherwise; that means if this is your
x 0, x 1, x 0 2 and so on, okay. Then you
269
00:53:22,099 --> 00:53:29,099
are inserting zeros here
and you are generating a new sequence. So,
x 1 whatever was x ten, sorry x 1 will be
270
00:53:43,670 --> 00:53:48,470
now, say x three or x four, depending on the
number of zeros that you are inserting; that
271
00:53:48,470 --> 00:53:55,470
is the factor L, okay. We will stop here,
today, next time we take up from this point,
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00:54:03,579 --> 00:54:09,759
okay. Thank you very much.