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Today, before we start our topic; that is
the relationship between continuous and discrete
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time signals, we will just briefly go through
some of
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the examples that we have taken up last time.
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If you remember, the last problem that we
took was a sequence which was symmetric, okay.
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It is like minus 1 plus 2 minus 3 again plus
2 and
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minus 1, it is symmetric about this central
point, okay. So, if you consider the signal,
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from the origin then we are asked to compute
some of
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these quantities; like this one and then angle
of X omega, and so on, integral minus p i
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to plus p i X omega, d omega and so on.
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Now, you can see first one; X e to the power
j 0 by putting omega equal to 0 what we get
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is summation of these terms, minus 1 plus
2 minus 3
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plus 2 minus 1 which gave us minus 1. Then
angle of X omega while computing this, there
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was a little bit of confusion and I deliberately
did not
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mention about the starting point.
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See, the first one is minus 1, if I pair them
minus 1 into 1 plus e to the power minus j
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4 omega, okay; so that can be written as,
if I take e to the
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power minus j 2 omega common, inside I will
have e to the power plus j 2 omega and e to
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the power minus j 2 omega which will be twice
cos of
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2 omega. Next, if I pair this and this it
will be, plus I will write these terms here.
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If I take two common, it will be e to the
power; I will call this as X 1 okay, X 1 omega
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this is the first part, e to the power minus
j omega plus e to
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the power minus j 3 omega that gives me, 2
into 2 into e to the power minus j 2 omega.
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Once again, I take e to the power minus j
2 omega
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common, so it will be e to the power plus
j omega, e to the power minus j omega which
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will be twice cos of that 2 I have taken out,
so cos of
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omega, okay.
And thirdly, it was 3 into cosine of sorry
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3 into e to the power minus j 2 omega. So,
if I add all the three terms, X e to the power
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j omega will be e
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to the power minus j 2 omega common and I
get inside the bracket twice cosine 2 omega
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plus minus plus 4 cos omega plus or minus
minus 3,
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that is all. Now, this is a real quantity,
so this gives only me the phase, so the phase
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theta omega will be minus 2 omega, okay.
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It is not 90 degrees, it is just minus 2 omega,
last time there was little slip. Now, if I
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have the sequence instead of starting from
here, sequence
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starting at this point; that is it is symmetric
about the origin then this will be starting
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with 3 into e to the power 0 and this will
be 2 into e to the
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power minus j omega, this will be 2 into e
to the power plus j omega.
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So this term, that is you are giving a shift
of the origin. So, this term will be multiplied
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by e to the power plus j 2 omega and hence
there will not
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be any angle associated with this. So, for
a symmetric sequence, for a symmetric sequence
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if the symmetry is about the origin
then we get
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angle theta is equal to 0, all right. Next
the third part was, if you remember we have
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asked you to compute minus p i to plus p i
X omega, d
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omega.
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Now, what is this? 1 by 2 p i into this; if
you remember, X n is equal to 1 by 2 p i minus
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p i to plus p i X omega e to the power j omega
n d
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omega is that all right? Now, if I put n is
equal to 0, this becomes just integral of
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X omega, d omega is that all right. So, that
is equal to x 0. So,
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what is this integral equal to 2 p i into
x 0?
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And you have been asked to compute; sorry
you have been asked to compute this integral,
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so that will be x 0. So, that is equal to
into 2 p i,
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okay. So that is equal to minus 2 p i, is
that all right, x 0, is minus 1. Next, it
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was what the fourth part? X at p i, all right.
If I put here X at p i, yes?
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How much is it? If you put omega is equal
to p i, what do you get from the series? What
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do you get from the series? X n e to the power
minus j
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p i n, is it not?
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And that is equal to; basically you are just
alternately changing the sign, e to the power
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minus j p i n means, when n is odd it is p
i, 3 p i, 5 p i
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when n is even it will be minus j 2 p i and
that is equal to 1. So, you just alternately
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change the signs. So the original sequence
that was given;
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minus 1 plus 2 into minus 1 minus 3 plus 2
into minus 1 minus 1, is that all right. So,
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that is equal to minus 1, minus 2, minus 3,
minus 2,
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minus 1, so that is minus 9.
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The last one, that we discussed, we did not
discuss that was asked was X omega square
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d omega, okay. Now, from Parseval’s theorem
that is
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the energy content of the signal; whether
you represent the signal in the time domain
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or the frequency domain will be same, and
that is equal to
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square of the components, okay, summation
of that, in the continuous domain it is integral.
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So that is will this one will be equal to
minus 1
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square plus 2 square plus 3 square plus 2
square plus 1, 4 plus 1, 5, plus 5, 10 plus
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9, 19, okay.
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Now this is actually, 1 by this is equal to
this into 2 p i, okay. So, 19 into 2 p i;
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that is 38 p i. Parseval’s theorem states
that, this is equal to 2 p i
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into sigma x n square, all right. Now, let
us take up another interesting problem, before
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we go over to the relationship. You are given,
x n this is
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a low pass filter; you are having an adder
here and this is y n. This low pass, low pass
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filter is having the characteristics like
this, minus p i to
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minus p i by 2 to plus p i by 2, it is 1 and
this is a characteristics of H e to the power
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j omega, okay, and ideal low pass filter.
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What will be y n like; if I give say a delta
input you are giving an impulse input, what
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will be the output, okay. So, let us take
this as some w n.
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So this is minus 1 to the power n into w n
and this is w n, okay. Now, what will be w
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e to the power j omega? It will be product
of the two
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frequency transforms, frequency transform
of delta n is 1, okay.
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So, if I call that as input X into H e to
the power and this is equal to 1; so, it will
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be H e to the power j omega okay. What will
be the frequency
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transform of this?
This is already known, same as h. What will
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be the frequency transform of this? If, I
call that as say; some W dashed n then what
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is the
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transform of W dashed?
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Now, minus 1 to the power n, I can write as
e to the power minus j p i n, is it not? Into
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w n, the frequency transform of this, is it
not? I am taking
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frequency transform of w dashed n which is
nothing but w n into e to the power minus
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j p i, okay. So, in the frequency domain,
if I multiply any
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function x or say f by e to the power minus
some quantity j into something, then what
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is the transform of that product?
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Will be the original function shifted by j
p i, all right; so this is nothing but H because
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W is equal to H for a delta input, so H e
to the power j
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omega minus p i. So, what will be Y e to the
power j omega? It is nothing but W plus W
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dashed, okay. So, this is H plus H e to the
power j
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omega minus p i, correct me if I am wrong,
is that all right?
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And, how much is this? If H omega is extending
from minus p i by 2 plus p i by 2 and after
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that after that it is all 0. This is omega.
This is 1,
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then just take any value. So, omega is equal
to a little above p i by 2, all right. Then
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this will be zero and this will be one. You
just see, what is w
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dashed, if this is H e to the power j omega,
what will be the shape of e to the power j
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omega minus p i? It I will start from p i
by 2, 2, 3 p i by 2, is
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that all right.
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It will be like this. So, if I add them together,
it will be always 1, okay. So, what will be
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Y n? This plus this, okay; if I keep on changing
this
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periodic and it is coming continuously as
1, so y n is delta n. So, if I have an ideal
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low pass filter of ideal low pass filter of
band minus p i by 2 to
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plus p i by 2 then this kind of arrangement
will give me, an overall transfer function
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unity. I give delta n input, delta n output,
okay.
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Okay, before we take up any more examples,
let us will take up some more interesting
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examples later on; let us get the relationship
between
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the continuous domain and discrete domain
signals. In real life, signals are signals
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are continuous, mostly all right; whether
it is speech or it
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may be the temperature recorded or may be
the voltage of a particular node. You record
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anything that is basically, analogue or continuous.
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Now, you discretize it in the time domain
and then you process it in a computer. So,
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for processing again the discrete signal depending
on the
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memory available, we discretize in magnitude,
okay, representing it by some bits. So, when
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we are doing that it requires some time.
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We discretize in magnitude that means, we
measure them at discrete intervals of time
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then this value is to be represented in the
digital form.
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So, it will take some time depending on the
length of the memory. So, we allow that time;
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so till that time it is kept on hold at this
value all right,
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so we also require sample and hold circuit.
The device by which we discretize it is analogue
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2 digital converter or we call it, ADC.
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Then similarly from the discreet value; we
translate them into continuous value we call
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it digital to analogue conversion and apart
form that, we
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require whole circuits. These are not the
only elements when we are getting a signal
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and we are using that hold element then we
are getting the
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sampled value like this, okay.
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These values are processed and then we are
doing some kind of processing, depending on
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your requirement. And then you are getting
the
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output converted into continuous domain signal;
while doing that, what you are getting is
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a discrete even though it is a continuous
function but
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it is having sharp bends like this, okay.
You have to smoothen it all right. So, at
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the final stage you have a smoothing filter.
So, after processing digital processing, you
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have a DAC and then you have to have a smoothing
filter, okay; smoothing filter, you also call
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it
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reconstruction filter all right or reconstruction
filter, okay. Now before feeding it to teddy
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converter, you also take care of the signal;
will very soon
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see what should be the condition for the signal
for discretizing, before feeding it we also
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pre filter it or we call it anti-aliasing
filter.
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So, the signal is feed here pass through an
anti-aliasing filter or pre-filter and then
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we discretize it, then we process it, then
again we go for
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analogue conversion, then we have smoothing
filter, okay. These two filters are in the
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analogue domain and this is the digital domain;
this filter
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is in the digital domain. So, what is the
most fundamental condition for processing
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a signal with the with the help of a computer?
What should
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be the condition for anti-aliasing filter,
because rest of them are very standard one;
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Anti-aliasing filter.
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Now the signal discrete signal x n, is nothing
but x analogue. We are computing the values
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at regular intervals of T, okay. So, I will
write like this
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and the sampling frequency is 1 by T in the
radian domain 2 p i into f s; that is 2 p
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i by T, so many radians per second. What will
be its
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frequency transform? It is minus infinity
to plus infinity, x at e to the power minus
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j omega t d t, okay.
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And what is the discrete domain representation?
It will be summation x n; this we are writing
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in the discrete domain, we want to establish
a
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relationship between this and this, okay,
you want to establish a relation between this
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and this. Now, x a n T, I can represent in
two ways, I
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should put this bracket. Now, this can be
shown as a multiplication
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of the function x t by a function f t
by a function f t which is basically, a string
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of delta function with period T, is it not,
that means okay.
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If I call these as periodic, X p j omega,
okay; this is a periodic function, so if I
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take the Fourier transform of this, I call
this as so this multiplied by
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x t, I call this as x p t, all right. When
we are looking at it as a product of this
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periodic function and the function x t, okay,
I call it x p t. What is the
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Fourier transform of this? This is x p j omega,
that is Fourier transform of x p t, okay.
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Now this I can look at it as, there are two
ways of looking at it; one is a summation
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x a n T, x a n T is a magnitude, okay minus
e to the power j n
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omega T, is it not. This small omega, I am
writing as omega into T that is at only those
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times, if you are taking it, so it can be
written like this.
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And the other way it will be; okay let me
write for f t first, f t is this will discuss
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later on in details. See, if I have a delta
function, a string of delta
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functions
and so on.
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This is summation delta n minus n minus, okay
I can write this also in the discrete domain.
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So, the transform of this e to the power j,
I am just
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writing this as K t okay, I can write f t
like this, is it not? Whenever, T is equal
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to K t, I take only discrete values, okay.
Whenever T becomes
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equal to this then it I will be giving me
a pulse; so f t can be written like this,
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this is f t which is nothing but summation
of this, okay.
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So, 1 upon T, summation e to the power j omega
s K t in terms of something frequency all
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right. So, correspondingly this is f t, so
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correspondingly if f t such a function; X
p j omega will be 1 by T summation X a j omega
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minus K omega s, okay. Translation, any function
is
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multiplied by e to the power some exponential
function; e to the power j, this thing will
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be having in the frequency domain, a shift
of K into
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omega s, okay, one of those properties of
Fourier transform, shifting property.
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So, it is nothing but a periodic function
all right. A summation, summation of a function
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which is shifted by omega s, two omega s,
three omega
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s and so on, is that all right. What does
it mean?
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Suppose, X a is a function like this; in the
null of domain, what is X a j omega minus
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omega s minus 2 omega s minus 3 omega s? So,
this is
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omega s and again around that omega s, you
are having a replica of this, is it not? Again,
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after 2 omega s we are having replica of this,
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similarly on this side okay, is it not?
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00:32:44,260 --> 00:32:57,820
So, if I have corresponding to this, say the
time function is like this; this is the original
203
00:32:57,820 --> 00:33:05,490
time signal, all right who’s Fourier transform
will be this,
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00:33:05,490 --> 00:33:18,470
okay. And the entire set corresponds to this
one, if I scaled it down by 1 by T that represents
205
00:33:18,470 --> 00:33:25,240
basically X p j omega. So, if this was X then
this
206
00:33:25,240 --> 00:33:32,920
whole thing is scaled by a factor 1 by T will
be X p, this entire set is X p, is that all
207
00:33:32,920 --> 00:33:33,920
right?
208
00:33:33,920 --> 00:33:43,860
So this, this one corresponds to the Fourier
transform of this, and this one is 1 by T
209
00:33:43,860 --> 00:33:49,380
times; I have to multiply by a factor 1 by
T, that I will give me
210
00:33:49,380 --> 00:34:05,010
X p, okay. How have you got it? If I consider,
in the time domain you are multiplying X T
211
00:34:05,010 --> 00:34:09,740
by what? By a chain of delta function, all
right? There is
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00:34:09,740 --> 00:34:18,081
a train of pulses, impulses of magnitude 1
and you are multiplying by that, okay. So
213
00:34:18,081 --> 00:34:21,409
in the frequency domain you can take the products,
all
214
00:34:21,409 --> 00:34:34,419
right. So, what is the product? One is this
one to be multiplied, by no x omega?
215
00:34:34,419 --> 00:34:47,460
See, if I have a delta function; periodically
appearing then what will be its Fourier transform?
216
00:34:47,460 --> 00:34:56,159
Also at chain of pulses in the frequency domain,
if
217
00:34:56,159 --> 00:35:17,200
this is f p t, t then f omega F p omega will
also be pulses, okay chain up pulses like
218
00:35:17,200 --> 00:35:26,809
this, sorry. So, this one I can put j omega,
this one is
219
00:35:26,809 --> 00:35:39,660
convolved with this. You convolve, just X
t, X j omega with these pulses, you will get
220
00:35:39,660 --> 00:35:46,329
this, is it not; convolution of one pulse
with this one which
221
00:35:46,329 --> 00:35:49,780
will give me the same function.
222
00:35:49,780 --> 00:35:59,490
And with a train of pulses, you will get repetition
of this, is that clear? So, in the time domain
223
00:35:59,490 --> 00:36:02,400
when we are multiplying, in the frequency
domain
224
00:36:02,400 --> 00:36:06,869
domain will be taking convolution of the two
frequency domain representations. And that
225
00:36:06,869 --> 00:36:12,730
is how we have got it. And we have seen it
also from
226
00:36:12,730 --> 00:36:14,700
the summation expression.
227
00:36:14,700 --> 00:36:31,059
Now, question comes while taking convolution,
this has got nothing to do with the with your
228
00:36:31,059 --> 00:36:35,680
sampling as such; it is depending on the type
of
229
00:36:35,680 --> 00:36:43,500
signal that you are getting. So, this is a
base band of the signal. Original signal,
230
00:36:43,500 --> 00:36:48,710
analogue signal is having this much of band
width; minus
231
00:36:48,710 --> 00:37:08,029
omega m to plus omega m, okay you can call
it omega m by 2 also, okay. Now, if this spacing
232
00:37:08,029 --> 00:37:13,099
is sufficient this is omega s, then how much
is
233
00:37:13,099 --> 00:37:14,099
this?
234
00:37:14,099 --> 00:37:28,400
Omega s minus omega m by 2, okay. So, if omega
s minus omega; in many books, they write just
235
00:37:28,400 --> 00:37:31,609
minus omega m to plus omega m, okay, that
236
00:37:31,609 --> 00:37:36,220
is the normal convention, so let us also follow
that.
237
00:37:36,220 --> 00:37:47,589
Omega s minus omega m should be greater than
omega m. If, I want clearly distinguishable
238
00:37:47,589 --> 00:37:55,510
forms, okay that means omega s should be greater
239
00:37:55,510 --> 00:38:04,029
than twice omega m or omega m should be less
than omega s by 2; that means sampling frequency
240
00:38:04,029 --> 00:38:06,780
should be more than twice the maximum
241
00:38:06,780 --> 00:38:15,580
frequency, that is present in the original
continuous domain signal, this is the same
242
00:38:15,580 --> 00:38:22,070
theorem that we have already discussed, sampling
theorem
243
00:38:22,070 --> 00:38:25,049
sampling theorem or Shannon’s sampling theorem.
244
00:38:25,049 --> 00:38:36,700
Now, if we have an overlap; if supposed this
condition is violated then it is like this,
245
00:38:36,700 --> 00:38:43,769
this is omega m and omega s is omega s by
2 is coming
246
00:38:43,769 --> 00:38:55,220
inside
247
00:38:55,220 --> 00:39:08,039
and so on. So, now when you are taking the
summation, the overall sum will be
248
00:39:08,039 --> 00:39:17,160
like this. So, this is not a clear picture
from here; I cannot
249
00:39:17,160 --> 00:39:25,950
segregate the base band element all right,
that is that segment. Earlier, when this was
250
00:39:25,950 --> 00:39:29,270
repeating with a distinct gap in between;
I could have
251
00:39:29,270 --> 00:39:39,289
now used a low pass filter, I could have used
a low pass filter and filter out this one.
252
00:39:39,289 --> 00:39:47,019
If you take the inversion, you will get the
original analog time domain signal; that means
253
00:39:47,019 --> 00:39:51,390
if you are given a discrete domain signal,
if you take it
254
00:39:51,390 --> 00:40:01,329
is Fourier transform, all right, if you pass
it through a low pass filter then again take
255
00:40:01,329 --> 00:40:07,549
inverse transform, you get the analog signal.
But, now if I
256
00:40:07,549 --> 00:40:14,480
have a low pass filter, I will be having a
signal which is not a replica of the original
257
00:40:14,480 --> 00:40:20,450
signal, you see here there will be lot of
error, okay. So, to
258
00:40:20,450 --> 00:40:24,279
avoid this we impose this condition.
259
00:40:24,279 --> 00:40:40,980
Now in reality, you may not have a signal
with a definite value of sigma m or the magnitude
260
00:40:40,980 --> 00:40:47,490
goes on diminishing but at a very slow rate,
it is like
261
00:40:47,490 --> 00:40:58,849
this. So somewhere we truncate it. This frequency
is ensured to be less than omega s by 2, okay.
262
00:40:58,849 --> 00:41:08,510
Then you get
this is the base band and then
263
00:41:08,510 --> 00:41:18,730
over this, you have another repetition and
so on, but there is a separation. This is
264
00:41:18,730 --> 00:41:24,660
somewhat closer to the original signal then
this distortion;
265
00:41:24,660 --> 00:41:30,720
because here because of the overlap even much
before this here the distortion is only at,
266
00:41:30,720 --> 00:41:35,570
the tail only this portion is eliminated,
okay.
267
00:41:35,570 --> 00:41:43,420
But, if it is like this then the other tail
also comes in, so the corruption starts from
268
00:41:43,420 --> 00:41:48,760
here because the resultant is somewhere here,
you see
269
00:41:48,760 --> 00:41:58,650
somewhere here. So, the value gets corrupted
even much before this value, omega s by 2,
270
00:41:58,650 --> 00:42:03,450
okay. So, that the distortion is less, if
we do the pre
271
00:42:03,450 --> 00:42:14,529
filtering where we are restricting the band,
okay.
272
00:42:14,529 --> 00:42:29,130
Now, let us take an example; it will be very
interesting, to demonstrate the effect of
273
00:42:29,130 --> 00:42:31,819
sampling on signals.
274
00:42:31,819 --> 00:42:48,099
Suppose, you have X 1 t is equal to 10 cosine
2 p i into 100 t plus 30 sin 2 p i into 200
275
00:42:48,099 --> 00:43:04,220
t minus 20 sin 2 p i into 400 t plus 15 cosine
2 p i into
276
00:43:04,220 --> 00:43:24,180
600 t plus 15 sin 2 p i into 900 t plus 0.5,
sorry, 5 cosine 2 p i into 1100 t. Suppose,
277
00:43:24,180 --> 00:43:31,990
we sample it at a frequency of 1 KHz, what
I will be the
278
00:43:31,990 --> 00:43:44,809
discrete domain representation of this signal?
Will be 10 cosine 2 p i into 100 into t, 1
279
00:43:44,809 --> 00:43:48,309
kilo hertz is the sampling times, so sampling
280
00:43:48,309 --> 00:43:55,119
frequencies. So, T is equal to 0, 0, 1 second,
okay.
281
00:43:55,119 --> 00:44:13,069
So, you put n into capital T, so this will
become 0.2 p i n plus 30 sin 0.4 p i n, correct
282
00:44:13,069 --> 00:44:25,349
me if I am wrong, minus 20 sin 0.8 p i n plus
15 cosine.
283
00:44:25,349 --> 00:44:36,490
Now, this will become 1.2 p i n. Now, 1.2
p i, I can always write as; 2 p i minus 0.8
284
00:44:36,490 --> 00:44:44,789
p i into n, okay, so 2 p i n minus 0.8 p i
n cos of 2 p i minus
285
00:44:44,789 --> 00:44:57,289
0.8 p i n. So, it will become 0.8 p i n, if
you permit me to write in one step.
286
00:44:57,289 --> 00:45:09,569
And then, where it would be, this one and
then plus 15 sin 1.8 p i n, 1.8 p i n. So,
287
00:45:09,569 --> 00:45:14,380
that can be written as 2 p i n minus 0.2 p
i n. So, sin of this
288
00:45:14,380 --> 00:45:32,250
means, minus 15 sin 0.2 p i n okay plus 5
cos; this is 2.2, if I subtract 2 p i then
289
00:45:32,250 --> 00:45:44,930
it 2 will become 0.2 p i n, okay. So, check
into 5,5 so 10 plus 5,
290
00:45:44,930 --> 00:46:16,670
15 cosine 0.2 p i n, okay minus 15 sin 0.2
p i n then plus 30 sin 0.4 p i n. Again, minus
291
00:46:16,670 --> 00:46:27,560
20 sin 0.8 p i n plus 15 cosine 0.8 p i n.
292
00:46:27,560 --> 00:46:37,640
I can write this as; this is 15 and 15, so
15 root 2 cos and sin, so cosine 0.2 p i n
293
00:46:37,640 --> 00:46:47,359
then this plus 45 degrees or minus 45 degrees?
Plus p i by 4,
294
00:46:47,359 --> 00:47:02,650
is that all right? Then this and this okay,
30 I write this first 30 sin 0.4 p i n
295
00:47:02,650 --> 00:47:09,500
and then minus 20 and 15 same frequency. So,
this is 20 square
296
00:47:09,500 --> 00:47:18,910
plus 15 square under root 25 okay, plus 25
cosine, cos a, cos b minus sin a, sin b; so
297
00:47:18,910 --> 00:47:30,020
cosine 0.8 p i n plus theta where n theta
is 20 by 15,
298
00:47:30,020 --> 00:47:44,749
okay. So, I can write like this where n theta
is minus, sorry plus 20 by 15, okay. It is
299
00:47:44,749 --> 00:47:48,980
4 by 3.
300
00:47:48,980 --> 00:48:03,490
Now, one more signal; we take, x 2 t, x 2
t which will be 10 cosine, which is different
301
00:48:03,490 --> 00:48:19,749
from this minus 30 sin 2 p i into 800 t minus
20 sin 2 p i
302
00:48:19,749 --> 00:48:45,849
into 400 t plus 15 cosine 2 p i into 1400
t plus 15 sin of 2 p i into 1100 t plus 5
303
00:48:45,849 --> 00:48:55,260
cosine of 2 p i into 1900 t. Now, if I have
a function like this, what
304
00:48:55,260 --> 00:49:07,609
will be x 2 n? Once again by the similar conversions;
we get 10 cosine, this is 800 into sorry 800
305
00:49:07,609 --> 00:49:14,910
into p i, so, that give that will give you
1.8 p i.
306
00:49:14,910 --> 00:49:30,109
So, that is 0.2 p i, 2 p i minus 0.2 p i cos
of 2 p i minus theta is cos theta itself.
307
00:49:30,109 --> 00:49:41,330
And then minus 30 into sin this is 1.6 p i,
so that will be plus 30
308
00:49:41,330 --> 00:49:49,299
sin 0. 4 p i n, okay.
309
00:49:49,299 --> 00:50:15,059
Next, 20 sin this is 800, so 20 sin 0.8 p
i, okay 0.8 p i, plus 15 cosine this is 2.8,
310
00:50:15,059 --> 00:50:28,849
that is equal to 0.8 p i n plus 15 into 2200.
So, this is actually
311
00:50:28,849 --> 00:50:40,809
minus, this was minus I am sorry, so this
will become minus 15 into sin 1, 2 p i n plus
312
00:50:40,809 --> 00:50:56,789
5 cos, this one is 5 cos 0.2 p i n, is that
all right. One is
313
00:50:56,789 --> 00:51:04,960
900, other one is 1900 but both of them are
generating 0.2 p i n. So, this is 10 plus
314
00:51:04,960 --> 00:51:12,490
5, 15 cosine 0.2 p i n; you check with a previous
example,
315
00:51:12,490 --> 00:51:24,190
is it coming same? Plus 30 sin 0.4 p i n.
I leave it to you to, complete this, we will
316
00:51:24,190 --> 00:51:27,020
come as same as x 1 n.
317
00:51:27,020 --> 00:51:41,819
Now, you change the sampling rate to 2 kilohertz;
will find x 1 n is not equal to x 1 n or x
318
00:51:41,819 --> 00:51:48,890
1 n can be equal to x 2 n, I ask you to verify
for these
319
00:51:48,890 --> 00:51:57,950
two frequencies, what will be the sample sequences
x 1 n and x 2 n? Are they equal for some of
320
00:51:57,950 --> 00:52:05,160
these frequencies? Will you guess or are they
321
00:52:05,160 --> 00:52:12,009
different? That means, two signals might appear
to be identical under a certain condition
322
00:52:12,009 --> 00:52:16,359
of sampling, it might be different at some
of the
323
00:52:16,359 --> 00:52:22,299
sampling rates. I want all of you to verify
this by taking two different frequencies and
324
00:52:22,299 --> 00:52:26,970
once again translate the time domain description
to
325
00:52:26,970 --> 00:52:33,309
continuous domain description, to discrete
domain description like this, okay. Thank
326
00:52:33,309 --> 00:52:37,900
you very much. We shall take up some examples
in the
327
00:52:37,900 --> 00:52:41,299
next class, using the frequency transforms.