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We are taking up a few numerical problems on discrete time systems and their transforms.
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So, let us start with the first question,
the response of an LTI system, is given here
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for an input of 1, 0 minus 1; the output sequence
is 1, 0 minus 1, 0, 1, 0 minus 1, 0, it continues.
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You determine
the impulse response
of the system; determine the impulse response
of the system, okay so, let us see if we apply
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Z transform then what would be X z, what would
be X z, 1 minus z to the power minus 2. What
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will be the corresponding Y z? 1 minus z to
the power minus 2 plus z to the power minus
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4 minus z z to the power minus 6 and so on,
okay.
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So, this is an infinite sequence, this will
be giving me 1 plus z to the power minus 2,
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okay; z to the power z magnitude should be
less than 1, that is our assumption. So, what
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will be H z? Y z by X z, there is a noise
1 plus z to the power minus 2 into 1 minus
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z to the power minus 2; that will be 1 minus
z to the power minus 4.
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So, what will be the inverse of this, very
you are very familiar with this; it will be
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1, 0, 0, 0 minus 1 or plus 1? If it is minus
then in the series it is basically z to the
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power minus 4 with plus signs. So, it is a
sequence like this, okay. This will be the
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impulse response.
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Now, next example is there is a sequence given
2, root 2, 0, minus root 2, minus 2, minus
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root 2, 0, root 2. This is just one period
of a sequence; A sin omega naught n plus phi,
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can you determine A omega naught and phi?
One may detected by inspection but we will
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do it, will solve it by rigorous mathematical
technique.
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Let me see, first of all these set of values
first one is 2 next one is root 2, next one
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is 0, next one is root 2, next is minus root
2 of course then minus 2, then again minus
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root 2, then 0, then root 2, next it repeats
again at this point. So this is a period,
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okay. You sketch it
in the form of a sinusoidal function, okay.
Now, tell me what will be A from here, what
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is the expression for this?
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Okay, A is equal to 2 and then what will be
the function h n, you may be able to write
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it, will solve it because the sequence could
have been something else also, here it is
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very simple. So, this is 2 p i, okay, this
is 0, this is the radian frequency omega.
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So 2, is it a sign function? Cos function,
so in terms of sin, sin how much is omega
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naught? How much is omega naught, how much
is this?
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So, 2 p i by 8; this is 45 degrees, 90 degrees.
So, 2 p i by 8 n all right and how much is
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the phase shift? Plus or minus, how much is
the phase shift plus or minus? Plus p i by
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2, is it all right? So, it will be 2 sin p
i by 4 n plus p i by 2, okay. So, let us see
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how we solve this, analytically. Put n is
equal to 0, so A sin phi that is equal to
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2.
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Next, A sin omega naught plus phi is equal
to root 2, know this is equation 2. A sin
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twice omega naught plus phi is equal to 0,
okay, is that all right, so substitute from
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here. In the second equation, the second equation
is A sin, break it up omega naught, cosine
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phi plus A cosine omega naught, sin phi is
equal to root 2 and A sin phi is 2, okay.
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If I put A sin phi is equal to 2, this A sin
phi will become 2. So, A cos phi sin omega
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naught plus 2 into cos omega naught is equal
to root 2; so this is modified equation 2,
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what about the third equation? A sin 2 omega
naught cos phi plus A cos 2 omega naught,
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sin phi is equal to 0, okay; put it to this
side A will get cancelled, okay or okay retain
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A.
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A sin omega naught, cosine omega naught, cos
phi okay into 2 plus A sin phi is 2, 2 cos
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omega naught squared twice sorry, cos squared
omega naught minus 1 into 2 is A sin phi is
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2, is equal to 0. Now, from this 1 and this
1 we can eliminate A; if you try to substitute
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this here, will finally get A sin p i by 4
plus phi is equal to root 2. And from there
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another equation get; before this 2 root 2
cos omega naught equal to 2 from where we
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get cos omega naught is equal to 1 by root
2 and hence omega naught is equal to p i by
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4, okay.
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That has been substituted here, you will get
this relation first by elimination okay and
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substitute it here, it will be root 2. So
from there you can compute phi, okay. There
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is one relationship A sin phi is equal to
2 and this one, if you expand and then substitute
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this condition; will get 2 cot of phi plus
2 is equal to 2, you just expand it A sin
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p i by 4 is 1 by root 2 that 1 by root 2 I
shift here, cos phi by 4 is also 1 by root
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2.
So, the 1 by root 2 terms, if I put here that
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will become 2 and on the left hand side we
will get this, friends. Cot phi will come
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out to be 0, so phi is p i by 2. So, it will
be A sin n p i by 2 sorry, n p i by 4 plus
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p i by 2 which means of course cos sign, is
that all right. Let us take up another example.
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I think I should change the pen. You are asked
to determine Y n which is the convolution
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of
x n and h n, okay. x n is given as n, n greater
than equal to 0, equal to 0, n less than 0,
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instead I can put greater than 0 because when
n is equal to 0, it automatically become 0
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at x is equal to 0. So, this equal to sign,
in this particular case does not have any
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significance, it may be just greater than
and h n, what is this function? It is a ram
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function; h n is minus n to the power n, n
greater than equal to 0, equal to 0 for n
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less than 0.
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The purpose of defining any any function in
these two ranges, is to just a wide writing
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u n; otherwise you can write n into u n, we
do not have to put all these, okay similarly
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this one. So, just just sketch the functions.
One is like this 1, 2, 3 and so on, this is
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x n the other one is h n, all right which
will be alternately plus 1, minus 1, this
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is start with plus 1 n is equal to 0, it will
be plus 1, minus 1 and so on, okay.
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So, let us see what will the response corresponding
to this; this is h n. Now, using Z transform
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you are asked to find out response using Z
transform, okay. So, X z is sigma n z to the
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power minus n, all right. So, if you remember
this is z, d, d z of sigma z to the power
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minus n which is minus z, d d z of, for z
to the power of minus n, what is the sum?
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n 1 by 1 minus z inverse? So, that you can
reduce it to this form also, just multiplied
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throughout by z.
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So, take the derivative that gives me z minus
1 into 1, minus z, so divided by z minus 1
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whole square; z into derivative of this. So
that will be, minus z will get cancelled,
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so minus 1 and minus 1 will get plus z by
z minus 1 whole square, is it okay. What will
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be h n, H z, H z; will be 1 by 1 plus z inverse
which means, z by z plus 1, okay.
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So, Y z will be the product of x n and x z;
that will be z square by z minus 1 whole squared
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into z plus 1, is it so? So, take the product
and make a partial fraction. So, z minus 1
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whole squared into z plus 1, I can write this
as A by z plus 1 plus B by z minus 1, okay
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sorry sorry; I take y z by z whenever you
are having the function in terms of z, not
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z inverse in terms of z then it is better
to take y z by z as a partial fraction,
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Why? After getting A B C, I will multiply
z here then A z by z plus 1 is a standard
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form, B z by z minus 1, C z by z minus 1 whole
squared, we know the corresponding values,
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okay values of variance. Similar thing you
have done in the network synthesis also, remember
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RL circuit? You take z as p i s first, make
the partial fractions and then get z s. So,
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what will be A if we multiplied by z plus
1, evaluated at z equal to minus 1?
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So, if I multiply these by z plus 1 and divided
by z; so it becomes z by z minus 1 whole squared
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at z is equal to minus 1, so that gives me
minus 1 by 4, is it all right. Similarly,
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C let us evaluate C first; because that takes
all the multiple roots. C, it will be Y z
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by z, z minus 1 whole squared, so that gives
me z by z plus 1; to be evaluated at z is
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equal to plus 1.
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So, this will be 1 by 2 and it is this function
which is to be differentiated with respect
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to z, so for the evaluation of B and if there
are more number of roots then you start with
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the highest term, term with the highest number
of roots and then keep differentiating. So,
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B will be d d z of z by z plus 1 and evaluated
at z is equal to 1, so that gives me again,
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1 by 4. Is any other opinion about this for
evaluation of B? Subtract, A by z plus one
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and C by z minus one whole square, from this
side you can straight away get B by z plus
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1, you do not have to take the derivative.
Now, is it not all right?
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So, Y z will be A z by z plus 1 plus B z by
z minus 1 plus C z, z minus 1 whole squared.
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Therefore y n, if you take the derivative
or if you take the inverse of this; z by z
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plus 1, how much is it? A into minus 1 to
the power n plus B into 1 to the power n,
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okay plus C into n just one or two problems
earlier we have derived this, is it not; for
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that ram function. So, whole thing multiplied
by u n. And now substitute the values A is
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1 by 4 okay, into minus 1 to the power n;
you can write 1 minus by 4 to the power n,
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can I write like that?
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Should, should it be minus 1 by 4 to the power
n, no this is all right? Okay plus 1 by 4
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plus0.5 n whole thing into u n, is that okay?
A was minus 1 by 4, so there will be a minus
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sign here, is that all right; this is the
one we are talking about. Next, let us take
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another example on Z inverse, z by z square
plus 4 z plus 13; this is a very familiar
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problem. We know Z transform of a to the power
n sin b n, how much is that?
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z a sin b divided by z square, no if I write
in terms of z then it will be z square minus
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2 a cos b into z plus a square, if you remember.
So, straightaway matching the coefficients,
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a squared is 13, do you all agree? So, a is
equal to root 13. How much is a cos b? Minus
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2 a cos b is 4, so it will be minus 2. So,
what will be a sin b? 13 minus 2 square 4,
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9 under root; so 13 minus 4 under root so
it will be 3, okay. a square, this square
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plus this square is 13 and that is equal to
a square, a sin b is 3 is that all right?
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So tan b is minus 3 by 2, is anything wrong?
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Therefore, Z inverse z by z squared plus 4
z plus 13 will be I can put; 1 by 3 here and
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3 into z there, divided by z square plus 4
z plus 13. I can write this as z inverse of
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this and which is nothing but, I made it in
the form of z a sin b, a sin b is 3, so I
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put 3 here and taken 1 by 3 outside. So, this
is nothing but this term and the denominator
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is same, so what is the corresponding inverse
a to the power n? Sin b, n.
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So, root 13 to the power n and sin b n, is
okay; I will write sin b n where into of course
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u n, where b is equal to tan inverse minus
3 by 2, is it okay? Another example would
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like to take; determine
the region of convergence ROC of Y z where
Y z is given as sum of X 1 z and X 2 z. X
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1 z is given as z by z plus 1, X 2 z is equal
to z to the power minus 2 by z plus 1; this
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will have a ROC, z greater than 1, this is
also having an ROC z greater than 1, what
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will be the ROC for this function?
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The function converges for this part outside
the unit circle, for this part also there
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is outside the unit circle; so the total sum
will have a region of convergence normally,
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normally let us see, okay. So, add them together,
what do you get?
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Y z, z plus z to the power minus 2 by z plus
1, okay. If, I take z to the power minus 3
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outside minus 2 outside then I will get z
to the power 3 plus 1 by z plus 1 which gives
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me; z cube plus 1 will have z plus 1 as a
common, common factor, so z squared plus z
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sorry minus z plus 1, sorry this minus into
z to the power minus 2 which gives me 1 okay,
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minus z to the power minus 1 plus z to the
power minus 2. And what is ROC of this? When
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is it becoming infinitive? Only, at z is equal
to 0.
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So these convergence; except the origin, it
converges in the entire space so, apparently
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the two function x 1 z and x 2 z, though they
were having the region of convergence outside
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the unit circle, the function x 1 z was converging
only in this region, so was x 2 z, but the
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total sum converges except at this point in
the entire space, all right. Let us see, what
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will be the response, the impulse response
of corresponding to i z? What will be x 1
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n and x 2 n? If we add them together, what
will be this?
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z by z plus 1, what kind of sequence will
it give? Alternately changing the sign, okay
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and z to the power minus 2 by z plus 1, what
is it? 0, 0 and then 1 minus 1, 1 minus 1,
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is it 0, 0? Only one 0? 0, 0.
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So, this was z to the power minus 2. This
is z by z plus 1, thank you.
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This is z by z plus 1 will be first one will
be, second one; it will be 0, 0, 0 then 1
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because I have to take one more 0. So, if
you add them together; so it is a sequence
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first one is acceleratory, so is a second
one with a shift but shift of 1, 2, 3, so
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now it is acceleratory. So, what will be the
net result? It is a finite sequence. It is
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not an infinite sequence.
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So, it will converge okay. Sometimes, it is
a very good example; sometimes now a days
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to reduce noise of any system, we have anti-noise
that means you create a signal which will
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be anti-phase with the noise, noise signal.
So you create another noise, to cancel the
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parent noise. So, that you have silencing
effect, you do not have to pad these, you
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do not have, to have a wall, barrier wall
to prevent yourself from the disturbing noise.
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You can create it by having just some anti-noise
generators, all right.
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If you can know the source of the noise, actual
noise, its behaviour; if you can sense it
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with the sensor and suppose the noise source
is at a far off place and it takes so much
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time to propagate, all right, and you have
a sensor there, it transmits a signal immediately
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the moment the signal comes here in that time,
you have already started the anti-noise generator.
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So that, that will be cancelled with that;
so you will be getting a signal free from
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noise. So, that is a very challenging task,
scientists have working on it. So, this is
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a very good example of digital signal processing,
in a very practical situation.
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Let us take up another example, what will
be the ROC? What will be the ROC of Y z equal
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to H z into X z? It is you are having x n
and h n; convolved output is y n, so in the
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Z domain it will be like this where H z is
given as 1 by z squared plus 7 z plus 10,
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okay. And x n is given as delta n plus delta
sorry, 2 times delta n minus 1; that that
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means you are giving an input of 1 and 2.
This is 0, 1, this is x n, what will be the
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region of convergence for Y z?
So, X z you can compute from here; it will
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be 1 plus 2 z inverse. I can write this as,
Z inverse into 1 plus z, is that be, all right?
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So, Y z will be how much is it? Z inverse,
it should be 2 plus z, 2 plus z divided by
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this and you can factorize this; z plus 2
into z plus 5, so that gives me Z inverse
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00:37:56,790 --> 00:38:03,790
by z plus 5, okay. So, what is a ROC? One
one may take 1 z common outside, so that will
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become z to the power minus 2 divided by 1
plus 5 z inverse, is that all right?
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00:38:20,790 --> 00:38:27,790
So, ROC should be 5 inverse, should be less
than 1 or z should be greater than 5, okay.
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Let us come to another example; okay I will
take another page.
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Consider the signal; x n is equal to u n minus
3 u n minus 1, okay, u n minus 1. y n is equal
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00:39:29,210 --> 00:39:36,210
3 to the power n minus 2 to the power n u
n, what will be h n? So, let us compute X
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00:39:49,599 --> 00:39:56,599
z, it will be okay, z by z plus 1. z by minus
1 all right, minus 3 into u n minus 1, it
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00:40:21,480 --> 00:40:27,760
is z by z minus 1 shifted by one step; so
z inverse multiplied by this, so z inverse
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00:40:27,760 --> 00:40:28,740
will get cancelled with this z.
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00:40:28,740 --> 00:40:35,740
So, it will be 3 by z minus 1. So, we may
take z minus 1, z minus 3, is that so? How
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00:40:44,480 --> 00:40:51,480
much is 8 Y z? 3 to the power n. So, 1 by
1 minus 3 z inverse minus 1 by 1 minus 2 z
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00:41:06,400 --> 00:41:13,400
inverse; which means z by z minus 3 minus
z by z minus 2, correctly me if I am wrong,
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00:41:18,369 --> 00:41:25,369
is it all right? So, that gives me z minus
2 minus z minus 3, so plus 1 by z minus 3
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00:41:27,660 --> 00:41:34,660
into z minus 2 and z can be taken out, okay.
So, H z will be Y z by X z, is it all right?
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00:41:49,390 --> 00:41:56,390
So, z by z minus 3 into z minus 2 into z minus
1 z minus 3, is this all right?
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00:42:14,000 --> 00:42:21,000
So, that equal to z into z minus 1 by z minus
2 into z minus 3 whole squared; make partial
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00:42:29,950 --> 00:42:36,950
fractions and then take the inverse, can you
make the partial fractions, letâ€™s see.
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00:42:40,579 --> 00:42:47,579
A by z minus 2 plus B by z minus 3 plus C
by z minus 3 whole squared. So, A will be
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00:42:56,819 --> 00:43:03,819
multiplied by z minus 2 and then put z is
equal to 2; so it will be 2 into 1 divided
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00:43:08,920 --> 00:43:15,920
by 2 minus 3 whole squared that is 1, so that
gives me 2, you all getting that? C will be,
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00:43:22,470 --> 00:43:29,470
I multiplied by z minus 3 whole squared then
put z is equal to 3; so it will be 3 into
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00:43:30,349 --> 00:43:37,349
2 divided by 3 minus 2, that is 1, so that
is 6, okay. And B will be, d d z of after
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00:43:49,900 --> 00:43:56,900
multiplying by z minus 3 whole squared; that
is z into z minus 1 by z minus 2, derivative
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00:43:57,220 --> 00:44:04,220
of that, if I write z square minus z by z
minus 2 and derivative of this at z is equal
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00:44:07,760 --> 00:44:10,890
to 3, is that all right?
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00:44:10,890 --> 00:44:17,890
So, numerator will be z, 2 z minus 1 into
z minus 2 minus z square minus z by z minus
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00:44:28,890 --> 00:44:35,890
2 whole square evaluated at z is equal to
3. So, 2 z square minus z square; that is
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00:44:41,630 --> 00:44:48,630
z square, 2 z into 2, 4, 5 plus 1, so 4 is
it so? z minus 2 whole square evaluated at
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00:44:56,160 --> 00:45:03,160
3; so that gives me 9 minus 12. So, minus
3 divided by 1 square, so minus 3, okay. So,
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00:45:08,400 --> 00:45:15,400
you have got these three values. So, 2 by
z minus 2 plus sorry, minus 3 by was any mistake?
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00:45:29,170 --> 00:45:36,170
I should have factorised, I told you H z by
z, okay. So, 1 z should have been reduced,
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00:45:41,430 --> 00:45:42,440
okay.
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00:45:42,440 --> 00:45:49,440
So, A z by a Y z by z; first of all we should
compute H z by z, H z by z that is equal to
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00:45:57,500 --> 00:46:04,500
z minus 1 by z minus 3 into z minus 2 remains
3 whole square into z minus 2 which will be
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00:46:07,170 --> 00:46:14,170
A by z minus 2 plus B by z minus 3 plus C
by z minus 3 whole square, okay. And you can
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00:46:19,150 --> 00:46:26,150
compute A B C that, will be A is 1 all right
multiplied by z minus 2; put z is equal to
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00:46:31,220 --> 00:46:38,119
1, z is equal to 2, so it will be 2 minus
1, 1 okay.
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00:46:38,119 --> 00:46:45,119
Then C, z minus 3 whole square, so it will
be 2. And B will be, if I take the derivative
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00:46:49,300 --> 00:46:56,300
of z minus 1 by z minus 2; so it will be z
minus 1, z minus 2 into derivative of this
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00:46:58,950 --> 00:47:05,950
is 1 minus z minus 1 divided z minus 2 whole
square evaluate at z is equal to 3. So, z
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00:47:11,250 --> 00:47:18,250
will go this is minus 1 by z minus 2 whole
square, that is minus 1, is it so? So, H z
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00:47:26,819 --> 00:47:32,810
after getting these you multiply by z, so
it will be A z by z minus 2 which will be
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00:47:32,810 --> 00:47:39,810
z by z minus 2 minus z by z minus 3 plus 2
z by z minus 3, whole square.
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00:47:45,589 --> 00:47:52,589
So, what will be the corresponding H n? This
is 2 to the power n minus 3 to the power n
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00:47:58,930 --> 00:48:05,930
minus 2, is it 2 into n? z by z minus 3 whole
square, is it not a ram function? So, it is
225
00:48:16,530 --> 00:48:23,530
minus 1, 2 z by z minus 3 whole square, is
it not same?
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00:48:26,640 --> 00:48:33,640
What is the Z transform of a ram function?
Okay, n into a to the power n, how much is
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00:48:40,040 --> 00:48:47,040
it? z by z minus a whole square. So, what
is this? So, it will be minus 2 into it will
228
00:48:59,910 --> 00:49:06,910
be 3, it is 3, plus 3 it is minus. So, plus
3 into 3 to the power n into n and there is
229
00:49:14,500 --> 00:49:21,500
a multiplier 2; okay, whole thing into u n,
is that all right?
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00:49:25,520 --> 00:49:32,520
The lase one is consider the signal, x n equal
to minus 1, 2, minus 3, 2, minus 1 with DTFT,
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00:49:52,290 --> 00:49:59,290
X e to the power j omega, determine a X at
0, b angle of X at any frequency, c minus
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00:50:10,280 --> 00:50:17,280
p i to plus p i X omega, d omega; d X at p
i. And lastly, minus p i to plus p i; X omega,
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00:50:30,839 --> 00:50:37,839
magnitude square omega, okay. So, we will
take up one by one at least some of them,
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00:50:42,280 --> 00:50:45,170
rest of them I can leave it to you.
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00:50:45,170 --> 00:50:52,170
X at 0, how much is it? Okay, so all right
let us see, what is X e to the power j omega?
236
00:50:57,089 --> 00:51:04,089
First, it will be minus 1 plus 2 into e to
the power minus j omega plus, sorry minus
237
00:51:09,140 --> 00:51:16,140
3 into e to the power minus j 2 omega and
so on, all right. And I am putting omega equal
238
00:51:18,559 --> 00:51:25,559
to 0, except 0 means basically; I should have
written e to the power j 0, okay omega 0.
239
00:51:32,369 --> 00:51:39,369
So, it will be just sum of these terms. So,
how much is it? If, you add up minus 1, 2,
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00:51:44,150 --> 00:51:51,150
minus 3, 2, minus 1; it will be minus one,
okay. What will be the angle of this? Okay,
241
00:51:55,300 --> 00:52:02,300
let me write this completely, minus 3 into
e to the power minus then plus 2 into e to
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00:52:03,800 --> 00:52:10,800
the power minus j, 3 omega minus 1 into e
to the power j minus 4 omega. Now, the sequence
243
00:52:13,089 --> 00:52:20,089
that has been given to you, you see if you
plot is there any symmetry? Minus 1, okay
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00:52:22,760 --> 00:52:29,760
then plus 2 then minus 3, then again plus
2, then minus 1. So, it is a symmetric function,
245
00:52:36,589 --> 00:52:38,559
all right.
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00:52:38,559 --> 00:52:45,559
If I take this and this, this and this and
this; if I pair them then it will be very
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00:52:47,020 --> 00:52:54,020
simple; this is minus 1 and this one minus
1 into 1 minus e to the power minus j 4 omega,
248
00:53:06,160 --> 00:53:13,160
both are okay plus 2 into 1 plus e to the
power minus j 4 omega into e to the power
249
00:53:19,059 --> 00:53:26,059
minus j omega and minus 3 into e to the power
minus j 2 omega, okay. So, this one if I take
250
00:53:31,130 --> 00:53:38,130
e to the power minus j 2 omega common then
that will give me, twice cos 2 omega into
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00:53:42,180 --> 00:53:49,160
e to the power minus j 2 omega. Similarly,
here if I take e to the power minus j 2 omega
252
00:53:49,160 --> 00:53:56,160
out; so this will be e to the power minus
j 3 omega into twice cos 2 omega inside and
253
00:54:02,990 --> 00:54:09,990
there is a 2 here plus 3 into e to the power
minus j 2 omega, okay. So, it will be minus
254
00:54:18,319 --> 00:54:25,319
1 into e to the power minus j 2 omega, 3 omega,
sorry minus. Yes second term is; e to the
255
00:54:39,569 --> 00:54:46,569
power minus j, sorry, 1 and this I have taken
3 plus 1 minus 1, okay. Let me rewrite this,
256
00:55:06,640 --> 00:55:13,640
it will be clear; see minus 1 and this I have
taken common, okay.
257
00:55:15,809 --> 00:55:22,809
Actually it is 4 and 0, so it is minus 2 e
to the power minus j 2 omega into cos 2 omega,
258
00:55:32,069 --> 00:55:39,069
is it all right, first one. Then next one;
2 into e to the power minus j omega, 2 into
259
00:55:41,180 --> 00:55:46,829
e to the power minus j 3 omega, if I take
e to the power minus j 2 omega common. So,
260
00:55:46,829 --> 00:55:53,829
4 into e to the power minus j 2 omega, all
right into cos omega, is it all right.
261
00:56:06,049 --> 00:56:13,049
And then the next one is minus 3 into e to
the power minus j 2 omega all right; so if
262
00:56:17,020 --> 00:56:24,020
I add them together, it will be minus 2 cos
2 omega which is a real quantity plus 4 into
263
00:56:24,950 --> 00:56:31,950
cos omega which is a real quantity, minus
3 and multiplied by e to the power minus j
264
00:56:32,950 --> 00:56:39,950
2 omega. e to the power minus j 2 omega for
any omega, it is having an angle of 90 degrees
265
00:56:40,430 --> 00:56:47,430
minus 90 degrees and this quantity is real,
all right.
266
00:56:51,059 --> 00:56:58,059
So, this will have an angle of minus j, that
is 2 omega; j will go, so minus 2 omega. Its
267
00:57:05,670 --> 00:57:12,670
sign is basically 180 degrees, no sorry e
to the power minus j 2 omega; I made a I made
268
00:57:15,440 --> 00:57:22,440
a slip, e to the power minus j 2 omega, I
this is real, angle is 2 omega, no. There
269
00:57:35,319 --> 00:57:42,319
is some slip here, minus 1 actually minus
1 plus, we are searching for the angle, okay
270
00:57:59,109 --> 00:58:06,109
it will have minus 2 omega. I think we will
stop here for today, we will continue with
271
00:58:09,480 --> 00:58:12,760
the other two parts in the next class. Thank you very much.