Youtube subtitles download by mo.dbxdb.com
the following content is provided under a Creative Commons license
your support will help MIT OpenCourseWare continue to offer
high-quality educational resources for free
to make a donation or view additional materials from 100 to them IT courses
visit MIT OpenCourseWare at 0 CW
that MIT the DD you
there
ok
here
in
well in the last lecture we
introduced the the transform in this lecture
I'd like to go backwards that is i'd like to
introduce the in for C transform and
demonstrate some of its properties with a few examples
to begin with let me remind you
love busy transform relationship
as we talked about it in the last lecture the transform
XZ the sequence X a van
is given by some are x7
times Z to the mine ascent the inverse eat transform
of course is the relationship or
the said rules that allow s
to obtain exit Van the original sequence
from it see transform XZ their
are a variety of methods that can be used
for implementing the inverse eat transform some love them
are somewhat informal methods and
others one in particular which will talk about
is a somewhat formal method to begin with
an informal method and a method that in fact
most diverse tend to use the most often is a method
that week kid refer to as
the inspection method for computing the University transform basically the
inspection method
says if we have computed z transform
then from a sequence then by looking at that
see transform we can recognize more or less by inspection
the sequence that that z transform came from
for example we have worked
several times now the computation
busy transform up an exponential sequence
which is a right-sided sequence and gotten is the answer
or three see transform one over one minus
AZ the -1 with the region of convergence
with a magnitude see greater than the magnitude a consequently
having worked at example if
I now present you with this the transformer
and ask you what sequence generated
one-way doing that obviously is
by inspection by recognizing
that this sequence generates that sea transport
another example that we've worked several times
now is the exponential sequence
which is left side that is zero for and greater than 0
generating the same ratio polynomials
as in the right-sided exponential
but with the region of convergence which is the interior of a circle
rather than in this case with the regional convergence is an exterior
is the exterior a circle and in fact it was
this the comparison between these two see
transforms because this is the focus in the last lecture
on the importance the regional convergence so
this is one method which I emphasize again is
in fact one of the most common methods that is
just simply recognizing busy transform something that we've computer previously
from a sequence
an app obviously also as you work more examples
you build up a larger list
love sequences and the transforms which
pairs which you can recognize second method
for computing the in Bercy transform
is to recognize
is based on the fact that the
z transform relationship is basically
a power series in ze that is
if we look back here at
busy transform expression this is
essentially a power series relationship
which ex expand XZ empowers if -z
consequently given z transform
if we can expand it in terms of powers UVC
then we can pick of the coefficients and that expansion
as values the sequence accident
for example let's take the sequence
sorry busy transform exigency which is one over
one minus a seat at the -1
or we can rewrite it as easy over C minus a
now this of course is the transform
that we in z transform infers
we know how to compute by inspection
however we can also compute the inverse E transform
by expanding this expression in a power series
let's divide one by one minus 8c to the -1
and then recognize the coefficients in an expansion
as values the sequence X event
so carrying that out we have one divided into one
is one multiplying back we have one
minus a see to the -1
subtracting we then have AZ to the -1
dividing by one we then have a time
see to the -1 multiplying down again
we have AZ to the -1
minus a squared see to the -2
subtracting we have a squared
see -2 dividing by one
we have a squared the to the -2
and if we continue this will generate terms
one a time see the -1
a square time see the -2
AQ time see -3
et cetera this series will continue
with an infinite number of terms and
we can recognize then this
as the expansion affects ISI
in the form some x7 times due to the minus n
consequently we can recognize that
apparently exit then is one
and any 0 a any one
a square that any to AQ any 3
said course
there are lots of ways I'll expanding
a function like this in a power series
for example if I simply reverse the order these two terms
and carried out the division this isn't the series i'd get
what i'd get instead is the series
which is minus a to the minus one time see
minus a to the minus two times the square
-8 minus three times EQ Center
in that case if this was the expansion
then when I would have to recognize as the sequence corresponding to this
is the sequence which is 0 NE 0
it has no turns the form see to the minus a positive power
so apparently it's all 0 for and greater than 0
and its -8 -1 at any -1
-8 to the -2 any -2 et cetera
so in fact for this example I could expand
this either in this power series
or I could expand in this power Sears
well which power series is the right power series
we know in fact that given just this ratio of polynomials
there are several choices for
the sequence that corresponds to depending on the region of convergence
if I associate as the region of convergence
the magnitude of Z greater than the magnitude a
then its this power series that converges
and this power series diverges so for that region of convergence
that is with the magnitude see
greater than the magnitude a as the region of convergence
this is the power series and consequently
the answer is a to the N
times you have an which of course is consistent with
what we know already York with only consistent with the inspection
on the other hand if the region of convergence
was the magnitude of the less than the magnitude day
then its this power series that converges
and consequently in that case we get a left-sided sequence which again is
consistent
with what we know either from the inspection method or
acquittal we from the example work before
one obvious drawback to the power series
is the fact dead when we generate the inverse each transform this way
we get the values in the sequence
as individual terms that is
we generate XFN 19 clothes for
generated as a sequence for simple sequences of course we can recognize a
closed-form
form for that but a more complicated cases we can
alright third method for generating
the University transform is basically
an extension the inspection method which
allows us to take more complicated z transform expert expression
and expand it out interns love
terms that we can recognize
by inspection and this method
all refer to as the method based on
the partial fraction expansion
I assume that some love you are familiar
with the partial fraction expansion in general
for a rational function but in case you're not
or in case you rusty on it let me just quickly
review what the partial fraction expansion consists
let's consider general function
all a variable all taken
complex variable axe which is
a rational function in Acts
it is one polynomial in the numerator
divided by denominator polynomial
and let me remark incidently that shouldn't confuse this
acts with the sequence values x7 we
been referring to this is simply some arbitrary complex
number or complex variable
now I can in general
expand the rational function interns
I'll an expansion all
for which
at least under certain simple conditions
is a psalm of terms
over the form old a constant our security
divided by X minus exit K
where the X minus except K's are
the roots 0 the denominator polynomial
cue backs and these coefficients
or refer to as the residues
now to expand this rational function
in this form
without additional terms I need to
apply the restriction which I'll do for the purposes
this lecture that the
order the numerator polynomial is less than the order of the denominator
polynomial
and net in addition there are no
multiple order roots the denominator polynomial
in other words the denominator polynomial is
and order then the n roots
the denominator polynomial this thing
in the text there is the more general
discussion the partial fraction expansion
when the order the numerator is greater than or equal to the order
the denominator or if there are multiple order roots
the essential ideas of course don't change its just the mechanics
love implementing the expansion and the form expansion
so for this lecture will assume that the expansion simply
as terms this for
and this then is referred to as the partial fraction expansion
love FX now
how do we get the coefficients are securely
in the expansion well essentially by recognizing
that if we multiply
FX by 10
the denominator factors say
X minus sex so far and evaluate that product
at X equals Xserve are
then relating that to disarm
as i've indicated here
when K is equal to are
these two factors cancel out
when K is not equal to are then when I substitute X
equal sex so far in here then
this term vanishes so
basically then what happens inside this some
is that the term inside this
some is equal to 0 for K
not equal are and
it's equal to capital R so far that is the residue
at the at the pole X
equals x so far when K
is equal to or so
consequently is only one term left in this song which is
capital are so far and therefore
the residues that is capital or so far
can be obtained by
multiplying FX by X minus X a bar
evaluating the result index equals x a bar
and the resulting number is the residue Jeff Backus
at the poll XE back so far
that's for the case love
simple order polls that is first-order roots
the denominator polynomial and also for the case when you order the numerator
polynomial
is less than or equal to the order that the nominator polynomial
and just store away for later because I want to refer back to this point
that in particular when there are multiple order polls
the evaluation the residue somewhat more complicated
in particular it involves computing some derivatives
et cetera and the more the higher the multiplicity the poll
more complicated it is and that's a point that will sneak back at us
toward the end lecture now
this of course is for a general rational function
let's apply this now to the inverse the transport
well we know that busy transform
well let's restrict ourselves to the case with the transform
is a rational function Z or Z to the -1
and consequently we can
referring back to the general notion the partial fraction expansion
either think the variable Z
as what we were calling X over there
in which case we have an expansion love busy transform
in the form all some on K
a sip K divided by C-minus
little lace K where the little a suitcase the polls
and the capital a suitcase at the residues
alternatively if we interpret see to the -1 is an axe
then we can expand exits the in the form
love some other residues bck
divided by see to the -1
minus some call fission
little piece K or we can rewrite that as capital be
K divided by one minus AC K
see -1 so here where expanding
in terms love retreating
the transfer is a rational function in C here we're treating you see transform
as a rational function to the -1
and so what is the method consists of well the method consists
love expanding exigency in terms
simple terms like this and then using the inspection method
to recognize what the inverse eat transform is
for each of the individual terms
to get the final sequence then we add those up
that is expressed z transform is a somewhat simple
factories so the sequence the original sequence
we can express as the song all the infers
transform obtained by the expected inspection method
%uh each of the individual facts all right let's just work
simple example to show you what the mechanics are
this and how it works out let's take an example
where exigency is
one over one minus a half see to the -1
times one minus two quarters eat -1
with region of convergence which I'll take
as the magnitude of the greater than a hand
this has two poles of course want to see have the other is equals a quarter
and the region of convergence is outside the outermost poll
so what we expect to get is a right-sided sequence
and let's work this one by treating
variable acts in our previous discussion
as Z to the -1 and obviously it doesn't matter
which one we use although it turns out actually that pending on how you pick it
you either and up with the numerator polynomials orders greater than the
denominator polynomial
or the other way around
okay well so we want then
to expand this the transform
in terms all a
some factors and let's just rewrite this
to make things a little clearer by
multiplying denominator enumerator
by to by -2 rather
and then also by -4 we can simply rewrite this is
8 divided by Z -1 -2
times seat -1 -4 so that its it's more in the form are
X minus a constant times
X minus another
alright well so here then
we have busy transform that we want to expand their partial fraction expansion
in terms all treating see to the -1
as the complex variable X in the expansion
we have to polls two routes have the denominator polynomial
wanted to see -20 to the other at the
-1 equals four so we need to calculate the residues a teacher those routes
to calculate the residue at see to the -1
equals to what are the steps well
we multiply this by
-1 -2 that is X -2
where X were interpreting is it -1
and then evaluate the result
at see -1 people's
to well obviously
course this term
cancels out that
and we want to do that before we substitute see the -1 equals two
and then substituting see the -24 to
we get -2 left over in the denominator
divided into a and that's equal
to -4
so the residue then at the to the -20
to is equal to -4
similarly to calculate the residue
at the route see to the -1 equals four
we multiply by z2 -1 -4
and these two tires will cancel out
evaluate the resulting expression at the to the minus twenty calls for
so we have for minus-2 is to enter a is for
and the residue then at -1 equals four
the residue is equal to four
consequently the partial fraction expansion
for exigency for this sexy is
-4 -4
over Z to the -1 -2
plus 4 over see -1
-4 well finally let's just put his back into form
that is more the form that were used to
looking at this transform an in particular
multiplying this piece numerator and denominator by minus
half we get to over $1 -2 Habsi to the -1
here multiplying numerator and denominator by minus a quarter
we get -1 over $1 -2 quarter the to the -1
and now we've broken XZ into a form
that allows us to use the inspection method
that is we can recognize from ness
remember that the region of convergence was
outside circle so
we want to interpret the region converges to be the same
here so consequently for this piece
we end up with to
times half to the N
times you have an
in and
this piece we end up with -1
times quarter
to the an times you
so by breaking XZ into
some of compounds simple terms
we were then able to apply the inspection method
to each one of those to get
the resulting see transform
in a closed-form expression or at least the summer
some closed form expressions
alright so that then is basically an extension
the inspection when the inspection method by itself
doesn't work it's usually
the partial fraction expansion method dead
one uses perhaps also in conjunction
with some properties busy transform which is something that will be going
into more in the next lecture
now finally their is a
someone more formal way
evaluating the inverse eat transform and in some cases
it's important to actually use this to explicitly evaluate
University transform in other cases
it's important to make reference two more formal
infers transform relationship
to discussed properties at the University transforms
sometimes the formal in Bercy transform
is computationally useful other
cases its useful improving the rooms were developing properties
or something sort well this method
the final method and the most formal method
is the method that I refer to here
as contour integration
and let me basically handed to you
that is rather than go through their ovation let me just
simply state what the inverse the transform relationship is
in the text there is derivation
this and so
for the purposes the lecture itself let me just present
any objections now
so and then also
work through at least one example to show you with the mechanics are
another nice thing about videotape lectures are never any objections
okay well here it is this is the inverse transformer what does it say it says
the sequence x7 is
one over to Pike J times
contour integral which are returned to a second
up XZ times the to the end -1
easy if you're not used to looking at
an inverse the transform relationship like this let me just indicated it's
surprisingly easy
or it's surprising how easy it is verified in fact this is the University
transform relationship basically using the county residue
but this is it this is of course a complex function
this is a complex integral and
the important thing two straight now
is this contour see
the contour see is a
closed contour in the seaplane
which in circles the origin
so it's a closed contour in circles the origin is like a circle
that goes around the origin with another very important condition on it
which is dead it lies inside the region of convergence
affects see well that's not surprising because is
the regional convergence for Xfce is the only place we were allowed to speak in
values for Z
so obviously quote
we were not allowed to look at value citizen agran
except where XZ make sense that is except in the region convergence
so obviously then we have to evaluate this integral considering only values
the inside the region convergence
so controversy is a closed contour in the plane in circles the origin
and it lies inside the region converge
now for XZ a rational function see
which the cases we've been talking about
this entire in a grand is a rational function
and consequently a contour integral is this
hi can be simply evaluated
using the notion all residues
all-state and hopefully this is a property
kayak tour complex contour integrals that you're familiar with
that the integral love
a rational function that
the integral itself stoop I J times the sum up the residues
inside the contour and
back to fight GA cancels out this one over to
by J and consequently exit and is simply the sum of the residues
ecstasy I'm to the end -1
at the polls inside the controversy
the polls what the polls not XZ
only but the polls all the entire
in a grand that is exigency multiplied by C
to the end -1 important point which
it's important to keep in mind the integral
for the residues that we're looking at the residues not just affects see
by XZ multiplied by C to the end -1
another statement just tuck it away
is that this is valid
for both and greater than 0 and equal to 0
and and less we'll see in a minute
that for and greater than 0 it might be easier to evaluate
for in less than zero depending on the example
but talk away the fact that this expression is a valid expression
no matter whether and is positive or negative
or 40 alright
let's look at the mechanics I'll evaluating the inverse each transform
this way
let's return to the example exigency
equals one over one minus the half see to the -1
with region of convergence the magnitude see
greater than a hand and by the way by now
if is one thing you should have gotten outta
these lectures it certainly should be be able to recognize with the University
transform
this is an example that I guess we bounced around so many times
alright let's a.m. look at this the transform
first let's just look at Exit see
in seaplane is the seaplane
we have a whole as equals a hand
and we have a zero at the
0 here's the unit circle and in
green what i've indicated is the region convergence
now to do the contour integration it's not just text you see that we're looking
at
its XZ times see to the end -1
equals see to the N over C-minus Hannah
incidently what's the
contour what is the contour integration look like
then seaplane for this example
it's a closed contour inside the region of convergence
encircling the origin so for example
for example if the contour that looks like this
and incidently it is and I
don't think I stress this before it's a counter-clockwise
so this is the kind tourists see
its a counter-clockwise contour as i've drawn it
I didn't drive very good circle partly because I can't
but in fact but also emphasizes the fact that it doesn't have to be a circle
just has to be close inside the region of convergence
circling the origin and a counter-clockwise
alright how many Poles all
ecstasy see the end -1 are the air that is
how many places I there that we have to evaluate residues
well let's look first the ball at and greater than or equal to 0
there this doesn't introduce any polls
this introduces one poll equals a hand
so we only need to evaluate the residue yes
at this one poll that is at the a hand
so we want to compute the residue then
XZ times to the end -1 at
see equals a half
so that seat to the N oversee minus the hair
multiplied by C minus the hair evaluated at C equals a hair
these two cancel out
and we end up with simply have to the end
so far in greater than or equal to 0
the sequence XFN is just
have to the end so
thats part the result now we have to get
the inverse E transform for an less than 0
well for in less than zero again of course we have
one poll it's equal to a hand
but now remember that we had this factor
which was easy to the end
and for and negative then introduces
holes at the origin and in fact
it introduces in polls so
for and less than 0 we have not only this poll
but we have in polls at the 0
that's a problem because now we're gonna have to evaluate the residues
and each of these polls
and the more negative than you have and we're trying to get active and for
the worse it gets because the higher the multiplicity of the poll
art but look we've seen this example before
we know we know there
ex-offenders to have to the end bring greater than recall 20
and 80 friend lessons are as we knew that from other times we work this
example
so obviously there must be some easy way
to get the fact that x7 in fact that this example
0 for and less 0
all right let's take a look at away an easy way
here we have again the inverse each transform accident is one over to Pike J
times this contour integral and what simply make a substitution variables
let's make the substitution variables see
equal to P to the -1
in which case seat to the end -1 is equal to P
minus and plus one
which means incidently dead
with C equal to are to the J data
he this new variable is one over our
times to the minus J
so now we want to make this substitution variables in this
contour integral well what CDC
is minus p to the -2
just differentiating yes well I'll let you go through the algebra
on your own but if you take this substitution variables
stuff into that contour integral
and the result that we end up with is dead
exit then is equal to -1 over to fight J
minus coming from the fact that DC is minus
P to the -2 times the that minus
ex the one over P times people the minus 10 -1
DP and what happens to the contour
well the cont to our previously was
let's say are to the J let's take it as a circular Conn
so the contour now is one over our to the minus J
that means as one consequence that are all
counterclockwise contour turns into a clockwise contour
so that's what I've indicated year well we also have this minus sign
and we can use that minus sign to
changed direction were so if we change the contour
back to it counterclockwise were get rid of the minus sign
then finally we have the expression exit and is one over to fight J
times the contour integral capital ex one over
the p minus -1 pp
and what is this contour now well
if the old contour see with circular with the radius of our
this contour see prime is a circle
the radius one over our
well let's just look at what this substitution variable Z
21 over P did as a matter a fact
year I go straight to the seaplane
here I've illustrated the people
the unit circle in the seaplane when I replace see by one over P
turns again into the unit circle in the P plane
because the unit circle is where the magnitude sees one
if I take the reciprocal that the magnitude is still one
the important thing is what happened to this contour
well this kind to or got converted
if it was outside the unit circle into
new contour which is inside the unit circle
and in fact in general stuff that was in here
the inside the unit circle is going to end up
outside the unit circle in the P plane
and things that were outside the unit circle
in the seaplane are going to end up inside the unit circle
in the people this transformation simply takes
stuff outside the unit circle unfolded inside the unit circle
takes things inside the unit circle and fold them outside
sir and it's just simply a substitution variables
right now with this substitution variables let's return
to the example there we've been working
had our example
exigency is one over $1 -2 have seeded -1
magnitude see greater than a hand
we want to replace the by one over P so
we want ex the one over P
ex the one over P then simply replacing see the -1 by P
is one over $1 -2 half times p
if this was true for the bank to do see greater than a hand
then this is true the magnitude less than
so we get -2 divided by p
-2 if we look at this
in P plane
we have then appall in the P plane
opponents the plane was sequels Hannah so the poll
in the P plane is at he
to that's now outside the unit circle
the region of convergence now is inside the circle
radius to and where is the contour integration see prime
back now is inside circle with radius
to for example let's brought here
so this is now the contour see prime
alright we want finally to
evaluate the inverse transform
where the rivers P transform by
looking at the residues love X one over P
speed to the minus and -1
the residues that at its
roots at its polls that are
inside the contour integration see prime
well for in less than 0
we have one poll people's to
and for an negative this will contribute no polls
therefore there are no roots
I love this inside this kind to or
and consequently the contour integral is obviously 0
so accident is equal to 0 for and less
0 now for and greater than or equal to 0
we have one poll people's to
but then because it this factor p for the minus 10 -1
we have M plus one polls at people 0
so now it seems like bring greater than or equal to 0
we're back where we started from because now we have these multiple order polls
at people 0 what do we do about that
well we don't have to do anything about it because we already considered
the case for and greater than or equal to 0
in that case we didn't run into multiple order polls
the origin use this substitution variables Eagles won over P
basically avoid that problem and to get the answer
for and less than zero so consequently the answer that we get
which we finally know to be the right answer
is that x7 is equal to 0
for and less than zero and as we worked at before
exit then is equal to have to the yen
for an greater than or equal to 0 or
finally exit Van
is equal to half to the N
times you then
alright that's an example we've worked out now three times in three times we
got the same
answers must be right alright well
this rough set out the inverse
see transform several methods
for getting me in Bercy transform
obviously to become fluent with the inverse each transform
requires working a lot of examples
and you know where you can have all those examples to work
that is study guide and you'll find as I
stress previously dead as you work
more and more examples what you find in computing in Bercy transforms
is there more and more frequently your able to recognize the answer
well in the next lecture will
complete the discussion busy transform in particular
will among other things present some properties busy transform
which also by the way helpers in computing
in Bercy transforms and so in the next lecture
will be continuing is the topic the transform
and in fact including thank you Subtitles End: mo.dbxdb.com