hello friends welcome to my lecture on initial and final value theorems for z transforms
in practical applications often we need the values of u n for n equal to zero and
tends to infinity without actually requiring the complete knowledge of the sequence u n
we know that the sequence u n is related to the its z transform uz so by using that
relation relationship we can find the values of u n for n equal to zero and n tends
to infinity without actually knowing the sequence u n so suppose the z transform
of u n sequence is known it is given by uz then it tells out that the value of u n for
n equal to zero is given by limit of uz as z tends to infinity the proof is quite
simple we consider the definition of the z transform z is given by which is z transform
of the sequence u n is given b y sigma n equal to zero to infinity u n z to the power minus
n which when expressed as u naught plus u one z plus u two yz square and so on on taking
the limit as z tends to infinity gives us the value of naught as limit z tends to
infinity uz
so when we take the limit of this equality as z tends to infinity we get limit z tends
to infinity uz as u naught so this limit of uz as z tends to infinity gives us the
value of u naught now similarly we can see that limit z tends to infinity z times z minus
u naught is equal to u one so in order to see this we have uz equal to z of u n sequence
which is sigma n equal to zero to infinity u n z to the power minus n which is u naught
plus ui one by z plus u to by z square and so on
so from here we see that uz minus naught multiplied by z this is equal to u one
plus u two by z plus u three by z square and so on and so taking the limit as z tends to
infinity z times uz minus u naught gives us the value
of u one similarly from equation one we can see that uz minus u naught minus
u one by z uz minus u naught minus u one by z when it is multiplied by z square gives
us a u two plus u three by z plus u four by z square and so on ok so taking the limit
again as z tends to infinity ok we get limit z tends to infinity z square times uz minus
u naught minus u one by z equal to u two and so on so this how we can get the initial value
of the sequence u n by its z transform
so let us now talk about the final value theorem here we will get the value of u n as n
tends to infinity so if the z transform of the sequence u n is known say it is uz then
this result tells us that the final value of the sequence u n that is u n as n tends
to infinity is given by the limit of z minus one uz as z tends to one let us look
at the proof of this final value theorem so we can take the z trans[form] suppose
we take the z transform of the sequence u n plus one minus u n then by linearity property
we can write it as z of u n plus one minus z of u n and then we use the shifting property
here shifting to the right shifting of the sequence u n to left so z of u n plus
k is z to the power k into uz minus u naught minus u one by z minus u two by z square and
so on minus uk minus one z to the power k minus one
if we use this shifting property then z of u n plus one will be equal to z times u z
minus u naught so we will get z of u n plus one equal to z times uz plus u naught
so let us substitute the value of z of u n plus one here so z of u n plus one minus u
n is z times uz minus u naught minus uz and then we can take write this further as
z times z minus one minus u naught z
now by definition of z transform z transform of the sequence u n plus one minus u n is
equal to sigma n equal to zero to infinity u n plus one minus u n into z to the power
minus n as z tends to one then we will have limit z tends to one z minus one uz
minus u naught as limit z tends to one n equal to zero to infinity u n plus one minus
u n z to the power minus n because z minus one into uz minus u naught z is equal
to this so from this equation and this equation we arrive here as z tends to one
limit of this expression is equal to limit z tends to one this so when z tends to one
will it will tend to sigma n equal to zero to infinity u n plus one minus u n which can
be expressed as limit n tends to infinity the its partial sum and partial sum so
u one minus u naught minus u two minus u one and plus u one u n min u n plus one minus
u n and the this will reduce further to u n plus one limit n tends to infinity u n plus
one minus u naught which is equal to limit of u n as n goes to infinity minus u naught
because u naught is independent of n so then we can cancel and the left hand side
is limit z tends to one z minus one uz minus u naught because u naught is independent of
z so we can cancel u naught on both sides and then we get limit of u n as limit z tends
to one z minus one uz so this is the proof of the final value theorem let us use this
initial value theorem to find the initial values of the sequence u n when
u its z transform is known
so let say uz is known to us and we want to determine the value of u n for n equal to
two and n equal to three so let us see how we get this uz is equal to two z square
plus five z plus fourteen divided by z minus one to the power four ok and the initial
value by initial value theorem initial value theorem u naught the initial value of the
sequence u n naught is limit z tends to infinity z tends to infinity uz so this is equal
to now uz can be written as in the numerator we have polynomial of degree two in z and
the denominator has a polynomial in z of degree four so i can write it as one by z
square two plus five by z plus fourteen by z square divided by one minus one by z
to the power four on taking z square common from the numerator and z to the power four
common from the denominator
so this will get we will get this now as z tends to infinity his expression tends to
two this tends to one so his whole thing tends to two and this goes to zero so the
limit is zero so we get the value of u naught as zero let us now find the value of u one
u one is equal to limit z tends to infinity z times uz minus u naught so this is limit
z tends to infinity z times two z square plus five z plus fourteen by z minus one to
the power four minus zero so uz uz can be expressed in this form when we multiply this
form by z ok what we get is one by z so this z cancels with one z here and we get
one by z two plus five by z plus fourteen by z square divided by one minus one by z
to the power four again this expression tends to two and this then goes to zero so we get
the limit as zero
now let us find u two ok so u two is limit as z tends to infinity z square times uz
minus u naught minus u one by z u naught and u one both are zeros and when we multiply
uz by z square ok what we get is limit z tends to infinity two plus five by z plus fourteen
by z square five by z divided by one minus one by z to the power four and when z goes
to infinity this limit clearly is two so we get the value of u two as two we can similarly
find u three
so similarly u three three is equal to limit z tends to infinity z cube times
uz minus u naught minus u one by z minus u two by z square now we have obtained u naught
as zero u one as zero u two equal to two by z u two equal to two so we can see find
the limit so u three is equal to limit z tends to infinity z cube times uz is one by z square
two plus five by z five by z plus fourteen by z square divided by one minus one by z
to the power four minus this zero this zero this two by z square so two by z square
z cube can be multiplied so this is limit z goes to infinity z cube when multiplied
gives you two z plus five plus fourteen by z divided by one minus one by z to the power
four minus two and we can simplify it further to find the limit
so this will give you this z z cube when multiplied by this is two z so two z
times this ok now we can write it further as
one minus one by z to the power four can be expanded by binomial expansion so we have
one then four times minus one by z then six times minus one by z square then we have
four times minus one by z cube and then we have one by z to the minus one by z to
the power four so we can do this now we can see when you multiply two z here and what
does then happens so two z minus two z two z cancel out we have
then five five ok so we two z two z cancel out we get
five plus fourteen by z ok and here what we get his z this z cancel out we get minus four
into minus two so plus eight and then this z when multiplied here to z square what we
get is minus twelve into one by z and then we get here minus one by z cube so we get
plus eight times one by z square and then we get hm minus two times one over z to
the power three
so when z goes to infinity this goes to zero this goes to zero this this goes to zero
and we what we get is thirteen so this is equal to thirteen so this is how we get the
value of u three now so there is example one an example two we have take
an example of inverse z transform let us find inverse z transform of l n z over z plus one
so we shall apply the power series method to determine the inverse z transform let
us see we ln z over z plus one is equal to ln z plus one over z to the power minus
one and this is ln or i can say this is minus ln one plus one by z
now if mod of z is greater than one then i can write it as minus one over z minus
one over two times z one by z square plus one by three times one by z cube and so on
provided mod of z is greater than one so i can write it as minus one by z plus one
by two z square minus one by three z cube plus one by four z to the power four and so
on
now comparing this series in finite series with sigma [vocalized noise] we note that
here there is no term which is free from z so we get z inverse ok so if so z inverse
of ln z over z plus one this is equal to u n sequence ok then u n is equal to zero
when n is equal to zero and so here when n is equal to one the coefficient is minus
one when n is equal to two the coefficient is one by two so we get minus one to the power
n divided by n when n is equal to one two three and so on
this is the example two let us now go to the next slide and see how we find the z trans[verse]
inverse z transform of z over z plus one whole square we know that z transform of n in to
a to the power n is a z over z minus a whole square provided mod of z is greater
than mod of a so when mod of a is greater than mod of a we know that z transform of
n a to the power n is equal to a z over z minus whole square so taking a equal to minus
one we have z transform of z transform of n times minus one to the power n equal
to minus z divided by z plus one whole square provided mod of z is greater than mod of minus
one so which is one
and this gives us inverse transform of so using the linearity property of the z transform
what we get is minus z transform of n times minus one to the power n minus one is equal
to z over z plus one whole square when mod of z is greater than one or we can say that
inverse z transform of z over z plus one whole square this is equal to minus one to the
power n minus one into n
now this is the method where we have used the known result to get the inverse transform
there is a method which we call long division method we can find the inverse transform by
that using the using that long division method also so let us see how we that method
works so we shall apply the long division method what we will do here is that we have
z over z plus one whole square z over z plus one whole square is z over z square plus two
z plus one so let us divide z by z square plus two z plus one we have to divide z
by z square plus two z plus one in such a way that it will lead us to an finite series
in powers of one over z
so z square plus z plus one by we multiply this b y z to the power minus one so that
here when we multiply by z to the power minus one we get z here so that that z will cancel
with this z here so z to the power minus one we take then we multiply z to the power minus
one we take here so we get z plus two times z in to z minus one is two plus z to the power
minus one and what we get is when we subtract the sign changes so minus two minus z to the
power minus one now our our aim will be to eliminate minus two so we write here minus
two times z to the power minus two so when you multiply by minus two times z to the power
minus two to z square you get minus two and then we get minus four z to the power minus
one and then minus two times z to the power minus two again we subtract so [we/the] sign
changes and this cancels and we get four z to the power one minus one minus z to the
power minus one so we get three z to the power minus one plus two times z to the power minus
two
now we multiply by three times z to the power minus three so that when we multiply it to
z square it will give us three times z to the power minus one so three times z to the
power minus one and then we get six times z to the power minus two and we will get
three times z to the power minus three so like this we continue so will get here
four times z to the power minus two be the negative sign ok minus three times z to
the power minus three
so now our aim will be to multiply z square by such by an expression which will give us
minus four z to the power minus two so minus four z to the power minus four so minus four
z to the power minus two and then we get minus eight z to the power minus three and then
we get minus four z to the power minus four so this is plus plus this cancels and
we get this minus this is five times z to the power minus three plus four times z
to the power minus four so continuing this process ok so thus z
over z plus one whole square it will come out to be z to the power minus one then
minus two times z to the power minus two and then three times z to the power minus three
minus four times z to the power minus four and then next one will be plus five
times z to the power minus five and so on ok and when we compare it with that series
sigma u one z to the power minus n so this gives us then z inverse of z over z plus
one whole square this equal to this equal to u n equal to zero because there is no
term contain[ed] which is free from z so zero when n is equal to zero n minus one to the
power n minus one when minus one to the power n into n when n is equal to when
n is equal to one we will write here minus one
so this how we get the inverse transform of z over z plus one whole square with that
i would like to conclude my lecture
thank you very much for your attention