Introduction to Lattice Filter - Electrical Engineering Video Lecture - Electrical Engineering (EE)

Shall we continue with Computer Aided Design, and then will take over Lattice Filter design.
The Parks and McClellan algorithm that we
discussed earlier for equi ripple filters
was like this, we got say some function A
e e to the power j omega as summation of a
K x to the power K, so we called it P x where
x was cosine omega. And then we fix the two
frequencies omega p and omega s at their desired
values and delta 1, delta 2 are allowed to
vary, but the ratio is given it is normally
kept constant. And then we took the error
function if remember, if you write error function
we took D x minus P x. Where D x was basically
the desired function
and p x was the polynomial approximation,
polynomial in cos omega and W x was a weighting
function
and P x is not constrained in the transition
region.
That is we compute the error only in these
two zones, we compute the error only in these
two zones. Here we do not constrict the function
and W x we took as 1 by K and equal to 1 for
omega s phi and K was delta 1 by delta 2.
So, basically this is normalizing the error.
So minimax criteria we had minimization of
maximum value of E x modulus, so this we solved
by alternation theorem.
So, crux of that theorem
was for P x to be unique, that minimizes
the error is the maximum of the error we defined
as E bar, so maximum of E we are writing as
E bar. So, P x to be unique that minimizes
this E, E P x that is, that normalized error
should exhibit at least, mind you at least
Q plus 2, where Q was the order of the function,
Q plus 2 alternations, that is x 1, x 2, x
Q plus 2.
Such that, E P x i is equal to E P x i plus
1 with a negative sign and that is equal to
plus minus this value, for all values of i
this should be L plus 1 such points, the values
of i. So, this was our alternation theorem,
we found that W x into D x minus P x that
is minus 1 to the power i plus 1 delta, so
delta is to be optimized is optimum error.
So, this we wrote as this center set of equations,
we wrote as x 1, x 1 square, x 1 to the power
Q and then W cos we wrote w 1, small w 1 into
the coefficients a 0, a 1 up to how many such
coefficients will come Q and then delta. And
this will be equal to that function H d e
to the power j omega 1 and so on, H d e to
the power j omega Q plus 2, now obviously
the value of delta is unknown, the omega is
are also unknown.
So, there is alternative process involved,
how do you solve this? So you make some initial
guess I will just outline first how such a
problem is solved, you are having within a
bound, say may be from here, so this is fixed
omega stop band, this is fixed omega pause
band. So, these are the points of maxima or
minima in the error function, if remember
in the error function this will come as maxima
and minima and these frequencies are not known,
nor is delta known.
So, we first of all assume some values, so
this is omega 1, omega 2, omega 3 and so on,
I can write this also as one of those omega
0, so initially you are having an approximate
values of these maximum. Then what you do
try to fit it the polynomial, that polynomial
may not have the maximum value here, that
polynomial will be satisfying this.
So, you may get a polynomial, where your earlier
values where these. So they may not correspond
to the maximum to get my point, so may be
here, so your earlier points where these,
whereas the new maxima are these. So, what
you do, how do make an approximation for the
polynomial, see it is passing through zero
in between the maxima and the minima.
So, you have selected first some values at
random for the frequencies omega 1, omega
2, omega 3, so between omega 1 and omega 2,
there will be a zero processing, omega 2 and
omega 3 there will be a another zero processing.
So, you take even a linear interpolation you
can make, so by interpolation technique you
form the polynomial, then what you do take
the derivative, find out the values of the
maximum and the corresponding frequencies,
you can do that from a polynomial of say order
n there will be n minus 1 number of maxima
and minima.
You can find out those frequencies and you
can also find out the values of the function
at those values, now find out all of them
may not be equal, so take the maximum one
then, take delta. Once again those frequencies
you take up as the next approximation, find
out once again the zero processing and it
is an iterative process, when you find all
of them are converging to a single delta that
is delta is not modified much, then you stop.
So, I will write down the steps, make an initial
guess
of x 1, x 2, up to x Q plus 2, x means off
course cosine omega 1, cosine omega 2 and
so on, calculate the optimum value of delta
on extremal set. Next
interpolate through Q plus 1 points to obtain
that polynomial A i omega or P x we can talk,
calculate error E omega or E x whatever you
call it and determine the local maxima, maxima
and minima, that is local maxima such that...
Check whether the extremal points changed,
if there is a change, then go back to step
2, step 2 is what calculate the optimum value
of delta on extremal set. So, basically this
is the desired function and this is the designed
filter this is the equation, where x i was
cosine omega i, so we are taking a different
points basically this is i i.
So, after guessing, so
you can put those iterative equations in this
form, we write delta equal to sigma b k H
d e to the power j omega k divided by sigma
b k minus 1 to the power K plus 1 by W omega
k, k varying from 1 to Q plus 2 again Q plus
2. S from here you compute delta where b k
is 1 by x k minus x i phi i equal to 1 to
Q plus 2 i not equal to K, this is Lagrange
interpolation technique, mind you x i is cosine
omega.
Therefore, A e e to the power j omega will
be that is P of cosine omega is sigma d k
by x minus x k into C k divided by sigma d
k by x minus x k, again k varying from 1 to
Q plus 1 and C k equal to H d e to the power
j omega i minus, omega k. And d k is equal
to phi 1 by x k minus x i, i varying from
1 to Q plus 1 i not equal to k, thus basically
b k by x k minus x Q plus 2, because b k was
this product where I was varying from 1 to
Q plus 2.
And here for d k it is varying from 1 by Q
plus 1, so in terms of b k this is one more
x k minus x Q k plus 2 left here, so once
you know C k and d k we can calculate these
polynomial, so it is like this you fix the
maxima minima frequencies to start with, you
just assume some values. But, omega p and
omega s will be one of the, I mean these two
values are fixed and then try to find out
the polynomial, again calculate the maxima
and minimum points those frequencies and also
the maximum value. And that should be greater
than delta, that again change the polynomial
and keep on trying this, next we shall see
what will be the approximation of the filter
order, how do calculate the approximate value
of the filter order.
Specially, for FIR filters various suggestions
have been made, so from these values omega
s, omega p, delta s and delta p these are
specified, what is the order of the filter.
So, Kaiser suggested this relationship approximately
N is equal to minus 20 log of root over of
delta p delta s minus 13 divided by 16.6 omega
s minus omega p divided by 2 phi, and that
is 20 and root over of this.
So, that will give me minus 10 log product
of delta p delta s minus 13 divided by 14.6
by 2 phi. So that gives me… and this I can
write as delta omega delta omega means the
stop band and pause band gap
for a very narrow band, narrow pause band,
this may not be very effective.
So, Parks has suggested N equal to minus 20
log of delta s plus 0.22 by 2.324 delta omega
this is more effective, so instead of minus
20 log of delta p delta s, it is approximated
as this. And for a wide band N is minus 20
log of delta p plus 5.94 divided by 27 into
delta omega by 2 phi, so these two modifications
were suggested by parks.
There are more accurate relationships, which
is very commonly adopted and given by Herman
it is like this, D infinity a function of
delta p delta s minus F delta p delta s omega
s minus omega p by 2 phi whole squared divided
by omega s minus omega p by 2 phi where D
infinity a 1 log of delta p to the base 10
of course, squared plus a 2 log of delta p
plus a 3 into log of delta s to the base 10
minus a 4 log of delta p to the base squared
plus a 5 log of delta p plus a 6, so it is
like this.
D infinity delta p delta s is one quadratic
involving log delta p to the base 10 multiplied
by log delta s, another quadratic again in
the same form without log of this and F delta
p delta s is equal to b 1 plus b 2 log of
delta p minus log of delta s. A 1is 0.005309,
a 2 is equal to 0.07114, a 3 equal to minus
0.4761, a 4 is 0.00266, a 5 is 0.5941 and
a 6 0.4278, b 1 equal to 11.01217, b 2 equal
to 0.51244, this is valid for delta p greater
than delta s.
If it is the other way round, if delta s is
greater than delta p, then you just interchange
delta p and delta s get this expression, so
this is very commonly used in filter design.
If time permits later on we take up one or
two tutorial problems on this filter design.
So, this is all about the computer aid design
that we have studied so far. Now, I will take
up lattice filter. Now, so far we have seen
a direct form, which is taken from the difference
equation or from the transfer function, lattice
structure we had studied in signals and networks,
networks synthesis.
Lattice structure is like this, let me take
a simple network. Suppose we given input here
this is z a, z b, z b, z a and we take the
output here, we will find this can be written
as z a, z b, z b and z a, 1 1 dashed 2 2 dashed,
this is a lattice structure. So, unlike a
lattice structure, latter develops by alternatively
putting series and shunt elements, this is
a lattice structure and in a lattice you have
cross connections. Now, in adaptive filtering
speech processing modeling of the signals
by filters will be very convenient with lattice
structure, there many users.
So, in a lattice structure we have like this.
Hence forth I will not put z inverse as a
single element I will show in the path itself
have multiplied z inverse, this is in signal
flow graph we use this symbol. So z inverse
this is the input
and so on, this is continuing from here. I
will write these symbols, suppose this is
e 0 n and this is e 0 n tilde, then in one
path you are putting a delay elements.
And then this is minus K 1, minus K 1 they
are multiplied these signals are multiplied
by the lattice constants K 1, K 2 and so on,
minus K 1 minus K 1, minus K 2 minus K 2.
So, at their next stage it is minus K n minus
K n, so at this point it is e 1 tilde n and
e 1 n, similarly here it is e 2 n and e 2
tilde n and so on. The initial value of these
two signals there are two channels through
which you are sending the signals, so the
initial values are same, so e 0 n is e 0 tilde
n and that is equal to x n itself.
Now, will this be an FIR filter? What is this
structure? Is it an FIR structure, of what
order? if it is an FIR of what order, let
us see whether it is an FIR or not. See there
is no feedback, so the signal is transmitting
only in the forward direction. At the end
suppose I give an impulse input, I give an
impulse input, then that impulse go straight
here without any delay here it is delayed
by one step, then again goes to this here
it is through two delays, so like that there
will be n such steps.So, there will be n delays
and after that there is no response, so it
is an FIR of degree n.
Now let us see what this relationship will
be like at any stage e i n can you tell me
what will be the relationship, I want to get
a recursive relation e i n, in terms of these
two signals in the previous steps. So, correct
me if I am wrong and check whether you are
also getting this, are you getting this say
for example, e to n is minus K 2 times e 1
tilde n, check once again e to n, let us take
e 2 n is equal to minus K 2 into e 1 tilde
n no, there is a delay z inverse.
So, n minus 1 there is a delay, so in the
second channel the signal is successively
delayed by one step, whereas in the forward
channel it is e 1 n, so e 1 n, so e 2 n will
be e 1 n minus e 1 tilde n delayed by one
steps. So, e i minus 1 tilde n minus 1 multiplied
by the corresponding constant K i, so that
will be e i n, check whether this is all right.
Now, you tell me what would be the relation
for e i tilde n at the next stage what will
be these value e 2 tilde n it is e i minus
1 into n into minus K i, this signal is multiplied
by K i and then plus e i minus 1 tilde is
it n. So, e 2 tilde n is e 1 tilde n delayed
by one step minus K 2 into e 1 n, so this
is the general form at the higher stage.
So what will be h at N th stage, what will
be these value, if I write h 0, h 1, h 2,
h n as the impulse response, what is the h
n, this is what I am writing as h capital
N, what is that h N, after Nth interval what
will be the output for an impulse input, this
an impulse what will be the output after Nth
interval it will be minus k n. The one of
which suffers Nth delays, will be the one
which will be coming through all these and
lastly take up this.
So, if the signal is coming through these,
there is only 1 delay, 2 delays, 3 delays
and so on, Nth delay and then multiplied by
minus k n that is the last signal, so h N
is minus k n that convinced. So, we are interested
in finding out H z, I will write Y z by X
z, so this is A z a polynomial of this type,
now what will be A i z, it is E i z by E 0
z, from here if I take the z transform, what
do I get.
E i z here is equal to E i minus 1 z minus
K i E i minus 1 tilde, then z into z inverse
and from this 1 E i tilde z equal to minus
K i E i minus 1 z, correct me if I am wrong,
E i minus 1 tilde into z inverse. Now, there
are two relationships, two recursive relationships
involving i minus 1 and for both the channels
and you have to find out what will be E i,
it will be better to see for yourself some
simple, starting from i is equal to 1, 2,
3, 4 by simple induction what these relationships
will be like. So, E 0 z is same as E 0 tilde
z is equal to x z is it not, this is equal
to this equal to x, z inverse and E i z inverse
is minus K i E i minus 1 z in I think this
is, so let us see with E 1.
What would be E 1 z, E 1 z is equal to E 0
z minus K 1 I will use this. So let me put
the paper here minus K 1 E 0 tilde, is same
as E 0 into z inverse. So that is equal to
1 minus K 1 z inverse into E 0 z, that is
1 minus K 1 z inverse in to x z, is it not
E 0 z is x z. So, A 1 z which is the transfer
function E 1 z by x z is 1 minus K 1 z inverse,
what will be E 1 tilde z, E 1 tilde tilde
z is minus K i minus K 1 E 0 z plus z inverse
E 0 tilde z, which is same as E 0 z, so it
will be minus K 1 plus z inverse into E 0
z, because E 0 tilde is same as E 0. Therefore,
what will be the corresponding transfer function
A 1 tilde z minus K 1 plus z inverse, can
you write A 2 and A 2 tilde z, then will generalize
something from here.
A 2 z it would be better if you write first
E 2 and then derive A 2 z, so this would be
E 1 z minus K 2. Now, once again taking this
E 2 z is equal to E 1 z minus K 2 into E 1
tilde z into z inverse. So, E 1 z minus K
2 into what was E 1 tilde z earlier, we have
derived, if minus K 1 plus z inverse E 0 z
and what was E 1 z, 1 minus K 1 z inverse
into E 0 z minus K 2 into minus K 1 plus z
inverse E 0 z, can I write this in some form
of this function.
Because, my E 0 z is basically x z, so my
aim is to calculate A 2 z, which is E 2 z
by E 0 z, so can I write this if I take z
common, this is z inverse into 1 minus K 1
z. So, what was 1 minus K 1 z inverse is appearing
as 1 minus K 1 z and K 2 z to the power minus
1, so can I write this as this is what, this
one was A 1 z. So, I can write this as A 1
z minus K 2 z inverse A 1 z inverse, z replaced
by z inverse can I write like this.
So, generalizing this we can write A i z as
A i minus 1 z minus K i z to the power minus
i A i minus 1 z inverse, if this is A 1 this
is also A 1, so A i minus 1 A i minus 1, z
is replaced by z inverse. In this bracketed
quantity, if I replace z by z inverse I get
this, and there will be accumulation of z
inverse there, so K i z i inverse z to the
minus i, it should be z to the minus 1, a
friend suggest that it should be minus 1.
What is it should it be z to the minus i or
z to the minus 1 i minus of 1, try with A
3, A 4 and then generalize it leave it as
an exercise, similarly A tilde let us see
A 2 tilde we have computed A 2 z.
A 2 tilde z is how much, so E 2 tilde, if
you write first E 2 tilde z, A 1 z inverse
is z to the power minus 1 into 1 minus K 1
z is equal to z to the power minus 1 into
A 1 z inverse. A 1 tilde is A 1 z inverse
z to the power minus 1, what it be E 2 z,
E 2 tilde z is if you remember E 2 will be
minus K 1 minus K 2 E 1 z minus K 2 E 1 z
plus E 1 tilde z z to the power minus 1.
And now let us substitute minus K 2 into E
1 z was 1 minus K 1 z inverse into E 0 z plus
E 1 tilde z E 1 tilde z is minus K 1 plus
z inverse E 0 z into z inverse. So, that gives
me minus K 2 into 1 minus K 1 z inverse plus
this I can write as z to the power minus 2
1 minus K 1 z inverse into E 0 z. Therefore
what will be A 2 tilde z, so 1 minus K 1 z
inverse is common, 1 minus there is some mistake
here this one, so it is 1 minus K 1 z inverse
is what, 1 minus K 1 z inverse is A 1 z.
So, it is minus K 2 into A 1 z plus z to the
power minus 2 into A 1 z inverse are you getting
this. So generalize this A i tilde z, it should
be z to the power minus i A i minus 1 z inverse
minus K i A i A i minus 1, A i minus 1 strike
this and write minus K i A i minus 1 z. Now,
based on these two relationships that is the
iterative relations that you have got, the
recursive relation that you have got for A
i z and A i tilde z, we have got what is that
A i z.
A i z as A i minus 1 z minus K i z to the
power minus i A i minus 1, so actually we
are interested in only A i, not A i tilde,
A i tilde has come as a byproduct for solving
this you needed A i tilde. So, finally, this
is the relationship for A i with respect to
it is previous values, now from here will
find out the polynomial coefficients. So,
given a polynomial what will be lattice constants
or given a lattice structure what will be
the polynomial, so in the next class will
take up from this. How to determine? Finally
A z at the N th stage, given the lattice constants
K 1, K 2 up to K n or given a polynomial how
do you determine the lattice constants.
Thank you very much.