Effects of Quantization - Electrical Engineering Video Lecture - Electrical Engineering (EE)

In the last class, we were discussing about Jury's criteria for stability. We saw with
one example 4 z 4 plus 3 z cube plus twice
z square plus z plus 1, we arrange the rows
and we found that the system is stable. Let
us take another example, this was meeting
all the criteria. Let us have the denominator
polynomial as z to the power 4 plus 6 z cube
plus 3 z square plus 4 z plus 5 is the system
stable, first you apply the condition D 0
is 1 plus 6 plus 3 plus 4 plus 5, so it is
positive.
What about D at minus 1 minus 1 to the power
4 that is minus 1 minus 6 minus 4 and 5 plus
3 plus 1, so that gives me minus 1, so this
condition is evaluated you need not go for
further trial, you do not have to make the
rows. If in the first two test you find the
conditions are not made, then it is unstable
will take up a few problems in the tutorial
class and see how to apply in jury's criteria
for stability. Today, we shall be considering
the Effect of Quantization.
Before we go for quantization, I will just
give you the basic idea about t to a conversion,
you have x a t that is an analog signal, then
you have sample and hold, then actually A
to D conversion takes place, so you get the
converted value x n. Now, the analog signal
may be like this, if you just sample it, it
is not an instantaneous process you sample
it for then
this sample value will be represented by binary
digits by any particular code.
And then till that particular value is approximately
closed that sample value it will not stopped,
so it will require continuously a fixed input,
if the input keeps on changing then it will
be disturbed. So, this level has to be maintained
for a period t, so this is the sampling period
and within this time the computation has to
be completed, the conversion from that exact
value here to the approximate value that binary
digit has to be computed within this time.
So, that is how the sampler time is fixed
and then the next value comes, so it will
stay there, then this value will stay here,
then this value will stay here, this value
will stay here and so on. So, there will be
some errors of course, we are quantizing the
quantity say 5.692347, now if I in a decimal
system, if I quantize it there are two types
of reducing this one is suppose I have three
decimal place approximation, then I will just
truncate it.
So, it will be 5.692, so the error is always
0.34, 0.0003447, if I have say 647, then also
if I truncate it then this would be the error,
so truncated value is always less than the
actual value, it can be at the most equal.
So, the error is always positive, error is
one side it actually error is actual value
to be subtracted from the final result, so
it is 5.692 minus 5.692647, so error is minus
0.000647, so it is always in one side. But,
if you round it off, then it will be a 5.693,
so it can be higher, so round of error can
be both positive and negative, whereas truncation
error will be one side it will be always less,
the error will be always negative.
So, this hold circuit is having a value 1
for this period and 0 otherwise, so the whole
circuit output is basically like this, one
may write x D to A converted value as x n
h naught t minus n T, so h naught t is the
impulse response of the hold circuit, of the
0 order hold.
Now, shall take up a little bit of simple
mathematics that you all know. How to represent
the numbers in binary from by different conventions.
So, DSP algorithm will be use for special
purpose DSP chips or the programs that will
be quantizing either the signal or the parameters.
Now, there are three places where this quantization
is coming, so error will be generated at three
points, one is input, input first of all you
have to scale it to that particular level
and then the input has to be quantized.
So, the input error, now input is a signal
which really does not alter the system property,
it will give you some output say instead of
5.2, suppose you are giving an input 5, so
had everything else been ideal, then the output
is just proportionately reduced. So, input
error is going to give you error in the output,
but it is not going to affect the property
of the system, then or the quality of the
output, then you have product you are having
multiplication inside.
So, you are having a finite bit length for
the representation of any quantity say 8 bit,
so an 8 bit number is multiplied by another
8 bit number, it may go to 16 bit, then you
are truncating it, so this is going to cause
more of errors. Product of finite length quantities
or variables, that is you can multiply by
a multiplier or just two quantities, two variables
may be multiplied, finite length quantities.
So, the product will have this gives a larger
bit length, see you have to truncate it the
third one, which is the most important one
is the parameter quantization. Parameter quantization,
you can see if we change the parameter the
roots may change H z we say B z by A z, we
may write b 0 plus b 1 z inverse and so on,
1 minus a 1 z inverse minus a 2 z to the minus
2 and so on.
Normally, we put a negative sign because it
is an I mean this coefficients will come to
this and if they are transferred you can write
plus and minus it really does not matter,
many books they follow this as a standard
form. Any way these b 0, b 1, a 1, a 2, etcetera;
these are the constants which if you quantize
will be changing the distribution of roots,
you must have studied in control systems root
locus technique. So, they are the last coefficient.
Again if you keep on changing the roots keep
on changing, now here you are having multiple
changes, changes in all the coefficients,
so you can imagine all the roots will be changing
in a very complex way and sometimes they may
go out of the unit circle. So, that may reduce
the stability of the system, it is not only
going to the unstable is one, even if they
change their positions the response is rustically
affected that will see very soon.
So, round off noise which may be positive
or negative, they may cause limit cycle oscillation
and noise, random noise; now representation
of numbers you may write a number in this
form, where C is the coefficient associated
with a particular radix r. So, in a decimal
system we have r is equal to 10 in a binary
system, we have r is equal to 2, so 25.564,
you write as C i 10 to the power i, so that
can be written as 2 into 10 to the power 1
plus 0 into 10 to the power 0 plus 5 into
10 to the power minus 1 plus 6 into 10 to
the power minus 2 plus 4 in to 10 to the power
minus 3. If r is equal to 2, you write 110.010
that represents 1 into 2 to the power 2 plus
1 into 2 to the power 1 plus 0 into 2 to the
power 0 plus 0 into 2 to the power minus 1
plus 1 into 2 to the power minus 2 plus 0
into 2 to the power minus 3. So, if you work
it 2 to the power 4 plus to 6 0 this is 146.25.
0, which one 5 into 10 to the power minus
1
0 into, not 20 5 into 10 to the power 0, thank
you very much, I was writing something I was
writing 20 in my mind, so this is 6.25.
So, for conversion from decimal to binary
we have say floating point arrangement like
this 30.275 how much is it, in the binary
system, so what we do this part we divide
by 2 and take the remainder 0. Then, that
gives me 15 divided by 2 the remainder is
1, the quotient is 7, 7 divided by 2 the remainder
is 1, then 3 divided by 2 remainder is 1,
1 divided by 2 remainder is 0, so we have
1 and quotient is 0.
So, we take it from this way will be 11110
and the fractional part we keep on multiplying
by 2 and starts removing the integral part,
so 0.275 multiplied by 2. And we go downward
is equal to 0.5, so this is 0, 0.55, 0.55
multiplied by 2, so that is 1.10, so that
is 0.10 into 2 how much 0.2 0, 0.2 into 2
0.4 that is 0, 0.4 into 2, 0.8 into 2 is 1.6,
so 1. And then 0.6 into 2 is 1.2 is 1, again
0.2, so if it is 0.2 you are coming to this
same place, so it will continue again 0 0
like that 1, 1, and so on. So, 30.275 will
be 11110 point will 0 1, go downward 00011001100
will it be two zeros are three zeros check.
Student: Two zeros.
Two zeros 1100, so it is like a your recording
decimal, so it is not only in decimal system
also in binary system, you can have a number
which may not end it may be endless.
So, there are three types of representation
in the binary system, one is sign magnitude,
one is one's complement and the third one
is two's complement, some of you have studied
this in digital electronics. So, the benefit
of those who have not gone through this course,
I will just briefly go through this will not
go in to the details of this representation.
So, the negative number is always given by
a 1 and the positive number were 0 in the
MSD more significant date, so say minus 1.75
will be 11.110000, 1.75 is 0 1, so this 1
is for the sign this 0 is for positive sign
0.110000, 0.75 is written like this.
So, this is half to the power 1 half to the
powers 2 and so on, so that gives me this,
so with b bits
you have 2 to the power b minus 1 number of
numbers can be represented, so this is a very
simple form.
In 1's complement form
for positive numbers it is same, for positive
numbers it is same as the earlier one that
is sign magnitude form, for negative numbers
say 0.875, if you write it will be 0.111000
in the binary form, so for minus 0.875 you
just complement each bit, so it will be 1.000111.
Then, 0 can be represented in this 1's complement
form as 00.0000 say we have 6 bit, I can write
also minus 0 that will be 11.1111. So, this
and this both will be representing 0's, in
the 1's complement form, in two's complement
form, let me write on this side.
Say the same number 8 7 5 is positive number
is once again same, so minus 0.875 you just
take the complement and then add the 1 at
the last bit, so it was 1.000111, so add 1,
so that gives me 001 plus 1000, then 1 0 0,
so it will be 1.001000. So, this will be the
representation in the binary form for 2's
complement, this is at as good as this is
same as subtracting 0.875 from 2, so 2 means
10.000000 and 0.875 is 0.111000 and you are
subtracting this, so that will give you 1.00100,
so this is same as this. Addition of
two fix point numbers, if it is 0.5 which
is 0.100 in the binary form and 0. Say 2 5,
0.125, which is 0.001 in the binary form if
you add them together, if the total does not
exceed 1 there is no problem.
So, 0.625 is the addition or 0.5 plus 0.125
in the decimal system, and it will be 0.100
plus 0.001, so that gives me 0.101 this sign
bit is not changing, but if the sum exceeds
1 that is rather exceeds the limit of 3 bits,
here the 3 bit representation if it exceeds
that, then we have problems. That carry will
be causing over flow say 0.5 plus 0.6 to 5,
if we had this is 0.100 and 0.6 to 5 is 0.101,
if you add this is 100.1, which is causing
a negative sign basically. So, this will be
causing an over flow which is here confuse
with a sign bit, so one point actual sum is
1.125, so 1.125 cannot be represented in a
3 bit system.
The subtraction
of two fix points number, this we can do by
2's complement say 0.5, we want to subtract
0.25 from here, so that gives me 0.25, so
0.5 in 2's complement it is 0.100 whether
it is a positive number there is no problem.
For a negative number 0.25 is how much and
then complement it will be 1.101, so 1.101.
We have to add 1 for 2's complement this is
1's complement, this 2's complement 1 1 1
0 1 and 1, so 1.110 add them straight 0 1
0 0 1 and then neglect this.
So, you drop this, so 0.010 this is for 0.25,
so this carry is dropped, otherwise 1.110,
let us say 1.110 what will be 2's complement,
0.010 you can do by a simple subtraction also.
The 2's complement is simpler you just take
the negative quantity in 2's complement, positive
quantity in remains as it is, so add them
together that gives you the result. Multiplication
this is somewhat involved we see that, suppose
we have a 2 bit representation of 2 numbers,
the product is a 4 bit number, when you multiply
specially you get this problem. So, if you
have 2 bit number for the integral part plus
2 bit number in the fraction part, then the
product may give you 4 bit part in the integral
part and 4 bit part in the fraction part.
So, in a floating representation we write
a number in this form, where M is written
half and 1, M is a mantissa, so C is either
positive or negative, for example 4.5 we should
write as 2 to the power how much, if I write
2 to the power 2, then I will get a quantity
m which will be more than 1. So, it should
be always next higher number which is some
power of 2, next higher number is 8, so 8
into something 8 into 4.5 by 8, so it is 2
to the power 3 multiplied by 0.5625.
So, it will be 2 to the power this 3 is 011
multiplied by 0.5625 if you write in a 4 bit
representation
it will be like this 0.5 plus 0.25 and so
on. Similarly, 1.5 will be next number is
2, so 2 to the power 1 into 1.5 by 2, so that
is 2 to the power 1 into 0.75 and that is
2 to the power 1, so it will be 001 into 0.750
0.1100. 6.5 2 to the power 3 into 0.8125,
so that gives me 2 to the power 001, to the
power 3 011 into 0.1101, 0.625 it will be
2 to
the power 0 into 0.625 is to be 2 to the power
000 into 0 0.1010. So, like this any number
you can express, in terms of decimal binary
forms.
Let us now multiply 1.5 into 1.25, so 1.5
is 2 to the power 1 into 0.75, that is 2 to
the power 001 into 0.1100. So 2 to the power
C 1 into M 1, 1.25 is 2 to the power 1 into
0.625 that is 2 to the power 001 into 0.1010,
so that is 2 to the power C 2 in to M 1, 2
to the power C 1, C 1 is equal to C 2 M 2.
So, multiplication will give me 2 to the power
C 1 that is common plus C 2, 2 to the power
C 1 plus 2 to
the power C 2 got it, into M 2. So, 2 to the
power 010 into 0 product of these 2, so 0.1100
0.1010 what is the product, 0000 1100, then
0000 1100, if I add 111. Now, 1 2 3 4 5 6
7 8, 1 2 3 4 7 8, so 1 0, so 0.0111 1 1 2
4, if it is a 4 bit representation then you
have to truncate.
Now, if you have to add two floating point
numbers for addition or subtraction for that
matter, for addition of two floating point
numbers, see when you add 10 with 7, basically
you put a 0 here. So, both of them are brought
to the same level of representation same number
of digits, so you have to bring them to the
same number of digits and then only you can
add.
So, 3.0 plus 0.125 what will be the addition
of these two, so 3.0 is 2 to the power 2 into
0.75, so 2 to the power 010 into 0.110000,
0.125 is 2 to the power 2 you have to bring
the same index into 0.125 by 4, so 0.03125
which gives me 2 to the power 010 into 0.0000100.
So, if you add 3.0 plus 0.125 that gives me,
correct me if I am wrong 0.125 by 4, so that
gives me this, so this will be 2 to the power
010 into product of this two...
Will become… this is sum, so this will give
me 0.1100 1100100, so will take it up in further
details later on if time permits, otherwise
I leave at this, you can read it off from
the books for further details of this. Now,
quantization error as we are discuss in earlier,
in the direct form
the coefficients of the polynomials will be
appearing in the structure as constants as
a multiplier, is it not. In the other forms
that is if you have cascaded form or parallel
form the constant of these are not going to
affect all the roots, they are going to affect
only that particular pair.
For example, if I have say A z, B z by A z
as product of some b i 0 plus b i 1 z inverse
plus b i 2 z to the power minus 2 divided
by 1 minus a i 1 z to the power minus a i
2 z to the power minus 2 and i varying from
say 1 to K. That means, there are many bike
wards I have got in cascade, I could have
also written this as some like a constant
K 1 by 1 minus a i 1 z inverse minus a i 2
z to the power minus 2 plus a 1 K 2 minus
a 2 1 z inverse minus a 2 2 z to the power
minus 2, and so on, I could have put in parallel
forms.
So, here the two 0's and two poles only will
be affected by quantization of these coefficients,
whereas in the direct form all the coefficients
that is B z by A
z say b 0 plus b 1 z inverse plus b 2 z to
the power minus 2 and so on. If you take the
polynomial as it is and quantization of this
will be affecting all the poles and 0's, here
quantization of a coefficients, suppose forget
about others just one coefficient is changed
only this whole pair will be affected, others
will not be affected.
So, here we are restricting it within a certain
group, restricting the sensitivity of a particular
group of poles and 0's with the variation
of these coefficients. So, this is much more
robust, however in these case the poles are
restricted, 0's will be 0's are not known
they are in a very complex mode, because if
you add up K 1 in to a 2 1, a 2 2, K 2 into
a 1 1, a 1 2 and so on, all this terms will
appear. So, here only the poles will be affected
by rather the effect of variation of the poles,
these parameters will be restricted to poles,
but these parameters will not be a affecting
the 0's independently.
In normalized lattice, if you are having a
smaller bit representation, then this is a
very robust one, very useful for smaller bit
representation, but it needs lot of computation,
computation is more. In the other two cases
the number of elements required will be same
as a direct form a little bit of extra computation
will be involved. So, these three representation
that is normalize lattice or cascade or parallel
forms, they will be more robust compare to
a direct form. Now, will see the effect of
quantization, let us consider H z as K varying
from 1 to N, K varying from 0 to M, so H hat
z will be b k hat z to the power minus K,
b k hat is b k plus delta b k the error due
to quantization, actual value plus the error.
Let us consider A z, the poles we can write
this as 1 minus a k z to the power minus k,
K equal to 1 to N, this can be factorized.
And I can write this as 1 minus z i z inverse
i varying from, I will write j z j, j equal
to 1 to N, N there are its an Nth order polynomial
having N roots. Now, let us see the sensitivity,
let us see the change delta z i is sigma delta
z i by delta a k in to delta a k, that is
shift in the sensitivity of the i th root
with respect to any one constant, multiplied
by the change in that constant that will be
the small change due to a particular element,
particular parameter.
Similarly, if you sum up, then that will be
the total change in the position of a particular
root, each root is dependent on all the parameters
a 1, a 2, a 3, a 4, so you make a small change
in a 1, a small change in a 2 and so on. So,
collect the total change, so the total change
in that ith root will be so much, again K
varying from 1 to N. Now, delta A z by delta
z i multiplied by delta z i by delta a k is
equal to delta A z by delta a k z equal to
z i. So, change in this, a partial derivative
of this with respect to any root z I, and
if you compute that slope at z is equal to
z i, that multiplied by the sensitivity of
z i with respect to a k will be the sensitivity
of this that partial derivative of a z with
respect to the coefficient a k.
So, from here delta z i delta a k will be
this divided by this, if you expand this delta
z by delta a k, you can just see take the
derivative with respect to a k. So that will
give me what it is 1 minus a 1 z to the power
minus 1, so a 2 z to the power minus 2, and
so on. If I take a k z to the power minus
k minus a N z to the power minus N, if I take
the derivative with respect to a k it will
be z to the power minus k with a negative
sign.
So, this will be z to the power minus K, similarly
if you do this will get z to the power i and
then pi z i minus z j if you evaluate this,
j equal to 1 to N, j not equal to i and there
will be a negative sign here, so that will
make it positive. Now, you see this term what
does it mean? Suppose the poles we are now
considering only the denominator factor poles,
if the poles are very tightly clustered, there
will be complex conjugates know, there will
be always appearing in the form of complex
conjugate.
Now, these are the vectors z i minus z j,
say suppose this is the i th one, so these
are the z i minus z j quantities z i minus
z j, now if you take the products if this
z i shifts a little it is a quite a big change.
It is somewhere a 0.01 to 0.02, the difference
is say if it is at a distance of 5.1, if it
shifts to 5.2, there is not much of a change
5.1 to 5.2, but the distance between the two
poles if there all in close cluster, it may
be from 0.01 to 0.02 it is doubled, do you
get my point.
So, the denominator which is the product of
such vectors, finally will give you a big
change, so they are very, very sensitive specially
when they are in close cluster same is the
case with the 0's, this is a narrow band,
band pass filter basically. If you are having
close cluster here that will correspond to
a low pass filter, narrow band low pass filter;
low pass filter roots are somewhere close
to the real axis and band pass filters here
away from the real axis, now let us see what
would be the effect on this.
We have say for a quadratic representation,
we may have
something like this, you can write
a quadratic form in this fashion also, that
is 1 by 1 minus 2 r cos theta z inverse minus
z to the power minus 2 r plus r square z to
the power minus 2. Suppose, we have the bike
ward representation like this, 1 minus 2 r
cos theta z to the power minus 1 plus r square
z to the power minus 2 it is in a quadratic
form, if the roots are complex.
If the roots are simple on the real axis,
then there is no problem the roots are here,
if they are complex then we can represent
by this form, so the roots are at r e to the
power j theta and r e to the power minus j
theta. Suppose, we quantize these two quantities,
then what are the possible values, so will
take it up in the next class, the time is
just over, what will be the effect on quantization
of these quantities. Thank you very much,
I will take it up on.