Welcome to the next video in the
binary series. In the previous
video we learned how to convert
numbers to floating point
notation. In this video I will
show you how to do the other way
around, so had to convert a
floating point notation number
back into ordinary decimal
numbers. So let's say that you
are given an 8 bits floating
point notation, something like
1001. 1011 OK,
now the first step that you
would do in here is break
this code into it
constituents break this
code into its parts.
Remember that the first bit
is the sign bit. That's
going to tell you if it's a
positive or a negative
number. The next 3 digits.
Is your exponent?
And this exponent, these 3
bits twos complement notation
will tell you how many places
you need to move the radix
point either in to the right
or to the left and the last
part. The last four bits.
Is your normalized?
Month is self.
This is the number that you will
have to apply the exponent two
and move the radix point into
the right. Place to be able to
decode the binary number so.
Step one. Sign because the sign
equals to one. We have got a
negative number here.
Then Step 2.
Exponent Our exponent is
001. Well, that's a positive
number because it starts with
zero and it's positive one. So
my exponent is positive one, so
I will need to move my
normalized mantissa one place
into the positive direction that
three. My normalized mantissa is
0.1011. Apply the exponent
two. It move the radix .1
in the positive direction.
So the number the binary number
that is given to me here is
one data .011. Now I need to
decode it back into decimal.
This is one that's the radix
point. This is 1/2. This is 1/4
and this is an 8. So what I have
in here is.
1 + 1/4 plus an 8. Remember,
how can we act together?
Fractions need to make them to
be the same denominators, so it
will be 1 + 2 eight +18,
which is altogether one and
three eights. And remember,
we had one to start with,
so 10011011 in eight bits
floating point notation is
the same as minus one and
three eights in decimal.
Let's look at the
next example, which
is 11011100 again.
The first step is to identify
the sign. Now the sign is 1
again. So we have got a
negative number.
Then Step
2. Find
the exponent.
Remember the exponent is 101
now. This is in three Bits 2's
complement, so the 1st digit
tells me that this is a negative
number. So this is a negative
exponent, so it went through the
inversion process. So what I
need to do to find out what
positive equivalent is in here
is to redo that inversion
process, so copy the number
until you copy the one.
Then invert everything else.
So that zero becomes one. This
one becomes a 0, so this will be
the positive equivalent of this
negative number, and the
positive equivalent in here
using the place values.
This is equal to
R. Three, so this original
number is negative three, so our
exponent is negative 3.
And that 3.
Using the mantissa.
Now the normalized mantissa is
the last four bits. Remember,
it's always starts with 0.1100,
but what that Azar's done?
The exponent was negative three,
so the computer was stored to
move it 3 places to the negative
direction. So what does that
mean? One place to place
three places. Filling the zeros.
So the original number
was a very small
number O radix .0001100.
So what is this number?
Use the place values again this
is 1 radix point or half or
quarter on 8R16R32 and the rest
of them are zero, so we don't
really need to bother about
them. So this number here is
basically the sum of the 16th
and the 32.
Again, how can I add
fractions together? I need to
make them to be the same
denominator, so how can I
make 30 twos from sixteens? I
just need to double it.
So it's altogether 3 / 32.
And remember I had a negative
number because my assignment was
one so one.
1011100 As an 8 bits floating
point is exactly the same as
minus 3 / 32.
Let's do one more example of
this floating point notation and
let's look at what number is 0.
101
1011
Step one, find
the sign. The
first bit is 0, so this
is a positive number.
Then comes Step
2. Using
the exponent.
Now our exponent.
Is 101.
Which is a negative number.
So again, going through the
inversion process copied the
number until your Capital One,
then invert everything else.
And. This is again.
Positive three so this is still
negative. 3 The difference here
now will be that my month is a
is slightly different, so let's
look at the last step and deal
with the mantissa.
So I'm normalized mantissa
is 0.1011. The computer
was told to move
this mantissa or the
radix point. Three in the
negative direction, so it would
need to move it 123.
So 0.00.
So the original
number was 0.0001011.
The accompanying place
values are one
radix point 1/2
or quarter an
8 or 16
or 3264 and
128th. So I need to
talk together. Now is 1 / 16 +
1 / 64 + 1 / 128. Now
again to be able to add them
together I need to get them all
to be the same denominator. Now
this is a common denominator
because I can get 228 from
doubling all of them but at
different numbers. So from.
16 to get 228 what I've done. I
doubled 123 times. So that is
basically 16 * 8 going to give
me the 128. So what this is
telling me that 8 / 128 is the
same as 1 / 1664 and 120. There
is only one doubling point in
here, so that is just 2 / 128.
Plus 1 / 128, so
this is altogether giving me
8 + 2 is 10
plus one is 11 /
128 so 01011011 in eight
bits within point.
Is the same as 11 / 128?
Remember this time it was zero,
so this is a positive number and
for positive numbers we don't
write out the positive sign.
Now we went through the 8 bits
floating point notation and you
might find it still a little bit
confusing. That's absolutely
fine, because this is like an
orchestra you pulling together
everything that you've learned
about binary numbers. So you
putting together the two bits
complement notation, some normal
mathematical knowledge about
moving decimal places applied 2
in binary numbers and so on. So
this is probably one of the most
difficult questions that you can
get in binary numbers, but once
you understand them then you
will really know.
What needed to be known
about this system, and
you know how the computer
looks at the binary code.
And now it is your turn to try
these conversions. You will find
the answers to these questions
shortly after the questions
appear. So these are the
practice questions.
And here are the answers.