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.