How to Convert binary 11110011 to decimal if the number is represented as........?

Question

A. Unsigned 8-bit number

B.Signed 8-bit Excess (biased)

C. Signed 8-bit two's complement

Three questions above i require urgent help with thanks to all that can offer help

Answer 1

A) unsigned is easiest:

1*(2^7) + 1*(2^6) + 1*(2^5) + 1*(2^4) + 0*(2^3) + 0*(2^2) + 1*(2^1) + 1*(2^0)

= 1*(128) + 1*(64) + 1*(32) + 1*(16) + 0*(8) + 0*(4) + 1*(2) + 1*(1)

= 128 + 64 + 32 + 16 + 2 + 1

= 243

For B) you usually need some number to bias by, say Excess-N where N is some number we bias by, then we solve the number and subtract the biasing N. Let's pretend N=128, then you simply subtract 128 from the SIGNED representation of the number (since you specify SIGNED):

First, the signed version of the number is solved similarly to A - but we only solve for the first 7 digits, the 8th is the sign-bit. so, dropping the 8th bit (or sign bit):

11110011 => 1110011 (note: sign bit was = 1)

And solving the same way as before:

= 1*(64) + 1*(32) + 1*(16) + 0*(8) + 0*(4) + 1*(2) + 1*(1)

= 64 + 32 + 16 + 2 + 1

= 115

No since the sign bit (the 8th bit we dropped earlier) was 1, we say it's a negative number (if sign bit had been 0, we'd say it was a positive number):

= -115

and lastly, we subtract our N from Excess-N (we assumed 128) so:

= -115 - N

= -115 - 128 = -243

For C) Since the sign bit is 1, we have a negative number..keep this in mind.

The method for solving a negative two-compliment is to invert all bits (so 1's become 0's and 0's become 1's), and then add 1 to the last bit. It sounds complicated, but it's not:

so flip it:

11110011 => 00001100

then add 1:

00001100 + 00000001 = 00001101.

Now, just solve as usual:

= 00001101

= 1*(2^3) + 1*(2^2) + 0*(2^1) + 1*(2^0)

= 8 + 4 + 1

= 13

BUT remember the sign bit? It was a 1, so we have a negative number:

= -13