Sala común de la asignatura Tecnología de Computadores. Plan 2023 (2024-25, Grado en Ing del Software, Informática, de Computadores y Doble Grado Matemáticas + Informática, Todos los grupos)
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Give the upper limit of the range of values of integer representation two's complement with 7 bits? (highest integer value representable in decimal)
We consider the division with rest algorithm for unsigned integers in 4 bits.
You give the values of A[4:0] and Q[3:0] and the followed operation in each iteration.
Dividend (numerator) = 1510 (11112) / Divisor (denominator) (M) = 710 (01112)
In each row, you provide the values of A[4:0] and Q[3:0] in binary wih all needed bits just after the opeartion given in the field "Operation in the same row.
If the operation is "No operation", you copy the values pf A[4:0] and Q[3:0] of the former row as they did not change.
|
| Iteración | A[4:0] | Q[3:0] | Operation |
| 2 | 2 | Initialization | |
| 0 | 2 | 2 | |
| 2 | 2 | | |
| 2 | 2 | | |
| 2 | 2 | | |
| 1 | 2 | 2 | |
| 2 | 2 | | |
| 2 | 2 | | |
| 2 | 2 | | |
| 2 | 2 | 2 | |
| 2 | 2 | | |
| 2 | 2 | | |
| 2 | 2 | | |
| 3 | 2 | 2 | |
| 2 | 2 | | |
| 2 | 2 | | |
| 2 | 2 | |
Quotient in binary, 4 binary digits:
2 and in decimal: 10Rest in binary, 4 binary digits:
2 and in decimal: 10Por favor, responda a todas las partes de la pregunta.Consider the multiplication aalgorithm for C2 multiplication based on add and shift for operands of 4 bits.
Give the iterative values of Z[8:4] and Z[3:0] and the performed operation.
X = -510 (10112) * Y = -410 (11002)
Nota:
| In each row, you give the value of Z[8:4] and Z[3:0] in binary with all necessary bits just after the iteration operation.
In the field "Operation" you may also put "No operation" and you copy the values of Z[8:4] and Z[3:0] from the last row to show they do not change.
|
| Iteración | Z[8:4] | Z[3:0] | Operation |
| 2 | 2 | Initialization | |
| 0 | 2 | 2 | |
| 2 | 2 | | |
| 1 | 2 | 2 | |
| 2 | 2 | | |
| 2 | 2 | 2 | |
| 2 | 2 | | |
| 3 | 2 | 2 | |
| 2 | 2 | |
After the procedure, give the 8 bits as a result of the multiplication (8 binary digits):
2Give the result in decimal: 10We consider the division with rest algorithm for unsigned integers in 4 bits.
You give the values of A[4:0] and Q[3:0] and the followed operation in each iteration.
Dividend (numerator) = 1510 (11112) / Divisor (denominator) (M) = 710 (01112)
In each row, you provide the values of A[4:0] and Q[3:0] in binary wih all needed bits just after the opeartion given in the field "Operation in the same row.
If the operation is "No operation", you copy the values pf A[4:0] and Q[3:0] of the former row as they did not change.
|
| Iteración | A[4:0] | Q[3:0] | Operation |
| 2 | 2 | Initialization | |
| 0 | 2 | 2 | |
| 2 | 2 | | |
| 2 | 2 | | |
| 2 | 2 | | |
| 1 | 2 | 2 | |
| 2 | 2 | | |
| 2 | 2 | | |
| 2 | 2 | | |
| 2 | 2 | 2 | |
| 2 | 2 | | |
| 2 | 2 | | |
| 2 | 2 | | |
| 3 | 2 | 2 | |
| 2 | 2 | | |
| 2 | 2 | | |
| 2 | 2 | |
Quotient in binary, 4 binary digits:
2 and in decimal: 10Rest in binary, 4 binary digits:
2 and in decimal: 10Given a bit string which represents a real number in IEEE 754 single precicion:
SCCCCCCCCMMMMMMMMMMMMMMMMMMMMMMM
01111110110100000000000000000000- Mantisa: Sign-Magnitud S: Bit 31 -signM: Bits 22:0 - normalised mantissa (1.M with implicit 1 in the representation)- Exponent: Excess 127C: Bits 30:23 - characteristic (exponent represented in excess 127)The represented value is given by V = (-1)S x 1.M x 2C-127 if it is not a special case.
Determine the represented real value in decimal with at most 5 significative digits (different from 0). For instance use decimal notation with a decimal comma, e.g. -2,625 or you can also use the scientific notation, e.g. -5,5432e-23.
Given a bit string which represents a real number in IEEE 754 single precicion:
SCCCCCCCCMMMMMMMMMMMMMMMMMMMMMMM
01111110110100000000000000000000- Mantisa: Sign-Magnitud S: Bit 31 -signM: Bits 22:0 - normalised mantissa (1.M with implicit 1 in the representation)- Exponent: Excess 127C: Bits 20:23 - characteristic (exponent represented in excess 127)The represented value is given by V = (-1)S x 1.M x 2C-127 if it is not a special case.
Determine the represented real value in decimal with at most 5 significative digits (different from 0). For instance use decimal notation with a decimal comma, e.g. -2,625 or you can also use the scientific notation, e.g. -5,5432e-23.
A number is represented in floating point according to the following format
S CCCC MMMMMM1 0100 011100- Mantisa: Sign-MagnitudS: Bit 10 signM: Bits 5:0 - mantisa normalized as 1.M, so using an implicit 1 - Exponent: Exces 8C: Bits 9:6 - characteristic for the exponent in exces 8)
So the represented number is V = (-1)S x 1.M x 2C-8
Give the represented number in decimal considering only 5 significative digits (nonzero) Example 1.2345
Given the simple precision IEEE 754 representation:
- 32 bits.
- Mantisa of 24 bits, fixed point (sign in bit 31, magnitud in bits 22:0) with 1 implícit.
- Characteristic of 8 bits (bits 30:23) represents the exponent in excess 127.
- Binary string: SCCCCCCCCMMMMMMMMMMMMMMMMMMMMMMM
represent real number 1,4 in this format
Given the single precision IEEE 754 representation:
SCCCCCCCCMMMMMMMMMMMMMMMMMMMMMMM
11000001111100000000000000000000- Mantisa: Signed-MagnitudS: Bit 31 -signM: Bits 30:23 - fractional normalized mantisa (1.M with implicit 1)- Exponent: Excess 127C: Bits 22:0 - characeristic representing the exponent in excess 127The represented value is given by = (-1)S x 1.M x 2C-127 as long as we are not dealing with a special case.
Determine the represented number in decimal notation. In the nswer, consider only 5 significative digits, i.e. digits that differ from 0.
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