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Chapter 3: 46 (page 241)

[20] Write down the bit pattern assuming that we are using base 30 numbers in the fraction instead of base 2. (Base 16 numbers use the symbols 0–9 and A–F. Base 30 numbers would use 0–9 and A–T.) Assume there are 20 bits, and you do not need to normalize. Is this representation exact?

Short Answer

Expert verified

The bit pattern is 3D7.T.

Yes, the representation is exact

Step by step solution

01

Define the concept.

The base-30 number system uses the base 30, the symbols 0 to 9, and A to T, where the decimal number system uses the base 10 and the binary number system uses the base 2.

For example, in the base-30 number system, “A” is equivalent to the value of the decimal number system “10” and that is also equivalent to the value of the binary number system “01010.”

In the base-30 number system, “B” is equivalent to the value of the decimal number system “11” and that is also equivalent to the value of the binary number system “01011.”

In the base-30 number system, “C” is equivalent to the value of the decimal number system “12” and that is also equivalent to the value of the binary number system “1100.”

In the base-30 number system, “D” is equivalent to the value of the decimal number system “13” and that is also equivalent to the value of the binary number system “01101.”

In the base-30 number system, “E” is equivalent to the value of the decimal number system “14” and that is also equivalent to the value of the binary number system “01110.”

In the base-30 number system, “T” is equivalent to the value of the decimal number system “29” and that is also equivalent to the value of the binary number system “11101.”

In the base-30 number system, “3” is also equivalent to the value of the binary number system “0011.”

02

Determine the calculation

The required bit pattern is written below by using the base-30 number system in the fraction, assuming there are 20 bits, and without normalization:

00011 01101 00111.11101

3D7.T is presented in a 20-bit base-30 number system:

000110110100111111013D7T

The presentation is exact.

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Most popular questions from this chapter

IEEE 754-2008 contains a half precision that is only 16 bits wide. The left most bit is still the sign bit, the exponent is 5 bits wide and has a bias of 15, and the mantissa is 10 bits long. A hidden 1 is assumed. Write down the bit pattern to represent -1.5625×10-1assuming a version of this format, which uses an excess-16 format to store the exponent. Comment on how the range and accuracy of this 16-bit floating point format compares to the single precision IEEE 754 standard.

Based on your answers to 3.32 and 3.33, does

3.984375×10-1+3.4375×10-1+1.771×103=3.984375×10-1+3.4375×10-1+1.771×103?

Calculate 1.666015625×100×1.9760×104+1.666015625×100×-1.9744×104 by hand, assuming each of the values is stored in the 16-bit half-precision format described in Exercise 3.27 (and also described in the text). Assume 1 guard, 1 round bit, and 1 sticky bit, and round to the nearest even. Show all the steps, and write your answer in both the 16-bit floating-point format and in decimal.

Calculate the product of-8.0546875×100and-1.79931640625×10-1by hand, assuming A and B are stored in the 16-bit half precision format described in Exercise 3.27. Assume 1 guard, 1 round bit, and 1 sticky bit and round to the nearest even. Show all the steps; however, as is done in the example in the text, you can do the multiplication in human readable format instead of using the techniques described in Exercises 3.12 through 3.14. Indicate if there is overflow or underflow. Write your answer in both 16-bit floating point format described in Exercise 3.27 and also as a decimal number. How accurate is your result? How does it compare to the number you get if you do multiplication on a calculator?

[30] <§3.4> Using a table similar to that shown in Figure 3.10, calculate 74 divided by 21 using the hardware described in Figure 3.11. You should show the contents of each register on each step. Assume A and B are unsigned 6-bit integers. This algorithm requires a slightly

different approach than that shown in Figure 3.9. You will want to think hard about this, do an experiment or two, or else go to the web to figure out how to make this work correctly. (Hint: one possible solution involves using the fact that Figure 3.11 implies the remainder register can be shifted either direction.)
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