
a)
Explanation of Solution
IEEE-754 floating point single precision:
IEE-754 floating point single precision has 32 bits.
- One bit for sign, 8 bits for exponent, and 23 bits for significant bits.
Storing “12.5” using IEEE-754 single precision:
Step 1: Converting decimal to binary number:
Step (i): Divide the given number into two parts, integer and the fractional part. Here, the integer part is “12” and the fractional part is “.5”.
Step (ii): Divide “12” by 2 till the quotient becomes 1. Simultaneously, note the remainder for every division operation.
Step (iii): Note the remainder from the bottom to top to get the binary equivalent.
Step (iv): Consider the fraction part “.5”. Multiply the fractional part “.5” by 2 and it continues till the fraction part reaches “0”.
Step (v): Note the integer part to get the final result.
Thus, the binary equivalent for “12.5” is
Step 2: Normalize the binary fraction number:
Now the given binary fraction number should be normalized. To normalize the value, move the decimal point either right or left so that only single digit will be left before the decimal point.
Step 3: Convert the exponent to 8 bit excess-127:
To convert the exponent into 8-bit excess-127 notation, the exponent value should be added with 127. After addition, it is converted into binary equivalent.
Converting
Step 4: Convert the significant to hidden bit:
To convert the significant to hidden bit the leftmost “1” should be removed.
Step 5: Framing the number “12.5” in 32 bit IEEE-754 single precision
Sign bit(1 bit) | Exponent bit(8 bits) | Significant bit(23) |
0 | 10000010 | 10010000000000000000000 |
Thus, the number “12.5” in 32 bit IEEE-754 single precision is represented as “
b)
Explanation of Solution
Storing “-1.5” using IEEE-754 single precision:
Step 1: Converting decimal to binary number:
Step (i): Consider the fraction part “0.5”. Multiply the fractional part “.5” by 2 and it continues till the fraction part reaches “0”.
Step (ii): Note the integer part to get the final result.
Thus the binary equivalent for “1.5” is
Step 2: Normalize the binary fraction number:
Now the given binary fraction number should be normalized. To normalize the value, move the decimal point either right or left so that only single digit will be left before the decimal point.
Step 3: Convert the exponent to 8 bit excess-127:
To convert the exponent into 8-bit excess-127 notation, the exponent value should be added with 127. After addition, it is converted into binary equivalent.
Converting
Step 4: Convert the significant to hidden bit:
To convert the significant to hidden bit the leftmost “1” should be removed.
Step 5: Framing the number “-1.5” in 32 bit IEEE-754 single precision
Sign bit(1 bit) | Exponent bit(8 bits) | Significant bit(23) |
1 | 01111111 | 10000000000000000000000 |
Thus, the number “-1.5” in 32 bit IEEE-754 single precision is represented as “
c)
Explanation of Solution
Storing “.75” using IEEE-754 single precision:
Step 1: Converting decimal to binary number:
Step (i): Consider the fraction part “.75”. Multiply the fractional part “.75” by 2 and it continues till the fraction part reaches “0”.
Step (ii): Note the integer part to get the final result.
Thus the binary equivalent for “.75” is
Step 2: Normalize the binary fraction number:
Now the given binary fraction number should be normalized. To normalize the value, move the decimal point either right or left so that only single digit will be left before the decimal point.
Step 3: Convert the exponent to 8 bit excess-127:
To convert the exponent into 8-bit excess-127 notation, the exponent value should be added with 127. After addition, it is converted into binary equivalent.
Converting
Step 4: Convert the significant to hidden bit:
To convert the significant to hidden bit the leftmost “1” should be removed.
Step 5: Framing the number “.75” in 32 bit IEEE-754 single precision
Sign bit(1 bit) | Exponent bit(8 bits) | Significant bit(23) |
0 | 01111110 | 10000000000000000000000 |
Thus, the number “.75” in 32 bit IEEE-754 single precision is represented as “
d)
Explanation of Solution
IEEE-754 floating point single precision:
IEE-754 floating point single precision has 32 bits.
- One bit for sign, 8 bits for exponent, and 23 bits for significant bits.
Storing “26.625” using IEEE-754 single precision:
Step 1: Converting decimal to binary number:
Step (i): Divide the given number into two parts, integer and the fractional part. Here, the integer part is “26” and the fractional part is “.625”.
Step (ii): Divide “26” by 2 till the quotient becomes 1. Simultaneously, note the remainder for every division operation.
Step (iii): Note the remainder from the bottom to top to get the binary equivalent.
Step (iv): Consider the fraction part “.5”. Multiply the fractional part “.625” by 2 and it continues till the fraction part reaches “0”.
Step (v): Note the integer part to get the final result.
Thus the binary equivalent for “26.625” is
Step 2: Normalize the binary fraction number:
Now the given binary fraction number should be normalized. To normalize the value, move the decimal point either right or left so that only single digit will be left before the decimal point.
Step 3: Convert the exponent to 8 bit excess-127:
To convert the exponent into 8-bit excess-127 notation, the exponent value should be added with 127. After addition, it is converted into binary equivalent.
Converting
Step 4: Convert the significant to hidden bit:
To convert the significant to hidden bit the leftmost “1” should be removed.
Step 5: Framing the number “26.625” in 32 bit IEEE-754 single precision
Sign bit(1 bit) | Exponent bit(8 bits) | Significant bit(23) |
0 | 10000011 | 10101010000000000000000 |
Thus, the number “12.5” in 32 bit IEEE-754 single precision is represented as “
Want to see more full solutions like this?
Chapter 2 Solutions
ESSENTIALS OF COMPUTER ORGAN..-TEXT
- I need help fixing the minor issue where the text isn't in the proper place, and to ensure that the frequency cutoff is at the right place. My code: % Define frequency range for the plot f = logspace(1, 5, 500); % Frequency range from 10 Hz to 100 kHz w = 2 * pi * f; % Angular frequency % Parameters for the filters - let's adjust these to get more reasonable cutoffs R = 1e3; % Resistance in ohms (1 kΩ) C = 1e-6; % Capacitance in farads (1 μF) % For bandpass, we need appropriate L value for desired cutoffs L = 0.1; % Inductance in henries - adjusted for better bandpass response % Calculate cutoff frequencies first to verify they're in desired range f_cutoff_RC = 1 / (2 * pi * R * C); f_resonance = 1 / (2 * pi * sqrt(L * C)); Q_factor = (1/R) * sqrt(L/C); f_lower_cutoff = f_resonance / (sqrt(1 + 1/(4*Q_factor^2)) + 1/(2*Q_factor)); f_upper_cutoff = f_resonance / (sqrt(1 + 1/(4*Q_factor^2)) - 1/(2*Q_factor)); % Transfer functions % Low-pass filter (RC) H_low = 1 ./ (1 + 1i * w *…arrow_forwardMy code is experincing minor issue where the text isn't in the proper place, and to ensure that the frequency cutoff is at the right place. My code: % Define frequency range for the plot f = logspace(1, 5, 500); % Frequency range from 10 Hz to 100 kHz w = 2 * pi * f; % Angular frequency % Parameters for the filters - let's adjust these to get more reasonable cutoffs R = 1e3; % Resistance in ohms (1 kΩ) C = 1e-6; % Capacitance in farads (1 μF) % For bandpass, we need appropriate L value for desired cutoffs L = 0.1; % Inductance in henries - adjusted for better bandpass response % Calculate cutoff frequencies first to verify they're in desired range f_cutoff_RC = 1 / (2 * pi * R * C); f_resonance = 1 / (2 * pi * sqrt(L * C)); Q_factor = (1/R) * sqrt(L/C); f_lower_cutoff = f_resonance / (sqrt(1 + 1/(4*Q_factor^2)) + 1/(2*Q_factor)); f_upper_cutoff = f_resonance / (sqrt(1 + 1/(4*Q_factor^2)) - 1/(2*Q_factor)); % Transfer functions % Low-pass filter (RC) H_low = 1 ./ (1 + 1i * w *…arrow_forwardI would like to know the main features about the following three concepts: 1. Default forwarded 2. WINS Server 3. IP Security (IPSec).arrow_forward
- map the following ER diagram into a relational database schema diagram. you should take into account all the constraints in the ER diagram. Underline the primary key of each relation, and show each foreign key as a directed arrow from the referencing attributes (s) to the referenced relation. NOTE: Need relational database schema diagramarrow_forwardWhat is business intelligence? Share the Business intelligence (BI) tools you have used and explain what types of decisions you made.arrow_forwardI need help fixing the minor issue where the text isn't in the proper place, and to ensure that the frequency cutoff is at the right place. My code: % Define frequency range for the plot f = logspace(1, 5, 500); % Frequency range from 10 Hz to 100 kHz w = 2 * pi * f; % Angular frequency % Parameters for the filters - let's adjust these to get more reasonable cutoffs R = 1e3; % Resistance in ohms (1 kΩ) C = 1e-6; % Capacitance in farads (1 μF) % For bandpass, we need appropriate L value for desired cutoffs L = 0.1; % Inductance in henries - adjusted for better bandpass response % Calculate cutoff frequencies first to verify they're in desired range f_cutoff_RC = 1 / (2 * pi * R * C); f_resonance = 1 / (2 * pi * sqrt(L * C)); Q_factor = (1/R) * sqrt(L/C); f_lower_cutoff = f_resonance / (sqrt(1 + 1/(4*Q_factor^2)) + 1/(2*Q_factor)); f_upper_cutoff = f_resonance / (sqrt(1 + 1/(4*Q_factor^2)) - 1/(2*Q_factor)); % Transfer functions % Low-pass filter (RC) H_low = 1 ./ (1 + 1i * w *…arrow_forward
- Task 3. i) Compare your results from Tasks 1 and 2. j) Repeat Tasks 1 and 2 for 500 and 5,000 elements. k) Summarize run-time results in the following table: Time/size n String StringBuilder 50 500 5,000arrow_forwardCan you please solve this without AIarrow_forward1. Create a Vehicle.java file. Implement the public Vehicle and Car classes in Vehicle.java, including all the variables and methods in the UMLS. Vehicle - make: String model: String -year: int + Vehicle(String make, String, model, int, year) + getMake(): String + setMake(String make): void + getModel(): String + setModel(String model): void + getYear(): int + set Year(int year): void +toString(): String Car - numDoors: int + numberOfCar: int + Car(String make, String, model, int, year, int numDoors) + getNumDoors(): int + setNumDoors (int num Doors): void + toString(): String 2. Create a CarTest.java file. Implement a public CarTest class with a main method. In the main method, create one Car object and print the object using System.out.println(). Then, print the numberOfCar. Your printing result must follow the example output: make Toyota, model=Camry, year=2022 numDoors=4 1 Hint: You need to modify the toString methods in the Car class and Vehicle class!arrow_forward
- CHATGPT GAVE ME WRONG ANSWER PLEASE HELParrow_forwardHELP CHAT GPT GAVE ME WRONG ANSWER Consider the following implementation of a container that will be used in a concurrent environment. The container is supposed to be used like an indexed array, but provide thread-safe access to elements. struct concurrent_container { // Assume it’s called for any new instance soon before it’s ever used void concurrent_container() { init_mutex(&lock); } ~concurrent_container() { destroy_mutex(&lock); } // Returns element by its index. int get(int index) { lock.acquire(); if (index < 0 || index >= size) { return -1; } int result = data[index]; lock.release(); return result; } // Sets element by its index. void set(int index, int value) { lock.acquire(); if (index < 0 || index >= size) { resize(size); } data[index] = value; lock.release(); } // Extend maximum capacity of the…arrow_forwardWrite a C program using embedded assembler in which you use your own function to multiply by two without using the product. Tip: Just remember that multiplying by two in binary means shifting the number one place to the left. You can use the sample program from the previous exercise as a basis, which increments a variable. Just replace the INC instruction with SHL.arrow_forward
- Database System ConceptsComputer ScienceISBN:9780078022159Author:Abraham Silberschatz Professor, Henry F. Korth, S. SudarshanPublisher:McGraw-Hill EducationStarting Out with Python (4th Edition)Computer ScienceISBN:9780134444321Author:Tony GaddisPublisher:PEARSONDigital Fundamentals (11th Edition)Computer ScienceISBN:9780132737968Author:Thomas L. FloydPublisher:PEARSON
- C How to Program (8th Edition)Computer ScienceISBN:9780133976892Author:Paul J. Deitel, Harvey DeitelPublisher:PEARSONDatabase Systems: Design, Implementation, & Manag...Computer ScienceISBN:9781337627900Author:Carlos Coronel, Steven MorrisPublisher:Cengage LearningProgrammable Logic ControllersComputer ScienceISBN:9780073373843Author:Frank D. PetruzellaPublisher:McGraw-Hill Education





