COMPUTER SCIENCE ILLUMIN.-TEXT
7th Edition
ISBN: 9781284156010
Author: Dale
Publisher: Jones & Barlett
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Expert Solution & Answer
Chapter 2, Problem 1E
Program Description Answer
A number is an abstract mathematical unit which is subjected to the laws of arithmetic.
Hence, the correct option is “A”.
Expert Solution & Answer
Explanation of Solution
Number:
- A number is actually a mathematical object used for counting, labeling and measuring.
- There are many types of numbers such as
- Natural numbers
- Whole numbers
- Negative numbers
- Rational numbers
- Irrational numbers
- Positive numbers
- Numbers are mainly used for arithmetic operations such as addition, subtraction, multiplication, division and so on.
- The study of properties of number is known as number theory.
Explanation for incorrect options:
b. Natural number:
A natural number is a type of number. Natural numbers are positive integers or whole numbers such as 1, 2, 3 and so on and sometimes 0 as well.
Hence, the option “B” is wrong.
c. Integer number:
An integer number is also a type of number. It is a whole number that can be positive, negative or zero.
Hence, the option “C” is wrong.
d. Negative number:
A negative number is also a type of number. It is a real number which is less than zero.
Hence, the option “D” is wrong.
e. Rational number:
Rational number is also a type of number. It is any number that can be expressed as fraction or as a quotient.
Hence, the option “E” is wrong
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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 *…
Chapter 2 Solutions
COMPUTER SCIENCE ILLUMIN.-TEXT
Ch. 2 - Prob. 1ECh. 2 - Prob. 2ECh. 2 - Prob. 3ECh. 2 - Prob. 4ECh. 2 - Prob. 5ECh. 2 - Prob. 6ECh. 2 - Prob. 7ECh. 2 - Prob. 8ECh. 2 - Prob. 9ECh. 2 - Prob. 10E
Ch. 2 - Prob. 11ECh. 2 - Prob. 12ECh. 2 - Prob. 13ECh. 2 - Prob. 14ECh. 2 - Prob. 15ECh. 2 - Prob. 16ECh. 2 - Prob. 17ECh. 2 - Prob. 18ECh. 2 - Prob. 19ECh. 2 - Prob. 20ECh. 2 - Prob. 21ECh. 2 - Prob. 22ECh. 2 - Prob. 23ECh. 2 - Prob. 24ECh. 2 - Prob. 25ECh. 2 - Prob. 26ECh. 2 - Prob. 27ECh. 2 - Prob. 28ECh. 2 - Prob. 29ECh. 2 - Prob. 30ECh. 2 - Prob. 31ECh. 2 - Prob. 32ECh. 2 - Prob. 33ECh. 2 - Prob. 34ECh. 2 - Prob. 35ECh. 2 - Prob. 36ECh. 2 - Prob. 38ECh. 2 - Prob. 39ECh. 2 - Prob. 40ECh. 2 - Prob. 41ECh. 2 - Prob. 42ECh. 2 - Prob. 43ECh. 2 - Prob. 44ECh. 2 - Prob. 45ECh. 2 - Prob. 46ECh. 2 - Prob. 47ECh. 2 - Prob. 1TQCh. 2 - Prob. 2TQCh. 2 - Prob. 3TQCh. 2 - Prob. 4TQ
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