COMPUTER SCIENCE ILLUMIN.-TEXT
7th Edition
ISBN: 9781284156010
Author: Dale
Publisher: Jones & Barlett
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Chapter 2, Problem 40E
Program Plan Intro
Hexadecimal subtraction:
The rules for subtracting two hexadecimal numbers are given below:
- Read both the numbers to find the number of bits.
- If the number of bits in the first number is less than that of second number, then pad the required zeroes to the left side.
- Numbers are represented in bytes (1 byte = 8 bits).
- Align the number in correct order after padding. The hexadecimal number that is followed by a minus sign should be subtracted from the other hexadecimal number.
Explanation of Solution
b.
“9988 – AB”:
- Start with the numbers from right to left.
- Subtract “B” from “8”. The equivalent hexadecimal value for “B” is “11”.
- Here, “8” is less than “11”. Borrow “16” from adjacent left side and add it to “8”. Since the adjacent left side number is “8”, it becomes “7” after borrowing.
- Add “16” to “8” and then subtract “11” which give “13” as the result. The equivalent hexadecimal digit for the number “13” is “D”. Hence write D to the result.
- Subtract A from “7”. The equivalent hexadecimal value for “A” is “10”.
- Here “7” is less than “10”...
- Subtract “B” from “8”. The equivalent hexadecimal value for “B” is “11”.
Explanation of Solution
c.
“A9F8 – 1492”:
- Start with the numbers from right to left.
- Subtract “8” from “2”. The equivalent hexadecimal digit for number “6” is “6”. Write “6” to the result.
- Subtract “9” from “F”. The equivalent hexadecimal value for “F” is “15”.
- Subtract “15” – “9” which gives “6”. The equivalent hexadecimal digit for the number “6” is “6”. Hence write “6” to the result.
- Subtract “4” from “9”. The equivalent hexadecimal value for number “5” is “5”...
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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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