PROGRAMMABLE LOGIC CONTROLLERS (LOOSE PA
5th Edition
ISBN: 9781264206216
Author: Petruzella
Publisher: MCG
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Chapter 15.3, Problem 3RQ
Program Plan Intro
Timers predefined structure:
Timer can perform the operations that are given below:
- Turning on and off the outputs after the completion of time delay.
- Turning on or off the output for a specific time.
- Determining the time when an output is in “on” or “off” condition.
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My 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 *…
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Chapter 15 Solutions
PROGRAMMABLE LOGIC CONTROLLERS (LOOSE PA
Ch. 15.1 - Prob. 1RQCh. 15.1 - Prob. 2RQCh. 15.1 - Prob. 3RQCh. 15.1 - Prob. 4RQCh. 15.1 - Prob. 5RQCh. 15.1 - Prob. 6RQCh. 15.1 - Prob. 7RQCh. 15.1 - Prob. 8RQCh. 15.1 - Prob. 9RQCh. 15.1 - Prob. 10RQ
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- What 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_forwardTask 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_forward
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