EBK LOGIXPRO PLC LAB MANUAL FOR PROGRAM
5th Edition
ISBN: 8220102803503
Author: Petruzella
Publisher: YUZU
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Question
Chapter 3, Problem 1P
Program Plan Intro
Programmable Logic Controller (PLC):
- PLC is a specialized computer used in industrial segments to control machines and process.
- PLC is program that consists of a set of instructions which resembles the controlling functions needed to perform specific tasks.
- The function of PLC is similar to that of a relay and hence, in an industrial segment a relay is greatly replaced by a suitable PLC.
- When compared to general Personal Computer (PC), a PLC is available in small and tiny sizes.
- The basic architecture of a PLC consists of an input-output interface module and a small Central Processing Unit (CPU) that runs by controlling
programming language.
Hexadecimal number system:
- In hexadecimal number system, the numbers from “0” to “9” and letters from “A” to “F” are used.
- It means, in total, hexadecimal system has 10 numbers and 6 alphabets.
- Therefore, the base used for hexadecimal numbering system is 16.
Binary number system:
- In binary system, the numbers “0” and “1” are only used.
- It means, in total, binary system has only 2 numbers.
- Therefore, the base used for binary numbering system is 2.
Explanation of Solution
b.
Conversion of binary number to its equivalent hexadecimal code:
- Given, binary number is “00100101”.
- First, the given binary number is partitioned into two groups, such that each group contains only four bits.
- Now, each group is converted to its corresponding hexadecimal code.
- Here, “0010” is one such group and “0101” is the other group.
- The following diagram describes the hexadecimal notation of the binary number “0010”.
- The hexadecimal equivalent of decimal number “2” is “2”...
Explanation of Solution
c.
Conversion of binary number to its equivalent hexadecimal code:
- Given, binary number is “01001110”.
- First, the given binary number is partitioned into two groups, such that each group contains only four bits.
- Now, each group is converted to its corresponding hexadecimal code.
- Here, “0100” is one such group and “1110” is the other group.
- The following diagram describes the hexadecimal notation of the binary number “0100”.
- The hexadecimal equivalent of decimal number “4” is “4”...
Explanation of Solution
d.
Conversion of binary number to its equivalent hexadecimal code:
- Given, binary number is “00111001”.
- First, the given binary number is partitioned into two groups, such that each group contains only four bits.
- Now, each group is converted to its corresponding hexadecimal code.
- Here, “0011” is one such group and “1001” is the other group.
- The following diagram describes the hexadecimal notation of the binary number “0011”.
- The hexadecimal equivalent of decimal number “3” is “3”...
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Students have asked these similar questions
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 *…
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 *…
I would like to know the main features about the following three concepts:
1. Default forwarded
2. WINS Server
3. IP Security (IPSec).
Chapter 3 Solutions
EBK LOGIXPRO PLC LAB MANUAL FOR PROGRAM
Ch. 3 - Prob. 1RQCh. 3 - Convert each of the following decimal numbers to...Ch. 3 - Prob. 3RQCh. 3 - Prob. 4RQCh. 3 - Prob. 5RQCh. 3 - Prob. 6RQCh. 3 - Prob. 7RQCh. 3 - Prob. 8RQCh. 3 - Prob. 9RQCh. 3 - Prob. 10RQ
Ch. 3 - Prob. 11RQCh. 3 - Define the term sign bit.Ch. 3 - Prob. 13RQCh. 3 - Prob. 14RQCh. 3 - Prob. 15RQCh. 3 - Prob. 16RQCh. 3 - Prob. 17RQCh. 3 - Prob. 18RQCh. 3 - Prob. 19RQCh. 3 - Prob. 1PCh. 3 - Prob. 2PCh. 3 - Prob. 3PCh. 3 - Prob. 4PCh. 3 - Prob. 5PCh. 3 - Prob. 6PCh. 3 - Prob. 7PCh. 3 - Prob. 8PCh. 3 - Prob. 9PCh. 3 - Prob. 10P
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