EBK LOGIXPRO PLC LAB MANUAL FOR PROGRAM
EBK LOGIXPRO PLC LAB MANUAL FOR PROGRAM
5th Edition
ISBN: 8220102803503
Author: Petruzella
Publisher: YUZU
Question
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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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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).
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