Electric Motor Control
10th Edition
ISBN: 9781133702818
Author: Herman
Publisher: CENGAGE L
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Chapter 53, Problem 1SQ
To determine
Explain the reason for the usage of gear motors in many industrial machines, instead of low rpm induction motors.
Expert Solution & Answer
Explanation of Solution
Discussion:
The reasons for the usage of gear motors in many industrial machines instead of low rpm induction motors are as follows:
- The gear motor is a machine which provides a direct power drive from a single unit and it is a speed reducing machine.
- The gear motors are extremely small size and provide efficient power drive. The low-speed induction motors are heavy, large in size, and very expensive for the same horse power as gear motors.
- The gear motors are available in different types of standard motors. For example, the gear motors are available as slip-ring induction motors and squirrel-cage induction motors.
- The gear motors have numerous applications in the field of industrial machines over the low rpm induction motors.
Conclusion:
Thus, the reason for the usage of gear motors in many industrial machines instead of low rpm induction motors is explained.
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help on this question about induction motors?
The MATLAB code is going well but the last part in bandpass, the legend that is supposed to tell the color of both lower and upper-frequency cutoff does not align with each other. As such I need help
My Matlab 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
R = 1e3; % Resistance in ohms (1 kΩ)
C = 1e-6; % Capacitance in farads (1 μF)
L = 0.1; % Inductance in henries (chosen for proper bandpass response)
% Compute cutoff frequencies
f_cutoff_RC = 1 / (2 * pi * R * C); % RC low-pass/high-pass cutoff
f_resonance = 1 / (2 * pi * sqrt(L * C)); % Resonant frequency of RLC
Q_factor = (1/R) * sqrt(L/C); % Quality factor of the circuit
% Band-pass filter cutoff frequencies
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));
% Define Transfer Functions
H_low =…
The MATLAB code is going well but the last part in bandpass, the legend that is supposed to tell the color of both lower and upper-frequency cutoff does not align with each other. As such I need help
My Matlab 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
R = 1e3; % Resistance in ohms (1 kΩ)
C = 1e-6; % Capacitance in farads (1 μF)
L = 0.1; % Inductance in henries (chosen for proper bandpass response)
% Compute cutoff frequencies
f_cutoff_RC = 1 / (2 * pi * R * C); % RC low-pass/high-pass cutoff
f_resonance = 1 / (2 * pi * sqrt(L * C)); % Resonant frequency of RLC
Q_factor = (1/R) * sqrt(L/C); % Quality factor of the circuit
% Band-pass filter cutoff frequencies
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));
% Define Transfer Functions
H_low =…
Chapter 53 Solutions
Electric Motor Control
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