Mixed Discipline Systems

Mixed Discipline Systems

Dr. Nhut Ho

ME584

Chp7 1

Agenda

• Electro-mechanical Systems– DC Motor Speed Control

– DC Motor Position Control

• Fluid-Mechanical Systems– Hydraulic Position Servo

– Electro-hydraulic Position Servo

– Pneumatic Position Servo

• Active Learning: Pair-share Exercises

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Electro-mechanical Systems

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DC Motor Speed Control (Open-Loop Control)

Without Feedback

Open-loop DC motor speed control

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DC Motor Speed Control (Open-Loop Control)

The model for the motor is based on ideal permanent-magnet DC motor

The motor is driven by an electronic power source, modeled as ideal voltage source driving motor through resistance R0 . Assuming overall amplifier gain G:

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Chp7 5

DC Motor Speed Control (Open-Loop Control)

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Chp7 6

DC Motor Speed Control (Open-Loop Control)

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DC Motor Speed Control (Closed-Loop Control)

With Speed Feedback

• Purpose of feedback control is to allow output to track input and to compensate for any error from command input and the actual output.

• Summing junction subtracts feedback signal from input signal to obtain an error signal which is usually amplified to drive actuator.

• Gain of tachometer is h

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DC Motor Speed Control

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DC Motor Speed Control

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Chp7 10

DC Motor Speed Control

Transfer function is first-order function of input and disturbance

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DC Motor Speed Control

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Static gain, disturbance sensitivity, time constant of closed-loop system are modified by

A reduction in static gain, not desirable, can be compensated by increasing input. A reduction in disturbance sensitivity means motor speed will not vary as much. A reduction in time constant means system will respond faster.

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Chp7 12

DC Motor Position Control

Set point/desired position is θd (with gain c)

Feed back position sensor is potentiometer, output voltage linearly proportional to angular position, θ, with sensitivity h

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DC Motor Position Control

The motor speed m is reduced to an output shaft θ

by a gear speed ration Rs

Neglecting inertia of gear train, assuming viscous damping B in the drive train, torque balance gives

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Chp7 14

DC Motor Position Control

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DC Motor Position Control

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General Approach to Model Electro-Mechanical Systems

1. Apply motor and amplifier equations

Tm=KimEm=Kωm

Em=Ge-R0im2. Write force or torque balance equation

F=ma or T=Jα

3. Combine equations to obtain transfer function, and compute static and dynamic characteristics

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Example 1

A DC motor is connected to 125 volts DC and is used to spin a grindstone. The diameter of the grindstone disk is 200 mm, and the thickness is 15 mm. The density of the disk material is 3000 kg/m3 . The free speed of the motor (with maximum voltage applied and no torque loading) is 1000 rpm. The voltage source that is used to drive the motor has an output impedance R0 of 2 Ω(i.e.,em=e0-R0im ). Neglecting the inertia and friction of the motor itself, derive a transfer function for the dynamic response of the speed of the motor as a function of the input voltage. Calculate the time response using Laplace transform techniques for a step input of 125 volts DC with zero initial conditions.

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Example 1Chp7 19

Example 1

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Example 1

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Example 2

Simulate the system with a step input of 125 V with zero initial condition

Matlab code:num = [8*125];den=[9.92e-3 1];sys=tf(num,den);% t = ?t= [0:.0001:.06];[y,t]=step(sys,t);plot(t,y); xlabel(‘Time (s)’); ylabel(‘Speed (rpm)’)

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Pair-Share Exercise: Example 3

Shown is a DC torque motor connected to a mechanical load through a gear reduction. The motor produces a torque in linear proportion to the current delivered from a constant-current source. The mechanical load is a disk with inertia J, and a translational spring and dashpot connected to the edge of the disk. Write the modeling equations for this system, and derive a transfer function for the angular position of the disk as a function of the input current.

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Example 3

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Example 3

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Pair-Share Exercise: Example 4

Simulate the step input of the system assuming zero initial conditions:

Matlab code:num = [2700];den=[1 1.25 2701];sys=tf(num,den);t=[0:.05:3];[y,t]=step(sys,t);plot(t,y); xlabel(‘Time (s)’); ylabel(‘position (rad)’)

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Pair-Share: Example 5A small DC motor is used to spin a disk with an op-amp as shown. In this application, we do not want the disk to spin up very fast, so a dynamic filter is added to the op-amp. The op-amp acts as an ideal voltage source with an output impedance R0. Write the modeling equation for this system, and derive a state-space representation of the system using the op-amp output e0 and the shaft speed ω as state variables. Is this a stable system?

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Example 5

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Example 5

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Example 5

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Fluid-Mechanical Systems

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Fluid-Mechanical Systems

• Wide range of applications in precision motional control at high power levels

– Flight controls (ailerons, rudder, stabilizer)

– Automotive (power steering) and industrial

• Provide high power actuation in small volume, high responsiveness

• Configuration

– Servo valve capable of controlling pressure and flow

– Mechanical/Electrical feedback with actuator output

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Hydraulic Position Servo

• Servo valve provides metered ΔP0 and Q0 in response to position x of spool valve

• Actuator moves M and experiences load FL

• Mechanical level feedback: input u and actuator position z change, causing servo valve to modulate ΔP0 and Q0 to reduce error x of servo system

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Hydraulic Position Servo

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Chp7 36

Electro-Hydraulic Position Servo

Electro-hydraulic position servo control system

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Electro-Hydraulic Position Servo

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Chp7 38

Pneumatic Position Servo

Chp7 39

Pneumatic Position Servo

Servo made of three-way valve connected to spring-loaded actuator with mechanical lever feedback

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Pneumatic Position Servo

• Decreasing y causes increase in δP0 and output position δz

• If δu held constant, feedback lever causes valve to return to steady-state position

• If FL causes decrease in z, valve will be actuated, pressure to actuator will increase a force to oppose FL

Chp7 41

Pneumatic Position Servo

• In this system, actuator stroke starts at an unactuated position (z0,y0,u0) when supply pressure is off and increases to null position when supply pressure is activated

• Examine small variations δz,δy,δz in positions around null operation:z= z0 +δzy= y0 +δyu= u0 +δu

• Nonlinear system difficult to analyze, linearization to obtain transfer function

Chp7 42

Pneumatic Position Servo

Nonlinear characteristics of three-way flapper-nozzle valve

Chp7 43

Pneumatic Position Servo

Linear Analysis

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Chp7 44

Pneumatic Position Servo

Linearized pressure-flow characteristics of three-way flapper-nozzle valve

Chp7 45

Pneumatic Position Servo

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Chp7 46

Pneumatic Position Servo

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Chp7 47

Example 7

In what follows, use the basic models for flows in the flapper-nozzle valve to derive the normalized model for the flapper-nozzle valve. Plot the output pressure δP0 versus the normalized valve clearance α with no output flow. Show that the point α = 0.7 is a good trade-off from among the maximum gain, the minimum mean output pressure, and the maximum linear modulation of the output pressure. Show that the null output pressure is 0.671 δPs. Linearize this model. The parameters for flow in the flapper-nozzle valve are:

Chp7 48

Example 7

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Chp7 49

Example 7

In your derivation of the output flow normalized to the maximum supply flow, you should find a natural grouping of terms such as the following:

For symmetric operation, the null position of the valve is , and the valve can stroke 0y y

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Chp7 50

Example 7

For this model to be correct, the circumferential (or curtain) flow area of the nozzle should be smaller than the area of the supply orifice; that is,

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Example 7: Solution

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Example 7: Solution

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Example 7: Solution

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Example 7: Solution

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Example 7: Solution

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Example 7: Solution

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Example 7: Solution

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Example 7: Solution

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Example 7: Solution

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Example 7: Solution

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Example 7: Solution

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