MECH 3140 SYSTEM DYNAMICS AND CONTROLS
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MECH 3140 SYSTEM DYNAMICS AND CONTROLS
EXAM NO. 2
April 13, 2022
1. (20 points) Consider the DC motor and rack and pinion steering assembly shown below. Assume the DC motor is attached to the steering wheel (not shown in schematic) which has inertia Is . Assume the inductance of the motor is negligible as shown in the circuit diagram on the right. The motor has torque constant kT, back emf constant kb, shaft inertia Im and tortional damping coefficient bm. The rack (mass m) and pinion (inertia Ip and radius R) has spring and damping coefficients as seen in the schematic.
a) Derive the equation(s) of motion for the system
b) Bonus(+3 Point): Put the system in state spaceform
2. ( 15 points) For the differential equations with given Bode plot solve for x(t). You do not have to solve for initial condition coefficients.
+ 2ẋ + 4x = 4sin(t)
3. ( 10 points) Consider the system described by the following differential equation. + 2ẋ + 4x = 4u(t)
Below is the response of the system for a unit step input (i.e., u(t>0)=1)
Given a new input u(t) = sin(t) (for t ≥ 0), sketch the steady state response (i.e. sketch x(t) for t ≥ 10S).
4. (20 points). Consider a system that can be modeled with the following differential equation: a + bẋ + Cx = u(t)
Several experiments were run to determine the coefficient of the differential equations. All the results below were attained using the same system. Determine the coefficients of the model (i.e. a, b, and c). Note the change in scale on the time axes.
5. (25 points) Consider the given 2nd order mass-spring-damper dynamic system shown below with control input F and zero initial conditions:
2 + 4ẋ + 20x = F
a. If F = Kp(xd − x) + Kd (ẋd − ẋ), what range of values for Kp and Kd would result in a stable response?
b. Now design a controller such that the steady-state position error is 0 (eSS = 0) when the desired position is constant and such that the rise time is tT = 0.18 S and the percent overshoot is Mp = 15%.
c. Prove that the controller designed in (b) yields eSS = 0?
6. (12 Points) Determine the frequency of each plot (including units).
a._________________ |
._____________________ |
c.________________ |
.___________________ |
e.________________ |
.___________________ |
7. (10 Points) Find the gain and phase for the following differential equation: 1ẋ + 2x = 3u̇ + 4u
Mechanical Modeling Steps 1) DOF and System Order 2) Constitutive coordinates and equations 3) FBD 4) Sum Forces and Moments as needed 5) Repeat 2 through 4 as needed 6) Simplify to one equation of motion per degree of freedom 7) Solve EOM if needed |
Mass in translation m = ∑Fx |
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Electrical Modeling Steps 1) DOF and System Order 2) Constitutive coordinates and equations 3) KCL and KVL as needed 4) Repeat 2 and 3 as needed 5) Simplify to one equation of motion per degree of freedom 6) Solve EOM if needed |
Electrical Elements |
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Resistors Inductor Capacitor |
vr = IR |
Linearization
Small Angles cos(e) ≈ 1 sin(e) ≈ e |
Taylor Series f(x) ≈ f(x0) + |
x0 (x − x0) |
Solving Differential Equations
Eigenvalue = s
Characteristic Equation = 0
General Solution x(t) = xh (t) + xp (t)
Specific Solution - Given initial conditions solve for unknown homogeneous constant (i.e. C1 )
Real Eigenvalue Homogeneous Solution |
xh (t) = C1est |
Complex Eigenvalues (Underdamped – 0 < 3 < 1) Homogeneous Solution xh (t) = C1e−3业n t sin(业d t + C2) |
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Imaginary Eigenvalues (Undamped – b=0) Homogeneous Solution xh (t) = C1 sin(业n t) + C2cos(业n t) |
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Particular Solution Follows form of input F(t) = C => xp (t) = C ∗ GDC F(t) = A sin(业t) => xp (t) = A(gain)sin(业t + p) |
1st Order Step Response
Time Constant |
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1 C |
2 C |
3 C |
4 C |
4.6 C |
5 C |
1 C = |s| |
63% |
86% |
95% |
98% |
99% |
99.3% |
2nd Order Step Response (Underdamped – 0 < 3 < 1)
Underdamped Eigenvalue Locations
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s2 + 23业ns + 业n(2) = 0 |
Mp = exp (− |
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ts = tr = t = p |
4.6 3业n 1.8 业n 几 业d |
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3 = 0.1 3 = 0.3 3 = 0.5 3 = 0.7 3 = 0.9 |
M = 75% Mp = 35% Mp = 15% Mp = 5% Mp = 0.2% |
Frequency Response
2022-11-01