ELE_292_F23_PreLab9

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Syracuse University *

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292

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Mechanical Engineering

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Dec 6, 2023

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ELE 292 – Linear Systems Lab PreLab 9 – Mechanical 2 nd Order Systems Name: Goals: 1) Estimate the Stokes’ drag of a pendulum, using the transfer function and an estimate of the time constant. 2) Simulate the unit-impulse response (angle) of the pendulum. 3) Simulate the response (angle and voltage) of the pendulum when given an initial angle. The pendulum In lab we are going to test a pendulum. Consider the transfer function from section 3.4.2 in the text. H ( s ) = φ τ = 1 ml 2 s 2 + s b ml 2 + g l Our pendulums in lab have l = 0.27 meters, m = 0.038 [kg], and we have no idea what b is! Remember, b, represents the friction in the system. Our friction mostly comes from the potentiometer used to measure the angle. What is a potentiometer? A potentiometer is a resistor! The picture below shows what one looks like. Our pendulum has one of these mounted at the top of the arm which holds the pendulum. The potentiometer itself is the rotary bearing which allows the pendulum to swing. What goes on in the inside the potentiometer case you see above is shown in the diagram below.

What is the formula that gives you V out as a function of the angle in radians? Type the formula here: 340*(pi/180)= 5.934 Guess b and plot the unit impulse response Consider the arbitrary transfer function H ( s ) = A s 2 + ps + q The quadratic equation tells us the poles are located at the frequencies s 1,2 = − p± √ p 2 − 4 q 2 Assuming an underdamped system, the real part of the pole frequency is -p/2. Comparing this to the transfer function for the pendulum tells us the real part of the pole frequencies for the pendulum is α = − p 2 = − b 2 ml 2

This makes the time constant for the decaying oscillations equal to (The time constant is positive even if α is negative.) τ = 1 α = 2 m l 2 b We know it takes about five time-constants, 5τ, for the oscillations to completely decay. Let’s assume the pendulum swings back and forth for about 20 seconds before it stops. Use this estimate to calculate a guess for the value of the Stokes’ drag, b. b = -1.3851000000000002 Now we have all the variables in the transfer function for the pendulum; m, l, b, and g. Enter the transfer function H(s) for the angle, φ, over torque, τ, as an lti object in a Jupyter notebook. Use a time array that goes from 0 to 20 seconds and plot the impulse response. Paste the unit impulse response here: The unit impulse is a huge amount of torque to apply to our pendulum. 1 [Nms] is like pushing the pendulum with 8 pounds of force for 1/10 of a second. The angles you observe in the graph above are not practically real angles. But in linear systems, it doesn’t really matter, nothing ever saturates if the system is linear. Starting the pendulum at an initial angle If you hold the pendulum up at an initial starting angle and then let go of it, it will swing. This problem is a problem where there is no applied torque to the pendulum, but

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the pendulum swings because of the initial condition. You can’t write a transfer function for this problem because of the initial condition, however, you can write the angle, φ, as a function of s. In the equation below, φ 0 is the initial angle the pendulum starts at. All angles are specified in radians. φ ( s )= φ 0 s + bφ 0 ml 2 s 2 + s b ml 2 + g l Create a new lti object for the angle, φ(s). Use the lti object and the impulse method to determine the angular response when the initial angle starts out at 45° (π/4 radians). Paste a plot of the 45° angular response here: For this problem, you should see real observable angles! Computing the voltage Vout The angle you got in the previous part can be converted to a voltage using the formula you found at the beginning of this prelab. Use that formula to find Vout versus time for the response you just calculated. When you make this calculation the voltage, Vout, should have both positive and negative values. But the actual voltage coming out of the potentiometer should be between 0 and 3.3V. To offset the voltage, simply add 1.65V to the previously calculated voltages. Now the resulting values should all be between 0 and 3.3V. Plot the response due to an initial angle of 45° below. This is an estimate of what we expect to see in lab. Paste a plot of the 45° voltage response here: