HW 08_2023 Fall HW Solution

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0000000000000000000000Name: __________________________________________ Section: _______________________ MEEN 260 Mechanical Measurements Homework 8: Laplace Transform & Dynamic Systems Assigned: Monday 10 October 2023 Due: Monday, 6 November, 2023, 11:59 pm Notice: Upload your assignment to Canvas, as a single PDF file. The purpose of this assignment is to help you analyze the time response of dynamic systems. Homework Exercise Problem 1) (10 points) Find the Laplace transform of Solution L { e − at f ( t ) } = ∫ 0 ∞ e − at f ( t ) e − st dt = F ( S + a ) F ( S ) = L { t e − 3 t } = 1 ( S + 3 ) 2 Problem 2) (10 points) Find the solution x(t) of the differential equation Where a and b are constants. Solution L { x ( t ) } = X ( S ) L { ˙ x ( t ) } = SX ( S ) − x ( 0 ) L { ¨ x ( t ) } = S 2 X ( S ) − Sx ( 0 ) − ˙ x ( 0 ) . The Laplace transform of the given differential equation becomes S 2 X ( S ) − Sx ( 0 ) −˙ x ( 0 ) + 3 ( SX ( S ) − x ( 0 ) ) + 2 X ( S ) = 0 .

Substituting the given initial conditions into the preceding equation yields [ S 2 X ( S ) − aS − b ] + 3 ( SX ( S ) − a ) + 2 X ( S ) = 0 . Thus, X ( S ) = aS + b + 3 a S 2 + 3 S + 2 = 2 a + b S + 1 − a + b S + 2 The inverse Laplace transform of X(S) produces, x ( t ) = L − 1 [ X ( S ) ] = L − 1 [ 2 a + b S + 1 ] − L − 1 [ a + b S + 2 ] = ( 2 a + b ) e − t − ( a + b ) e − 2 t ,t ≥ 0. Which is the solution of the given differential equation. Notice that the initial conditions a and b appear in the solution. Thus, x(t) has no undermined constants. Problem 3) (10 points) Consider the following 1 st order low-pass-filter system with a cut off frequency ω=4.0 rad/s, X(s) is the input, TF(s) is the transfer function of the filter, and Y(s) is the response output in a Laplace domain function. Here s=jω is the Laplace variable. The K is a constant. TF ( s ) = K s + 3 Find the K value when the input signal x(t)=sin(4×t) is applied? Here t is the time in seconds. (Answer should have at least two significant figures, for example, 1.2)

Problem 4) (10 points) Consider the following 1 st order dynamic system (thermocouple). X(s) is the input, TF(s) is the transfer function, and Y(s) is the response output in a Laplace domain function. Here s=jω is the Laplace variable. TF ( s ) = 1 s + 1 When the step input is given, calculate the response output y at t = 2s. Here t is the time in seconds. (Answer should have at least two significant figures, for example, 0.12) Problem 5) (20 points) The students are testing two thermocouple sensors with a different junction size. They measure the thermocouple output when they put the sensor into the hot chamber (at 3s) as seen in the graph below. Assuming the sensor output has a 1st order dynamic response as, τ ˙ y ( t ) + y ( t ) = x ( t ) . Here τ is a time constant of the sensor, t is the time in second, y(t) is the time response of the sensor output, and x(t) is the input. a) Is the time constant τ A of Sensor A larger than τ B of Sensor B? (Answer, Yes or No) - 5 points Yes (3 points or nothing, no partial credit) b) Describe the reason for the answer (a) with one sentence (less than 30 words). Do not include the equation. - 5 points The answers could be:

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● The output of sensor B showed a steeper slope (sensitivity) than that of Sensor A. ● The sensor B responses to the temperature change more quickly than the Sensor A. Problem 6) (30 points) Dr. Rasmussen is using a sensor with a 1 st order dynamic response, as shown below. a) Assume that the sensor’s time constant is τ = 0.2 sec . Assume that we are trying to measure a pressure signal that is oscillating at a frequency of 3 rad/sec, has an average pressure of 20 psi , and fluctuates with an amplitude of ¿ ± 7 psi . In other words, x ( t ) = 20 + 7sin ( 2 πt ) Will the measured signal y ( t ) be a reasonable approximation of the true signal x ( t ) ? Briefly justify your answer (1-3 sentences). Solution: The time constant of the sensor is 0.1 sec, while the sine wave has a period of T = 2 π 2 π = 1.0 sec . Thus the sensor is 10 times faster than the period of the sine wave. Thus the measured signal should be a reasonable approximation of the true signal. b) Assume that the sensor’s time constant is τ = 1 sec . Assume that we are trying to measure a pressure signal given by x ( t )= 2cos ( √ 8 t + ϕ ) , where ϕ =0 in rad. What will be the amplitude and phase of the measured signal y ( t ) ? Solution: The frequency response of the sensor is give as: H ( ω ) = 1 τjω + 1 Thus the output signal will be: y ( t ) = ( K ) ( 2cos ( √ 8 t + ϕ ) ) , in rad. where

K = | H ( ω ) | ω = √ 8 = | 1 ( 1 ) j ( √ 8 ) + 1 | = 1 √ 1 2 + √ 8 2 = 0.333 , ϕ = 1.23 ∈ rad ( 70.53 ∈ degree ) Thus the amplitude of the measured signal is 0.333: y ( t ) = ( 0.333 ) ( 2cos ( √ 8 t + ϕ ) ) = 0.666cos ( √ 8 t + 1.23 ) Problem 6) (10 points) Consider the following 1 st order dynamic system. X(s) is the input, TF(s) is the transfer function, and Y(s) is the response output in a Laplace domain function. Here s=jω is the Laplace variable. The K is the constant. TF ( s ) = K s + 4 Find the K value that the magnitude of the response output becomes 1 at ω=3.0 rad/s upon the impulse input. Y ( s ) X ( s ) = TF ( s ) = K s + 4 | TF ( jω ) | = K √ ( ω ) 2 + 4 2 | TF ( jω ) | = K √ ( 3 ) 2 + 4 2 = K 5 = 1.0 (ω=3.0 rad/s). K=5.0 Problem 7) (10 points) Consider the following 1 st order dynamic system. X(s) is the input, TF(s) is the transfer function, and Y(s) is the response output in a Laplace domain function. Here s=jω is the Laplace variable. The K is the constant. TF ( s ) = K s + 1 Find the K value that the response output y becomes 0.5 at t = 1 s upon the step input. Count to two decimal places (e.g., 0.01) Y ( s ) X ( s ) = TF ( s ) = K s + 1 Y ( s ) = K s + 1 × X ( s ) = K s + 1 × 1 s = K ( s + 1 ) s