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Diterbitkan olehCinta Januar Telah diubah "9 tahun yang lalu
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1 Dynamic Response: 1 st and 2 nd Order Systems March 24, 2004
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2 Pokok Bahasan Persamaan Differensial: Order satu dan order dua Basic Persamaan Differensial Penyelesaian persamaan Order satu Penyelesaian persamaan Order dua Contoh Soal
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3 Persamaan Differensial Order dari Persamaan Differensial adalah order tertinggi dari turunan dalam persamaan. Persamaan Differensial adalah Persamaan yang mengandung turunan atau differensial Derajat dari Persamaan Differensial adalah Pangkat dari order Persamaan Differensial
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4 Persamaan Differensial Order satu dan dua P. D. Order satu P. D. Order dua P.D. Homogen P.D. Non-Homogen Syarat awal (IVP) Syarat batas Konstante waktu Penyelesaian Umum Penyelesaian Khusus Driving Force Damping Ratio Natural Frequency Ringing Frequency Static Sensitivity
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5 Dimana Persamaan Differensial digunakan Beberapa contoh aplikasi dari Persamaan differensial antara lain : Gerakan dalam mekanika Rambatan panas Getaranl dynamics & seismology aerodynamics & fluid dynamics electronics & circuit design population dynamics & biological systems climatology and environmental analysis options trading & economics
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6 Beberapa Contoh Dapat Dibuka Pada Alamat Berikut : SOS Mathematics http://www.sosmath.com/diffeq/diffeq.html Wolfram Research – Math World http://mathworld.wolfram.com/OrdinaryDifferenti alEquation.html http://mathworld.wolfram.com/OrdinaryDifferenti alEquation.html Math Forum @ Drexel http://mathforum.org/differential/differential.html Internet Differential Equations Activities http://www.sci.wsu.edu/idea/
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7 Persamaan Differensial Contoh dari Persamaan Differensial First Order Equation Second Order Equation
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8 Persamaan Differensial Bentuk Umum Persamaan Differensial dapat ditulis sebagai berikut : Second Order Equation First Order Equation
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9 Persamaan Differensial Persamaan Diff.order satu dapat dinyatakan dalam bentuk M(x,y) dx + N(x,y) dy = 0 Persamaan.Diff. Separabel M(x) dx+ N(y) dy = 0 Persamaan Diff. Homogen
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10 Persamaan Differensial A differential equation that has a zero on the right-hand- side (RHS) of the equation is called a homogeneous equation. A differential equation that has a constant or a function of the independent variable on the RHS is called a non- homogeneous equation. The function on the RHS of the differential equation is know as the driving force, and represents the input for which the solution of the equation is the response.
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11 Persamaan Differensial Penyelesaian Umum ( total solution ) (y t ) terdiri dari dua bagian yaitu: Penyelesaian umum dari persamaan homogen C.F(complementary Function (y g ) Penyelesaian khusus( particular solution) (y p ), yaitu penyelesaian bila ada RHS P.U(total solution ) adalah jumlah dari C.F dan penyelesaian khusus
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12 Contoh: Dari persamaan differensial order satu Penyelesaian Umum dari soal adalah: Penyelesaian Umum ( C.F ): Penyelesaian khusus:
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13 Differential Equation Basics Finally, we need to solve for any constant(s) in the total solution by defining the system at some point(s) in time or space. These points are called initial conditions or boundary conditions, depending on when or where they are in the solution space.
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14 Contoh: Diberikan persamaan : Syarat awal ( initial value problem ): Penyelesaian Umum adalah: Total penyelesaian dinamakan penyelesaian khusus ( karena tidak memuat lagi konstante )
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15 Persamaan Differensial Banyaknya constante dalam penyelesaian umum ( yg tentunya sama dengan banyaknya IVP ) sama dengan order dari persamaan.
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16 Penyelesaian dari Persamaan differensial order satu Bentuk umum adalah : Second Order Equation First Order Equation
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17 Penyelesaian dari Persamaan differensial order satu For 1 st order systems, this general form can be rewritten as follows: The constant ( ) is known as the time constant of the 1 st order system.
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18 Solving 1 st Order Systems The general solution for a homogeneous 1 st order system:
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19 Solving 1 st Order Systems The value of the constant (A) can be determined by the initial condition or boundary condition. (Note that there is no particular solution for a homogeneous system.)
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20 Solving 1 st Order Systems The time constant ( ) measures how fast the 1 st order system responds. Specifically, at t= , the response, y(t), will have moved ~63.2% of the way between its initial value at t=0 and its final value at t= .
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21 Effect of Changing t = ; y( )=36.8
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22 Solving 1 st Order Systems A particular solution for a non-homogeneous 1 st order system can be found as follows:
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23 Particular Solution Example:
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24 Particular Solution Example:
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25 Solving 1st Order Systems So, to solve a 1 st order differential equation: 1) find the particular solution (if the problem is non-homogeneous) and solve for all constants 2) find the general solution (this will be of the same form for all 1 st order differential equations) 3) add the particular solution to the general solution to get the total solution 4) solve for the remaining constant using the initial condition or boundary condition.
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26 Some Example Problems:
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27 Example (1) Solution
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28 Some Example Problems:
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29 Example (2) Solution
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30 Some Example Problems:
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31 Example (3) Solution
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32 Some Example Problems:
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33 Example (4) Solution
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34 Solving 2 nd Order Systems Dynamic Response: 1st and 2nd Order Systems Differential Equation Basics 1st & 2nd Order Systems Homogeneous vs. Non-Homogeneous General vs. Particular Initial or Boundary Conditions Solving 1st Order Systems General Solution Particular Solutions Initial Conditions Physical Significance Solving 2nd Order Systems Mechanical Analogy Physical Significance Step Change Driving Force Varying the Damping Ratio Some Example Problems
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35 Solving 2 nd Order Systems Examples of some phenomena that can be modeled as 2 nd order systems: electrical capacitance–inductance circuit vibrating string on a musical instrument car shock absorber–spring assembly vibrating cantilever beam stereo speaker cone pressure transducer bungi cord
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36 Solving 2 nd Order Systems Recall that the general form of a linear differential equation can be written as: Second Order Equation First Order Equation
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37 Solving 2 nd Order Systems For 2 nd order systems, this general form can be rewritten in several different ways:
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38 Mass/Spring/Dashpot Analogy
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39 Solving 2 nd Order Systems In terms of measurable quantities:
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40 Physical Significance of Constants Driving Force [f(t)] the input force that drives the output motion of the system Static Sensitivity [K] a measure of the “amplification” of the driving force by the system in the absence of any dynamic response Natural Frequency [ n ] a measure of the frequency of the dynamic response of the system in the absence of any damping forces Damping Ratio [ ] a measure of the degree to which the dynamic response of the system damps out with time
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41 Physical Significance of Constants Damping Ratio [ ] =0:“undamped” <1:“underdamped” =1:“critically damped” >1:“overdamped” Ringing Frequency [ r ] a measure of the frequency of the dynamic response of the system in the presence of damping forces
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42 Example Problem: The following equation describes the response of a 2 nd order system. Determine the natural frequency, the damping ratio, and the static sensitivity of the system.
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43 Example Problem:
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44 Solving 2nd Order Systems We will only look at solutions to 2 nd order systems for one specific driving forces:
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45 Step Input Solution
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46 Step Input Solution (cont.)
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47 Example Problem
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48 Example Problem (underdamped)
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49 Example Problem (no damping)
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50 Example Problem (critically damped)
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51 Example Problem (over damped)
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52 Effect of Changing y(t)/K x 0 n t
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53 Solving 2 nd Order Systems Dynamic Response: 1st and 2nd Order Systems Differential Equation Basics 1st & 2nd Order Systems Homogeneous vs. Non-Homogeneous General vs. Particular Initial or Boundary Conditions Solving 1st Order Systems General Solution Particular Solutions Initial Conditions Physical Significance Solving 2nd Order Systems Mechanical Analogy Physical Significance Step Change Driving Force Varying the Damping Ratio Some Example Problems
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54 Example: 1 st Order Instrument A thermocouple is a transducer that converts temperature to voltage. As a probe, it can be considered a 1 st order instrument. A thermocouple is placed in ice water where it equilibrates to a temperature of 0 o C. The probe is then placed in boiling water, which can be assumed to be at 100 o C.
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55 Example: 1 st Order Instrument If it is observed that it takes 15 seconds for the thermocouple to reach a temperature of 99 o C, determine the following. What is the time constant for this probe? What would the time lag be in the recorded temperature if this probe were used to monitor the temperature of a water bath that is being rapidly heated from 0 o C to 100 o C at 2 o C per second?
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56 Solution:
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57 Solution (continued):
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58 Solution (continued):
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59 Example: 1 st Order Instrument If it is observed that it takes 15 seconds for the thermocouple to reach a temperature of 99 o C, determine the following. What is the time constant for this probe? What would the time lag be in the recorded temperature if this probe were used to monitor the temperature of a water bath that is being rapidly heated from 0 o C to 100 o C at 2 o C per second?
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60 Solution (continued):
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61 Solution (continued):
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62 Solution (continued):
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63 Example: 2 nd Order Instrument A strain gauge accelerometer is used to measure the acceleration of a vehicle. The natural frequency of the accelerometer is 1 kHz and the damping ratio is measured to be 0.5. Estimate the maximum instantaneous error as a function of time if the acceleration of the vehicle suddenly changes from zero to 10 m/s 2.
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64 Background Information A strain-gauge accelerometer is an instrument that can be used to measure the acceleration of a vehicle. It consists of a mass (m) connected to the instrument housing by a spring (k). The mass can move relative to the housing, and this displacement (y) is measured using a strain gauge. There is inherent damping ( ) due to the mechanical components in the device.
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65 Background Information Therefore, a strain-gauge accelerometer can be modeled as a second order instrument.
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66 Solution:
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67 Plotting The Solution:
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68 Zooming In: 0.05 error (0.5%)
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