Chapter 7: Time Domain Analysis


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1 Chapter 7: Time Domain Analysis Samantha Ramirez Preview Questions How do the system parameters affect the response? How are the parameters linked to the system poles or eigenvalues? How can Laplace transforms and transfer functions be used to analyze the time domain response of a system? How can linear algebra and the statespace model be used to analyze the time domain response? 1
2 Objective and Outcomes Objectives To understand firstorder responses To understand secondorder responses To understand how higherorder responses are composed from first and secondorder terms, and To understand the relation between the system poles or eigenvalues and the overall system response. Outcomes Identify the characteristics of firstorder responses Identify the characteristics of secondorder responses Identify the dominant poles of higherorder systems, Use the transfer function or statespace representation to determine a system s characteristic roots, and Predict overall response based on pole placement Transient Response of FirstOrder Systems 2
3 The Natural Response T dx + x = 0 x t = x(0)e t/t dt The Impulse Response T dx dt + x = A ሚδ(t) x t = A T e t/t 3
4 The Step Response Tsx s + x s = Ts + 1 x s = A x x t = A(1 e t/t ) Example 7.1 A simple torsion system is depicted. It is a disk mounted on a bearing and is excited by an input torque. The rotational inertia can be readily attained by measuring the mass and diameter of the disk. The bearing damping coefficient, on the other hand, is not something that is commonly advertised or supplied. However, this can be readily determined using experimental data and a model of the system. Given the measured step response plotted in the following figure, what are the rotational inertia, J, and damping coefficient, β? 4
5 The Ramp Response T dx + x = At dt x t = A(t T Te t T) Complex Numbers 5
6 5.2.1 Complex Numbers z = x + jy z = x 2 + y 2 z ҧ = x jy θ = tan 1 y x Transient Responses of SecondOrder Systems 6
7 Prototypical SecondOrder System x ሶ = p m p ሶ = kx b p + F(t) m The Natural Response Prototypical secondorder differential equation with no input x ሷ + 2ζω n x ሶ + ω 2 n x = 0 Associated characteristic equation in the sdomain Characteristic roots s 2 + 2ζω n s + ω n 2 = 0 s 1,2 = 2ζω n ± 4 ζω n 2 4ω n 2 2 = ζω n ± ω n ζ 2 1 7
8 Undamped Natural Response (ζ=0) Differential Equation x ሷ + ω 2 n x = 0 Characteristic Equation Characteristic Roots Response s 2 + ω n 2 = 0 s 1,2 = ±jω n x t = x 0 cosωt Purely Imaginary Roots Underdamped Natural Response (ζ<1) Roots Response s 1,2 = ζω n ± ω n ζ 2 1 = ζω n ± ω n 1 1 ζ 2 = ζω n ± jω n 1 ζ 2 Complex Conjugate Roots Damping Frequency 8
9 Critically Damped Natural Response (ζ=1) Characteristic Roots s 1,2 = ζω n ± ω n ζ 2 1 Response = ζω n ± ω n = ω n x t = x(0)(1 + ω n t)e ω nt Repeated Negative Real Roots Overdamped Natural Response (ζ>1) Characteristic Roots Response s 1,2 = ζω n ± ω n ζ 2 1 = ω n (ζ ζ 2 1 Distinct Negative Real Roots 9
10 SecondOrder Natural Responses SecondOrder Natural, Underdamped Response x(0) 1 ζ 2 e ζω nt T 2π p= ω n 1 ζ 2 ) =2π ω d From Logarithmic Decrement ζ = 4π n 1 ln x 1 x n 1 n 1 ln x 2 1 x n T = 1 ζω n 10
11 Example 7.2 Given the natural response, how could you determine the damping coefficient, β, and the shaft rigidity, κ? The disk is the same as from Example 1. The Natural Frequency, Damping Ratio, and Pole Placement 11
12 SecondOrder Response of Varying Damping Ratio SecondOrder Response of Varying Natural Frequency 12
13 The Impulse Response x ሷ + 2ζω n x ሶ + ω 2 n x = A ሚδ(t) Undamped Underdamped x t = A ω n sinω n t x t = A ω n e ζω nt sinω d t Critically Damped Overdamped x t = 2ω n x t = Ate ω nt A ζ 2 1 e ω nζ 1 t + e ω nζ 2 t where ζ 1 = ζ + ζ 2 1 and ζ 2 = ζ ζ 2 1 SecondOrder Impulse Response 13
14 The Step Response x ሷ + 2ζω n x ሶ + ω 2 n x = A1(t) Undamped Underdamped Critically Damped x t = A 2 ω (1 cosω nt) n x t = A 2 ω 1 ζ e ζω nt n 1 ζ sinω dt + cosω d t 2 x t = A 2 ω 1 e ωnt ω n e ωnt t n Overdamped x t = A 2ω n 2 ζ ζ 2 1 e ω nζ 1 t e ω nζ 2 t e ω nζ 1 t e ω nζ 2 t + 2 SecondOrder Step Response 14
15 Characteristics of an Underdamped Step Response Maximum (percent) overshoot M p = x t p x( ) A/ω n Delay Time Peak Time Rise Time Settling Time t s = 4T = 4 ζω n (2% Criteria) The Ramp Response x ሷ + 2ζω n x ሶ + ω 2 n x = At Undamped Underdamped Critically Damped Overdamped x t = A e ζω nt ω3 n ω n x t = A 2ω n 3 ζ ζ x t = A ω n 3 (ω nt sinω n t) 1 ζ 2 ω dcosω d t + ζω n sinω d t +ζω n cosω d t ω d sinω d t + ω n t 2ζ x t = A ω n 3 ω nt + ω n e ω nt t 2e ω nt 2 ζ 2 1 ζ 2e ω nζ 1 t + ζ 1 e ω nζ 2 t 2 ζ ζ 2 e ω nζ 1 t + ζ 1 e ω nζ 2 t 2ζ + 2ω n t 15
16 SecondOrder Ramp Response Transient Responses of Higher Order Systems 16
17 Example 7.3 A twodisk system is depicted, and the system parameters are provided in Table 7.1 in your textbook. Determine the lowerorder components that contribute to the overall responses ϴ 1 (t) and ϴ 2 (t). Plot and compare the individual contributions to the overall responses. θ 1 (s) T(s) = J 2 s 2 + β 2 s + κ 2 + κ 1 a 0 s 4 + a 1 s 3 + a 2 s 2 + a 3 s + a 4 θ 2 (s) T(s) = κ 1 a 0 s 4 + a 1 s 3 + a 2 s 2 + a 3 s + a 4 a 0 = J 1 J 2 a 1 = J 1 β 2 + J 2 β 1 a 2 = J 1 κ 2 + J 1 κ 1 + J 2 κ 1 + β 1 β 2 a 3 = β 1 κ 2 + β 1 κ 1 + β 2 κ 1 a 4 = κ 1 κ 2 Example 7.3 Use MATLAB to find poles >> J1 = ; >> J2 = J1; >> K1 = 7; >> K2 = 2; >> B1 = 0.01; >> B2 = 0.02; >> num1 = [J2 B2 (K1+K2)]; >> num2 = K1; >> den = [J1*J2 (J1*B2+J2*B1)... (J1*(K1+K2)+J2*K1+B1*B2)... (B1*(K1+K2)+B2*K1) K1*K2]; >> G1 = tf(num1,den); >> G2 = tf(num2,den); >> pole(g1) Responses are combinations of two secondorder underdamped responses ans = i i i i 17
18 Example 7.3 Response An Introduction to PoleZero Analysis When a zero is in close proximity to a pole Except for oscillatory or undamped responses, stable responses will tend to attenuate Remember that underdamped secondorder responses will attenuate to within 2% in 4T 18
19 PoleZero Analysis using MATLAB Function damp(sys) pzmap(sys) sgrid poly(a) Description Computes natural frequencies and damping ratios Plots polezero map of dynamic systems Generates splane grid lines for a polezero map Computes characteristic polynomial of matrix Example 7.4 Determine the dominant pole of the system. >> m = 10; b = 20; k = 60; >> sys2 = tf([m b 2*k],... [m^2 m*b 3*m*k b*k k^2]); >> Z2 = zero(sys2) Z2 = i i >> P2 = pole(sys2) P2 = i i i i >> pzmap(sys2); sgrid 19
20 Example 7.4 Response Example 7.4 >> damp(sys2) Eigenvalue Damping Frequency 2.89e e+00i 1.85e e e e+00i 1.85e e e e+00i 1.85e e e e+00i 1.85e e+00 (Frequencies expressed in rad/seconds) >> pzmap(sys2) >> sgrid(0:0.1:1,1:4) >> impulse(sys2) 20
21 PoleZero Analysis in the State Space The poles are the roots of the denominator To find the poles, we seek the eigenvalues or the roots of the characteristic equation. det(si A) ቚ = det pi A = 0 s=p The zeros are the roots of the numerator To find the zeros, we seek the roots of det A zi B C D = 0 Example
22 SteadyState Analysis in the State Space Example
23 Summary The natural response (also known as the unforced response) is the dynamic response that results when there are no inputs but the initial conditions are generally not zero. Natural and impulse responses for firstorder systems exponentially decay. The time constant, T, is the amount of time required for the response to decay to 36.8% of the initial value. The responses decay to under 2% in 4T. The firstorder step response asymptotically approaches the steadystate value. The time constant,t, is the time required to reach 63.2% of the steadystate value. Within 4T the step response is within 98% of the final value. The firstorder ramp response (increases at a constant rate that runs parallel to the input. At steadystate, there is a constant time delay between the input and response equal to the time constant, T. Also, at steadystate, the difference between the response and the input is AT, where A is the rate. Secondorder systems can be categorized based on the amount of damping: Undamped, ζ = 0 Underdamped, 0 < ζ < 1 Critically damped, ζ = 1 Overdamped, ζ > 1 Summary Continued Undamped responses oscillate forever. Underdamped responses will oscillate, but attenuate to the steadystate value. Critically damped and overdamped responses do not oscillate, but rather asymptotically approach steadystate. The roots of the characteristic equation are referred to as the poles. Poles are generally complex. The placement of secondorder poles in the realcomplex plane is correlated with the damping ratio (ζ), undamped natural frequency (ωn), and the damped frequency of oscillation (ωd): Undamped, purely imaginary roots Underdamped, complex conjugate roots Critically damped, real repeated negative roots Overdamped, real distinct negative roots The secondorder natural and impulse responses oscillate about, attenuate to, or asymptotically approach zero. The unit step responses for secondorder systems oscillate about, attenuate to, or asymptotically approach one. The ramp responses of secondorder systems oscillate about the ramp or settle to a steadystate response that runs parallel to the ramp with a constant error. 23
24 Summary Continued Higherorder responses are just combinations of first and secondorder responses. Transfer functions can be factored into first and secondorder terms. The secondorder terms have complex poles that have associated damping ratios and natural frequencies. Generally, the poles closest to the imaginary axis have dynamics that take longer to settle. If not negated by the presence of zeros, the poles closest to the imaginary axis will tend to dominate the overall response. MATLAB includes functions that facilitate transient analysis including commands to determine eigenvalues, damping ratios, and natural frequencies. It also includes functions for generating and annotating polezero maps. Though poles and zeros are commonly determined by factoring the numerator and denominator of the transfer function representation, they can also be readily determined using the statespace model. The poles, for example, are the eigenvalues of the A matrix in the statespace representation. For systems that reach a static condition at steadystate, the rate of change of the state vector, x, is 0. Thus, the steadystate condition, xss, can be determined algebraically from the steadystate input, uss. 24
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