First-Order and Second-Order Systems: Analysis and Response, Lecture notes of Law

Download First-Order and Second-Order Systems: Analysis and Response and more Lecture notes Law in PDF only on Docsity! MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING 2.151 Advanced System Dynamics and Control Review of First- and Second-Order System Response1 1 First-Order Linear System Transient Response The dynamics of many systems of interest to engineers may be represented by a simple model containing one independent energy storage element. For example, the braking of an automobile, the discharge of an electronic camera flash, the flow of fluid from a tank, and the cooling of a cup of coffee may all be approximated by a first-order differential equation, which may be written in a standard form as τ dy dt + y(t) = f(t) (1) where the system is defined by the single parameter τ , the system time constant, and f(t) is a forcing function. For example, if the system is described by a linear first-order state equation and an associated output equation: ẋ = ax + bu (2) y = cx + du. (3) and the selected output variable is the state-variable, that is y(t) = x(t), Eq. (3) may be rearranged dy dt − ay = bu, (4) and rewritten in the standard form (in terms of a time constant τ = −1/a), by dividing through by −a: −1 a dy dt + y(t) = − b a u(t) (5) where the forcing function is f(t) = (−b/a)u(t). If the chosen output variable y(t) is not the state variable, Eqs. (2) and (3) may be combined to form an input/output differential equation in the variable y(t): dy dt − ay = d du dt + (bc− ad) u. (6) To obtain the standard form we again divide through by −a: −1 a dy dt + y(t) = −d a du dt + ad− bc a u(t). (7) Comparison with Eq. (1) shows the time constant is again τ = −1/a, but in this case the forcing function is a combination of the input and its derivative f(t) = −d a du dt + ad− bc a u(t). (8) In both Eqs. (5) and (7) the left-hand side is a function of the time constant τ = −1/a only, and is independent of the particular output variable chosen. 1D. Rowell 10/22/04 1 Example 1 A sample of fluid, modeled as a thermal capacitance Ct, is contained within an insulating vacuum flask. Find a pair of differential equations that describe 1) the temperature of the fluid, and 2) the heat flow through the walls of the flask as a function of the external ambient temperature. Identify the system time constant. T C T a m b q w a l l s R t f l u i d C t C t R t T C T a m b T r e f h e a t f l o w Figure 1: A first-order thermal model representing the heat exchange between a laboratory vacuum flask and the environment. Solution: The walls of the flask may be modeled as a single lumped thermal resistance Rt and a linear graph for the system drawn as in Fig. 1. The environment is assumed to act as a temperature source Tamb(t). The state equation for the system, in terms of the temperature TC of the fluid, is dTC dt = − 1 RtCt TC + 1 RtCt Tamb(t). (i) The output equation for the flow qR through the walls of the flask is qR = 1 Rt TR = − 1 Rt TC + 1 Rt Tamb(t). (ii) The differential equation describing the dynamics of the fluid temperature TC is found directly by rearranging Eq. (i): RtCt dTC dt + TC = Tamb(t). (iii) from which the system time constant τ may be seen to be τ = RtCt. The differential equation relating the heat flow through the flask is dqR dt + 1 RtCt qR = 1 Rt dTamb dt . (iv) This equation may be written in the standard form by dividing both sides by 1/RtCt, RtCt dqR dt + qR = Ct dTamb dt , (v) and by comparison with Eq. (7) it can be seen that the system time constant τ = RtCt and the forcing function is f(t) = CtdTamb/dt. Notice that the time constant is independent of the output variable chosen. 2 v m F ( t ) m B mBF ( t ) BK V ( t ) B K V ( t ) v = 0r e fv = 0r e f R C + - V ( t ) V ( t ) V = 0r e f R C LRI ( t ) I ( t ) V = 0r e f LR Figure 4: Time constants of some typical first-order systems. Example 2 A water tank with vertical sides and a cross-sectional area of 2 m2, shown in Fig. 5, is fed from a constant displacement pump, which may be modeled as a flow source Qin(t). A valve, represented by a linear fluid resistance Rf , at the base of the tank is always open and allows water to flow out. In normal operation the tank is filled to a depth of v a l v e R f t a n k C f cp ( t ) Q ( t )o u t Q ( t ) p = pr e f R C ffi n Q ( t )i n a t m Figure 5: Fluid tank example 1.0 m. At time t = 0 the power to the pump is removed and the flow into the tank is disrupted. If the flow through the valve is 10−6 m3/s when the pressure across it is 1 N/m2, determine the pressure at the bottom of the tank as it empties. Estimate how long it takes for the tank to empty. 5 Solution: The tank is represented as a fluid capacitance Cf with a value: Cf = A ρg (i) where A is the area, g is the gravitational acceleration, and ρ is the density of water. In this case Cf = 2/(1000× 9.81) = 2.04× 10−4 m5/n and Rf = 1/10−6 = 106 N-s/m5. The linear graph generates a state equation in terms of the pressure across the fluid capacitance PC(t): dPC dt = − 1 RfCf PC + 1 Cf Qin(t) (ii) which may be written in the standard first-order form RfCf dPC dt + PC = RfQin(t). (iii) The time constant is τ = RfCf . When the pump fails the input flow Qin is set to zero, and the system is described by the homogeneous equation RfCf dPC dt + PC = 0. (iv) The homogeneous pressure response is (from Eq. (11)): PC(t) = PC(0)e−t/Rf Cf . (v) With the given parameters the time constant is τ = RfCf = 204 seconds, and the initial depth of the water h(0) is 1 m; the initial pressure is therefore PC(0) = ρgh(0) = 1000 × 9.81 × 1 N/m2. With these values the pressure at the base of the tank as it empties is PC(t) = 9810e−t/204 N/m2 (vi) which is the standard first-order form shown in Fig. 3. The time for the tank to drain cannot be simply stated because the pressure asymptot- ically approaches zero. It is necessary to define a criterion for the complete decay of the response; commonly a period of t = 4τ is used since y(t)/y(0) = e−4 < 0.02 as shown in Table 1. In this case after a period of 4τ = 816 seconds the tank contains less than 2% of its original volume and may be approximated as empty. 1.2 The Characteristic Response of First-Order Systems In standard form the input/output differential equation for any variable in a linear first-order system is given by Eq. (1): τ dy dt + y = f(t). (13) The only system parameter in this differential equation is the time constant τ . The solution with the given f(t) and the initial condition y(0) = 0 is defined to be the characteristic first-order response. 6 The first-order homogeneous solution is of the form of an exponential function yh(t) = e−λt where λ = 1/τ . The total response y(t) is the sum of two components y(t) = yh(t) + yp(t) = Ce−t/τ + yp(t) (14) where C is a constant to be found from the initial condition y(0) = 0, and yp(t) is a particular solution for the given forcing function f(t). In the following sections we examine the form of y(t) for the ramp, step, and impulse singularity forcing functions. 1.2.1 The Characteristic Unit Step Response The unit step us(t) is commonly used to characterize a system’s response to sudden changes in its input. It is discontinuous at time t = 0: f(t) = us(t) = { 0 t < 0, 1 t ≥ 0. The characteristic step response ys(t) is found by determining a particular solution for the step input using the method of undetermined coefficients. From Table 8.2, with a constant input for t > 0, the form of the particular solution is yp(t) = K, and substitution into Eq. (13) gives K = 1. The complete solution ys(t) is ys(t) = Ce−t/τ + 1. (15) The characteristic response is defined when the system is initially at rest, requiring that at t = 0, ys(0) = 0. Substitution into Eq. (14) gives 0 = C + 1, so that the resulting constant C = −1. The unit step response of a system defined by Eq. (13) is: ys(t) = 1− e−t/τ . (16) Equation (16) shows that, like the homogeneous response, the time dependence of the step response depends only on τ and may expressed in terms of a normalized time scale t/τ . The unit step char- acteristic response is shown in Fig. 6, and the values at normalized time increments are summarized in the fourth column of Table 1. The response asymptotically approaches a steady-state value yss = lim t→∞ ys(t) = 1. (17) It is common to divide the step response into two regions, (a) a transient region in which the system is still responding dynamically, and (b) a steady-state region, in which the system is assumed to have reached its final value yss. There is no clear division between these regions but the time t = 4τ , when the response is within 2% of its final value, is often chosen as the boundary between the transient and steady-state responses. The initial slope of the response may be found by differentiating Eq. (16) to yield: dy dt ∣∣∣∣ t=0 = 1 τ . (18) The step response of a first-order system may be easily sketched with knowledge of (1) the system time constant τ , (2) the steady-state value yss, (3) the initial slope ẏ(0), and (4) the fraction of the final response achieved at times equal to multiples of τ . 7 0 1 2 3 4 5 0 1 2 3 4 5 t T i m e t y ( t )r Ra mp re sp on se Figure 8: The ramp response of a first-order system described by τ ẏ + y = ur(t). The characteristic responses yu(t) are by definition zero for time t < 0. If there is a discontinuity in yu(t) at t = 0, as in the case for the characteristic impulse response yδ(t) (Eq. (19)), the derivative dyu/dt contains an impulse component, for example d dt yδ(t) = 1 τ δ(t)− 1 τ2 e−t/τ (26) and if q1 6= 0 the response y(t) will contain an impulse function. 1.3.1 The Input/Output Step Response The characteristic response for a unit step forcing function, f(t) = us(t), is (Eq. (16)): ys(t) = ( 1− e−t/τ ) for t > 0. The system input/output step response is found directly from Eq. (25): y(t) = q1 d dt ( 1− e−t/τ ) + q0 ( 1− e−t/τ ) = q0 [ 1− ( 1− q1 q0τ ) e−t/τ ] . (27) If q1 6= 0 the output is discontinuous at t = 0, and y(0+) = q1/τ . The steady-state response yss is yss = lim t→∞ y(t) = q0. (28) The output moves from the initial value to the final value with a time constant τ . 1.3.2 The Input/Output Impulse Response The characteristic impulse response yδ(t) found in Eq. (19) is yδ(t) = 1 τ e−t/τ fort ≥ 0 10 Input u(t) Characteristic Response Input/Output Response y(t) for t ≥ 0 u(t) = 0 y(t) = y(0)e−t/τ u(t) = ur(t) yr(t) = t− τ ( 1− e−t/τ ) y(t) = [ q0t + (q1 − q0τ) ( 1− e−t/τ )] u(t) = us(t) ys(t) = ys(t) = 1− e−t/τ y(t) = [ q0 − ( q0 − q1 τ ) e−t/τ ] u(t) = δ(t) yδ(t) = 1 τ e−t/τ y(t) = q1 τ δ(t) + ( q0 τ − q1 τ2 ) e−t/τ Table 2: The response of the first-order linear system τ ẏ + y = q1u̇+ q0u for the singularity inputs. with a discontinuity at time t = 0. Substituting into Eq. (25) y(t) = q1 dyδ dt + q0yδ(t) = q1 τ δ(t) + ( q0 τ − q1 τ2 ) e−t/τ , (29) where the impulse is generated by the discontinuity in yδ(t) at t = 0 as shown in Eq. (26). 1.3.3 The Input/Output Ramp Response The characteristic response to a unit ramp r(t) = t is yr(t) = t− τ ( 1− e−t/τ ) and using Eq. (21) the response is: y(t) = q1 d dt [( t− τ ( 1− e−t/τ )) us(t) ] + q0 ( t− τ ( 1− e−t/τ )) us(t) = [ q0t + (q1 − q0τ) ( 1− e−t/τ )] us(t). (30) 1.4 Summary of Singularity Function Responses Table 2 summarizes the homogeneous and forced responses of the first-order linear system de- scribed by the classical differential equation τ dy dx + y = q1 du dt + q0u (31) for the three commonly used singularity inputs. The response of a system with a non-zero initial condition, y(0), to an input u(t) is the sum of the homogeneous component due to the initial condition, and a forced component computed with zero initial condition, that is ytotal(t) = y(0)e−t/τ + yu(t), (32) where yu(t) is the response of the system to the given input u(t) if the system was originally at rest. 11 The response to an input that is a combination of inputs for which the response is known may be found by adding the individual component responses using the principle of superposition. The following examples illustrate the use of these solution methods. Example 3 A mass m = 10 kg is at rest on a horizontal plane with viscous friction coefficient B = 20 N-s/m, as shown in Fig. 9. A short impulsive force of amplitude 200 N and duration 0.01 s is applied. Determine how far the mass travels before coming to rest, and how long it takes for the velocity to decay to less than 1% of its initial value. Solution: The differential equation relating the velocity of the mass to the applied v m F ( t ) m mBF ( t ) v = 0r e f B 0 . 0 1 F ( t ) 2 0 0 t Figure 9: A mass element subjected to an impulsive force. force is m B dvm dt + vm = 1 B Fin(t) (i) The system time constant is τ = m/B = 10/20 = 0.5 seconds. The duration of the force pulse is much less than the time constant, and so it is reasonable to approximate the input as an impulse of strength (area) 200× .01 = 2 N-s. The system impulse response (Eq. (29) is vm(t) = 1 m e−Bt/m (ii) so that if u(t) = 2δ(t) N-s the response is vm(t) = 0.2e−2t. (iii) The distance x traveled may be computed by integrating the velocity x = ∫ ∞ 0 0.2e−2tdt = 0.1 m. (iv) The time T for the velocity to decay to less than 1% of its original value is found by solving vm(T )/vm(0) = 0.01 = e−2T , or T = 2.303 seconds. Example 4 A disk flywheel J of mass 8 Kg and radius 0.5 m is driven by an electric motor that 12 and with the values given (J = 1 Kg-m2 and BR = 0.1 N-m-s/rad) 10 dΩJ dt + ΩJ = 10Tin(t), (ii) The torque input shown in Fig. 12 may be written as a sum of unit ramp and step singularity functions Tin(t) = 0.2ur(t)− 0.2ur(t− 100)− 200us(t− 125). (iii) The response may be determined in three time intervals (1) Initially 0 ≤ t < 100 when the input is effectively Tin(t) = 0.2ur(t), (2) for 100 ≤ t < 125 seconds when the input is Tin(t) = 0.2ur(t)− 0.2ur(t− 100), and (3) for t ≥ 125 when Tin(t) = 0.2ur(t)− 0.2ur(t− 100)− 20us(t− 125). From Table 2 the response in the three intervals may be written 0 ≤ t < 100 s: ΩJ(t) = 2 [ t− 10 ( 1− e−t/10 )] rad/s, 100 ≤ t < 125 s: ΩJ(t) = 2 [ t− 10 ( 1− e−t/10 )] −2 [ (t− 100)− 10 ( 1− e−(t−100)/10 )] rad/s, t > 125 s: ΩJ(t) = 2 [ t− 10 ( 1− e−t/10 )] −2 [ (t− 100)− 10 ( 1− e−(t−100)/10 )] rad/s, −200 [ 1− ( 1− e−(t−125)/10 )] The total response is plotted in Fig. 13. Example 6 The first-order electrical circuit shown in Fig. 14 is known as a “lead” network and is commonly used in electronic control systems. Find the response of the system to an input pulse of amplitude 1 volt and duration 10 ms if R1 = R2 = 10, 000 ohms and C = 1.0 µfd. Assume that at time t = 0 the output voltage is zero. Solution: From the linear graph the state variable is the voltage on the capacitor vc(t), and the output is the voltage across R2. The state equation for the system is dvc dt = −R1 + R2 R1R2C vc + 1 R2C Vin(t) (i) 15 W ( t )J 0 5 0 1 0 0 1 5 0 2 0 0 0 5 0 1 0 0 1 5 0 2 0 0 2 5 0 T i m e ( s e c ) t Figure 13: Response of the rotary flywheel system to the torque input profile Tin(t) = 0.2ur(t) − 0.2ur(t− 100)− 20us(t) N-m, with initial condition ΩJ(0) = 0 rad/s. R C + - V ( t ) V ( t ) V = 0 R C 1 R 2 V ( t )oi n R 1 2 r e f Figure 14: Electrical lead network and its linear graph. and the output equation is vo(t) = vR2 = −vc + Vin(t), (ii) The input/output differential equation is R1R2C R1 + R2 dvo dt + vo = R1R2C R1 + R2 dVin dt + R1 R1 + R2 Vin. (iii) with the system time constant τ = R1R2C/(R1 + R2) = 5× 10−3 seconds. The input pulse duration (10 ms) is comparable to the system time constant, and therefore it is not valid to approximate the input as an impulse. The pulse input can, however, be written as the sum of two unit step functions Vin(t) = us(t)− us(t− 0.01) (iv) and the response determined in two separate intervals (1) 0 ≤ t < 0.01 s where the input is us(t), and (2) t ≥ 0.01 s, where both components contribute. 16 The input/output unit step response is given by Eq. (27), vo(t) = R2 R1 + R2 − ( R2 R1 + R2 − 1 ) e−t/τ = R2 R1 + R2 + R1 R1 + R2 e−t/τ = ( 0.5 + 0.5e−t/0.005 ) for t ≥ 0. (v) At time t = 0+ the initial response is vo(0+) = 1 volt, and the steady-state response (v0)ss = 0.5 volt. The settling time is approximately 4τ , or about 20 ms. The response to the 10 ms duration pulse may be found from Eqs. (iv) and (v) by using the principle of superposition: vpulse(t) = vo(t)− v0(t− .01). (vi) In the interval 0 ≤ t < 0.01, the initial condition is zero and the response is: vpulse(t) = ( 0.5 + 0.5e−t/0.005 ) , (vii) in the second interval t ≥ .01 , when the input is Vin = us(t)− us(t− .01), the response is the sum of two step responses: vpulse(t) = ( 0.5 + 0.5e−t/0.005 ) − ( 0.5 + 0.5e−(t−0.01)/0.005 ) = 0.5 ( et/.005 − e−(t−.01)/.005 ) = 0.5et/0.005 ( 1− e2 ) = −3.195e−t/.005 V. (viii) The step response (Eq. (v)) and the pulse response described by Eqs. (vii) and (viii) are plotted in Fig. 15. 2 Second-Order System Transient Response Second-order state determined systems are described in terms of two state variables. Physical second-order system models contain two independent energy storage elements which exchange stored energy, and may contain additional dissipative elements; such models are often used to represent the exchange of energy between mass and stiffness elements in mechanical systems; be- tween capacitors and inductors in electrical systems, and between fluid inertance and capacitance elements in hydraulic systems. In addition second-order system models are frequently used to rep- resent the exchange of energy between two independent energy storage elements in different energy domains coupled through a two-port element, for example energy may be exchanged between a mechanical mass and a fluid capacitance (tank) through a piston, or between an electrical induc- tance and mechanical inertia as might occur in an electric motor. Engineers often use second-order system models in the preliminary stages of design in order to establish the parameters of the energy storage and dissipation elements required to achieve a satisfactory response. Second-order systems have responses that depend on the dissipative elements in the system. Some systems are oscillatory and are characterized by decaying, growing, or continuous oscillations. 17 and for x2(t): d2x2 dt2 + 4 dx2 dt + 7x2 = 2u. (iv) By inspection of either Eq. (iii) or Eq. (iv), ω2 n = 7, and 2ζωn = 4, giving ωn = √ 7 rad/s, and ζ = 2/ √ 7 = 0.755. 2.0.2 Generation of a Differential Equation in an Output Variable The output equation y = Cx + Du for any system variable is a single algebraic equation: y(t) = [ c1 c2 ] [ x1 x2 ] + [d] u(t) = c1x1(t) + c2x2(t) + du(t). (41) and in the Laplace domain Y (s) = ( C(sI−A)−1B + D ) U(s) = 1 det [sI−A] (Cadj(sI−A) + det [sI−A]D) The determinants may be expanded and the resulting equation written as a differential equation: d2y dt2 − (a11 + a22) dy dt + (a11a22 − a12a21) y = q2 d2u dt2 + q1 du dt + q0u (42) or in terms of the standard system parameters d2y dt2 + 2ζωn dy dt + ω2 ny = q2 d2u dt2 + q1 du dt + q0u (43) where the coefficients q0, q1, and q2 are q0 = c1 (−a22b1 + a12b2) + c2 (−a11b2 + a21b1) + d (a11a22 − a12a21) q1 = c1b1 + c2b2 − d (a11 + a22) q2 = d. (44) Notice that the left hand side of the differential equation is the same for all system variables, and that the only difference between any of the differential equations describing any system variable is in the constant coefficients q2, q1 and q0 on the right hand side. Example 8 A rotational system consists of an inertial load J mounted in viscous bearings B, and driven by an angular velocity source Ωin(t) through a long light shaft with significant torsional stiffness K, as shown in the Fig. 16. Derive a pair of second-order differential equations for the variables ΩJ and ΩK . 20 W ( t ) W = 0r e f JB K i n F l y w h e e lB e a r i n gT o r s i o n a l s p r i n g M o t o r ( v e l o c i t y s o u r c e ) K B J W ( t )J W ( t )i n Figure 16: Rotational system for Example 8. Solution: The state variables are ΩJ , and TK , and the state and output equations are [ Ω̇j Ṫk ] = [ −B/J 1/J −K 0 ] [ ΩJ TK ] + [ 0 K ] Ωin. (i) [ ΩJ ΩK ] = [ 1 0 −1 0 ] [ ΩJ TK ] + [ 0 1 ] (ii) In this case there are two outputs and the transfer function matrix is H(s) = C [sI−A]−1 B + D = Cadj [sI−A]B + D det [sI−A] =   K/J s2 + (B/J)s + K/J s2 + (B/J)s s2 + (B/J)s + K/J   (iii) The required differential equations are therefore d2ΩJ dt2 + B J dΩJ dt + K J ΩJ = K J Ωin. (iv) and d2ΩK dt2 + B J dΩK dt + K J ΩK = d2Ωin dt2 + B J dΩin dt . (v) The undamped natural frequency and damping ratio are found from either differen- tial equation. For example, from Eq. (v) ω2 n = K/J and 2ζωn = B/J . From these relationships ωn = √ K J and ζ = B/J 2 √ K/J = B 2 √ KJ . (vi) 2.1 Solution of the Homogeneous Second-Order Equation For any system variable y(t) in a second-order system, the homogeneous equation is found by setting the input u(t) ≡ 0 so that Eq. (43) becomes d2y dt2 + 2ζωn dy dt + ω2 ny = 0. (45) 21 The solution, yh(t), to the homogeneous equation is found by assuming the general exponential form yh(t) = C1e λ1t + C2e λ2t (46) where C1 and C2 are constants defined by the initial conditions, and the eigenvalues λ1 and λ2 are the roots of the characteristic equation det [sI−A] = λ2 + 2ζωnλ + ω2 n = 0, (47) found using the quadratic formula: λ1, λ2 = −ζωn ± ωn √ ζ2 − 1. (48) If ζ = 1, the two roots are equal (λ1 = λ2 = λ), a modified form for the homogeneous solution is necessary: yc(t) = C1e λt + C2te λt (49) In either case the homogeneous solution consists of two independent exponential components, with two arbitrary constants, C1 and C2, whose values are selected to make the solution satisfy a given pair of initial conditions. In general the value of the output y(0) and its derivative ẏ(0) at time t = 0 are used to provide the necessary information. The initial conditions for the output variable may be specified directly as part of the problem statement, or they may have to be determined from knowledge of the state variables x1(0) and x2(0) at time t = 0. The homogeneous output equation may be used to compute y(0) directly from elements of the A and C matrices, y(0) = c1x1(0) + c2x2(0), (50) and the value of the derivative ẏ(0) may be determined by differentiating the output equation and substituting for the derivatives of the state variables from the state equations: ẏ(0) = c1ẋ1(0) + c2ẋ2(0) = c1 (a11x1(0) + a12x2(0)) + c2 (a21x1(0) + a22x2(0)) . (51) To illustrate the influence of damping ratio and natural frequency on the system response, we consider the response of an unforced system output variable with initial output conditions of y(0) = y0, and ẏ(0) = 0. If the roots of the characteristic equation are distinct, imposing these initial conditions on the general solution of Eq. (46) gives: y(0) = y0 = C1 + C2 dy dt ∣∣∣∣ t=0 = 0 = λ1C1 + λ2C2. (52) With the result that C1 = λ2 λ2 − λ1 y0 and C2 = λ1 λ1 − λ2 y0. (53) For this set of initial conditions the homogeneous solution is therefore yh(t) = y0 [( λ2 λ2 − λ1 ) eλ1t + ( λ1 λ1 − λ2 ) eλ2t ] (54) = y0 λ1λ2 λ2 − λ1 [ 1 λ1 eλ1t − 1 λ2 eλ2t ] . (55) 22 z = 1 z = 0 . 1 0 . 2 0 . 5 N o r m a l i z e d t i m e y ( t ) / y ( 0 ) w tn0 5 1 0 2 01 5 0 0 . 5 1 . 0 - 0 . 5 - 1 . 0 No rm ali ze d r es po ns e Figure 18: Normalized initial condition response of an underdamped second-order system as a function of the damping ratio ζ. at an arbitrary time t is compared with the response at time t + Tp, an amplitude decay ratio DR may be defined as: DR = y(t + Tp) y(t) provided y(t) 6= 0 = e−ζωn(t+2π/ωd) e−ζωnt = e−2πζ/ √ 1−ζ2 (68) The decay ratio is unity if the damping ratio is zero, and decreases as the damping ratio increases, reaching a value of zero as the damping ratio approaches unity. Unstable System (ζ < 0): If the damping ratio is negative, the roots to the characteristic equation have positive real parts, and the real exponential term in the solution, Eq. (46), grows in an unstable fashion. When −1 < ζ < 0, the response is oscillatory with an overall exponential growth in amplitude, as shown in Figure 19, while the solution for ζ < −1 grows as a real exponential. Example 9 Many simple mechanical systems may be represented by a mass coupled through spring and damping elements to a fixed position as shown in Figure 20. Assume that the mass has been displaced from its equilibrium position and is allowed to return with no external forces acting on it. We wish to (1) find the response of the system model from an initial displacement so as to determine whether the mass returns to its equilibrium position with no overshoot, (2) to determine the maximum velocity that it reaches. In addition we wish (3) to determine which system parameter we should change in order to guarantee no overshoot in the response. The values of system parameters are m = 2 kg, K = 8 N/m, B = 1.0 N-s/m and the initial displacement y0 = 0.1 m. 25 w t N o r m a l i z e d t i m e z = - 0 . 1 y ( t ) / y ( 0 ) n0 5 1 0 2 01 5 0 2 4 6 8 - 2 - 4 - 6 No rm ali ze d r es po ns e Figure 19: A typical unstable oscillatory response of a second-order system when the damping ratio ζ is negative. Solution: From the linear graph model in Figure 20 the two state variables are the v m F ( t ) m B mBF ( t ) v = 0r e f K K y Figure 20: Second-order mechanical system. velocity of mass x1 = vm, and the force in the spring x2 = FK . The state equations for the system, with an input force Fin(t) acting on the mass are: [ v̇m ḞK ] = [ −B/m −1/m K 0 ] [ vm FK ] + [ 1/m 0 ] Fin(t). (i) The output variable y is the position of the mass, which can be found from the consti- tutive relation for the force in the spring FK = Ky and therefore the output equation: y (t) = (0) vm + ( 1 K ) FK + (0)Fin(t). (ii) The characteristic equation is det [λI−A] = det [ λ + B/m 1/m −K λ ] = 0, (iii) or λ2 + B m λ + K m = 0. (iv) 26 The undamped natural frequency and damping ratio are therefore ωn = √ K m , and ζ = B 2mωn = B 2 √ Km . (v) With the given system parameters, the undamped natural frequency and damping ratio are ωn = √ 8 2 = 2 rad/s, ζ = 1 4× 2 = 0.125. Because the damping ratio is positive but less than unity, the system is stable but underdamped; the response yh(t) is oscillatory and therefore exhibits overshoot. The solution is given directly by Eq. (65): yh(t) = y0 e−ζωnt √ 1− ζ2 cos(ωdt− ψ), (vi) and when the computed values of ωd and ψ are substituted, ωd = ωn √ 1− ζ2 = 2 √ 1− (.125)2 = 1.98 rad/s, and ψ = tan−1 0.125√ 1− (.125)2 = 0.125 r, the response is: yh(t) = 0.101e−.25t cos(1.98t− 0.125) m. (vii) The response is plotted in Fig. 21a, where it can be seen that the mass displacement response y(t) overshoots the equilibrium position by almost 0.1 m, and continues to oscillate for several cycles before settling to the equilibrium position. The velocity of the mass vm(t) is related to the displacement y(t) by differentiation of Eq. (vi), vm(t) = d dt yh(t) = − y0ωn√ 1− ζ2 e−ζωnt sinωdt. (viii) The velocity response is plotted in Figure 21b, where the maximum value of the velocity is found to be -0.17 m/s at a time of 0.75 s. In order to achieve a displacement response with no overshoot, an increase in the system damping is required to make ζ ≥ 1. Since the damping ratio ζ is directly proportional to B, the value of the viscous damping parameter B would have to be increased by a factor of 8, that is to B = 8 N-s/m to achieve critical damping. With this value the response is given by Eq. (61): y(t) = 0.1 ( e−2t + 2te−2t ) (ix) The critically damped displacement response is also plotted in Figure 21a, showing that there is no overshoot. As before, the velocity of the mass may be found by differentiating the position response v(t) = 0.1 ( −2e−2t + 2e−2t + 4te−2t ) = 0.4te−2t (x) 27 The particular solution may be found using the method of undetermined coefficients, we take yp(t) = K and substitute into the differential equation giving ω2 nK = 1 (74) or ys(t) = C1e λ1t + C2e λ2t + 1 ω2 n . (75) The constants C1 and C2 are chosen to satisfy the two initial conditions: ys(0) = C1 + C2 + 1 ω2 n = 0 (76) dys dt ∣∣∣∣ t=0 = C1λ1 + C2λ2 = 0, (77) which may be solved to give: C1 = λ2 ω2 n (λ1 − λ2) , C2 = λ1 ω2 n (λ2 − λ1) . (78) The solution for the unit step response when the roots are distinct is therefore: ys(t) = 1 ω2 n [ 1− ( λ2 λ2 − λ1 eλ1t + λ1 λ1 − λ2 eλ2t )] (79) = 1 ω2 n [ 1− λ2λ1 λ2 − λ1 ( 1 λ1 eλ1t − 1 λ2 eλ2t )] (80) It can be seen that the second and third terms in Eq. (80) are identical to those in the homogeneous response, Eq. (55), so that the solution may be written for the overdamped case as: ys(t) = 1 ω2 n [ 1− 1 τ2 − τ1 ( τ2e −t/τ2 − τ1e −t/τ1 )] for ζ > 1, (81) where τ1 = −1/λ1 and τ2 = −1/λ2 are time constants as previously defined. For the underdamped case, when λ1 = −ζωn + jωn √ 1− ζ2 and λ1 = ζωn − jωn √ 1− ζ2, from Eq. (65) the solution is: ys(t) = 1 ω2 n [ 1− e−ζωnt √ 1− ζ2 cos(ωdt− ψ) ] for 0 < ζ < 1. (82) where as before the phase angle ψ = tan−1 ( ζ/ √ 1− ζ2 ) . When the roots of the characteristic equation are identical (ζ = 1) and λ1 = λ2 = −ωn, the homogeneous solution has a modified form, and the total solution is: ys(t) = C1e λt + C2te λt + 1 ω2 n . (83) The solution which satisfies the initial conditions is: ys(t) = 1 ω2 n [ 1− eλt + λteλt ] = 1 ω2 n [ 1− e−ωnt − ωnte−ωnt ] for ζ = 1. (84) 30 In all three cases the response settles to a steady equilibrium value as time increases. We define the steady-state response as yss = lim t→∞ ys(t) = 1 ω2 n . (85) The second-order system step response is a function of both the system damping ratio ζ and the undamped natural frequency ωn. The step responses of stable second-order systems are plotted in Figure 22 in terms of non-dimensional time ωnt, and normalized amplitude y(t)/yss. N o r m a l i z e d t i m e y ( t ) / y s s 0 2 . 5 5 7 . 5 1 0 1 2 . 5 1 5 1 7 . 5 2 0 0 0 . 2 5 0 . 5 0 . 7 5 1 1 . 2 5 1 . 5 1 . 7 5 2 z = 0 . 1 0 . 2 0 . 5 0 . 7 0 7 1 . 0 1 . 5 2 . 0 5 . 0 nw t No rm ali ze d r es po ns e Figure 22: Step response of stable second-order systems with the differential equation ÿ + 2ζωnẏ + ω2 ny = u(t). For damping ratios less than one, the solutions are oscillatory and overshoot the steady-state response. In the limiting case of zero damping the solution oscillates continuously about the steady- state solution yss with a maximum value of ymax = 2yss and a minimum value of ymin = 0, at a frequency equal to the undamped natural frequency ωn. As the damping is increased, the amplitude of the overshoot in the response decreases, until at critical damping, ζ = 1, the response reaches steady-state with no overshoot. For damping ratios greater than unity, the response exhibits no overshoot, and as the damping ratio is further increased the response approaches the steady-state value more slowly. 31 Example 10 The electric circuit in Figure 23 contains a current source driving a series inductive and resistive load with a shunt capacitor across the load. The circuit is representative of motor drive systems and induction heating systems used in manufacturing processes. Excessive peak currents during transients in the input could damage the inductor. We therefore wish to compute response of the current through the inductor to a step in the input current to ensure that the manufacturers stated maximum current is not exceeded during start up. The circuit parameters are L = 10−4 h, C = 10−8 fd, and R = 50 ohms. Assume that the maximum step in the input current is to be 1.0 amp. C L R L o a d I ( t )S I ( t ) V = 0r e f L R cS Figure 23: A second-order electrical system. Solution: From the linear graph in Figure 23 the state variables are the voltage across the capacitor vC(t), and the current in the inductor iL(t). The state equations for the system are: [ v̇C i̇L ] = [ 0 −1/C 1/L −R/L ] [ vC iL ] + [ 1/C 0 ] Is. (i) The differential equation relating the current iL to the source current Is is found by Cramer’s rule: det [ S 1/C −1/L S + R/L ] {iL} = det [ S 1/C −1/L 0 ] {Is} (ii) or d2iL dt2 + R L diL dt + 1 LC iL = 1 LC Is, (iii) and the undamped natural frequency ωn and damping ratio ζ are: ωn = 1√ LC = 106 rad/s (iv) ζ = (R/L) 2/ √ LC = R 2 √ C L = 0.25. (v) The system is underdamped (ζ < 1) and oscillations are expected in the response. The differential equation is similar to the standard form and therefore has a unit step response in the form of Eq. (2.2.2): iL(t) = ( ω2 n ) 1 ω2 n [ 1− e−ζωnt √ 1− ζ2 cos(ωdt− ψ) ] (vi) = 1− 1.033e−0.25×106t cos ( 0.968× 106t− .2527 ) (vii) 32 where us(t) is the unit step function. For the standard system defined in Eq. (70) with f(t) = t, the forced differential equation is d2yr dt2 + 2ζωn dyr dt + ω2 nyr = t. (94) When the roots of the characteristic equation are distinct, the ramp response is found by integrating Eq. (80), that is yr(t) = 1 ω2 n ∫ t 0 [ 1− λ1λ2 λ2 − λ1 ( 1 λ1 eλ1t − 1 λ2 eλ2t )] dt = 1 ω2 n [ t− λ1λ2 λ2 − λ1 ( 1 λ2 1 [ eλ1t − 1 ] − 1 λ2 2 [ eλ2t − 1 ])] = 1 ω2 n [ t− λ1λ2 λ2 − λ1 ( 1 λ2 1 eλ1t − 1 λ2 2 eλ2t ) − λ1 + λ2 λ1λ2 ] (95) For an overdamped system with real distinct roots, λ1 = −ζωn + √ ζ2 − 1ωn and λ2 = −ζωn −√ ζ2 − 1ωn, the ramp response may be found from Eq. (95) directly, or by making the partial substitutions for ζ and ωn: yr(t) = 1 ω2 n t− 1 2ωn √ 1− ζ2 ( τ2 1 e−t/τ1 − τ2 2 e−t/τ2 ) − 2ζ ω3 n . (96) which consists a term that is itself a ramp, a pair of decaying exponential terms, and a constant offset term. When the system is underdamped with complex conjugates roots, Eq. (95) may be written: yr(t) = 1 ω2 n t + e−ζωnt ω3 n ( 2ζ cosωdt + 2ζ2 − 1√ 1− ζ2 sinωdt ) − 2ζ ω3 n (97) which consists of a ramp function, a damped oscillatory term and a constant offset. When the roots are real and equal (ζ = 1) the response is found by integrating Eq. (84): yr(t) = 1 ω2 n ∫ t 0 [ 1− e−ωnt − ωnte−ωnt ] dt = 1 ω2 n [ t + 2 ωn e−ωnt + te−ωnt − 2 ωn ] (98) 2.2.5 Summary of Singularity Function Responses The characteristic responses of a linear system to the ramp, step, and impulse functions are sum- marized in Table 3. 2.3 Second-Order System Transient Response The characteristic response defined in the previous section is the response to a forcing function f(t) as defined in Eq. (69). The response of a system to an input u(t) may be determined directly by superposition of characteristic responses. The complete differential equation d2y dt2 + 2ζωn dy dt + ω2 ny = q2 d2u dt2 + q1 du dt + q0u (99) in general involves a summation of derivatives of the input. The principle of superposition allows us to determine the system response to each component of the forcing function and to sum the 35 Damping ratio Input f(t) Characteristic Response y(t) 0 ≤ ζ < 1 f(t) = ur(t) yr(t) = 1 ω2 n [ t + e−ζωnt ωn ( 2ζ cosωdt + 2ζ2 − 1√ 1− ζ2 sinωdt ) − 2ζ ωn ] f(t) = us(t) ys(t) = 1 ω2 n [ 1− e−ζωnt √ 1− ζ2 cos(ωdt− ψ) ] f(t) = δ(t) yδ(t) = e−ζωnt ωn √ 1− ζ2 sin (ωdt) ζ = 1 f(t) = ur(t) yr(t) = 1 ω2 n [ t + 2 ωn e−ωnt + te−ωnt − 2 ωn ] f(t) = us(t) ys(t) = 1 ω2 n [ 1− e−ωnt − ωnte−ωnt ] f(t) = δ(t) yδ(t) = te−ωnt ζ > 1 f(t) = ur(t) yr(t) = 1 ω2 n [ t + ωn 2 √ 1− ζ2 ( τ2 1 e−t/τ1 − τ2 2 e−t/τ2 ) − 2ζ ωn ] f(t) = us(t) ys(t) = 1 ω2 n [ 1− ωn 2 √ ζ2 − 1 ( τ1e −t/τ1 − τ2e −t/τ2 )] f(t) = δ(t) yδ(t) = 1 2ωn √ ζ2 − 1 ( e−t/τ1 − e−t/τ2 ) Notes: 1. The damped natural frequency ωd = √ 1− ζ2ωn for 0 ≤ ζ < 1. 2. The phase angle ψ = tan−1 ( ζ/ √ 1− ζ2 ) for 0 ≤ ζ < 1. 3. For over-damped systems (ζ > 1) the time constants are τ1 = 1/ ( ζωn − √ ζ2 − 1ωn ) , and τ2 = 1/ ( ζωn + √ ζ2 − 1ωn ) . Table 3: Summary of the characteristic transient responses of the system ÿ + 2ζωnẏ + ω2 ny = f(t) to the unit ramp ur(t), the unit step us(t), and the impulse δ(t). 36 individual responses. In addition, the derivative property tells us that if the response to a forcing function f(t) = u(t) is yu(t), the other components are derivatives of yu(t) and the total response is y(t) = q2 d2yu dt2 + q1 dyu dt + q0yu. (100) As in the case of first order systems, the derivatives must take into account discontinuities at time t = 0. Example 11 Determine the response of a physical system with differential equation d2y dt2 + 8 dy dt + 16y = 3 du dt + 2u to a step input u(t) = 2 for t ≥ 0. Solution: The characteristic equation is λ2 + 10λ + 16 = 0 (i) which has roots λ1 = −2 and λ2 = −8. For this system ωn = 4 rad/s and ζ = 1.25; the system is overdamped. The characteristic response to a unit step is (from Table 3): ys(t) = 1 ω2 n [ 1− ωn 2 √ ζ2 − 1 ( τ1e −t/τ1 − τ2e −t/τ2 )] (ii) where τ1 = 1/2, and τ2 = 1/8, or ys(t) = 1 16 [ 1− 8 3 ( 1 2 e−2t − 1 8 e−8t )] = 1 16 − 1 12 e−2t + 1 48 e−8t (iii) The system response to a step of magnitude 2 is therefore y(t) = 2 [ 3 dys dt + 2ys ] = 2 [ 3 ( 1 6 e−2t − 1 6 e−8t ) + 2 ( 1 16 − 1 12 e−2t + 1 48 e−8t )] = 1 4 − 2 3 e−2t + 11 12 e−8t (iv) For systems in which q2 6= 0 a further simplification is possible. The system differential equation may be written in operational form y(t) = q2S 2 + q1S + q0 S2 + 2ζωnS + ω2 n {u} (101) 37 and with the system parameter values this equation gives Kr = 5.83 N/m. (vii) With this value of Kr the system parameters are ζ = 1 and ωn = 3.41 rad/s. From Table 3 the unit characteristic step response is ys(t) = 1 ω2 n [ 1− e−ωnt − ωnte−ωnt ] (viii) and the impeller response to a step of 100 rad/s is ΩJ(t) = 100q0 ω2 n [ 1− e−ωnt − ωnte−ωnt ] = 50 [ 1− e−3.41t − 3.41te−ωnt ] . (ix) The response in fan speed is similar to the non-dimensional form shown in Fig. 22, and is plotted in Fig 27. Note that the steady-state speed is 50 rad/s, which is one half of the motor no-load speed of 100 rad/s. 2. The differential equation relating the torque TK to the source velocity is d2TK dt2 + 2ζωn dTK dt + ω2 nTK = Kr dΩs dt + KrBr J Ωs (x) which contains both the input Ωs and its derivative. Then TK(t) = 100 [ Krte −ωnt + KrBr ω2 nJ ( 1− e−ωnt − ωnte−ωnt )] = 50 ( 1− e−3.41t + 8.22te−3.41t ) N-m, (xi) which is plotted in Fig. 27. Notice that in this case, although the system is critically damped (ζ = 1), the response overshoots the steady-state value. This behavior is common for output variables that involve the derivative of the input in their differential equation. 40 0 0 . 5 1 1 . 5 2 2 . 5 3 0 2 0 4 0 6 0 8 0 1 0 0 T i m e ( s e c s ) t T ( t )K W ( t )J W ( t )J T ( t )K An gu lar ve loc ity (ra d/s ) a nd to rqu e ( N- m) Figure 27: Step response of shaft coupling torque TK , and fan angular velocity ΩJ . 41