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MIT 2 004 - Lecture Notes

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MIT OpenCourseWare http://ocw.mit.edu 2.004 Dynamics and Control II Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.Ks + 2 s 5s + bU ( s )V ( s ) Y ( s )u ( t )v ( t )y ( t )sjwxxob i n c r e a s i n g- b� Massachusetts Institute of Technology Department of Mechanical Engineering 2.004 Dynamics and Control II Spring Term 2008 Lecture 221 Reading: Nise: 4.1 – 4.8 • 1 The Time-Domain Response of Systems with Finite Zeros Consider a system: K(s + b)Gs = , s2 + 2s + 5we have seen that we can consider this as two cascade blocks Then if the response of the a system 1/D(s) is v(t), then dv y(t) = + bv(t)dt and as the zero (at s = −b) moves deeper into the l.h. s-plane,, the relative contribution of the derivative term decreases and the system response tends toward a scaled version of the all pole response v(t). In general, the presence of the derivative terms in the response means that: 1copyright cD.Rowell 2008 22–10 1 2 3 4 5 6 700.20.40.60.811.21.41.61.82Step ResponseTime (sec)Amplitudez = −1z = −2z = −3all−pole• The response is faster (shorter peak-time TP and rise-time TR). • Greater overshoot in the response (if any). A zero may cause overshoot in the response of an over-damped second-order system. Example 1 The following MATLAB step response compares the response for the under-damped system 5 G(s) = s2 + 2s + 5 with similar unity-gain systems with zeros at s = −1, −2, −3: 5(s + 1) 5/2(s + 2) 5/3(s + 3) G(s) = , G(s) = , G(s) = s2 + 2s + 5 s2 + 2s + 5 s2 + 2s + 5 Note the increase in the overshoot, and the decrease in TP as the zero approaches the origin. 22–20 0.5 1 1.5 2 2.500.20.40.60.811.21.41.6Step ResponseTime (sec)Amplitude12(s + 1)s + 7s + 12s + 7s + 121222Example 2 The following MATLAB step response compares the response for the unity-gain overdamped system 12 G(s) = s2 + 7s + 12 with two real poles at s = −3 and s = −4 with the similar system with a zeros at s = −1: 12(s + 1) G(s) = s2 + 7s + 12 Note the overshoot caused by the zero, but that the overshoot is not oscillatory. Clearly the rise-time TR is much shorter for the system with the zero. 2 The Time-Domain Response of Systems where the Order of the Numerator equals the Order of the Denominator Consider systems of the form bnsn + bn−1sn−1 + . . . + b1s + b0G(s) = ansn + an−1sn−1 + . . . + a1s + a0 22–3b - as + a1++y ( t )u ( t )U ( s )Y ( s )d i r e c t f e e d - t h r o u g h f r o m i n p u t t o o u t p u twhere the degree of the numerator equals that of the denominator. In such systems it is possible to do polynomial division and write the transfer function as N(s) N�(s)G(s) = = K + D(s) D(s) where N�(s) is a polynomial of degree less than that of D(s). For example, a system with transfer function s + a G(s) = s + b may be written b − a G(s) = 1 + , s + a which may be represented in block-diagram form showing a direct feed-through of the input into the output. In other words, when the order of the numerator is the same of the denominator the input will appear directly as a component of the output. The step-response ystep(t) of this system will therefore be b − a �1 − e−at�ystep(t) = us(t) + a where us(t) is the unit-step (Heaviside) function. Note: • That ystep(0+) = 1, that is there is a step transient in the response (which does not occur if the order of N(s) is less than that of D(s)). • The steady-state step response yss = b/a, and if b > a then yss > 1, while if a > b yss < 1. The following MATLAB plot shows the step responses for the two systems s + 6 s + 2 G(s) = (b > a), and G(s) = (a > b) s + 4 s + 4 with step responses ystep(t) = 1 + 2 �1 − e−4t� and ystep(t) = 1 − 2 �1 − e−4t� 4 4 22–4Step ResponseTime (sec)Amplitude0 0.5 1 1.500.20.40.60.811.21.41.61.82s + 6s + 4s + 2s + 4Initial transient at t=0G(s) = G(s) = Find the step response of the following electrical circuit: +-CRR12vVo1mF1 0 kW2 0 kWExample 3 The transfer function is Vo(s) s + 1/R1C G(s) = = V (s) s + (R1 + R2)/R1R2C and with the values shown Vo(s) s + 100 50 G(s) = = = 1 − . V (s) s + 150 s + 150 The step response is therefore 50 �1 − e−150t� =2 1 e−150t ystep = 1 − + 150 3 3 which is plotted below: 22–5−0.005 0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.0400.20.40.60.811.2Step ResponseTime (sec)AmplitudeExample 4 Find the step response of the following third-order system: 2s3 + 17s2 + 13s + 12 G(s) = s3 + 7s2 + 6s + 5 3s2 + s + 2 = 2 + s3 + 7s2 + 6s + 5 showing a direct feed-through term of amplitude two. From Maple-Syrep, the step response is ystep(t) = 2.4−0.5307e−6.157t+0.1307e−0.4213t cos(0.7966t)−0.2667e−0.4213t sin(0.7966t) from which ystep(o+) = 2, and yss = 2.4. The step response is plotted below. 22–60 2 4 6 8 10 12 14 1600.511.522.53Step ResponseTime


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