自动化专业英语Chapter41b.docx

1、自动化专业英语Chapter41bChapter 4 Response Classical Method 4-1 INRODUCTION Having represented control systems using block diagrams as well as state variables, we turn our attention to system response, i.e. how does a system respond as a function of time when subjected to various types of stimuli? Here we

2、are interested in the system output without regard to the behavior of variables inside the control system. When this is the case, we can work with the system transfer function. If we desire C(s) we can work with C(s)/R(s) and specify R(s) and obtain the output. On the other hand, if we need E(s) we

3、should work with E(s)/R(s) and specify R(s). In any event it is important to recognize that when the response to a single input is required without regard to the behavior of variables inside a control system, we speak of applying the classical approach. This can be most readily achieved by employing

4、 techniques. This technique involves representing the output (or desired variable) as the ratio of two polynomials and then expanding the expression in partial fractions. The constants of the partial fraction are calculated by the residue theorem. The output in the time domain is the obtained by tak

5、ing the inverse Laplace transform. A detailed discussion of Laplace transforms is given in Appendix A and should be reviewed by those that do not have a good working knowledge in the use of Laplace transforms. In general, the input excitation to a control system is not known ahead of time. However,

6、for purposes of analysis it is necessary that we assume some simple types of excitation and obtain system response to at least these types of signals. In general, there are three types* of excitations used in obtaining the response of linear feedback control systems. They are the step input, ramp in

7、put, and the parabolic input. These are typical test or reference inputs. In practice, the input is generally never exactly specifiable.Step input A step input consists of a sudden change of reference input at t = 0. Mathematically it is The function shown in Fig. 4-1a is not defined for t = 0. The

8、Laplace transform of the step input is A/s.Ramp Input (Step Velocity) A ramp input is a constant velocity and is represented as The function is shown in Fig. 4-1b and has a Laplace transform of A/s2.Parabolic Input (Step Acceleration) In this case the input is a constant acceleration, * In many cont

9、rol systems the input may be a sinusoidally varying signal. When this is so, and we know the system is linear, then the output also consists of a sinusoidally varying signal but having a different magnitude and a phase shift which may be functions of the input frequency. We shall consider this in mo

10、re detail in a later chapter. The function is shown in Fig.4-1c and has a Laplace transform of 2A/s3.Fig.4-1 Three test signals for linear feedback control systems In studying the system response of a feedback control system there are three things we wish to know, viz. the transient response, the st

11、eady state or forced response, and the stability of the system. The transient solution yields information on how much the system deviates from the input and the time necessary for the system response to settle to within certain limits. The steady state or forced response gives an indication of the a

12、ccuracy of the system. Whenever the steady state output does not agree with the input, the system is said to have a steady state error. By stability we mean that the output does not get uncontrollably large.4-2 TRANSIENT RESPONSE Consider a closed loop system shown in Fig.4-2. The output and the err

13、or transfer functions are (4-1a) (4-1b)The transient response of the system, be it the error E or the output C, depends upon the roots (also called zeros) of the characteristic equation (4-2) The zeros of the characteristic equation are also the poles of the transfer functions given by Eqs. (4-1a) a

14、nd (4-1b). These poles are known as the closed loop poles. It is interesting to note that the transient response does not depend upon the kind of input but depends only on the zeros of the characteristic equation. Now if the forward and feedback transfer functions are defined as ; then the character

15、istic equation becomes and the zeros are obtained from which correspond to the poles of The right-hand side of the above equation is a ratio of two polynomials where the degree of the denominator is equal to or higher than the order of the numerator. Let us assume that the degree of the denominator

16、is n and of the numerator is v, then if we factorize Eq.(4-1a) it can be written as (4-3)Note that we began with the system transfer function to obtain this expression. It is perhaps interesting to observe that had we begun with a state representation, then which would have been factored to obtain t

17、he right-hand side of Eq.(4-3). When we are interested only in the system transients we need not be concerned with the form of the input since the transients are a function of only the characteristic roots. It is therefore convenient to set R(s) = 1. (Since R(s) = 1 when r(t) is an impulse, it follo

18、ws that the transients may be obtained by applying an impulse to the input of a system.) When the input is included, the transients will not only include the response due to the characteristic roots, but terms in the image of the input and its derivatives. Only these terms will survive as t and yiel

19、d the steady state performance. For obtaining the total transient response we will include the input term. Returning now to Eq.(4-3) we now assume that the input r(t) is a unit step, then R(s) = 1/s and the output becomes (4-4)Let us now assume that of the n distinct poles, 2k poles are complex* and

20、 the remaining poles are real. If we denote the conjugate of sm and Km by and, then Eq. (4-4) may be expanded in partial fractions and written as (4-5) * Complex poles appear as conjugates. where If we denote sm = -, then the output in the time domain is obtained by taking the inverse Laplace transf

21、orm of Eq.(4-5), (4-6)where is the phase contribution of the constant Km. Notice the second term of Eq.(4-6) is obtained by combining two terms. If C(s) has m poles that are equal (i.e. repeat), then (4-7)whereand (4-8)where goes from 1 to m. In general, the response of a system contains terms of th

22、e type given in Eq.(4-6) as well as Eq.(4-8). The important fact here is that the form of the transient response is a function of the location of the closed loop poles, which are identical to the zeros of the characteristic equation, on the s-plane. For real, simple poles the time response is simply

23、 an exponential which decays if the pole is in the left half s-plane and increases with time if the pole is in the right half s-plane. The rate of this decay or increase is dependent upon the magnitude of pole. Poles closer to the imaginary axis are referred to as dominant poles since the decay due

24、to them takes longer. For complex poles the response is oscillatory with the magnitude varying exponentially with time. Again, if the real part is in the left half s-plane, the magnitude decreases with time. If the real part is positive, then the magnitude increases exponentially with time. Finally,

25、 if the poles are real and of multiplicity m, then the time response is of the form. We have not shown the response if the poles are multiple and complex. It is left for you to show that for complex multiple poles the response is of the form Our ideas of this section are consolidated and shown graph

26、ically in Fig.4-3. We note that for well-behaved systems, i.e. systems exhibiting a stable response, it is reasonable to require that the closed loop poles of the control system be located in the left half s-plane. If the poles exist on the imaginary axis they must be simple. Otherwise, the control

27、system responds in such a way that the magnitude of the output becomes uncontrollably large.Fig.4-3 Transient response as a function of the closed loop poles on the s-plane.EXAMPLE 4-1 The forward loop of a unity feedback control system is given by G(s) = K/s(s2+19s+118). Obtain e(t) if K = 240 and

28、the input is r(t) = t.For K = 240 and r(t) = tExpanding this in partial fractions we obtain where s1 = 0, s2 = - 5, s3 = - 6 and s4 = - 8. Now we evaluate the constants, The inverse Laplace transform yieldsEXAMPLE 4-2 The forward loop of a unity feedback control system is given by G(s) = K/s(s+6). I

29、t is desired to vary K from 8 to 13 for the case of a unity step input. Obtain the output for K= 8 and 13 and determine the value of K above which the system exhibits oscillatory behavior. The overall transfer function is given by Substituting for G(s) and simplifying yields The roots of the denomin

30、ator are s1 and s2 whereFor K 9 the roots are real and for K 9 the roots are complex. The system therefore will exhibit oscillations for K 9. (See next example.) When K = 8 the roots becomes1 = - 2, s2 = - 4and for a step inputThe constants are The output for K = 8 becomesC(t) = 1 - 2e- 2t + e- 4tWh

31、en K = 13 the roots become s1 = -3 + j2, s2 = -3 j2and for a step inputThe constants become We note that K2 is the complex conjugate of K3. Whenever the roots are complex conjugates we will find that the constants are also complex conjugates. The output for K = 13 becomes Noting that The output is s

32、een to have an exponentially damped oscillatory term superimposed on a constant term. EXAMPLE 4-3 Obtain the output for K = 9 for the system described in Example 4-2. Assume the input is a unit step. For K = 9, the overall transfer function becomes Since we have repeated roots the partial fraction expansion becomesThe