Automatic Control Principle Question, Urgent 50

It is difficult for you to do this question directly for undergraduates, because it is a third-order problem, and the textbooks used by undergraduates generally talk about the dynamic and static performance of analysis are usually second-order questions, and even the postgraduate examination questions of Tsinghua University are second-order time domain analysis questions, because the second-order questions can be used to calculate dynamic and static parameters with ready-made formulas, as long as there is a transfer function, you can calculate the rise time, peak time, overshoot and other dynamic performance, but the third-order I turned various textbook reference books do not have a clear way to do the questions. Even the fifth edition of Hu Shousong (with more than 700 pages) of the version of the third-order time domain analysis is directly done with MATLAB, so, alas, it can only be said that the landlord does not have to be too entangled in this kind of deviation, I sent out the logarithmic frequency characteristic diagram and the time domain diagram, all of which are made with MATLAB, the input of the time domain diagram is a unit step, but it is a bit oscillating, I don't know if there is a problem, the amplitude and frequency must be right.

If you can draw an approximate curve, first calculate that the handover frequency is w1=1, w2=10, and the slope of the low-frequency asymptote is -20db dec, which is composed of an integral link and two inertial links, and several links are connected by the superposition method. The inertia link decreases by 20db dec every time it passes a handover frequency, and you can try to draw it after (1,40db) and (10,20db).

Summary. As shown in the figure above, according to the l(w) diagram, the cut-off frequency is between 1 and 5, so for the second term in the denominator, wc is larger than 1, so the important component is wc, and this link takes it. In the third link, since (wc 5) is compared with 1, the value of 1 is large, so 1 is taken in this link.

Finally, make the overall formula 1

Please send me the content of the question, thank you for helping you answer.

Because the phase angle margin of the original system is not enough to 50, the series advance correction option (2) is used

According to Grandpa Lu's steps, you can refer to his example questions.

To do this kind of problem, the steps are very fixed, the open-loop gain k is determined by the steady-state accuracy requirements, and then the BODE diagram of the system before the correction is drawn, according to the system requirements of the problem, take a look at the index, and find that the original truncation frequency is lower than the required ruler, and the phase angle margin is low, oh, at this time, the first thing to consider is to use the advanced correction.

You can solve it by referring to this method.

The pre-corrected a is greater than 1.

Approximate cut-off frequency.

As shown in the figure above, according to the L(W) diagram, the cut-off frequency is between 1 and 5, so for the second term in the denominator, wc is larger than 1, so the weight of the Nakai family is wc, and this link takes it. In the third link, because (wc 5) and 1 are compared, the disadvantage value of 1 is large, so 1 is taken in this link. Finally, make the overall formula 1

Principles of automatic control" Northwestern Polytechnical University Lu Jing Eyes Tide (to black edge) - Beep Xiye noisy ridge like Bilibili].

Let the signal x between g1g2 and the signal y y of auspicious and bright h1 be burned.

x=(r+y)g1,y=(x+c*h3+y*h2))h1,c=x*g2+g3*r.

It is easy to eliminate x and y to obtain the transfer function.

The meaning of the title is that the input signal is a velocity signal of 12t and a steady-state error of 2

So k=12 2=6

The rest is solved according to the example steps in the book.

That is, on the amplitude and phase characteristic diagram:

The starting point is infinity from amplitude and the amplitude angle is -90°

In this case, w=0 , and in this case, the real part is -3 4, and the imaginary part is -, so the amplitude is , and the amplitude angle is arctan(2-w w) Therefore, the amplitude angle is -90°

The endpoint to the amplitude is 0 and the amplitude angle is -270°

At this time, the real part is 0 and the imaginary part is also 0, then the amplitude is 0, and the amplitude angle is arctan(2-w w), so the amplitude angle is -270°

It looks like this.

The starting point is the point where the amplitude of the line is infinity at -90°.

I can only pass a **, so I simply spelled it, intercepted it again, tried it, and I saw it clearly! Hope it helps!

It's all basic.

The equation of the integral is logarithmic amplitude = 20lgk-20lgw, 20lgk-20lg10 = 0, and k=10 is obtained

Transfer function 10 [s(s ;

w=, knowing the slope of a point, write the equation of the middle segment of the straight line, and find wc;

According to the transfer function, wc to find the phase angle margin...

1. Feedback control, also known as deviation control, is carried out by the difference between the input quantity and the feedback quantity.

2. The closed-loop control system is also known as the feedback control system.

3. The main mathematical models used in classical control theory are differential equations, transfer functions, structural block diagrams and signal flow diagrams.

4. The automatic control system can be divided into constant value control system, follow-up control system and program control system according to the change law of input.

5. The basic requirements for the automatic control system can be summarized into three aspects, namely: stability, rapidity and accuracy.

6. The mathematical model of the control system depends on the system structure and parameters, and has nothing to do with the external action and initial conditions.

7. The two transfer functions are the links of g1 (s) and g2 (s) respectively, which are connected in parallel mode, and their equivalent transfer functions are g1 (s) + g2 (s), and connected in series mode, and their equivalent transfer functions are g1 (s) * g2 (s).

8. The forward channel transfer function of the system is g(s), and the transfer function of its positive feedback is h(s), then its closed-loop transfer function is g(s) (1- g(s)h(s)).

9. If the forward channel transfer function of the unit negative feedback system is g(s), then the closed-loop transfer function is g(s) (1+ g(s)).

10. In a typical second-order system, =, the system is said to be in the best state of the second-order project, and the overshoot is .

11. The Rouse criterion is used to judge the stability of the system, and the data in the first column of the Rouse table are all positive, then the system is stable.

12. The sufficient and necessary condition for the stability of linear systems is that the real part of the root of all closed-loop eigenequations is negative, that is, they are all distributed in the left plane of the s-plane.

13. The steady-state error of the follow-up system is mainly higher than that of the given signal, and the steady-state error of the constant-value system is mainly higher than that of the disturbance signal.

14. For systems with steady-state errors, the systematic error will become zero in the series proportional integration link in the forward channel.

15. The steady-state error of the system is divided into two types: given steady-state error and disturbance steady-state error.

16. For a system with steady-state error, the steady-state error will decrease if the system gain is increased.

17. For a typical second-order system, the greater the inertial time constant t, the worse the rapidity of the system.

18. Using the frequency domain analysis method, the larger the traversal frequency, the smaller the corresponding time domain index TS, that is, the better the rapidity.

19. The minimum phase system refers to the open-loop pole and open-loop zero point of the system that do not exist in the right half plane of the s.

20. According to the different positions of the calibration device in the system, the system calibration can be divided into four types: series calibration, feedback correction, compensation correction and compound correction.

21. For linear systems, the larger the phase margin, the better the relative stability of the system.