I. The concept of PWM and duty-free ratio
①PWM: Also known as pulse width modulation technology, is an analog control mode, the PWM waveform as shown below.
② duty cycle: refers to the proportion of the high level in a period. For example, a duty cycle of 50% means that the high level accounts for half of the period.
Two, PWM fixed frequency speed regulation principle
In the PWM speed regulation syste

"Electromechanical Drive Control"--DC motor speed Regulation simulation operationHu Hengxiang Machine Zhuo 1301 U201310563First, the PID of each adjustment link introductionIn the PID adjustment, the function of each part is as follows:Proportional regulation : Proportional control is one of the simplest control methods. The output of its controller is proportional to the input error signal. is proportional

The principle of PWM voltage regulation and speed regulation for DC motorThe expression of the DC motor Speed N is: n=u-ir/kφThe speed control method of DC motor can be divided into two types: the excitation control method of regulating excitation flux and the armature control method of regulating armature voltage. The excitation control method at low speed by the limit of magnetic pole saturation, at high

By the original DC motor speed control example can be seen in the stability of the current is not good, to achieve a long time stability, overshoot is large, steady-state error is not small enough, the oscillation is obvious.The original controller only proportional control, very coarse, when the gain is low, the steady state error is large, when the gain becomes larger, it will cause the motor current and acceleration oscillation.After considering the decision with PID adjustment, three adjustm

divert my attention. There is a good word in idealistic thought, "If you don't look at the moon, the moon will not exist." "Tamper with unauthorized, it becomes:" You do not look irritable, irritability does not exist. ”All things in the world are rooted, there are branches to depend on. The source of irritability comes from the conflict between our stated goals and the realities of something. Take tonight this irritability me, the established goal is not realistic cooperation began to irritabi

DC Motor Speed Regulation?for a given model, just modify the Controller part of the simulation purposes can be achieved, the previous code only proportional link, no integral and differential links, so need to increase these two links, here set kp=8 , Ki=1 , kd=60 , the resulting simulation waveform isIt can be seen that the acceleration and deceleration time is very short, rapid reaction, the overshoot is not small, the speed curve has a very low dev

China Insurance Regulatory Commission regulates Internet Insurance Regulation: whenever possibleOn September 11, Chen Wenhui, vice chairman of the China Insurance Regulatory Commission, said at the opening ceremony of sunshine Internet Financial Innovation Research Center, National Financial Research Institute of Tsinghua University, for internet finance (http://www.csix.cn /), the attitude of the Regulatory layer can be opened up as much as possible.

Read the Lujiaxin blog, feel good waves ah .... I'm sorry.Brush the question when the time is not enough, each algorithm is written only one or two passes, so how can progressDon't always complain that time is too little, you do not work hard >_Well, look at the problem.NF[a][b][t] Indicates that there is no shortest answer to M a~b in this interval.Enumerate breakpoints (very important idea), when T==1, apparently on the right side of the left are DP a bit; (think clearly)Then, see if the left

] What is the end result? Is the length of the F-list, which is the M-value after the final change. If we look at the loop one by one, there is the time complexity of O (M), but because of the monotonicity, we can find this x using the binary lookup algorithm, so the time complexity here is O (LOGM), because Mwe got a faster algorithm. Consider how to find a specific subsequence? (Hint: It's still a "record" decision, and it's OK to record it when X is found.) )The following:#include #includeusi

-SPAN-17" class= "Mi" >i− 1, J) (jai ) /span> {MAX(F(I−1,J),F(I,J−AI)+ai ) Span id= "mathjax-span-60" class= "Mi" >ai j [ Sum/2]) Span id= "mathjax-span-75" class= "Mo" > So what's the final answer? By definition, it should be f (n, [SUM/2]).Time complexity and spatial complexity are still O (n * [SUM/2]), which can also optimize the spatial complexity of one dimension.In summary, the time complexity of both methods are O (n * sum) level, usually sum is not very l

one-dimensional space complexity, so as to achieve O (n). optimized pseudo-code: forj =0to n DoF[j]=jendfor fori =1to M Do Last= f[0] f[0] =I forj =1to n DoTemp=F[i,j] F[i,j]= min (last + Same (i,j), temp +1, f[j–1] +1) Last=Temp endforendforNote: Our order for I actually update J is by small arrival, so we need to save "old" f[i-1,j–1].Exercises#include #include#includeusing namespacestd;Chara[1010],b[1010];intf[1010][1010];intMain () {CIN>>A; CIN>>b; intm=strlen (a); intn=strlen (b); for(int

Of course, first of all, the speed of the PID adjustment, talk about the speed of the feedback and speed sensor connectionThe current still has a certain impact, but the adjustment of Ki and KD eliminates most of the steady-state errors and adjusts the softness of the initial rise current.In the current loop compound adjustment, the same PIDWhen the current loop plays a relatively small value, the effect of a good current is basically no oscillation, but the linear part of the rotational speed b

-workers" refers to the same position, with pay, with the proceeds; the so-called "equal remuneration" means that the same remuneration, including wages, insurance and benefits, should be enjoyed under the premise of working together.(c) The trade unions that are dispatched to participate in the employment of the workers may also organize their own unions. Therefore, the employment Unit to this provision should be sent through the dispatching agreement and the specific agreement, to be implement

Title Link: http://poj.org/problem?id=1458This is a template problem of the longest common subsequence;#include Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced. Poj-1458-common subsequence-the longest common sub-sequence of motion regulation

Test instructions: input n, m, p then n Cow Tower coordinates, m-haystack coordinates, p-order opportunity.Each instruction can move up or down One direction, allowing all Nouta to be moved one unit at a time.Then every haystack/Ox tower, there are 1 weights.The maximum value of the weighted value and the maximum weight of the moving instruction sequence (minimum dictionary order).F[i][j][k] Represents the first I, moving sequence x axis coordinates of J,y axis K, the maximum weight.Then do it b

Test instructions: There are a number of numbers that can then be used to cost me a number +i or-I.Now ask you to put the sequence into a non-ascending or non-descending, asking the minimum price.ExercisesThe first thing that can be proved is that the last number of the best spent can be a number now:Card: If the two number is adjusted in the middle of the two number, then you can change the two number to one of the number, and the price is obviously equal.This is good to do after it comes out.W

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# Include Question:
First, a vertex can be split into multiple new vertices, thus having the foundation of the motion regulation on the graph.That is, F [I] indicates the minimum cost of point I being eliminated. It can be updated by Split points.
However, this is post-efficient, so we use spfa to handle it.Spfa post-effect dynamic regulation
Every time we update the dynamic value of point A,

namespacestd;intk[10000]={0};intMainintargcChar Const*argv[]) { intM; CIN>>M; intAns =0;//The last man was the only person in the last Joseph Ring. for(inti =1; I 1; ++i) {cin>>K[i]; } //starting from the penultimate Joseph Ring, a total of M-Joseph rings for(inti =2; I i) {ans= (ans+k[m-i+1])% i;//at this time, I have a number of people in the last game k[m-i+1] only to the ANS//using 0 To identify the initial position can effectively avoid the problem of 0 after taking the remai

[j])M (i, j) = Max (M (I-1, J), M (I, j-1)) (when a[i]! = B[j])The first is a good proof, that is, when a[i] = B[j]. Can be disproved, assuming M (i, J) > m (i-1, j-1) + 1 (M (i, J) cannot be less than M (i-1, j-1) + 1, the reason is obvious), then can be introduced m (I-1, j-1) is not the longest of this contradictory result.The second one has some trick. When a[i]! = B[j], it is still a rebuttal, assuming M (i, J) > Max (M (I-1, J), M (I, j-1)).From the counter-disprove hypothesis, M (i, J) >

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