Train travel between cities will increase as trains are...

Train travel between cities will increase as trains are developed that travel at high speeds, making the travel time from city center to city center equivalent to airline travel time. The Japanese National Railway has a train called the Bullet Express that travels between Tokyo and Osaka on the Tokaido line. This train travels the 320 miles in 3 hours and 10 minutes, an average speed of 101 $\mathrm{mph}[20]$. This speed will be increased as new systems are used, such as magnetically levitated systems to float vehicles above an aluminum guideway. To maintain a desired speed, a speed control system is proposed that yields a zero steady-state error to a ramp input. A thirdorder system is sufficient. Determine the optimum system transfer function $T(s)$ for an ITAE performance criterion. Estimate the settling time (with a $2 \%$ criterion) and overshoot for a step input when $\omega_n=10$.

Key Concepts

ITAE Performance Criterion

The ITAE (Integral of Time-weighted Absolute Error) performance criterion is a method used in control system design to evaluate and optimize the transient behavior of the system. It emphasizes minimizing the weighted error over time, particularly penalizing errors that persist for longer durations, leading to designs with faster settling and reduced oscillations.

Steady-State Error and Ramp Input

Steady-state error refers to the difference between the system output and the desired input as time goes to infinity. For a ramp input, which increases linearly with time, the design goal is often to achieve zero steady-state error. This ensures that the system output accurately tracks the increasing input without lag, which is critical in applications where precise position tracking or speed control is required.

Third-Order System

A third-order system in control theory is characterized by a transfer function with three energy storage elements, typically resulting in three poles in the s-domain. Such systems offer a balance between design simplicity and performance, and in many applications, they are sufficient to meet specified dynamic performance criteria such as rate of response, overshoot, and stability margins.

System Transfer Function

The system transfer function is a mathematical representation that describes the input-output relationship of a control system in the Laplace domain. It encapsulates all the dynamics of the system, including the natural frequency, damping ratios, and gains. It serves as the basis for analyzing system behavior and designing controllers to meet specified performance criteria.

Settling Time

Settling time is a key time-domain performance metric that measures the time required for the system's response to remain within a specified error band (commonly a 2% or 5% band) around the final value. It reflects how quickly a system stabilizes after a disturbance or a step input, indicating the speed of the transient response.

Overshoot

Overshoot is the measure of how much the system response exceeds its final steady-state value during the transient period. It is an important performance indicator, particularly in underdamped systems, because excessive overshoot can lead to instability or undesirable performance in control applications where precise and safe operation is critical.

Transcript

00:01 So here we need to separate it into three different sections of the problem, essentially.

00:06 We have the acceleration part, the constant speed.

00:13 The second part would be the constant speed, and then the third part would be the deceleration.

00:17 So we can find the maximum speed of the train.

00:22 This is going to be 95 kilometers per hour.

00:24 Let's just convert real quick.

00:26 1, multiplied by 1 meter per second for every 3 .6 kilometers per hour.

00:32 This is going to be equal to 26 .39 meters per second.

00:38 At this point, we're going to, we're going to evaluate the acceleration phase.

00:45 So we can say that v final, so let's say part 1, v final equals the initial plus at.

00:55 We know that the initial velocity is going to be equal to zero.

00:58 So t is going to be equal to 26 .39, our final velocity and our maximum velocity of the train, divided by our acceleration of the train.

01:12 So 1 .1 meters per second squared.

01:15 This is giving us 23 .99 seconds.

01:19 So let's not round until the very end.

01:22 Let's find the displacement during this part.

01:26 So we can say that delta x would be equal to v.

01:29 X initial t plus one half a t squared we know that vx initial is zero so we can eliminate that term and this is going to be equal to one half times 1 .1 meters per second squared multiplied by 23 .99 seconds quantity squared this is going to equal 316 .5 meters now we need to find the we need to evaluate now part the deceleration phase.

02:03 So here we can say that for the deceleration phase, we'll say acceleration for part one.

02:20 And then for part three essentially, because this comes after the constant speed phase, we can say deceleration.

02:34 Let's find how long it takes to decelerate.

02:40 So we can say that v final equals v initial plus a t.

02:44 We know that here now v final is zero because we're coming to a stop and t would be equal to negative 26 .39 meters per second divided by our maximum deceleration.

02:57 So this is not the acceleration.

03:01 This would be negative 2 .0 meters per second squared.

03:06 This is equaling approximately 13 point two zero seconds and then let's find the distance traveled during this phase so delta x would be equal to v x initial t plus one half a t squared this would be equal to 26 .39 meters per second multiplied by 13 .20 seconds plus one half times negative 2 .0 .0 seconds plus 1 half times negative 2 .0 .0.

03:43 Meters per second squared multiplied by 13 .20 seconds quantity squared.

03:51 And we find that here, delta x is going to equal 174 .1 meters.

04:00 Now, let's get a new workbook.

04:03 We want to find the total elapsed time.

04:06 So t of acceleration plus t of deceleration would be simply equal to 23 .99 seconds plus 13 .20 seconds, and this is equaling 37 .19 seconds.

04:22 And then for the total distance, this would simply be equal to 316 .5 plus 174 .1.

04:36 This is equaling 491 meters.

04:41 Now, for part a, we want to find, rather, we know that the stations are spaced 1 ,800.

04:50 Meters apart or 1 .8 kilometers...