Question

please use MATLAB, screenshot the code and the plots. Answer all parts.

Laboratory description: The speed control of a high-speed train is represented by the system shown in Figure 1. The transfer function of the train dynamics is as follows. G(s) = 2 + 10s + 20 Laboratory assignment: Part A. For the open-loop system described by the following block diagram, assume that the input r(t) is a unit step. (1) Compute the steady-state system output Yss. (2) Find the steady-state error egs (3) Make a plot for the system output y(i) by using MATLAB sa +10s + 20 Figure 1 A high speed train system Part B. Consider the closed loop control system shown below 0 D(6) | 2 + 10 + 20 Figure 2 PID closed loop control system

Answer 1

Suppose we have a simple mass-spring-damper system.

"Mt9 | m

The governing equation of this system is

(3)

më + bc + kc = F

Taking the Laplace transform of the governing equation, we get

(4)

ms-x(s) + b3X(s) +kX() = F(8)

The transfer function between the input force and the output displacement then becomes

(5)$$$ \frac{X(s)}{F(s)} = \frac{1}{ms^2 + bs + k} $$$

Let

    m = 1 kg
    b = 10 N s/m
    k = 20 N/m
    F = 1 N

Substituting these values into the above transfer function

(6)$$$ \frac{X(s)}{F(s)} = \frac{1}{s^2 + 10s + 20} $$$

The goal of this problem is to show how each of the terms, $$K_p$$ , $$K_i$$ , and $$K_d$$ , contributes to obtaining the common goals of:

• Fast rise time
• Minimal overshoot
• Zero steady-state error

Open-Loop Step Response

Let's first view the open-loop step response. Create a new m-file and run the following code:

s = tf('s');
P = 1/(s^2 + 10*s + 20);
step(P)

Step Response 0.05 0.045 0.04 0.035 0.03 Amplitude 0.02 0.015 0.01 0.005 0 0. 5 1 2 2.5 3. 1.5 1 Time (seconds)

The DC gain of the plant transfer function is 1/20, so 0.05 is the final value of the output to a unit step input. This corresponds to a steady-state error of 0.95, which is quite large. Furthermore, the rise time is about one second, and the settling time is about 1.5 seconds. Let's design a controller that will reduce the rise time, reduce the settling time, and eliminate the steady-state error.