Cruise Control System With Time Delay — Simulation & Power Modeling

Abstract — Cruise Control System With Time Delay

possible to create models to study the relationship between control structure, process response, and control action. The purpose of this notebook is to introduce for the analysis of block diagrams made up of linear transfer functions. Let's imagine you own a car with a mass of 1,500 kg (including passengers) that is traveling down a level stretch of highway at a desired speed of 100 kilometers per hour. You wish to design a cruise controller that will adjust the throttle position $u$ within the limits 0 to 1 in order to maintain constant speed.

Analytical & State-Space Control Methodology

Figure 2: Power Converter Circuit Schematic & Electrical Interfacing for Cruise Control System With Time Delay The disturbance to the system is a change in road grade measured in percent. (A one percent grade corresponds to 1 foot of rise in 100 feet of horizontal travel, or a tangent of 0.01). The dynamics are represented by three transfer functions denoting the response of the engine torque to throttle position, We assume that a one percent increase in grade introduces 150 Newtons of force in opposing the car acceleration, that is Chapter 2: Semiconductor Dynamics: MOSFET & IGBT Switching Characteristics & Losses The controller is assumed to be a proportional-integral control written as Internal model control is a version of feedback control that incorporates an explicit process model. A simple block diagram of internal model control is given by There are two designated inputs, the reference $w$ and disturbance $y_d$, and three labeled signals $u$, $y$, and $d$ that appear relevant to closed-loop performance. Here's the solution for all six closed-loop transfer transfer functions: < 5.1 Getting Started with Transfer Functions | Contents | Tag Index | 5.3 Creating Bode Plots > Example: Solution to the Cruise Control Problem Using State Space Control design using pole placement Reference input The state equations for this problem are: where m=1000 kg b=50 N*sec/kg u=500 N v=velocity y=output The design criteria are: Rise time < 5 sec Overshoot < 10% Steady state error < 2% To see the original problem setup , see Cruise Control Modeling page. Control design using pole placement The schematic of a full-state feedback system is shown below. where K=Control matrix U=-Kv=input R=Reference Recall from the State-Space Tutorial page, we should use the technique called "pole placement" to obtain the desired output. Poles of a closed-loop system can be found from the characteristic equation: the determinate of [sI-(A-B*K)] matrix. If desired poles can be placed into the system by designing right control matrix (K), then the desired output can be obtained. In this tutorial, poles will be chosen first, then use Matlab to find the corresponding control matrix (K). Now, we need to determine where to place poles for our system. Since our [sI-(A-B*K)] matrix is 1x1, we have only one pole to place. Let the pole to be at -1.5 (arbitrary). Just as in the State-Space Tutorial, the Matlab function called place will be used to find the control matrix K . Create an new m-file and enter the following commands. m=1000; b=50; t=0:0.1:10; u=500*ones(size(t)); A=[-b/m]; B=[1/m]; C=; D=; x0=; p1=-1.5; K= place (A,B,[p1]) A1=A-B*K; lsim (A1,B,C,D,u,t,x0); Running this m-file in the Matlab command window should give you the control matrix and the following step response. As you can see, the rise time is satisfactory, but the steady-state error is too large. Reference input Once again from the State-Space Tutorial , scaling factor called Nbar (the schematic is shown below) should be used to eliminate the steady-state error. Unlike the example in the Tutorial, the command rscale is not applicable for our system. Nbar needs to be determined manually.

Simulink Waveforms & Power Quality Findings

After several trial-and-error runs, the Nbar equals 30 provided the desired step response. Copy the following commands to an m-file and run it in the Matlab command window. You should get the step response shown below. m=1000; b=50; t=0:0.1:10; u=500*ones(size(t)); A=[-b/m]; B=[1/m]; C=; D=; x0=; p1=-1.5 K= place (A,B,[p1]); Nbar=30; A1=A-B*K; lsim (A1,B*Nbar,C,D,u,t,x0); As you can see, the steady-state error has been eliminated. The rise time is less than 5 seconds and the overshoot is, in fact, zero. All the design requirements are satisfied. User feedback State Space Examples Cruise Control | Motor Speed | Motor Position | Bus Suspension | Inverted Pendulum | Pitch Controller | Ball and Beam Cruise Control Examples Modeling | PID | Root Locus | Frequency Response | State Space | Digital Control: RL Tutorials Basics | Modeling | PID | Root Locus | Frequency Response | State Space | Digital Control | Examples 8/26/97 DK Example: Solution to the Cruise Control Problem Using State Space Control design using pole placement Reference input The state equations for this problem are: where m=1000 kg b=50 N*sec/kg u=500 N v=velocity y=output The design criteria are: Rise time < 5 sec Overshoot < 10% Steady state error < 2% To see the original problem setup , see Cruise Control Modeling page. Control design using pole placement The schematic of a full-state feedback system is shown below. where K=Control matrix U=-Kv=input R=Reference Recall from the State-Space Tutorial page, we should use the technique called "pole placement" to obtain the desired output. Poles of a closed-loop system can be found from the characteristic equation: the determinate of [sI-(A-B*K)] matrix. If desired poles can be placed into the system by designing right control matrix (K), then the desired output can be obtained. In this tutorial, poles will be chosen first, then use Matlab to find the corresponding control matrix (K). Now, we need to determine where to place poles for our system. Since our [sI-(A-B*K)] matrix is 1x1, we have only one pole to place. Let the pole to be at -1.5 (arbitrary). Just as in the State-Space Tutorial, the Matlab function called place will be used to find the control matrix K . Create an new m-file and enter the following commands. m=1000; b=50; t=0:0.1:10; u=500*ones(size(t)); A=[-b/m]; B=[1/m]; C=; D=; x0=; p1=-1.5; K= place (A,B,[p1]) A1=A-B*K; lsim (A1,B,C,D,u,t,x0); Running this m-file in the Matlab command window should give you the control matrix and the following step response. As you can see, the rise time is satisfactory, but the steady-state error is too large. Reference input Once again from the State-Space Tutorial , scaling factor called Nbar (the schematic is shown below) should be used to eliminate the steady-state error. Unlike the example in the Tutorial, the command rscale is not applicable for our system. Nbar needs to be determined manually.

Scope & Practical Applications

• High-efficiency DC-DC converters and isolated resonant topologies (LLC / DAB)
• Sinusoidal and Space Vector PWM inverters for grid integration & industrial motor drives
• MOSFET / IGBT gate driver optimization with active Miller-clamp shoot-through protection
• Power quality conditioning with Total Harmonic Distortion (THD) strictly compliant with IEEE 519
• Complete Simscape .SLX model files, parameter calculation scripts, and IEEE dissertation reports