control system lab report
DESCRIPTION
It contain all performances of control system lab,which were verified on Matlab .TRANSCRIPT
Lab Report
Control System
Submitted to:
Mr. Usman
Submitted by:
Junaid Mukhtar
2012-MC-139
Lab Session 1
Introduction of MATLAB and its application in control
system design and analysis.
MATLAB:
MATLAB (matrix laboratory) is a numerical computing environment
and fourth-generation programming language. Developed by MathWorks,
MATLAB allows matrix manipulations, plotting of functions and data,
implementation of algorithms, creation of user interfaces, and interfacing with programs written in other languages, including C, C++, Java, and Fortran.
Although MATLAB is intended primarily for numerical computing, an
optional toolbox uses the MuPAD symbolic engine, allowing access to symbolic
computing capabilities. An additional package, Simulink, adds graphical
multidomain simulation and Model-Based Design fordynamic and embedded systems.
Application of Matlab:
Mathematical operations
Vector
Matrices and Matrix operations
Solving systems of linear equations
Plotting
Curve fitting, Interpolation
Solving Ordinary Differential Equations
Compiling MATLAB and Toolboxes: MATLAB Compiler supports the full MATLAB language and most MATLAB
toolboxes, with some exceptions. For details on compliable features and products.
Control System Toolbox:
Control System Toolbox™ provides industry-standard algorithms and
apps for systematically analyzing, designing, and tuning linear control systems.
You can specify your system as a transfer function, state-space, pole-zero-gain,
or frequency-response model. Apps and functions, such as step response plot
and Bode plot, let you visualize system behavior in time domain and frequency
domain. You can tune compensator parameters using automatic PID controller
tuning, Bode loop shaping, root locus method, LQR/LQG design, and other
interactive and automated techniques.
Three Steps:
There are three major steps which are followed in Control System design.
(1) Building Models:
In building models there are two systems
SISO system
MIMO system
SISO Systems:
SISO means "single input single output" systems e.g. dc motor speed
control. In which we give step input voltage to motor and resulting speed control
of dc motor rotor.
MIMO Systems:
MIMO means" multiple input multiple outputs" systems e.g. jet transport
air craft. All standard model representations are supported, including;
Transfer function
State Space Model
Zeros, poles and gains etc.
Control System Toolbox provides commands for:
Performing arithmetic on linear models
Building complex block diagrams by connecting simple models in
series, parallel, or feedback
Discretizing continuous-time models
Computing low-order approximations of high-order models
Consider a block diagram:
(2) Analyzing Models:
We can analyze the model by LTIview (Linear time invariant) and
GUI (Graphical user interface). In which we can analyze the model
behavior by plotting different graphs e.g.
Step Response
Impulsive Response
Bode diagram
Nyquist Plot
By comparing the time and frequency responses of several linear models
at once we can analyze the system behavior for different parameters.
(3) Designing Compensator:
How we build compensators using functions from the Control System Toolbox.
We can build SISO system by
Roots locus method
Bode diagram (compensator)
Pole placement technique (for MIMO)
Controller design can be validated by verifying rise time, overshoot, settling time,
gain and phase margins, and other requirements.
Lab Session 2
To illustrate control system characteristics with the help of
MATLAB by introducing a control to speed Techo-generator
system.
DC Motor:
The DC motor is a power actuator device that delivers energy to a load, as
shown in Figure2.18 (a); a sketch of a DC motor is shown in Figure2.18 (b). The
DC motor converts direct current (DC) electrical energy in to rotational mechanical
energy. A major fraction of the torque generated in the rotor (armature) of the motor
is available to drive an external load.
The transfer function of the DC motor will be developed for a linear
approximation to an actual motor, and second order effects, such as hysteresis
and the voltage drop across the brushes, will be neglected. The input voltage
may be applied to the field or armature terminals. The air gap flux of the motor
is proportional to the field current, provided the field is unsaturated, so that
The torque developed by the motor is assumed to be related linearly to
and the armature current as follows:
Then we have, in Laplace transform notation,
Where = is a constant armature current, and is defined as the motor constant. The
field current is related to the field voltage as,
The motor torque is equal to the torque delivered to the load. This relation may be
expressed as,
As,
The block diagram model of the field controlled DC motor is shown in Figure2.19.
Alternatively, the transfer function may be written in terms of the time constants of the
motor as,
Where, and, , and
Armature controlled DC motor:
The armature controlled DC motor uses the armature current as the control
variable. The stator field can be established by a field coil and current or a permanent
magnet. When a constant field current is established in a field coil, the motor torque is
Where is a function of the permeability of the magnetic material. The armature current
is related to the input voltage applied to the armature by
Where is the back electromotive force voltage proportional to the motor speed.
Therefore, we have
Where the transform of the angular speed and the armature is current is,
Open loop control of DC motor:
Ra=1; Km=10; J=2; f=0.5; Kb=0.1; b=0.5; num1=[1];
den1=[J,b];
sys1=tf(num1,den1);
num2=[Km*Kb/Ra]; den2=[1];
sys2=tf(num2,den2)
sys_o=feedback(sys1 ,sys2);
sys_o=-sys_o
[yo,T]=step(sys_o);
plot(T,yo) title('Open-Loop Disturbance Step Response') xlabel('Time (s)'),ylabel('\omega_o'), grid yo(length(T))
Output:
Transfer function:
-1
---------
2 s + 1.5
ans = -
0.6650
Closed loop control of DC motor
(Techogenerator system):
clc; clear all; Ra=1; Km=10; J=2; f=0.5; Kb=0.1; b=0.5; Kt=8; for(i=1:10); Ka=i; num1=[1]; den1=[J,b]; sys1=tf(num1,den1); num2=[((Ka*Km*Kt*Ka)+(Ka*Km*Kt*Kb))/(Ra*Ka)]; den2=[1]; sys2=tf(num2,den2); sys=feedback(sys1 ,sys2); step(sys); hold on; end
Output:
Transfer function:
1
----------
2 s + 808.5
Comments:
As we can see from the graph that in closed loop system the response meet with
x-axis but in open loop system the response meet with x-axis after a little bit of
time. It takes more time to touch real axis for open loop system as compared to
closed loop system.
Open loop system response is quicker than the system response of closed loop
system.
Lab Session 3
To determine the performance of a second order system
& adjust the system’s performance with the variation of
parameter.
Second Order System:
Let us consider a single loop second order system and determine its
response to a unit step input. A closed loop feedback control system is shown in
Figure. The closed loop system is,
With a unit step input, we obtain,
For which the transient output, as obtained from the Laplace transform is,
Where, , and ,
Damping ratio (ζ):
The damping ratio plays a fundamental role in closed loop system
performance. As seen in the design formulas for settling time, percent overshoot,
peak time, and rise time, the damping ratio is a key factor in determining the
overall performance. Infect, for second order systems, the damping ratio is the
only factor determining the value of the percent overshoot to a step input. As it
turns out, the damping ratio can be estimated from the response of a system to a
step input
Over-damped (ζ > 1): The system returns (exponentially decays) to
equilibrium without oscillating. Larger values of the damping ratio ζ return
to equilibrium more slowly.
Critically damped (ζ = 1): The system returns to equilibrium as quickly as
possible without oscillating. This is often desired for the damping of
systems such as doors.
Under-damped (0 < ζ < 1): The system oscillates (at reduced frequency
compared to the un-damped case) with the amplitude gradually
decreasing to zero.
Un-damped (ζ = 0): The system oscillates at its natural resonant
frequency ( ).
Part 1: Effect of Damping ratio “Zeta” on system
performance
clc t=[0:0.1:12];num=[1]; zeta=0.3:0.1:1.2; zeta(9)=2;
for(i=1:9) sys(i)=tf(1,[1 2*zeta(i)1]); end
[y1,T1]=step(sys(1),t);
[y2,T]=step(sys(2),t);
[y3,T3]=step(sys(3),t);[y4,T4]=step(sys(4),t); [y5,T5]=step(sys(5),t);[y6,T6]=step(sys(6),t); [y7,T7]=step(sys(7),t);[y8,T8]=step(sys(8),t); [y9,T9]=step(sys(9),t)
plot(T1 ,y1 ,T2,y2,T3,y3,T4,y4,T5,y5,T6,y6,T7,y7,T8,y8,T9,y9); xlabel('\omega_n t'), ylabel('y(t)')
legend('underdamped:zeta<1','overdamped:zeta>1','criticallydamped:zeta=0','un
damped:zeta=1 ')
Graph:
Assignment: clc
t=[0:0.1:12]; a=0.5:0.5:3; for(i=1:length(a)) sys(i)=tf([a(i)*a(i)],[1 2*0.4 a(i)*a(i)]); end ltiview %import system “sys” in ltiview % or [y1,T1]=step(sys(1),t);[y2,T2]=step(sys(2),t); [y3,T3]=step(sys(3),t);[y4,T4]=step(sys(4),t); [y5,T5]=step(sys(5),t); % plot(T1 ,y1 ,T2,y2,T3,y3,T4,y4,T5,y5); xlabel('\omega_n t'), ylabel('y(t)') title('\omega_n t=0.5, 1.0, 1.5. 2.0, 2.5'), grid
Comments:
As we can see from the first graph by decreasing the damping ratio (ζ) the over
shoot is less.
From second graph, it is clear that by increasing the value of natural frequency ( ) the
response is quicker but over shoot is increase.so varying value of wn and zeta both effect
the system response.so these two are the important factor which effect P.O,Tr,Ts.
Perameters of system is totally dependent of these two variables.(wn,zeta)
Lab Session 4
To determine the steady state error of a mobile robot steering
control.
Part 1: Steady state response to a triangular wave input (kind of
ramp input)
clc; K1=1; K2=2; K=1; a=1/10; numg=[K1*K K*K2]; deng=[a 1 0]; sysg=tf(numg,deng); [sys]=feedback(sysg, [1 ]); t=[0:0.1:8.2]'; v1=[0:0.1:2]; v2=[2:-0.1:-2]; v3=[-2:0.1:0]; u=[v1 v2 v3]; [y,T]=lsim(sys,u,t); plot(T,y,t,u,'-'), xlabel('Time (s)'), ylabel('\theta (rad)'), grid
Output:
Assignment:
clc syms s d1 d2 ; a=input('enter numinator:') b=input('enter denumerator:') M=d2*s^2+d1*s+1 m1=diff(M,s) m2=diff(m1,s) m3=diff(m2,s) m4=diff(m3,s) %%%%% putting s=0 in M and its derrivatives m0=limit(M,s,0) m10=limit(m1,s,0) m20=limit(m2,s,0) m30=limit(m3,s,0) m40=limit(m4,s,0) %%%%% finding M2 and M4.................. M2=(-(m0*m20)/2)+((m10*m10)/1)-((m20*m0)/2) M4=((m20*m20)/(4)) D=tf(a,b); %%D=((1)/(114))*s^3+((11.4)/(114))*s^2+((14)/(114))*s+1 d1=diff(D,s) d2=diff(d1,s) d3=diff(d2,s) d4=diff(d3,s) %%%%% putting s=0 in D and its derrivatives d0=limit(D,s,0) d10=limit(d1,s,0) d20=limit(d2,s,0) d30=limit(d3,s,0) d40=limit(d4,s,0) %%%%% finding D2 and D4.................. D2=(-(d0*d20)/2)+((d10*d10)/1)-((d20*d0)/2) D4= -2*(d10*d30/6)+((d20*d20)/(4)) f=D2-M2 g=D4-M4 %%%%%% calculating d1 and d2 from (f and g)equation [d1,d2]=solve(f,g) d1=sym2poly(abs(d1(1))) d2=sym2poly(abs(d2(1))) den11=1+d1*s+d2*s^2 p=sym2poly(den11); num=1; den=p; X=sym2poly(D) disp('orignal transfer') sys1=tf(num,X) disp('approximate 2nd order system') sys=tf(num,p) t=0:.01:4; [y,t]=step(sys1,t); plot(t,y); hold on [y1,t1]=step(sys,t);
plot(t1,y1);
Output:
Assignment: (General code) function RouthApprox=Routh_Approximation(G,r) Function RouthApprox=Routh_Approximation(G,r)
% it Computes the r-th order Routh Approximation of a given n-th order % transfer function G, with 1<=r<=n.
clc; G=tf([3 1 2 4],[1 3 4 8 5 4]) r=3; R=Routh_Approximation(G,r)
% system Approximation if mod(n,2) A=[den1(1:2:end);den1(2:2:end)]; B=[num1(1:2:end);num1(2:2:end)]; else A=[den1(1:2:end);den1(2:2:end) 0]; B=[num1(1:2:end);num1(2:2:end) 0]; end gam(r)=0;del=gam; gam(1)=A(1,1)/A(2,1); if gam(1)<=0 disp('System Unstable. Routh Approximation does not exist'); RouthApprox=0; return end for i=3:r+1
for j=1:(size(A,2)-1) A(i,j)=A(i-2,j+1)-gam(i-2)*A(i-1,j+1); end gam(i-1)=A(i-1,1)/A(i,1); if gam(i-1)<=0 disp('System Unstable. Routh Approximation does not exist'); RouthApprox=0; return end end
del(1)=B(1,1)/A(2,1); for i=3:r+1 for j=1:(size(A,2)-1) B(i,j)=B(i-2,j+1)-del(i-2)*A(i-1,j+1); end del(i-1)=B(i-1,1)/A(i,1); end
P_1=0; P_2=del(1);
Q_1=1; Q_2=[1 gam(1)];
if r==1 P=P_2; Q=Q_2; end
for i=3:(r+1) if i>3 P=del(i-1)*[1 zeros(1,(i-2))]+conv([1 0 0],P_1)+[0 gam(i-1)*P_2]; else P=del(i-1)*[1 zeros(1,(i-2))]+[0 gam(i-1)*P_2]; end Q=conv([1 0 0],Q_1)+[0 gam(i-1)*Q_2]; P_1=P_2;P_2=P; Q_1=Q_2;Q_2=Q; end [P,Q]=tfdata(tf(P,Q)+D_fact,'v'); P=P.*(abs(P)>1e-6); Q=Q.*(abs(Q)>1e-6); RouthApprox=tf(P,Q)+D_fact; t=0:.001:2; [y,t]=step(G,t); plot(t,y); hold on [y1,t1]=step(R,t); plot(t1,y1);
Output:
Comment: In the graph of robot the green line is required response but the blue line is actual
response.
The difference between these responses is due to steady state error. So with
approximation we cant get exact the required system response but we just get a
system response near to original required response .This method give us easy
way to analyze the system with a simple approximate system transfer function.
Lab Session 5 Stability analysis of feedback system using MATLAB and also
to draw the ROUTH array.
Stability:
A stable system is a dynamic system with a bounded response to a bounded input.
Absolute stability:
We can say that a closed loop feedback system is either stable or it is not stable.
This type of stable/not stable characterization is referred to as absolute stability. A
system possessing absolute stability is called a stable system the label of absolute is
dropped.
Relative stability:
A closed loop system is stable; we can further characterize the degree of stability.
This is referred to as relative stability.
Routh-Hurwitz Stability criterion:
The Routh-Hurwitz criterion is a necessary and sufficient criterion for the stability
of linear systems. The method was originally developed in terms of determinants, but
we shall use the more convenient array formulation.
The Routh-Hurwitz criterion is based on ordering the coefficients of the
characteristic equation,
In to an array or schedule as follows
Further rows of the schedule are then completed as,
Where,
And so on. The algorithm for calculating the entries in the array can be followed on
a determinant basis or by using the form of the equation for .
The Routh-Hurwitz criterion states that the number of roots of q(s) with positive
real parts is equal to the number of changes in sign of the first column of the Routh
array. This criterion requires that there be no changes in sign in the first column for a
stable system. This requirement is both necessary and sufficient.
Part 1: Finding system’s stability:
clc; k=0:0.5:20; check=0; for i=1:length(k) q=[1 2 5 k(i)]; p=roots(q); re=real(p); for j=1:3 if re(j)>0 check=1; end end if check==1 break end end k(i)
Output:
ans =
10.5000
Assignment:
K=[0:0.5:20]; for i=1:length(K) q=[1 2 4 K(i)]; p(:,i)=roots(q); end plot(real(p),imag(p),'x'), grid xlabel('Real axis'), ylabel('Imaginary axis');
Assignment: (General code)
clc;
disp(' ')
D=input('Input coefficients of characteristic equation');
l=length (D);
disp(' ')
disp('--------------------')
disp('poles of characteristic equation :')
roots(D)
---- Bulding array--------
if mod(l,2)==0
m=zeros(l,l/2);
[cols,rows]=size(m);
for i=1:rows
m(1,i)=D(1,(2*i)-1);
m(2,i)=D(1,(2*i));
end
else
m=zeros(l,(l+1)/2);
[cols,rows]=size(m);
for i=1:rows
m(1,i)=D(1,(2*i)-1);
end
for i=1:((l-1)/2)
m(2,i)=D(1,(2*i));
end
end
for j=3:cols
if m(j-1,1)==0
m(j-1,1)=0.001;
end
for i=1:rows-1
m(j,i)=(-1/m(j-1,1))*det([m(j-2,1) m(j-2,i+1);m(j-1,1) m(j-1,i+1)]);
end
end
disp('--------The Routh-Hurwitz array is:--------'),m
Temp=sign(m);a=0;
for j=1:cols
a=a+Temp(j,1);
end
if a==cols
disp(' ----> System is Stable <----')
else
disp(' ----> System is Unstable <----')
end
output:
Comments:
Routh-Hurwitz table is an effective way to see whether the system is stable or not.
It also helps us in determining the auxiliary roots of the system and rout locus of
the system.
With this we can also find the point where rout locus cut the imaginary axis.
With this we can analyze the system status that weather the system is stable or
unstable before implementing it. We can also find the asymptotic center and also
we can find poles location of our system. We can also find break away and break in points from the graph.
Lab Session 6
Obtaining a Root Locus plot & Design of Parameter of K for an Electro-
hydraulic Servo Mechanism using MATLAB Root
locus analysis:
The root locus is the path of the roots of the characteristic equation traced
out in the s-plane as a system parameter varies from zero to infinity.
Therefore it is necessary that,
And
Where, k=0,±1,±2,±…………
The simple second order system considered in the previous chapters is
shown in Figure. The characteristic equation representing this system is
Or,
The locus of the roots as the gain K is varied is found by requiring that
The gain K may be varied from zero to an in finitely large positive value. For a
second order system, the roots are,
Procedure of finding of Rout Locus:
Program:
p=[1 1];
q=[2 5 5 0]; sys=tf(p,q); sisotool(sys)
Comments:
Rout locus defines a path with which the root of a system moves by changing the
value of k.
Due to attraction of zero the root locus differ from their asymptotes.
Using sisotool we added two zeros to the system and got the desired desired the
stable
step response alonge with desired P.O%,Tr,Ts .
It is also helpful to us for getting response for many inputs, just like ramp input
parabolic input.
Lab session 7 PID Controller Design
In this lab session, we will consider the following unity feedback system:
The output of a PID controller, equal to the control input to the plant, in the time-domain is as follows:
First, let's take a look at how the PID controller works in a closed-loop system using the schematic shown above. The
variable ( ) represents the tracking error, the difference between the desired input value ( ) and the actual output ( ).
This error signal ( ) will be sent to the PID controller, and the controller computes both the derivative and the integral
of this error signal. The control signal ( ) to the plant is equal to the proportional gain ( ) times the magnitude of the
error plus the integral gain ( ) times the integral of the error plus the derivative gain ( ) times the derivative of the
error.This control signal ( ) is sent to the plant, and the new output ( ) is obtained. The new output ( ) is then fed back
and compared to the reference to find the new error signal ( ). The controller takes this new error signal and computes
its derivative and its integral again, ad infinitum.
The transfer function of a PID controller is found by taking the Laplace transform of Eq.(1).
= Proportional gain = Integral gain = Derivative gain
We can define a PID controller in MATLAB using the transfer function directly, for example:
we may use MATLAB's pid controller object to generate an equivalent continuous-time controller as follows:
C = pid(Kp,Ki,Kd)
C =
1
Kp + Ki * --- + Kd * s
s
with Kp = 1, Ki = 1, Kd = 1
Continuous-time PID controller in parallel form.
Automatic PID Tuning
MATLAB provides tools for automatically choosing optimal PID gains which makes the trial and error process described
above unnecessary. You can access the tuning algorithm directly using pidtune or through a nice graphical user
interface (GUI) using pidtool.
The MATLAB automated tuning algorithm chooses PID gains to balance performance (response time, bandwidth) and
robustness (stability margins). By default the algorthm designs for a 60 degree phase margin.
Let's explore these automated tools by first generating a proportional controller for the mass-spring-damper system by
entering the following commands:
The Characteristics of P, I, and D Controllers
A proportional controller ( ) will have the effect of reducing the rise time and will reduce but never eliminate
the steady-state error. An integral control ( ) will have the effect of eliminating the steady-state error for a constant
or step input, but it may make the transient response slower. A derivative control ( ) will have the effect of increasing
the stability of the system, reducing the overshoot, and improving the transient response.
The effects of each of controller parameters, , , and on a closed-loop system are summarized in the table
.
CL RESPONSE RISE TIME OVERSHOOT SETTLING TIME S-S ERROR
Kp Decrease Increase Small Change Decrease
Ki Decrease Increase Increase Eliminate
Kd Small Change Decrease Decrease No Change
Note that these correlations may not be exactly accurate, because , , and are dependent on each other. In
fact, changing one of these variables can change the effect of the other two. For this reason, the table should only be
used as a reference when you are determining the values for , and .
Example Problem
Suppose we have a simple mass, spring, and damper problem.
The modeling equation of this system is
Taking the Laplace transform of the modeling equation, we get
The transfer function between the displacement and the input then becomes
Let
M = 2 kg
b = 5 N s/m
k = 25 N/m
F = 4 N
Plug these values into the above transfer function
X(s) / F(s) = 1 / 2s^2+5s+25
The goal of this problem is to show you how each of , and contributes to obtain
Fast rise time
Minimum overshoot
No 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/(2*s^2 + 5*s + 25);
step(P)
Comment:
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 the steady-state error of 0.95, quite large indeed. Furthermore, the rise time is about one second, and
the settling time is about 3.5 seconds. Let's design a controller that will reduce the rise time, reduce the settling time,
and eliminate the steady-state error.
Proportional Control
From the table shown above, we see that the proportional controller (Kp) reduces the rise time, increases the overshoot,
and reduces the steady-state error.
The closed-loop transfer function of the above system with a proportional controller is:
X(s) /F(s) = Kp / 2s^2+5s+(25+Kp)
Let the proportional gain ( ) equal 300 and change the m-file to the following:
Kp = 300;
C = pid(Kp)
T = feedback(C*P,1)
t = 0:0.05:5;
step(T,t)
C = Kp = 300
P-only controller.
T =
300
----------------
2 s^2 + 5*s + 325
Continuous-time transfer function.
Comment:
The above plot shows that the proportional controller reduced both the rise time and the steady-state error, increased
the overshoot, and decreased the settling time by small amount.
Proportional-Derivative Control
Now, let's take a look at a PD control. From the table shown above, we see that the derivative controller (Kd) reduces
both the overshoot and the settling time. The closed-loop transfer function of the given system with a PD controller is:
X(s) /F(s) = (Kd*s+Kp) / 2s^2+(5+Kd)s+(25+Kp)
Let equal 300 as before and let equal 20
Kp = 300;
Kd = 20;
C = pid(Kp,0,Kd)
T = feedback(C*P,1)
t = 0:0.01:2;
step(T,t)
T =
20 s + 300
-----------------
2 s^2 + 25 s + 325
Comment:
This plot shows that the derivative controller reduced both the overshoot and the settling time, and had a small effect
on the rise time and the steady-state error.
Proportional-Integral Control
Before going into a PID control, let's take a look at a PI control. From the table, we see that an integral controller (Ki)
decreases the rise time, increases both the overshoot and the settling time, and eliminates the steady-state error. For
the given system, the closed-loop transfer function with a PI control is:
X(s) /F(s) = (Kp*s+Ki) / s^3+ 5s^2+(25+Kp)s+Ki
Let's reduce the to 30, and let equal 70.
Kp = 30;
Ki = 70;
C = pid(Kp,Ki)
T = feedback(C*P,1)
t = 0:0.05:5;
step(T,t)
Continuous-time PI controller in parallel form.
T =
30 s + 70
------------------------
s^3 + 5 s^2 + 55 s + 70
Continuous-time transfer function.
Comment:
We have reduced the proportional gain (Kp) because the integral controller also reduces the rise time and increases
the overshoot as the proportional controller does (double effect). The above response shows that the integral controller
eliminated the steady-state error.
Proportional-Integral-Derivative Control
Now, let's take a look at a PID controller. The closed-loop transfer function of the given system with a PID controller is:
After several trial and error runs, the gains = 350, = 300, and = 50 provided the desired response. To
confirm, enter the following commands to an m-file and run it in the command window. You should get the following
step response.
Kp = 350;
Ki = 300;
Kd = 50;
C = pid(Kp,Ki,Kd)
T = feedback(C*P,1);
t = 0:0.01:2;
step(T,t)
Continuous-time PID controller in parallel form.
Comment:
After several trial and error runs, the gains = 350, = 300, and = 50 provided the desired response.
Now, we have obtained a closed-loop system with no overshoot, fast rise time, and no steady-state error.
Which is our desired response.
General Tips for Designing a PID Controller
When you are designing a PID controller for a given system, follow the steps shown below to obtain a desired response.
1. Obtain an open-loop response and determine what needs to be improved
2. Add a proportional control to improve the rise time
3. Add a derivative control to improve the overshoot
4. Add an integral control to eliminate the steady-state error
5. Adjust each of Kp, Ki, and Kd until you obtain a desired overall response. You can always refer to the table shown
in this "PID Tutorial" page to find out which controller controls what characteristics.
Lastly, please keep in mind that you do not need to implement all three controllers (proportional, derivative, and integral)
into a single system, if not necessary. For example, if a PI controller gives a good enough response (like the above
example), then you don't need to implement a derivative controller on the system. Keep the controller as simple as
possible.