5. System Responses
Standard forms of first and second-order systems with no zeros and unity gain
are considered.
5.1 First-order systems
DEU-MEE 5017 Advanced Automatic Control
Assoc.Prof.Dr.Levent Malgaca, Fall 2015
5.1.1 Common step response characteristics
5.2 Second-order systems
5.2.1 Pole position for underdamped systems
5.3 Step response based second-order system specifications
For a second-order system there is no general expression fof TR. TR can
be derived emprically from the following figure.
TR 
0.8  2.5
n
TP is found by differentiating the step response, and equating to zero..
Ts  k 
k
k

  n
[ Feedback control systems book, page 130]
where k is determinied by the defined percentage (% 2, % 5, %0.1). k is ofeten chosen as 4.
t ss 
Tn
(%0.13)

Example k values for the defined percentage (depends on ).
>> ksi=0.7;eb=0.02;k=-log(eb*sqrt(1-ksi^2))
k=
4.2487
>> ksi=0.7;eb=0.05;k=-log(eb*sqrt(1-ksi^2))
k=
3.3324
>> ksi=0.7;eb=0.001;k=-log(eb*sqrt(1-ksi^2))
k=
7.2444
Example 5.1
Design a second-order system under feedback control that a peak time 0.5 sec, and a % 5
overshoot.
5.4 Higher order systems
Example 5.2
Example 5.3
Consider the transfer function in Example 5.2. Find the step response by Simulink and
Laplace transform and compare the results.
5.4 Fundemental properties of linear systems
The impulse and ramp responses can be found from the step response by using the
derivative and integral properties.
Example 5.4
For a transfer function given below, find the impulse and ramp responses
form the step response and compare the results with “step” command in
MATLAB.
G (s ) 
The step response:
The impulse response:
1
s  10
y step  0.1(1  e 10t )
y 
d
y step (t )  e 10t
dt
t
The ramp response:
y ramp   y step (t )dt  0.1(t  0.1(1  e 10t ))
0
MATLAB code:
clc;clear;
% Solution with step command
ng=[1];dg=[1, 10];
p=roots(dg);p0=p(1);cksi_v2;t=0:dt:ts;
g=tf(ng,dg);
ys=step(g,t);plot(t,ys,'b')
hold on
g=tf([ng 0],dg);
yi=step(g,t);plot(t,yi,'r')
g=tf(ng,[dg 0]);
yr=step(g,t);plot(t,yr,'g')
% Analytical solution
ysa=0.1*(1-exp(-10*t));plot(t,ysa,'ob')
yia=exp(-10*t);plot(t,yia,'or')
yra=0.1*(t+0.1*(1-exp(-10*t)));plot(t,yra,'og')
Example 5.5
Consider the standart form of second-order transfer function (=0.15,
n=2) given below.
a) Find the impulse and ramp responses form the step response and
compare the results with “step” command in MATLAB.
b) Calculate the perfomance specifications. (TR,TP,TS, % OV, TD). Compare
the results with those obtained with step command in MATLAB.
G (s ) 
4
s 2  0.6s  4
The step response:
The impulse response:
y step  ?
y 
d
y step (t )  ?
dt
t
The ramp response:
y ramp   y step (t )dt  ?
0
5.5 Effect of zeros on the time-domain responses
Example 5.6
5.6 Reduced order systems (Dominant poles):