1. KEM100031
MOHD. REFDI BIN MOHD. REDZUAN
KMEM 3122: Control Engineering
Exercise 3: Transient Response and LTI Viewer
1. General form of step response.
a)
G(s)  
s 12s 400
2
s 6 19.08i
Under damped response:
c(t) ABe-6t cos(19.08t )
b)
H(s)  
S2 90s 900
 11.49 2 78.54
Over damped response:
c(t) ABe11.46t Ce78.54t
c)
R(s) 
s2 30s 225
1,2 15
Critically damped response:
c(t) ABe5t Cte15t
d)
E(s) 
s2 625
1,2 25i
Undamped response:
c(t) ABcos(25t 

2. KEM100031
MOHD. REFDI BIN MOHD. REDZUAN
2.
a)
n 20 ζ0.3
Under damped response because 0 < ζ < 1
b)
n 30 ζ1.5
Over damped response because ζ > 1

3. KEM100031
MOHD. REFDI BIN MOHD. REDZUAN
c)
n 15 ζ1.0
Critically damped response because ζ =1
d)
n 25ζ0
Undamped response because ζ = 0

4. KEM100031
MOHD. REFDI BIN MOHD. REDZUAN
3.
From MATLAB, the values are:
Tp 0.6778
%OS 28.0597
Ts 2.1333
Tr 0.3050
The values obtained are different since MATLAB calculations take into account more decimal
places compare to LTI Viewer.

5. KEM100031
MOHD. REFDI BIN MOHD. REDZUAN
4.
From MATLAB,
n 65
0.9231
Tp 0.1257
%OS 0.0531
Ts 0.0667
Tr 0.0498