3.0 Time Response
OBJECTIVE
• Influence of poles toward the time response
• Routh-Hurwitz Criteria used for determining the stability of a system
• Transient response for a system
• Parameter used for steady state response for a system
First Order System
Standard form for a first order transfer function is
G( s) 
Where K is the dc gain and

C ( s) K

R(s) s  1
Eqn. (3.1)
is the time constant
For a unit step of R(s)  1 s
C (s) 
1 K A B
 
s s  1 s s  1
Its response
c(t )  A  B' et 
A, B,
and
B'
are constant. For K  1
and
 1 a
then
c(t )  1  e  at
For a=1,
» num=[1];den=[1 1];[y,x,t]=step(num,den,0:0.01:10);plot(t,y);grid
» xlabel('t (saat)');ylabel('Amplitud');title('Sambutan Unit Langkah bagi Tertib Pertama');
(i) Time constant,  when the final value 1 e1 =0.6321.
(ii) Rise time, tie range for response of 0.1 to 0.9 from final value
tr 
2.31 0.11 2.2


a
a
a
(iii) Settling time, time range when the response is 0.98-1.02 of the final value.
ts 
4

Example
A process control for controlling the height of a water tank, h (cm) is done through controlling the
in-flow rate, qi (The out-flow rate, cm3.s-1). qo (cm3.s-1) is related to the
A process control for controlling the height of a water tank, h (cm) is done through controlling the
in-flow rate, qi (cm3.s-1). The out-flow rate, q (cm3.s-1) is related to the
o
(a) Obtain the system block diagram
2
(b) If R  4  10 s.cm-2 and C  6000 cm2, determine the open-loop transfer function
(c) Determine the system closed-loop transfer function, dc gain and time constant.
qi (cm3.s-1)
0.5
t
Solution
(a) Taking Laplace transform of C
dh
 qi  qo
dt
sCH ( s)  Qi ( s)  Qo ( s)
Rearrange
H ( s)
1

Qi ( s )  Qo ( s ) sC
By converting the Laplace transform of
Qo ( s ) 1

H ( s) R
qo  h R
Qi (s ) +
1
sC
Qo (s)
1
R
(b) This gives the open loop transfer function as
1 1
.
sC R
1

sCR
Go ( s ) 
Using
Go ( s ) 
R  4  10 2
1
240s
dan C  6000
H (s )
(c) While the closed loop transfer function is given as
1
H ( s)
sC

Qi ( s ) 1  1 . 1
sC R
R

sCR  1
Comparing with standard first order transfer function of Eqn. (3.1) G ( s ) 
Thus, the dc gain
And time constant
K  R  0.04
  RC  240 s
K
s  1
Second Order System
a
C ( s)
 2 0
R( s ) s  b1s  b0
 b1  b12  4b0
For impulse response where 1,2 
2
K n2
C ( s)

R( s) s 2  2 n s   n2
Standard form
Eqn. (3.2)
Where K is the dc gain,
 is the damping ratio
 n is the undamped natural frequency

exponentia l decay frequency
natural frequency
Roots of the denominator
s  2 n s   n2  0
2
Eqn. (3.3)
will give the poles of the second order system
s1,2  n  n  2 1
Eqn. (3.4)
Example
For the transfer function below, obtain the damping ratio, undamped natural frequency and its dc
gain for a unit step input.
(a)
C ( s)
25
 2
R( s) 2s  20s  36
(b) G ( s ) 
s4
s 2  5s  20
Penyelesaian
(a) Standard form of second order transfer function is
K n2
C ( s)

R( s) s 2  2 n s   n2
Normalized the transfer function
C ( s)
12.5
 2
R( s ) s  10s  18
Compare with the standard form
 n2  18
Giving the natural undamped frequency of
 n  4.24 rad.s-1
Damping ratio of
2n  10

10
 1.18
2  4.24
For a step input, the dc gain is given by
K dc  c(t  )  lim s.
s 0
25
1 25
.

 0.69
2s 2  20s  36 s 36
(b) Comparing with standard form of second order system of Eqn (3.2)
Undamped natural frequency n  20  4.47
5
Damping ratio of

 0.56
2  4.47
rad.s-1
For a step input the dc gain is given by
K dc  c(t  )  lim s.
s 0
s4 1 4
.   0.2
s 2  5s  20 s 20
Under damped
When
0   1
Its transfer function is.
K n2
C ( s)

R( s) s   n  j d s   n  j d 
where
d  n 1  2
Pole position
j

n

  d   n

  kos
j d  j n 1   2

 j d   j n 1   2
MP
1.0
t
tr
tp
ts
Underdamped step response of a second order system
Peak time,
tp
The time taken for the first peak of second order response
tp 

d
Percentage of overshoot, M p
Percentage different of the maximum peak from the final value
Mp 
 
c t p  c 
c 
100%




   1 2 


1  e
 1

 100%
1




Settling time, t s
Time taken for the final value to be within 2% of the final value
c(t s )  0.98  1 

e  n t s
1  2
 ln 0.02 1   2
ts 
 n
4
ts 

 n
Example:
Consider a second order RLC circuit with current, i that varies in time, t is subjected to a step voltage,
v and is represented as a differntial equation below. Assume the initial value of the current is zero.
d 2 i (t )
di (t )
8

15
 16i (t )  10v (t )
dt 2
dt
(a) Obtain the transfer function of the system.
(b) Determine damping ration, undamped and damped natural frequencies for a unit step input.
solution:
(a) Using the diffrential property of Laplace transform
8s 2 I ( s )  15sI ( s )  16 I ( s )  10V ( s )
Rearrange the equation give the transfer function as
I ( s)
10

V ( s)
8s 2  15s  16
or
I ( s)
1.25
 2
V ( s ) s  1.875s  2
(b) Comparing with standard form of
K n2
I ( s)

V ( s) s 2  2 n s   n2
Nutral undamped natural frequency
1.25
 0.63
2
And the damping ratio is obtained from
Dc gain of
n  2  1.41
rad.s-1
K
2 n  1.875 Which gives
And the damped natural frequency
 d   n 1   2  1.414 1  0.66 2  1.06
rad.s-1
  0.66
»wn=2;zeta=0.5;num=[wn*wn];den=[1,2*zeta*wn,wn*wn];[y,x,t]=step(num,den,0:0.01:10);»
plot(t,y);grid;
» xlabel('t (saat)');ylabel ('Amplitud');title('Sambutan Unit Langkah bagi Tertib Kedua')
Over damped
 1
transfer function
C ( s)
 n2

R(s) s      2  1 s      2  1
n
n
n
n



»wn=2;zeta=5;num=[wn*wn];den=[1,2*zeta*wn,wn*wn];[y,x,t]=step(num,den,0:0.01:50);
» plot(t,y);grid
» xlabel('t (saat)');ylabel('Amplitud');title('Sambutan Unit Langkah bagi
Tertib Kedua');
Critical damped
  1 , to produce transfer function of
 n2
C ( s)

R(s) s   n 2
»wn=2;zeta=1;num=[wn*wn];den=[1,2*zeta*wn,wn*wn];[y,x,t]=step(num,den);
grid; xlabel('t (saat)');ylabel('Amplitud');title('Sambutan Unit Langkah bagi
Tertib Kedua');
Undamped
  0 , transfer function becomes
 n2
C ( s)

R(s) s 2   n2
>>wn=2;zeta=0;num=[wn*wn];den=[1,2*zeta*wn,wn*wn];step(num,den)
grid;xlabel('t (saat)');ylabel('Amplitud');title('Sambutan Unit Langkah bagi Tertib Kedua');
Pole position of a second order
The movement of the poles will determine the time response behaviour of the system.
We will study these behaviours
(1) Real poles moving horizontally
j
-3
-1
Satah s

(2) Complex poles on the jw-axis moving vertically
j
5j
j
-j
-5j
s-plane

(3) Complex poles moving vertically
j
10j
s-plane
5j
3j

-3j
-5j
-10j
(4) Complex poles moving vertically
j
-3
-1
s-plane

Dominant pole
Poles that dominate the response over the others
Consider
A
Y ( s)
A
A
 1  2  3
R(s) s  20 s  1 s  2
y(t )  A1e20t  A2et  A3e2t
A
Y (s) A2

 3
R( s ) s  1 s  2
Example
Conider the response of two transfer function of
G1 ( s) 
1.5
1

s  10 s  1
and G2 ( s) 
0.9
1

s  10 s  1
We can consider the second order of G2 (s) is dominated by the pole at -1, hence the
second order model can be resperented with a first order model of G3 ( s) 
Wherelse the model of G1 (s) can’t be approximated by a first order model
1
s 1
Stability
Routh-Hurwitz Stability Criteria
If a polynomial is given by
T (s)  an s n  an 1s n 1  .....  a1s  a0  0
where an , an 1....., a1, a0
(i)
are constants and n  1...
necessary condition for stability are:
All the coefficients of the polynomial are of the same sign. If not, there are poles
on the right hand side of the s-plane
.
(ii) All the coefficient should exist accept for the
a0
For the sufficient condition, we must formed a Routh-array,
sn
s n 1
an
an  2 an  4
an  6
an 1
an  3
an  5
an  7
b1
b2
b3
b4
c1
c2
c3
c4






s3
h1
h2
s2
i1
i2
s1
j1
s0
k1
s n2
s n 3









  
  
  
  
an an  2
a
a
a a a a
b1   n 1 n 3  n 1 n  2 n n 3
an 1
an 1
an an  4
a
a
a a a a
b2   n 1 n 5  n 1 n  4 n n 5
an 1
an 1
an 1 an 3
b
b2
ba a b
c1   1
 1 n 3 n 1 2
b1
b1
an 1 an 5
b
b3
ba a b
c2   1
 1 n 5 n 1 3
b1
b1
j1  
h1
i1
h2
i2
i1

i1h2  h1i2
i1
k1  i2
Routh-Hurwitz Criteria states that the number of roots of characteristic equation is the
same as the numberof sign changed of the first column.
Case 1: No zero on the first column
After the Aray has been tabled, all the elements on the first column is not equal to zero.
If there is no sign changed, all the poles are in the LHP. While the number of poles on
the RHP is equal to number of sign change on the first column of the Routh’s array.
Example:
Consider a fourth order characteristic equation
D( s)  2s 4  s 3  12s 2  8s  2  0
Solution:
Form theRouth’s array
s
4
2
12
s
3
1
8
12  16
 4
1
2
s
2
s1
s
0
2
 32  2
 8.5
4
2
There are two sign change on rows 2 and 3. Hence, there two poles on the RHP
(Right-half f s-palne).
MATLAB solution
>>roots([2 1 12 8 2])
ans = 0.0885 + 2.4380i 0.0885 - 2.4380i
-0.3385 + 0.2311i -0.3385 - 0.2311i
Case 2: Coeffiecient of the first column is zero but not the others.
Change the zero element by a small number positive number, . The number of
pole on the RHP will depend on the number of sign change.
Example:
Consider a fifth order characteristic equation
D( s)  s 5  2s 4  3s 3  6s 2  5s  3  0
Solution:
Form the Routh’s array
5
1
3
5
s4
2
6
3
s
s3
s
2
66
 0,
2

6  7
42  49  6 2
12  14
0
3
s
3

1
s
10  3
 3 .5
2
is a small positive number there are two sign change at row 3 and 4   
and also at row 4 and 5     . Hence, there two poles on the RHP.
MATLAB Solution
» roots([1 2 3 6 5 3])
ans = 0.3429 + 1.5083i 0.3429 - 1.5083i
-1.6681
-0.5088 + 0.7020i -0.5088 - 0.7020I
Case 3: All the coefficients on a row are zeros.
Form an auxillary equation from the row above it and replace the coefficient of the
row with the differentiated coefficient of the auxillary equation. For this case, ifthere
is no sign change, the characteristic equation has a pair of poles with opposite
sign of real component or/and a pair of conjugate poles on the imaginary axis.
Example:
Consider this fifth order characteristic equation
D( s)  s 5  7 s 4  6s 3  42s 2  8s  56  0
Formed Routh array
s
5
s4
s3
1
6
8
7
42
56
42  42
0
7
56  56
0
7
84
28
s2
1
21
9.3
s
s0
56
56
Form the auxillary equation on the second row: P( s)  7 s  42s  56
dP( s )
Differentiate the equation:
 28s 3  84 s As there is no sign change,
ds
there is a pair of conjugate poles on the axis and/or a pair of poles with opposite
sign of real component. To be sure we can use MATLAB
4
» roots([1 7 6 42 8 56])
ans = -7.0000
0.0000 + 2.0000i 0.0000 - 2.0000i
0.0000 + 1.4142i 0.0000 - 1.4142i
2
Use of Routh Hurwitz Criteria
Main use is to determine the position of the poles, which in turns can determine the stability of the
response.
j
STABIL
TAK STABIL

Example
A closed-loop transfer function is given by
C (s)
K
 2
R ( s ) s s  s  1 s  2   K


Determine the range for K for the system to be always stable and its oscillating
frequency before it becomes unstable.
Solution:
Charactristic equation is
ss 2  s  1s  2  K  0
Expand the equation
s 4  3s 3  3s 2  2s  K  0
Form the Routh’s array
s4
1
3
3
3
2
s
92 7

3 3
s2
14 3  3K 14  9 K

73
7
1
s
s0
K
K
K
For nosign chage on the first column:
Refering to row 4
14  9 K
0
7
which gives K  14 9
and row 5
K 0
Hence its range
0  K  14 9
 7 3 2  14 9  0
Oscillating frequency 2 3
rad.s-1.
Steady state
R(s)
E(s)
+
Y(s)
G (s )
-
B(s)
H (s)
From the diagram
E ( s) 
1
R( s )
1  G( s) H ( s)
Consider
G (s)  K
H ( s) 
Ta1 s  1Ta 2 s  1...Tai s  1
s n Tb1 s  1Tb 2 s  1...Tbj s  1
and
Tx1 s  1Tx 2 s  1...Txk s  1
T
y1
s  1T y 2 s  1...T yl s  1
Use the final value theorem and define steady state error, ess that is given by
ess  lim e(t )  lim sE (s)
t 
s0
Unit step
1
Unit step input, R ( s ) 
s
From
E ( s) 
1
R( s)
1  G ( s) H ( s)
Steady state error,
s
1
s0 1  G ( s ) H ( s ) s
ess  lim
We define step error coefficient, K s  lim G(s) H (s)
s0
Thus, the steady state error is
e ss 
1
1 KS
By knowing the type of open-loop transfer function,
G ( s) H ( s)
we can know step error coefficient and thus the steady state error
K s  lim G(s)H (s)
s0
 lim K
s0
Ta1s  1Ta2 s  1...Tai s  1 T s  1T
s n Tb1s  1Tb 2 s  1...Tbj s  1 T s  1T
x1
y1
s  1...Txk s  1
y 2 s  1...Tyl s  1
x2
For open-loop transfer function of type 0: K s  K
ess 
1
1 K
1
0
1 
1
For open-loop transfer function of type 2: K s  , ess 
0
1 
For open-loop transfer function of type 1: K s  , ess 
Example:
A first order plant with time constant of 9 sec and dc gain of 5 is negatively feedback with unity gain,
determine the steady state error for a unit step input and the final value of the output.
Solution:
The block diagramKofs the system is
.
R(s)
+
-
5
9s  1
Y(s)
As we are looking for a stedy state error for a step input, we need to know
Knowing the open-loop transfer function, then
K s  had G ( s) H ( s)  had
s 0
And steady state error of
1
1
1
ess 


1 KS 1 5 6
Its final value is
y ss  1  1 6  5 6
s 0
5
5
9s  1
Ks
Unit Ramp
As in the above section, we know that r ( t )  t , while its Laplace form is R( s ) 
E ( s) 
1
R( s )
1  G( s) H ( s)
1
s2
Thus, its steady state error is
ess  lim s
s 0
1
1
1  G( s) H ( s) s 2
Define ramp error coefficient,
K r Kr  lim sG(s) H (s)
Which the steady state error as
1
ess 
Kr
s0
Just like for the unit step input we can conclude the steady state error for a unit ramp through the type
of the open-loop transfer function of the system.
For open-loop transfer function of type 0:
Kr  0
For open-loop transfer function of type 1:
Kr  K
For open-loop transfer function of type 2:
Kr  
Example:
A missile positioning system is shown.
 m (s)
(i) Find its closed-loop transfer function ( s )
i
3
(ii) Determine its undamped natural frequency and its damping ratio if K  10
(iii) Determine the steady state error, if the input is a unit ramp.
(iv) Cadangkan satu kaedah bagi menghapuskan ralat keadaan mantap untuk (iii).
Compensatot
i
+
-
K
DC motor
0.01
s(0.4s  1)
m
Solution:
(a) By Mason rule, the closed-loop transfer function is
0.01
 m ( s)
0.01K
0.025K
s(0.4 s  1)


 2
2
 i ( s ) 1  K . 0.01
0.4 s  s  0.01K s  2.5s  0.025K
s(0.4 s  1)
K.
0.01K
0.4 s 2  s  0.01K
0.025K
 2
s  2.5s  0.025K

,
3
(b) If K  10
 m ( s)
25
 2
 i ( s ) s  2.5s  25
Comparing with a standard second order transfer function
 m ( s)
K n2
 2
 i ( s) s  2 n s   n2
Comparing
 n2  25
Thus undamped natural frequency
n  5
rad.s-1
and
2 n  2.5
damping ratio of
  0.25
(c) To determine the ramp error coefficient, we must obtain its open-loop transfer function
Go ( s )  K .
0.01
s(0.4s  1)
As it is a type 1, the system will have a finite ramp error coefficient, putting
Go ( s ) 
10
s(0.4 s  1)
K r  lim sGo ( s)  lim s.
s0
s0
10
 10
s(0.4s  1)
Hence steady state error of
ess 
1
 0.1
Kr
K  10 3
Unit Parabola
Its time function
r (t )  t 2 ,while its Laplace R( s)  2
3
s
,thus its steady state error is
s
2
3
s 0 1  G( s) H ( s) s
ess  had
Define parabolic error coefficient, K pa
K pa  had s 2G(s) H (s)
s 0
Similarly we can determine its steady state error by knowing the type of the open-loop transfer function
For open-loop transfer function of type 0:
K pa  0
For open-loop transfer function of type 1:
K pa  0
For open-loop transfer function of type 2:
K pa  K
In summary we can make a table of the staedy state error for the above input
Unit step
Type 0
1
1 Ks
Type 1
0
Type 2
0
Unit ramp
Unit parabolic


1
Kr
0

2
K pa