1. 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

2. First Order System
Standard form for a first order transfer function is
Eqn. (3.1)
Where K is the dc gain and is the time constant
For a unit step of
Its response
are constant. For
and
then
and
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');

3. (i) Time constant,  when the final value =
(ii) Rise time, tie range for response of 0.1 to 0.9 from final value
(iii) Settling time, time range when the response is of the final value.

4. (cm3.s-1). The out-flow rate, (cm3.s-1) is related to the
Example
A process control for controlling the height of a water tank, h (cm) is done through controlling the
in-flow rate,
(The out-flow rate, cm3.s-1).
(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,
(cm3.s-1). The out-flow rate,
(cm3.s-1) is related to the
(a) Obtain the system block diagram
(b) If
s.cm-2 and
cm2, determine the open-loop transfer function
(c) Determine the system closed-loop transfer function, dc gain and time constant.

5. (cm3.s-1)
0.5 t

6. Solution
(a) Taking Laplace transform of
Rearrange
By converting the Laplace transform of

7. + -
(b) This gives the open loop transfer function as
Using
dan

8. (c) While the closed loop transfer function is given as
Comparing with standard first order transfer function of Eqn. (3.1)
Thus, the dc gain
And time constant s

9. Second Order System For impulse response where Standard form
Eqn. (3.2)
Where K is the dc gain,
 is the damping ratio
is the undamped natural frequency
Roots of the denominator
Eqn. (3.3)
will give the poles of the second order system
Eqn. (3.4)

10. (a) Standard form of second order transfer function is
Example
For the transfer function below, obtain the damping ratio, undamped natural frequency and its dc
gain for a unit step input.
(a)
(b)
Penyelesaian
(a) Standard form of second order transfer function is
Normalized the transfer function
Compare with the standard form
Giving the natural undamped frequency of
rad.s-1

11. (b) Comparing with standard form of second order system of Eqn (3.2)
Damping ratio of
For a step input, the dc gain is given by
(b) Comparing with standard form of second order system of Eqn (3.2)
Undamped natural frequency
rad.s-1
Damping ratio of
For a step input the dc gain is given by

12. Under damped     Its transfer function is. When where
Pole position    

13. Underdamped step response of a second order system

14. Peak time,
The time taken for the first peak of second order response
Percentage of overshoot,
Percentage different of the maximum peak from the final value

15. Settling time,
Time taken for the final value to be within 2% of the final value
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.
(a) Obtain the transfer function of the system.
(b) Determine damping ration, undamped and damped natural frequencies for a unit step input.

16. Rearrange the equation give the transfer function as
solution:
Using the diffrential property of Laplace transform
Rearrange the equation give the transfer function as or

17. (b) Comparing with standard form of
Nutral undamped natural frequency
rad.s-1
Dc gain of
And the damping ratio is obtained from
Which gives
And the damped natural frequency
rad.s-1

18. »wn=2;zeta=0. 5;num=[wn. wn];den=[1,2. zeta. wn,wn
»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')

19. Over damped transfer function
»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');

21. Critical damped
, to produce transfer function of
»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');

23. Undamped , transfer function becomes
>>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');

26. 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
Satah s -3 -1

28. (2) Complex poles on the jw-axis moving vertically
s-plane 5j j -j
-5j

30. (3) Complex poles moving vertically
10j
s-plane 5j 3j
-3j
-5j
-10j

32. (4) Complex poles moving vertically
s-plane -1 -3

34. Dominant pole Poles that dominate the response over the others
Consider
Example
Conider the response of two transfer function of
and
We can consider the second order of
is dominated by the pole at -1, hence the
second order model can be resperented with a first order model of
Wherelse the model of
can’t be approximated by a first order model

36. Stability Routh-Hurwitz Stability Criteria
If a polynomial is given by
where
are constants and
necessary condition for stability are:
(i)
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
For the sufficient condition, we must formed a Routh-array,

38. Routh-Hurwitz Criteria states that the number of roots of characteristic equation is the
same as the numberof sign changed of the first column.

39. 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

40. Solution:
Form theRouth’s array 2 12 1 8
There are two sign change on rows 2 and 3. Hence, there two poles on the RHP
(Right-half f s-palne).

41. MATLAB solution
>>roots([ ])
ans = i i
i i
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

42. ,  and also at row 4 and 5 Solution: Form the Routh’s array1 3 5 2 6
, 
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.

43. MATLAB Solution
» roots([ ])
ans = i i
i I
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

44. Formed Routh array 1 6 8 7 42 56 21
9.3 28 84

45. Form the auxillary equation on the second row:
Differentiate the equation:
As there is no sign change,
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
» roots([ ])
ans =
i i
i i

46. 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
TAK STABIL
STABIL 
Example
A closed-loop transfer function is given by
Determine the range for K for the system to be always stable and its oscillating
frequency before it becomes unstable.

47. 1 3 K 2 Solution: Charactristic equation is Expand the equation
Form the Routh’s array 1 3 K 2

48. For nosign chage on the first column:
Refering to row 4
which gives
and row 5
Hence its range
Oscillating frequency
rad.s-1.

49. Steady state R(s) E(s) Y(s) + - B(s) From the diagram Consider and
Use the final value theorem and define steady state error,
that is given by

50. Unit step Steady state error, Unit step input, From
We define step error coefficient,
Thus, the steady state error is
By knowing the type of open-loop transfer function,
we can know step error coefficient and thus the steady state error

51. For open-loop transfer function of type 0:,
For open-loop transfer function of type 2: ,

52. 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 diagram of the system is .
R(s)
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
And steady state error of
Its final value is

53. Unit Ramp , while its Laplace form is
As in the above section, we know that
Thus, its steady state error is
Define ramp error coefficient,
Which the steady state error as
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:
For open-loop transfer function of type 1:
For open-loop transfer function of type 2:

54. - + Example: A missile positioning system is shown.
(i) Find its closed-loop transfer function
(ii) Determine its undamped natural frequency and its damping ratio if
(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 DC motor + -

55. (a) By Mason rule, the closed-loop transfer function is
Solution:
(a) By Mason rule, the closed-loop transfer function is ,
(b) If
Comparing with a standard second order transfer function

56. Comparing
Thus undamped natural frequency
rad.s-1
and
damping ratio of
(c) To determine the ramp error coefficient, we must obtain its open-loop transfer function
As it is a type 1, the system will have a finite ramp error coefficient, putting
Hence steady state error of

57. Unit Parabola Its time function ,while its Laplace
,thus its steady state error is
Define parabolic error coefficient,
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:
For open-loop transfer function of type 1:
For open-loop transfer function of type 2:

58.  Unit step Unit ramp Unit parabolic Type 0 Type 1 Type 2
In summary we can make a table of the staedy state error for the above input
Unit step
Unit ramp
Unit parabolic
Type 0 
Type 1
Type 2