1. 4.0 ROOT LOCUS OBJECTIVE ~ Rules on sketching the root locus.
~ Determination of root from the characteristic equation by using graphical
solution.
~ Rules on sketching the root locus.
~ Analysis of closed-loop using root locus

2. Root locus concept R(s) Consider + E(s) Y(s) KG(s) B(s) H(s)
Open-loop transfer function If
and
where
and

3. Closed-loop transfer function
~ Number and position of zeros for open-loop and closed-loop are the same
~ Position of poles for the closed-loop depend on the position of poles, zeros and K.
Characteristic equation is If .
Let
and
Its open-loop transfer function .
and its closed-loop as
~ Position of poles for the closed-loop depend on the position of poles, zeros and K of the open-loop.
transfer function.

4. As K varies the closed-loop poles follow similarly and form a locus
As K varies the closed-loop poles follow similarly and form a locus. For that we define
Root locus is a locus of charateristic equation as K varie from 0 to .
Refering to a characteristic equation
Let say
Re-arrange
-1 is a complex number
where

5. Magnitude condition Re-arrange where
is the magnitude from a test point to open-loop poles .
and
the magnitude from a test point to open-loop poes
From the magnitude condition, we can determine K. Let s be the test point, the magnitude are
and
from open-loop zeros and poles respectively.
Hence the magnitude condition is

6. s-plane s m1 x n1 mm nn x x x

7. Angle condition Revisiting the complex number of -1 Its angle Expand
where
is the angle from a test point to open-loop poles
and
where
is the angle from a test point to open-loop poles

8. and is the angle from the test point s to the open loop zeros
From the angle condition, we can determine the angle contribution by s the test
point, the angle are
and
from open-loop zeros and poles respectively.
Hence the angle condition is

9. satah s
 s O   O  

10. Procedure for drawing a Root Locus
(1) Start with the characteristic equation.
(2) The locus is symmetrical at the real axis.
Consider
Characteristic equation is If
» K=3;a=1;roots([1 2*a 4+K])
ans = i i

11. satah s  
That shows the locus is symmetrical at
-axis.

12. (3) The number of locus depends on the number of open loop poles-
Open-loop transfer function
Closed-loop transfer function
Its characteristic equation
Number of locus is reflected by the order of characteristic equation which is the same as the
number of open-loop poles of the system.

13. , then the characteristic equation becomes
(4) Root locus begin from open loop poles and end at open loop zeros or symmetrical lines
Consider closed-loop transfer function, If
, then the characteristic equation becomes
which are open-loop poles.
, characteristic equation is If . If
, the equation becomes
, which are open-loop zeros.
Locus will begin from the open-loop poles,
, and end at the open-loop zeros,
If the number of open-loop zeros are less than the open-loop poles, then the locus will end at
asymptote lines given by angle

14. where m and n are the order of numertor and denominator of the open-loop transfer function
respectively.
And the line intersect the real-axis at
where
and
are the sum of all poles and zeros of open-loop transfer function
respectively.

15. - Example Consider a P-compensator
Find the open-loop zero and poles. Obtain the closed-loop poles at
and
Consequently, shows where the locus end.

16. Solution
which give
an open-loop zero at –2
open-loop poles at 0, –5 and –8
Its characteristic equation
substitute this to the characteristic equation
As we know that the locus start at
, give closed-loop poles at 0, –5 and –8, which is the same
as open-loop poles.

17. To find the closed-loop poles at
We rearrange the characteristic equation
becomes If
, the root of the characteristic equation is –2, which is the same as the open-loop zero.
As there is 3 open-loop poles, there will be 3 loci, one will end at open-loop zero of -2 and the other
two will end at the asymptote lines, with angles at
where
with intersection of the real-axis at

18. satah-s   o 

19. will give the following table:
Example
Consider the following closed-loop transfer function If
, find the closed-loop poles at
and 100.
Finally, trace the loci for
Solution:
As the charactristic equation is
. By substituting the above values to the equation
will give the following table:

20. Gain, K
Closed-loop pole 3 10 -5 21 40 i i
100 i i

21. s­-plane 
j8.8882 
j4.3589   
-5.0 
-j4.3589 
- j8.8882

22. For
we can trace the locus as it move from 0 to ∞ as shown below
Satah s
j8.8882
j4.3589 x x
-j4.3589 
- j8.8882

23. (5) Locus on the real-axis is
It should fulfill he angle condition.
Consider
s-plane H G F D C B A E x x x x

24. Take a test point at A-H and use the angle conditionNo B
-180
Yes C
= -360 D
= -180 E
= -360 F
= -180 G
= -360 H
= -540
Test point

25. (6) Breakaway point of locus from the real-axis
It is the point where the root locus orignating from the open poles meet at the real-axis and breakaway
Consider back the closed-loop transfer function of
If we plot a graph of the closed-loop position against the gain K. We notice that at the point of -5,
before the closed-loop poles become complex, the gain is maximum for any real poles. After this the
locus will breakaway from the real-axis If
We can determine this by taking the differential of the characteristic equation,

26. The characteristic equation is
Example:
A unity feedback system with an open-loop system as given below
Determine the maximum gain K before the system starts to osccilate.
Solution:
The characteristic equation is
Rearrange
Differentiate to obtain the maximum gain

27. Out of the two values, we choose -0.93 as –5.74 is not on the locus.
which gives
Out of the two values, we choose as –5.74 is not on the locus.
use the magnitude condition
Using
, maximum K before the starts to oscillate

28. (7) Stability boundary of the locus
We use the Routh-Hurwitz criteria determine the maximum gain before instability and its
frequency. This can be obtained through

29. (8) Angle of departure and arrival
For complex poles, the locus will leave the poles from an angle called angle of departure.
By selecting a test point very near to the open-loop pole, ,
the angle of departure,
is given by or
 = (Sum of angle from all open-loop zeros to ) - (Sum of angle from all open-loop poles to ) -r

30. Find the angle of departure of the complex poles
Example:
Find the angle of departure of the complex poles
satah-s    O  

31. As for the above diagram, the angle of departure from the complex pole is
while the conjugate pair, its angle of departure is
, the locus will end here and the angle of arrival, , given by an
For complex open-loop zero
angle condition as or
 = r -
(Sum of angle from all open-loop zeros to )
+ (Sum of angle from all open-loop
poles to )

32. Its characteristic equation is
Example:
A plant of transfer function
is feedback with a unity feedback. Sketch a
root locus for
Solution:
Its characteristic equation is
As the number of open-loop poles is 3, there will be 3 loci and the loci start at 0,-1 dan –2.
The locus on the real-axis

33. Satah-s    -2 -1
As there is no zeros, the locus will end at asymtote lines,
where
This gives asymtote angles at
, with real-axis crossing at

34.    -2 -1 satah-s The loci will meet and depart at breakway point
The loci will meet and depart at breakway point
Rearrange
Differentiate for the maximum value of K

35. The probable root is –0.4 as it is on the locus
Its root
The probable root is –0.4 as it is on the locus
The crossing at the boundary of stability 1 2 3 K .
and
Range of gain for stability
. The natural frequency is determined by taking
, replacing to
gives

36. » [k,p]=rlocfind(num,den)
» num=[1];den=[ ]; %Takrifkan rangkap pindah gelung buka
» rlocus(num,den); %Kerahan
» grid;title('LONDAR PUNCA')
We can determine the gain at particular points through
» [k,p]=rlocfind(num,den)

37. Determination of gain   
Can be determined using magnitude condition.
satah-s s
m1 n1 m m3   O 
Gambarajah 5.13: Magnitud sifar dan kutub daripada titik uji.
Take a test point, s the gain K is

38. Pole Determination
If the order of the charateristic equation is n, there will be n poles. If n-1 poles are
known we know the last pole by comparing the coefficient.
Consider
Which can be factored to
where its poles are
. Expanding the factor form
Equating the coefficient
will give the n pole as

39. + - Example For a third order system with
, two of the poles are –2 and –3, determine its third pole. + -
Solution:
Characteristic equation
Comparing
and
which give
hence the third pole is