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Problem 5.23PP
Suppose that in Fig.
= , , !.— ^ and Dc(s) =
j(j2 + 2 s + 5) s + 2
Without using Matlab, sketch the root locus with respect to K of the characteristic equation for the 
ciosed-loop system, paying particuiar attention to points that generate multiple roots. Find the 
value of K at that point, state what the location of the mulitple roots is, and how many multiple 
roots there are.
Figure Unity feedback system
- • a»)
Step-by-step solution
step 1 of 8
Consider the general form of characteristics equation
i+ o , w c w = o (1)
I KSubstitute
1 + ii: -
7 1 —Z— G( j ) and -------for D^(ff)in Equation (1).
(■s + 2 j+ 5 j j+ 2
1
(2 )
+ 2 j+ 5 j
Consider the roots of the general form of an equation by the root locus method.
(3)
Where,
The roots of JV (j)= 0 are called the zeros of the problem.
The roots of D ( j) = 0 are the poles.
Consider the number of poles and zeros from the characteristics equation. 
Compare the Equation (2) and the Equation (3).
To find zeros put numerator JV(j) = 0 
Thus, there is no zero in the transfer function.
To find poles put denominator D{s) = 0 • 
j (j + 2 )(j '+ 2 j +5) = 0 ...... (4)
The roots of the equation (4) are 0 , -2 , —\+ 2 J and —l - 2 y .
Thus, the four poles are 0, - 2, —l + 2y'and —1—2y-
Step 2 of 8
Consider the formula for the centre of asymptotes.
(sum of finite poIes)-(sum of finite zeros) 
(numberof finite poles )-(nunU>er of finite zeros) 
( 0 - 2 - l + 2 y - l -2 y ) - (0 )
(4 ) - (0 )
= - I
Thus, the centre of asymptotes is O
Step 3 of 8
Consider the formula for the angle of asymptotes.
(5), l80°+360«(/-l) 
n —m
Where,
Number of poles is n 
Number of zeros is m 
/ = l,2 ,..ji-m
Substitute 1 for /, 4 for n and 0 for m in equation (5).
I80°+360°(1-1)
^ ” >1 A4 -0
= 45*
Substitute 2 for /, 4 for n and 0 for m in equation (5).
l80°+360°(2-l)
4 -0
= 135»
Substitute 3 for /, 4 for n and 0 for m in equation (5).
, 180»+360»(3-l)
4 -0
«225«
= -45»
Substitute 4 for /, 4 for n and 0 for m in equation (5).
180* *+360»(4-l)
4 -0
-315«
= -135»
Thus, the angie of asymptotes are [45^ . |135|, |-45°|and | - 135|.
Step 4 of 8
Consider the following formula for the departure angle ^ ^ fro m the pole at - l+ 2 y . 
= - t a n - ' j - tan-'^ I j - 90°+180°
— 116.6°-63.4°-90°+180° 
= -90°
Thus, the departure angle ^ ^ fro m the pole at —l°h2y is |-9Q^|.
Step 5 of 8
From equation (2), the characteristics equation is given below.
A(a)=a (a+2)(j’+ 2a + 5)+X 
A(a)=a‘+4a’+9a“+10a+X... (6)
Apply Routh-Hurwitz criteria to equation (6).
tS - l
50
1 ^ 9
4 ' ^10
6.5 K
10 -
13
K
Figure 1
Step 6 of 8
The system is stable if the equation satisfies the following condition.
• All the terms in the first column of the Routh’s array must be positive sign. 
From Figure 1, the stable condition is 0