Question

B3. Sketch the root locus of the control system shown in the figure below K s(s? + 65 +25)

Answer 1

For given Control system,

page - I k 81 8² +68+25) OʻLTF = 913) H1B) = k S12763475) Poles 0,-3+4j, -3-4 Total Number of bles=3=P Total Number of zeros =o=z since, Number of Poles > No. of zeros. (#p > #2) dep820 Kenace Hence, (P-Z) Numbers of Asymptotes will guide the RL brach to arrive at kao calculation of centroid lua) E(Reel Part of)- E(Reed part of TA #(P-Z) o+(-3)+(-3) -o - 6-0 3 (3-0) 11 -2

op pole 2 = -36.86° Calculation of Angre of Asymtotes Page-2 4A (2K+1)180 (P-2) Where, k=0,1,2,- (P-2-1). (260) +1)180 = 60° & Anco 3 (2(1) +1) 180 180° = - A K=1 3 (212)+1) 180° 3000 K=2 3 Now Let us Calculate Angle of Departure for Combley poles & Opole@ +180 to & Net Angre made by a poles (other than pole ②] on pole 2. Above real axis real axis. - Oppo = + 180 – (85 + tant Cys) – 98 below Oppole

pole Similarly for po = 180-(g: + towe (9 got 53.13 - 180° 9 - + 36.86 Page-3

• When RL branch is moving along asymptotes ,it cuts imaginary axis at some point.
• The point at which RL Branch cuts imaginary asix can be calculated using Routh Hurwitz criteria. At the point of intersection system will become marginal stable.

Charactstics eah page 4 It GHI 8) =0 s(3²+68+25) tk o 83 +682 +258 +K=0 25 13 12 6 K 150-k si O 6 so k Locles toch touch The point where Root jo axis, at that system becomes marginal stable, 15ook =) -o 6 K = 150. 68 + 150 =0 s - 150 - -25 6 = s=tis -

So, Root locus will cut jw axis at +5j and -5j respectively .

Now let us plot Root Locus -

jo 10 -36.86 6 js Pole @x -3tju 2 szo tt 8. kao -6 -4 G 2 TA=I-2 Pole ② x -4 -3-34 -is 7-10 page-5

7.5 6 2.5 -5 -2.5 2.5 6 -2.5 x -5 -7.5

Hence This above is the root locus plot of given transfer function.

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