Asymptote = × 180° q = 0, 1, 2………..(pz1) P=3, Z=0. q= 0, 1, 2. For q=0 Asymptote = 1/3 × 180° = 60° For q=1 Asymptote = × 180° = 180° For q=2 Asymptote = × 180° = 300° Asymptotes = 60°,180°,300°.

Centroid = = Centroid = 0.66

S3  1  1  0 
S2  2  K  0 
S1  0 
 
S0  K 
 
0<K<2. So, when K=2 root locus crosses imaginary axis S3 + 2S2 + 2S + 2 =0 For k Sn1 = 0 n : no. of intersection S21 = 0 at imaginary axis S1 = 0 = 0 K<4 For Sn = 0 for valve of S at that K S2 = 0 2S2 + K = 0 2S2 + 2 = 0 2(S2 +1) = 0 32 = 2 S = ± j

Asymptote = ×180°. q=0,1,………pz1 P=3, Z=0 q= 0,1,2. For q = 0 Asymptote = × 180° = 60. For q=1 Asymptote = × 180° = 180° For q=2 Asymptote = × 180° = 300° 
Centroid = = = 1 
So, calculating breakaway point. = 0 The characteristic equation is 1+ G(s) H (s) = 0. 1+ = 0 K = (S3+3S2+2s) = 3S2+6s+2 = 0 3s2+6s+2 = 0 S = 0.423, 1.577. So, breakaway point is at S=0.423 because root locus is between S= 0 and S= 1 
Characteristics equation is S3+3S3+2s+K = 0
For k Sn1= 0 n: no. of intersection with imaginary axis n=2 S1 = 0 = 0 K < 6 Valve of S at the above valve of K Sn = 0 S2 = 0 3S2 + K =0 3S2 +6 = 0 S2 + 2 = 0 S = ± j 
Asymptote = q = 0,1,…..(Þz1)
q=0,1,2,3. (PZ = 40) for q=0 Asymptote = ×180° =45° For q=1 Asymptote = ×180° =135° For q=2 Asymptote = ×180° =225° For q=3 Asymptote = ×180° =315° 
Centroid = Centroid = = = 0.75 6) As poles are complex so angle of departure is ØD = (2q+1) ×180 + ø ø = ∠Z –∠P. 
So, the breakaway point is calculated 1+ G(s)H(s) = 0
1+ = 0
(S2+S)(S2 +2S+2) + K =0 K = [S4+S3+2s3+2s2+2s2+2s] = 4S3+9S2+8S+2=0 S = 0.39, 0.93, 0.93. 
I + G(s) H(s) = 0 K+S4+3S3+4S2+2S=0

Centroid = = Centroid = 1. 
S4+4S3+6S2+4s+K = 0
≤ 0 
Asymptote = q=0,1,….(pz). For q = 0 Asymptote = 45 For q = 1 Asymptote = 135 For q = 2 Asymptote = 225 For q = 3 Asymptote = 315 
Centroid = ∑Real part of poles  ∑Real part of zero / P – Z = [311] – 0 / 4 – 0 Centroid = 1.25 
5. Let S = + jw. G( + jw) = K( + jw + 6)/( + jw)( + jw + 4 ) = + π tan1 w/ + 6  tan1 w/ – tan1 w / + 4 =  π taking tan of both sides. w/ + w/ + 4 / 1 – w/ w/ + 4 = tan π + w / + 6 / 1  tan π w/ + 6 w/ + w/ + 4 = w/ + 6[ 1 – w2 / ( + 4) ] (2 + 4)( + 6) = (2 + 4 – w2) 2 2 + 12 + 4 + 24 = 2 + 4 – w2 22 + 12 + 24 = 2 – w2 2 + 12 – w2 + 24 = 0 Adding 36 on both sides ( + 6)2 + (w + 0)2 = 12
