Root locus question

  • #1
I was told to plot a root locus(by hand) of the open loop transfer function:

H(s)G(s)= (s+2)/(s^2 + 2s + 3)

and for a value of k=1.33 determine the location of the root, then find the equivalent damping coefficient.

After several attempts i was able to draw the correct root locus but I am not sure how to approach the other part of the question. Could someone with knowledge of the root locus offer me some assistance? Any comment would be greatly appreciated.
 

Answers and Replies

  • #2
it should be in the numerator...

H(s)G(s)= k(s+2)/(s^2 + 2s + 3)
 
  • #3
118
1
form characteristic equation 1+G(s)H(s)=0
and substitute K=1.33
solve the equation u ll get value of roots at K=1.33
 
  • #4
118
1
1+G(s)H(s)=0
s2+2s+3+k(s+2) = 0
s2+3.33S+5.66=0
solve this you'll get roots at k=1.33
 
Last edited by a moderator:
  • #5
ok thanks... how would I go about finding the equivalent damping coefficient?
I am thinking to compare it to the standard second order equation and equate like terms and solve for zeta. Is that the right approach?
 
  • #6
118
1
ok thanks... how would I go about finding the equivalent damping coefficient?
I am thinking to compare it to the standard second order equation and equate like terms and solve for zeta. Is that the right approach?
that's exact approach
i think you'll get 0.699 zeta
 
Last edited by a moderator:
  • #7
thanks a lot... you really helped me to clear up some misconceptions i had... this will help in my preparation for my finals...
 
  • #8
118
1
you're welcome .:cool:
and all the beat for your finals. . .:smile:
 
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  • #10
hey reddvoid, I tried solving for zeta however i got 0.58. What I did was to equate the constants for the characteristic equation from the standard second order equation to solve for omega, then i equate the coefficients of the "s" term and substitute omega and solve for zeta. Dnt know why i am not getting 0.699
 
  • #11
118
1
S2 + 2zWnS + Wn2 = s2+3.33S+5.66
Wn2=5.66
Wn = sqrt(5.66) = 2.38
2zWn = 3.33
z=3.33 / (2*2.38)
z = 0.699

this is how i got 0.699
correct me if i am wrong . . .
 
  • #12
ok I see where i made the error in my calculations. Thanks a lot for making those clarifications.
 

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