# How to impose 2nd order approximation?

Discussion in 'Homework Help' started by woodmark75, Dec 11, 2014.

1. ### woodmark75 Thread Starter Member

Dec 11, 2014
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0
Plot the root locus of the system as a function of K1. By imposing the 2nd order system approximation, estimate; settling time, rise time & peak time with 20% overshoot.
Here's the system:
https://app.box.com/s/unjk9dm1goaftgititwq
(image wouldn't work)

G(s)=-0.125(s+0.452)/(s+1.25)(s^2+0.234s+0.0163)
• The closed loop tf I get is 4th order
• How do I then impose 2nd order approx?

2. ### Papabravo Expert

Feb 24, 2006
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You always start a root locus problem by plotting all of the poles and all of the zeros. You then locate the branches. From this you would estimate the significance of the various poles and zeros and if possible hypothesize a 2nd order approximation that behaved similarly. So what does the initial pole zero plot look like? Don't forget the poles and zeros at infinity.

3. ### woodmark75 Thread Starter Member

Dec 11, 2014
36
0
If I plot 'Root Locus' should I use 1 transfer function to represent entire system i.e. G(s)_closed loop?
• Gcl(s)= k1.(2/s+2).G(s)/1+k1(2/s+2)G(s)

4. ### Papabravo Expert

Feb 24, 2006
10,340
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That doesn't look right. Ask yourself what happens when K1 = 0, where are you. Then ask what happens as K1 goes to ∞, where do you end up?
On the real axis the branches start at a pole and end at a zero, possibly a zero at infinity. Each branch of the root locus has an odd number of poles to the right

Another hint. How do you bring the parameter K1 inside the loop? Don't you have to modify the transfer function of the feedback path?

Last edited: Dec 11, 2014
5. ### woodmark75 Thread Starter Member

Dec 11, 2014
36
0
I'm a real novice at this. I'm looking at old notes, should my transfer function for the root locus have the 'K1' as the numerator? If this is the case, how do I derive the denominator?

6. ### Papabravo Expert

Feb 24, 2006
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It actually needs to be in two places. IIRC you need to bring K1 inside the loop. To do this you need to modify the transfer function of the feedback path so that the result is identical to the original system.

In the original system you have:
$E(s)=-K1*C_P(s)-P(s)$
where E(s) is the error signal to the elevator actuator,Cp(s) is the Pitch Command, and P(s) is the Pitch output.
With a -K1 gain block in the forward path on the other side of the summing junction, what gain do you need in the feedback path so that E(s) to the elevator actuator is equal to E(s) from the original system.

$\frac{-1}{K1}$
Then
$E_1(s)=C_P(s) - \frac{-1}{K1}*P(s)$
and
$E_2(s) = -K1*E_1(s)= -K1*C_P(s)-P(s)$
which matches the original input to the elevator control.

Now you can use standard techniques to get the closed loop transfer function.

Last edited: Dec 11, 2014
7. ### woodmark75 Thread Starter Member

Dec 11, 2014
36
0
So for the new system:
https://app.box.com/s/w97s8mpbkgjwuv78klv5
• When I try to find the 'closed loop' tf of this it equates to my post in #3 above?
• G(s) closed loop = -K1*Cp(s)*G(s)/1+ -K1*Cp(s)*G(s)
• Isn't this how you derive 'closed loop' tf's?

8. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
Sorry that is wrong. I told you that you can't just move the Gain constant inside the loop without modifying the feedback path. You should verify this result. That the system transfer function is unchaged between the original drawing and the revised drawing.

Aside: I guess I should figure out how to crop images from .png files

9. ### woodmark75 Thread Starter Member

Dec 11, 2014
36
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When I now calculate my 'closed loop' transfer function;
• would this now be, numerator= -k1*2/(s+2)*G(s), denomenator=1+(-k1*2/(s+2)*G(s))*-1/k1
• would this not cancel out the 'k1'?
• As I haven't got a value for 'k1', only a sign '-ve', if k1 is cancelled in the denomenator this will only remove the effect of its sign?

Dec 11, 2014
36
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11. ### Papabravo Expert

Feb 24, 2006
10,340
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Yeah..that kinda changes everything. I was puzzling about the fact that in your original statement K1 has no effect on the denominator and thaus has no effect on the pole locations.

12. ### woodmark75 Thread Starter Member

Dec 11, 2014
36
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My appologies, I thought I was being smart jumping ahead to what I 'assumed' was the new configuration. My grasp of 'control' is so bad I should never make assumptions!
• with these new specs how should I go about plotting root locus
• & by imposing 2nd order approx, find settling time, rise time , peak time (PO=20)

Last edited: Dec 11, 2014
13. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
I already gave you the prescription in my earlier post.
1. You derive the closed loop transfer function
2. For the open loop transfer function, which should match the closed loop transfer function for K=0, you locate the finite poles and zeros
3. Also for the open loop transfer function, locate any complex conjugate pairs. Complex poles may head for the real axis, spilt, and go towards a real zero or a zero at infinity. They may also depart directly for the zeros at infinity.
4. You locate the branches on the real axis. A branch begins on a pole and ends on a zero, and there is always an odd number of poles to the right of a branch.

To do a 2nd order approximation you look at the sketch and determine if a single pair of poles can be imagined that has the same behavior.
Do that much and then show me what you get.

Last edited: Dec 11, 2014
14. ### Papabravo Expert

Feb 24, 2006
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The rules are:
1. The number of branches of the root locus equals the number of closed loop poles.
2. The root locus is symmetrical about the real axis.
3. On the real axis for K > 0 the root locus exists to the left of an odd number of real-axis, finite open-loop poles and/or finite open-loop zeros
4. The root locus begins at the finite and infinite poles of G(s)H(s) and ends at the finite and infinite zeros of G(s)H(s).
5. The root locus approaches straight lines as asymptotes as the locus approaches infinity. Further the equation of the asymptotes is given by the real axis intercept, $\sigma_a$ and angle, $\theta_a$
$\sigma_a=\frac{\sum \text{finite poles}-\sum \text{finite zeros}}{\text{\sharp finite poles}-\text{\sharp finite zeros}}$

$\theta_a=\frac{(2k+1)\pi}{\text{\sharp finite poles}-\text{\sharp finite zeros}}$

Last edited: Dec 12, 2014

Dec 11, 2014
36
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16. ### Papabravo Expert

Feb 24, 2006
10,340
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The shape is correct, but the particulars may or may not be. It is hard to read the values on your plot of figure out if the gain at the jω axis crossing is correct.

Dec 11, 2014
36
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18. ### woodmark75 Thread Starter Member

Dec 11, 2014
36
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• poles taken from denomenator of F(s)
• zero taken fron numerator of F(s)
• complex conjugate pair taken from solution to quadratic (s^2+0.234s+0.0163=0)
• one pole at s=-2
• one pole at s=-1.25
• one zero at s=-0.452
• complex conjugate pair at: -0.177 +/- 0.051j
• 2nd order approx:

19. ### woodmark75 Thread Starter Member

Dec 11, 2014
36
0
If anyone can help with determining the '2nd order system approximation' it would be really appreciated, as I'm totally stuck....

20. ### MrAl Distinguished Member

Jun 17, 2014
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515
Hi,

Try doing a search on this forum for that 2nd order approximation as i gave an idea how to go about doing this in another thread. There are a couple different ways though so if this is an assignment the instructor might want you to do it a certain way and allow certain assumptions.
If you cant find it i'll have to try to find it again