# Tag Info

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You could use Newton's method. Yeah I know, it's not the "root locus method", but hey, it works. Let $F(s)=3s^4+10s^3+21s^2+24s+30$ Clean it up by dividing everything by 3 so our largest s is alone Let $f(s)=\dfrac{F(s)}{3}=s^4+\frac{10}{3}s^3+7s^2+8s+10$ then $f'(s)=4s^3+10s^2+14s+8$ Now we have everything we need to start converging onto the ...

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These are time constants and the formula would more advantageously be written as $C(s) = K\frac{1+\tau s}{1+\alpha\tau s}$ where $\tau$ is a time constant and has a dimension of time while $\alpha$ is unitless. $K$ represents the gain of this compensator for $s=0$. In a first-order system, the pole is the inverse of the time constant and this ...

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Since you have the closed-loop system you need to put it into the from $1+\frac{k n(s)}{d(s)}=0\$ to use the standard root locus techniques. The roots depend on only the denominator of the closed-loop system. $$s^3+K s^2+7 s^2+7 K s+10 s+20 K =0$$ $$s^3+7 s^2+10 s+K (s^2+7 s+20) =0$$ $$1+\frac{K (s^2+7 s+20)}{s^3+7 s^2+10 s}=0$$ And from the root ...

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The characteristic equation is: $$s(s+1)(s+3)=-K$$ Hence, plug in any point on the locus, and that will give the corresponding value of K. It appears that your selected point on the locus is approximately: $s=-0.4+j0.7$, which will give $K\approx 1.8$ For an accurate answer to this particular problem ($\small \zeta=0.5$), write the CE as: $$s(s+1)(... 2 I'm not sure about your calculation but Matlab yields the correct result. In this problem, the common approach is to use Routh-Hurwitz criterion and search for a row of zeros that yields the possibility for imaginary axis roots. For convert the system to the closed-loop transfer function, hence$$ \frac{K}{s^4 + 10s^3 + 88s^2 + 256s + K} $$The Routh ... 2 As long as build-ability is not a design requirement, why not place two pairs of conjugate zeroes? Place one pair next to each pair of conjugate open-loop poles. Use SISOtool to play around with where exactly the zeroes should be placed in relation to the poles. If the zeroes are placed appropriately close to the poles, the residues on the closed-loop ... 1 With $\small T=1$, the Laplace transform of the sampler and ZOH is $\large\frac{1-e^{-s}}{s}$. This is partitioned into a $z$ term, $\small (1-z^{-1})=\large \frac{z-1}{z}$, and an $s$ term, $\large \frac{1}{s}$. Combine the $z$ term with $\small C(z)$ and the $s$ term with $\small G(s)$, then determine the z-transform of the resultant s-... 1 Ignoring the semantics of "root locus," you can certainly plot the roots of a polynomial as its coefficients change. The following script plots the set of poles of your system as $RL$ varies from 0 to 10000 with $IL$ fixed at 1: IL = 1; RLvec = linspace(0,10e3,1e5); rts = zeros(1e5,4); for k = 1:length(RLvec); RL = RLvec(k); D = [8*IL, 28*IL+8*RL+3,... 1 I used Octave and I got that the asymptotes goes at -1.4, just as you calculated. You should double check the Matlab result. I think you might be misunderstanding the meaning of the centroid, it is the point where the asymptote lines meet at the real line, it is not the breakaway point (where the poles leave the real line). Check slide 3 of these notes 1 s=tf('s'); GH=((s-4)*(s-3))/((s+1)*(s+2)); rlocus(GH) 1 Looking at the root locus with the damping factor and natural frequency isolines, we can better see where the poles move. For continuous-time systems, we have that the settling time is given by$$ T_s = \frac{\ln(0.05\sqrt{1-\zeta^2})}{\zeta \omega_0},$$As an approximation, to reach the smallest possible settling time you would want to maximize$$ \max_{...

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In order to determine $K$. We need to investigate: $1+KF(s)=0\implies 1+L(s)=0 \implies s(s+4)(s^2+6s+64)+K=0.$ We know that $s=j\omega$. Plugging this into the equation and collecting the real and imaginary part we obtain. $\left(10\omega^3-256\omega \right)+j\left(\omega^4-88\omega^2+K \right)=0+j\cdot 0$ By comparing the complex number on the ...

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MATLAB is plotting the root locus only for positive values of $K$. Your analytical calculation may be considering both positive and negative values of $K$ and this is why you end up with two pair of points. Keep just the one that is obtained with $K>0$

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First, drawing two more lines with angles 53⁰ and -53⁰ should be easy, just do another plot while keeping the root locus. The lines should be, $$y = \tan(53^\circ)x$$ and, $$y = -\tan(53^\circ)x.$$ Once you have those lines, set some $k$ to be the maximum value of the evans function and change that value until you find the one that lies on the line you ...

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The only thing you're misunderstanding is that those vertical root loci do, in fact, imply and ever-diminishing phase margin. They're describing a system that goes resonant, with a Q that increases (or $\zeta$ that decreases) as the gain goes up. That increasing Q is indicative of ever-diminishing margins. (And just to note, even though you seem to ...

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The answer is relatively simple: From stability considerations we know that the poles of the closed-loop function H(s)=N(s)/D(s) must not enter the right side of the complex s-plane. The poles of the closed-loop function are identical to the roots (zeros) of the denominator D(s) (which is identical to the characteristic polynominal of the system). ...

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In order to draw the root locus, you need to convert the open-loop system into the closed-loop system. You can do this \begin{align} T(s) &= \frac{ \frac{1}{(s+3)(s+4)} }{ 1+ \frac{1}{(s+3)(s+4)} \frac{K}{s(s+1)} } \\ &= \frac{s^2+s}{s^4+8s^3+19s^2+12s+K} \end{align} Now the gain K appears in the characteristic equation of the closed-loop \...

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No - the root locus is not drawn for the open-loop function. It is the main purpose of the root locus of a system with feedback to show if resp. for which constant gain values the CLOSED-LOOP function will have poles in the "forbidden" region (right half of the s-plane). Hence, you have to find the poles of the function H(s)=G(s)/[1+G(s)*H(s)] with G=1/(s+...

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When you are working with root locus, you are making a figure about your characteristic equation behavior. So, the root locus of a negative-feedback loop is going to be the same at any point of the loop. Notice: the name 'characteristic equation' is just because of that. In your case, we have: characteristic equation = \\$1 + \frac{1}{(s+3)(s+4)} \frac{K}{s(s+...

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