Root Locus and Steady-State Error with Rate Feedback: 22-Mec-A3 System Analysis and Control, May 2015, Q4

Question 4: Root Locus and Steady-State Error with Rate Feedback (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. From Fig. 2: forward path $\dfrac{K}{s(0.2s+1)(0.5s+1)}$; an inner (rate, tachometer) feedback path of transfer function $s/3$ taken from the output and subtracted at the inner summing junction; and an outer unity position feedback loop, with the error $E = R - C$ defined at the outer junction.

Find. (a) the root locus of the closed loop as $K$ varies, and the value of $K$ that gives $\zeta = 0.5$ for the dominant complex pair; (b) the steady-state errors to unit step and unit ramp inputs at that gain.

Figure 2 — position loop with an inner rate (tachometer) feedback path $s/3$ and outer unity feedback.

Approach. Reduce the two nested loops to a single characteristic polynomial, rearrange it into the equivalent single-parameter root-locus form (which reveals a zero created by the rate path), impose $\zeta = 0.5$ through Vieta's relations on the cubic, then compute $K_v$ from the reduced error transmittance.

Close the inner rate loop. With forward transfer $K/[s(0.2s+1)(0.5s+1)]$ and inner feedback $s/3$, the transmittance from the inner error to the output is $$\frac{C}{E}=\frac{K}{s(0.2s+1)(0.5s+1)+\dfrac{Ks}{3}}$$ The tachometer term adds directly to the denominator without changing the free integrator, so the loop remains type 1.
Close the outer loop and clear fractions. Setting $1 + C/E = 0$ and expanding $s(0.2s+1)(0.5s+1)=0.1s^3+0.7s^2+s$: $$0.1s^3+0.7s^2+s+\frac{Ks}{3}+K=0$$ Multiplying by 10 gives the characteristic equation in a convenient monic form: $$\boxed{s^3+7s^2+\left(10+\tfrac{10K}{3}\right)s+10K=0}$$
Recast as a single-parameter root locus. Grouping the $K$-dependent terms, $$s^3+7s^2+10s+\frac{10K}{3}\,(s+3)=0 \;\Longrightarrow\; 1+\frac{10K}{3}\cdot\frac{s+3}{s(s+2)(s+5)}=0$$ The rate feedback has therefore created an open-loop zero at $s=-3$ while leaving the open-loop poles at $s=0,\,-2,\,-5$ — that zero is the whole benefit of the tachometer, since it pulls the locus leftwards. Real-axis segments belong to the locus on $[-2,\,0]$ and $[-5,\,-3]$; with $n-m = 2$ there are two asymptotes at $\pm 90^\circ$ about the centroid $\sigma_a = \frac{(0-2-5)-(-3)}{3-1} = -2$; and $dK/ds=0$ gives a breakaway on the real axis at $s=-1.136$.
Impose $\zeta = 0.5$ using Vieta's relations. A damping ratio of 0.5 means the dominant pair lies on the $60^\circ$ rays, i.e. at $s=-a\pm ja\sqrt3$ (so that $\tan 60^\circ = \sqrt3$), with the third root at $-c$. Writing $b = 10K/3$, the characteristic polynomial is $s^3+7s^2+(10+b)s+3b$, and matching the elementary symmetric functions of its roots gives $$2a+c=7,\qquad 4a^2+2ac=10+b,\qquad 4a^2c=3b$$ Eliminating $b$ and $c$ leaves a single cubic in $a$: $$4a^3-14a^2+21a-15=0 \;\Longrightarrow\; a = 1.7132$$ Hence $c = 7-2a = 3.5735$ and $b = 4a^2c/3 = 13.985$.
Recover the gain and check the roots. Since $b=10K/3$, $$\boxed{K = \frac{3b}{10} = 4.196}$$ Substituting back gives the closed-loop roots $s = -1.713 \pm j2.968$ and $s = -3.573$, for which $\zeta = 1.713/3.427 = 0.500$ exactly and $\omega_n = 3.427\ \text{rad/s}$. The third pole sits at $-3.573$, only slightly to the left of the complex pair, so on its own it would not dominate — but it very nearly cancels against the rate-feedback zero at $s=-3$, and its residue is correspondingly small. The complex pair therefore governs the response, which justifies the phrase "dominating poles" in the question.

Question 4(a): computed root locus of $1+\frac{10K}{3}\frac{s+3}{s(s+2)(s+5)}$. Crosses are open-loop poles, the circle is the rate-feedback zero; the red markers are the $\zeta = 0.5$ design points at $K = 4.196$.

Identify the error transmittance for part (b). The error is defined at the outer junction, $E = R - C$, so the relevant open-loop transfer is the rate-closed inner loop: $$L_e(s)=\frac{K}{s\left[(0.2s+1)(0.5s+1)+\dfrac{K}{3}\right]}$$ There is still exactly one free integrator, so the loop is type 1: the step error is zero and the ramp error is finite.
Compute the velocity error constant. $$K_v=\lim_{s\to 0}sL_e(s)=\frac{K}{1+\dfrac{K}{3}}=\frac{3K}{3+K} =\frac{3(4.196)}{7.196}=1.7492\ \text{s}^{-1}$$ The identical value follows from the characteristic polynomial coefficients, $K_v = a_0/a_1 = 10K/(10+10K/3)$, which is a useful cross-check.
State the two steady-state errors. $$e_{ss}\big|_{\text{step}}=\frac{1}{1+K_p}=0 \quad (K_p=\infty),\qquad e_{ss}\big|_{\text{ramp}}=\frac{1}{K_v}=\boxed{0.5716}$$ The pedagogical point is visible in the algebra: without the tachometer the loop would have $K_v = K = 4.196$ and a ramp error of only 0.238. Rate feedback has divided $K_v$ by $1+K/3 = 2.399$, buying damping at exactly the cost of tracking accuracy. Meeting a tight damping specification and a tight ramp-error specification therefore requires lead or PI compensation, not a tachometer alone.

Quantity	Symbol	Value
Characteristic equation	—	$s^3+7s^2+(10+\tfrac{10K}{3})s+10K=0$
Equivalent open-loop poles / zero	—	$0,\ -2,\ -5$ / $-3$
Asymptote centroid	$\sigma_a$	$-2$
Real-axis breakaway	$s_b$	$-1.136$
Gain for $\zeta = 0.5$	$K$	4.196
Dominant closed-loop poles	$s_{1,2}$	$-1.713 \pm j2.968$
Third closed-loop pole	$s_3$	$-3.573$
Undamped natural frequency	$\omega_n$	3.427 rad/s
Velocity error constant	$K_v$	1.749 $ ext{s}^{-1}$
Steady-state error, unit step	$e_{ss}$	0
Steady-state error, unit ramp	$e_{ss}$	0.5716

Topic. Root-locus design of a type-1 servo with inner rate (tachometer) feedback, and the damping-versus-accuracy trade it imposes.

Key relations. Magnitude and angle conditions $|KL(s)|=1$, $\angle L(s) = 180^\circ$; asymptote centroid $\sigma_a=(\sum p_i-\sum z_j)/(n-m)$; breakaway from $dK/ds=0$; for $\zeta = 0.5$ the pole angle is $60^\circ$ from the negative real axis so $\omega_d = \sqrt3\,\sigma$; $K_v=\lim_{s\to0}sL(s)$ and $e_{ss,\text{ramp}} = 1/K_v$.

Why this works. Because the tachometer signal is proportional to $sC$, closing the inner loop adds a term proportional to $Ks$ to the characteristic polynomial. When the polynomial is rearranged so that $K$ multiplies a single factor, that added term appears as an open-loop zero — here at $s = -3$ — and a zero on the negative real axis bends the locus away from the imaginary axis, which is precisely how rate feedback stabilises a sluggish servo. The same $K$ that creates the zero also scales the constant term, so the velocity constant grows more slowly than $K$ itself, and the ramp error stops improving. Recognising that both effects come from the one added term is what turns a messy two-loop reduction into a one-line design conclusion.

Common pitfall. Treating the problem as an ordinary $1+KG(s)$ locus on the bare plant poles $0,-2,-5$ and forgetting that the rate path introduces a zero — that yields a jw-axis crossing and a far smaller design gain. The second trap is computing $K_v$ from the forward path alone, which gives 4.196 instead of 1.749 and understates the ramp error by a factor of 2.4.

Source. Ogata, Modern Control Engineering, 5th ed., §6.2 and Example 6–5 (velocity feedback); Nise, Control Systems Engineering, 8th ed., §8.4 and §7.3; Franklin, Powell and Emami-Naeini, 8th ed., §5.4.

Question 4 of 6: Root Locus and Steady-State Error with Rate Feedback

Question 4: Root Locus and Steady-State Error with Rate Feedback (25 marks)