Chapter 4

Root Locus and Nyquist Plots

4.1 Root Locus

Consider the single loop feedback system

+ -  K  C  F  P

The dynamics of the closed loop system is intimately related to locations of closed loop poles.

Closed loop transfer function = \frac{KCP}{1 + KCP F} , with K an adjustable gain parameter

Let

L(s) = C(s)P (s)F (s) = \frac{N(s)}{D(s)} ⇒ closed loop poles at D(s) + KN(s)

Closed loop poles are roots of the characteristic polynomial D(s) + KN(s). May assume no common roots in D, N, otherwise cancel common factors. For the rest of this chapter, take F = 1.

Since K is a freely adjustable parameter, and we often get better performance by increasing K, it is of interest to study variation of the closed loop poles as a function of K. Without loss of generality, may assume N(s), D(s) \underline{\underline{monic}} degree of D = n, degree of N = m, with n ≥ m. The (positive) root locus is the locus of the roots of D(s) + KN(s) as a function of K, for 0 ≤ K < ∞. The root locus equation

D(s) + KN(s) = 0

is equivalent to

L(s) = \frac{N(s)}{D(s)} = −\frac{1}{K}

This implies that a point s is on the root locus if and only if

argL(s) = (2k + 1)π

i.e., ∑ angle at zeros - ∑ angle at poles = (2k + 1)π (including multiplicity). This is referred to as the angle condition. The corresponding gain K is given by K = \frac{1}{|L(s)|}.

Properties:

1. There are n branches, each a continuous path.

2. Begin at poles. m branches ⟶_{K→∞} zeros, n − m branches → ∞. This is because each closed loop pole s_{0} satisfies \frac{1}{K} D(s_{0}) + N(s_{0}) = 0. If |s_{0}| bounded as K → ∞ ⇒ s_{0} tends to a zero. This accounts for m branches. For the remaining n − m branches, |s_{0}| is unbounded.

3. Symmetry with respect to real axis.

4. A point s on the real axis that is not a pole or zero is on the root locus if and only if it lies to the left of odd number of poles and zeros.

contrib = 0 s

s

From the above figures, we see that the only poles and zeros which make a net nonzero contribution to the angle of L(s) are those on the real axis to the right of the point s. Let a = # of zeros to the right of s, and b = # of poles to the right of s. By the angle condition, we must have (a − b)π = odd multiple of π. Since a − b is odd ⇔ a + b is odd, this proves property (4).

5. For n = m, all branches are bounded. For n > m, n − m branches tend to ∞ as K → ∞ along asymptotes radiating from centroid σ_{0} on the real axis

σ_{0} = \frac{∑ poles − ∑ zeros}{n − m}

with asymptotic angles \frac{(2k+1)π}{n−m}. n − m is called the relative degree of L(s).

Heuristic explanation:

\frac{s^{n} + a_{n−1}s^{n−1}... + a_{n}}{s^{m} + b_{m−1}s^{m−1} + ... + b_{m}} + K = 0

\frac{s^{n}}{s^{m}} (\frac{1 + a_{n−1}s^{−1} + ...}{1 + b_{m−1}s^{−1}...}) + K

s^{n−m} (1 + \frac{a_{n−1} − b_{m−1}}{s} + ...) + K = 0

s^{n−m} (1 + \frac{a_{n−1} − b_{m−1}}{s} ) ≃ −K

s (1 + \frac{a_{n−1} − b_{m−1}}{s} )^{\frac{1}{n−m}} ≃ K^{\frac{1}{n−m}} e^{i(2k+1) \frac{π}{n−m}}

s (1 + \frac{a_{n−1} − b_{m−1}}{s(n − m)} ) ≃ K^{\frac{1}{n−m}} e^{i(2k+1) \frac{π}{n−m}}

Let s = σ + iω. We have

σ + iω − σ_{0} ≃ K^{\frac{1}{n−m}} [cos ((2k + 1)\frac{π}{n − m}) + n sin ((2k + 1)\frac{π}{n − m})]

where

σ_{0} = −\frac{a_{n−1} − b_{m−1}}{(n − m)} = \frac{∑ poles − ∑ zeros}{n − m}

Letting α = \frac{π}{n−m}, we obtain

σ − σ_{0} = K^{\frac{1}{n−m}} cos(2k + 1)α

ω = K^{\frac{1}{n−m}} sin(2k + 1)α

Eliminating K^{\frac{1}{n−m}}, we get

ω = (σ − σ_{0}) tan(2k + 1)α

6. If branches of root locus meet at s, it is necessary that

N(s)D′(s) − D(s)N(s)′ = 0  (4.1)

This is because at s, we must have

D + KN = 0
D′ + KN′ = 0

Eliminating K gives the desired equation. Note that not all solutions of Eq. (4.1) will correspond to meeting points of root locus branches. Those meeting points on the real axis at which branches leave (enter) the real axis are called breakaway (breakin) points.

Simple examples of root locus sketches Root locus examines variation of roots of D(s) + KN(s). Any polynomial dependent on a single parameter can be considered in the same way.

Example

1 −α −5  0 \frac{6(s + α)}{s + 5}  \frac{1}{s(s − 1)} −

s^{3} + 4s^{2} − 5s + 6(s + α) = s^{3} + 4s^{2} + s + 6α
= s(s^{2} + 4s + 1) + 6α

s + 4s + 1 has roots

\frac{−4 ± \sqrt{16 − 4}}{2} = −2 ± \sqrt{3}

Considering 6α as K̄, the root locus for D̄(s) = s(s^{2} + 4s + 1) and N̄(s) = 1 gives locus of the closed loop poles as a function of K̄, hence of α.

4.2 Nyquist stability criterion

(i) Root locus need complete knowledge of open loop poles and zeros (knowing number of poles in RHP not enough)

(ii) Does not really determine stability of feedback system (in root locus, stability determined by Routh Hurwitz)

Graphical method using frequency response.

Based on the following observations

Consider a closed path D in the s-plane with no self-intersections, in the clockwise direction. Suppose D enclosed a point s_{0}. Then as s traverses D, net change in ∠(s − s_{0}) is −2π (i.e. 1 CW rotation)

G(s) = s − s_{0} maps into  G D s_{0}s_{0} becomes new origin in G

Consider a rational function G(s) = K \frac{∏(s − z_{i})}{∏(s − p_{i})}, with K > 0.

Principle of the argument for rational functions

Suppose G has no poles or zeros on D, and D encloses n poles and m zeros of G. Then the image of D under G, G, encircles the origin exactly n − m times c.c.w.

Proof: ∠G(s) = ∠K + ∑_{i} ∠(s − z_{i}) − ∑_{i} ∠(s − p_{i}). By above observation, if z_{i} is enclosed by D, then

net change in ∠(s − z_{i}) = −2π.

Net change in ∠G(s) = m(−2π) − n(−2π) = (n − m)2π ⇒ n − m c.c.w. encirclements.

Nyquist contour or path

\begin{matrix}∞ radius \\ D \end{matrix}

G is called the Nyquist plot. Assume no poles and zeros on D (can always be done) ⇒ Nyquist plot encircles 0 n − m times c.c.w. Now consider the feedback system

- +  F  P KC

Assume PCF has no pole-zero cancellations. Then the feedback system is stable ⇔ G(s) = 1 + KP (s)C(s)F (s) ≠ 0 in Res ≥ 0. Suppose there are n_{0} poles of PCF in Res ≥ 0. Since Poles of G = poles of PCF ⇒ G has n_{0} poles in Res ≥ 0 Then there are no zeros enclosed by D in Res ≥ 0 ⇔ Nyquist plot of G does not pass through 0 and encircles it n_{0} times c.c.w. ⇔ PCF does not pass through −\frac{1}{K} and encircles it n_{0} times c.c.w.

Example 1: 2nd order system (i)

\frac{1}{s(s + α)} = \frac{1}{αs(1 + \frac{s}{α})} α > 0

G(iω) = \frac{1}{iω(iω + α)} = \frac{1}{ω(−ω + iα)}

= \frac{(−ω − iα)}{ω(ω^{2} + α^{2})} = \frac{−1}{ω^{2} + α^{2}} − i\frac{α}{ω(ω^{2} + α^{2})}

∠G(iω) = −90^{o} − tan^{−1} \frac{ω}{α}

−\frac{1}{α^{2}}  for all K > 0 Always stable ∞ −∞

(ii)

P (s) = \frac{1}{s(1 − sT)}

P (iω) = \frac{1}{iω(1 − iωT)} = \frac{T}{1 + ω^{2}T^{2}} − i\frac{1}{ω(1 + ω^{2}T^{2})}

∠P (iω) = −90^{o} + tan^{−1} ωT

−\frac{1}{K}  0+ 0-

P (s) = −\frac{1}{T s(s − \frac{1}{T})}

For K > 0 give the negative root locus for the system. Hence number of closed loop poles in RHP = 1. (iii)

P (s) = \frac{1}{s(−1 + T s)}

P (iω) = \frac{1}{iω(−1 + iωT)} = \frac{−i(−1 − iωT)}{ω(1 + ω^{2}T^{2})} = −\frac{T}{1 + ω^{2}T^{2}} + i\frac{1}{ω(1 + ω^{2}T^{2})}

−\frac{1}{K} 0^{−} 0^{+}

Example 2:

G(s) = \frac{K}{s(1 + τ_{1}s)(1 + τ_{2}s)} τ_{1} > 0, τ_{2} > 0

G(iω) = \frac{K}{iω(1 + iτ_{1}ω)(1 + iτ_{2}ω)}
= \frac{−iK(1 − iτ_{1}ω)(1 − iτ_{2}ω)}{ω(1 + τ_{1}^{2}ω^{2})(1 + τ_{2}^{2}ω^{2})}
= \frac{−iK(1 − τ_{1}τ_{2}ω^{2} − i(τ_{1} + τ_{2})ω)}{ω(1 + τ_{1}^{2}ω^{2})(1 + τ_{2}^{2}ω^{2})}
= \frac{−K(τ_{1} + τ_{2})}{(1 + τ_{1}^{2}ω^{2})(1 + τ_{2}^{2}ω^{2})} − i\frac{K(1 − τ_{1}τ_{2}ω^{2})}{ω(1 + τ_{1}^{2}ω^{2})(1 + τ_{2}^{2}ω^{2})}

G(iω) = −\frac{τ_{1} + τ_{2}}{(1 + τ_{1}^{2}ω^{2})(1 + τ_{2}^{2}ω^{2})} − i\frac{(1 − τ_{1}τ_{2}ω^{2})}{ω(1 + τ_{1}^{2}ω^{2})(1 + τ_{2}^{2}ω^{2})}

0^{+} 0^{−} −\frac{1}{K}

ω^{2} = \frac{1}{τ_{1}τ_{2}}

−\frac{1}{K} < −\frac{τ_{1} + τ_{2}}{(1 + \frac{τ_{1}^{2}}{τ_{1}τ_{2}} )(1 + \frac{τ_{2}^{2}}{τ_{1}τ_{2}} )} = −\frac{τ_{1} + τ_{2}}{(1 + \frac{τ_{1}}{τ_{2}} )(1 + \frac{τ_{2}}{τ_{1}} )}

⇒ \frac{1}{K} > \frac{τ_{1} + τ_{2}}{(1 + \frac{τ_{1}}{τ_{2}} )(1 + \frac{τ_{2}}{τ_{1}} )}

\frac{(1 + \frac{τ_{1}}{τ_{2}} )(1 + \frac{τ_{2}}{τ_{1}} )}{τ_{1} + τ_{2}} > K > 0 for stability

Example 3:

G(s) = \frac{(s + 1)(s + 2)}{s^{3}}

G(iω) = −\frac{3}{ω^{2}} + i\frac{2 − ω^{2}}{ω^{3}}

∠G(iω) = 90^{o} + tan^{−1} ω + tan^{−1} \frac{ω}{2}

ImG = 0 ω = ±\sqrt{2}

At ω^{2} = 2, ReG(iω) = −\frac{3}{2}

−∞ −\frac{3}{2} 0- 0+

No encirclement if 0 > −\frac{1}{K} > −\frac{3}{2}, \frac{1}{K} < \frac{3}{2} or K > \frac{2}{3} for stability

4.3 Design Specifications

Design specifications for simple control systems are often developed based on the transient response of a standard second order system driven by a step input. For the standard second order system

G(s) = \frac{ω_{n}^{2}}{s^{2} + 2ζω_{n}s + ω_{n}^{2}}

the poles are given by

p_{1,2} = −ζω_{n} ± ω_{n}\sqrt{ζ^{2} − 1} ζ ≥ 1
= −ζω_{n} ± iω_{n}\sqrt{1 − ζ^{2}} 0 ≤ ζ < 1

This can be shown pictorially as

ω_{n}\sqrt{1 − ζ^{2}} −ζω_{n} p_{1}  ϕ  ω_{n}  ϕ = cos^{−1} ζ ζ < 1

The step response is given by: for ζ < 1,

y(t) = 1 − \frac{1}{\sqrt{1 − ζ^{2}}} e^{−ζω_{n}t} sin(ω_{n}\sqrt{1 − ζ^{2}}t + ϕ)
= 1 − \frac{ζ}{\sqrt{1 − ζ^{2}}} sin ω_{n}\sqrt{1 − ζ^{2}}te^{−ζω_{n}t} − cos ω_{n}\sqrt{1 − ζ^{2}}te^{−ζω_{n}t}

for ζ = 1,

y(t) = 1 − (1 + ω_{n}t)e^{−ω_{n}t}

for ζ > 1,

\begin{aligned}y(t) = & 1 − ^{\frac{ζ+\sqrt{ζ^{2}−1}}{2\sqrt{ζ^{2}−1}}}e^{−ω_{n}(ζ−\sqrt{ζ^{2}−1})t} \\ & +^{\frac{ζ−\sqrt{ζ^{2}−1}}{2\sqrt{ζ^{2}−1}}}e^{−ω_{n}(ζ+\sqrt{ζ^{2}−1})t} \end{aligned}

0.98 1.02 t_{r} 0.1 0.91  overshoot  2% peak time T_{P}

T_{p} = \frac{π}{ω_{n}\sqrt{1 − ζ^{2}}}

Graph of T_{p} vs. ζ

overshoot = e^{− \frac{ζπ}{\sqrt{1−ζ^{2}}}}

Graph of overshoot vs. ζ

If we let h_{0} to denote the overshoot, we can determine the damping ratio using

ζ^{2} = \frac{(logh_{0})^{2}}{π^{2} + (logh_{0})^{2}}

Typically, want small overshoot, small rise time (i.e. fast response) small settling time (not much ringing) t_{r} estimated using t_{R}, time to go from 0 to 1.

1 = y(t_{R}) = 1 − \frac{1}{\sqrt{1 − ζ^{2}}} e^{−ζω_{n}t} sin(ω_{n}\sqrt{1 − ζ^{2}}t + ϕ)

⇒ sin(ω_{n}\sqrt{1 − ζ^{2}}t_{R}) + ϕ) = 0
0 = sin ω_{n}\sqrt{1 − ζ^{2}}t_{R} cos ϕ + cos(ω_{n}\sqrt{1 − ζ^{2}}t_{R}) sin ϕ
= ζ sin ω_{n}\sqrt{1 − ζ^{2}}t_{R} + cos ω_{n}\sqrt{1 − ζ^{2}}t_{R}\sqrt{1 − ζ^{2}}

⇒ tan(ω_{n}\sqrt{1 − ζ^{2}}t_{R}) = −\frac{\sqrt{1 − ζ^{2}}}{ζ} = − tan(π − ω_{n}\sqrt{1 − ζ^{2}}t_{R})

⇒ ω_{n}\sqrt{1 − ζ^{2}}t_{R} = − tan^{−1} \frac{\sqrt{1 − ζ^{2}}}{ζ}

t_{R} = −\frac{1}{ω_{n}\sqrt{1 − ζ^{2}}} tan^{−1} \frac{\sqrt{1 − ζ^{2}}}{ζ} = \frac{π − ϕ}{ω_{n}\sqrt{1 − ζ^{2}}} = \frac{π − cos^{−1} ζ}{ω_{n}\sqrt{1 − ζ^{2}}}

Recall tan(180 − α) = − tan α

∴ π − ω_{n}\sqrt{1 − ζ^{2}}t_{R} = tan^{−1} \frac{\sqrt{1 − ζ^{2}}}{ζ} = ϕ

⇒ t_{R} = \frac{π − ϕ}{ω_{n}\sqrt{1 − ζ^{2}}}

Can estimate settling time t_{s} as follows: ζ < 1

|e(t)| = |\frac{e^{−ζω_{n}t}}{\sqrt{1 − ζ^{2}}} sin(ω_{n}\sqrt{1 − ζ^{2}}t + ϕ)|
≤ \frac{e^{−ζω_{n}t}}{\sqrt{1 − ζ^{2}}}

If ζ < 0.8, |e(t)| ≤ 1.7e^{−ζω_{n}t_{s}}. For |e(t)| ≤ 0.02, set t_{s} ≥ \frac{4.5}{ζω_{n}} = \frac{4.5}{σ} where σ is real part of pole.

Thus the overshoot requirement specifies the range of values for ζ, which in turn specifies the range of values of the angle ϕ. The settling time requirement specifies the magnitude of the real part of the poles. Combining the 2 requirements gives rise to a region in the complex plane. For example, if the overshoot is required to be less than 5%, then we want which translates into ϕ < 45^{∘}. If, in addition, we want the settling time to be less than 5 sec., we would need the poles to lie to left of the line Re s = −0.9. The desired region in the complex plane for the location of the 2 complex conjugate poles are shown below.

-0.9 45^{∘} Im  Re

For second order system

+ -  K  \frac{1}{s(s + τ)}  K > 0

Closed loop transfer function is \frac{k}{s^{2} + τs + k}, ω_{n} = \sqrt{K}, ζ = \frac{τ}{2\sqrt{K}}

\begin{matrix}⇒ & K ↑ ∞, ω_{n} ↑ ∞ \\ ζ ↓ 0 \end{matrix}

T_{p} = \frac{π}{ω_{n}\sqrt{1 − ζ^{2}}} = \frac{π}{\sqrt{K}\sqrt{1 − \frac{τ^{2}}{4K}}}

⇒ K ↑, T_{p} ↓ 0

Corresponds to arbitrarily fast response.