Control Systems I - ETH Z · Bode plots Potentially misleading results unless the system is open-loop stable and minimum-phase. Easy to represent uncertainty. Easy to draw, this is

Control Systems ILecture 8: Frequency Response

Readings: Astrom & Murray Ch. 8

Jacopo Tani

Institute for Dynamic Systems and ControlD-MAVT

ETH Zurich

November 9, 2018

J. Tani, E. Frazzoli (ETH) Lecture 8: Control Systems I 09/11/2018 1 / 40

Tentative schedule

# Date Topic

1 Sept. 21 Introduction, Signals and Systems2 Sept. 28 Modeling, Linearization

3 Oct. 5 Analysis 1: Time response, Stability4 Oct. 12 Analysis 2: Diagonalization, Modal coordinates5 Oct. 19 Transfer functions 1: Definition and properties6 Oct. 26 Transfer functions 2: Poles and Zeros7 Nov. 2 Analysis of feedback systems: internal stability,

root locus8 Nov. 9 Frequency response, Bode Plots9 Nov. 16 Analysis of feedback systems 2: the Nyquist

condition

10 Nov. 23 Specifications for feedback systems11 Nov. 30 Loop Shaping12 Dec. 7 PID control13 Dec. 14 State feedback and Luenberger observers14 Dec. 21 On Robustness and Implementation challenges


Standard feedback configuration

C (s) P(s)r e u y

−

Transfer functions make it very easy to compose several blocks (controller,plant, etc.) Imagine doing the same with the state-space model!

(Open-)Loop gain: L(s) = P(s)C (s)

Complementary sensitivity: (Closed-loop) transfer function from r to y

T (s) =L(s)

1 + L(s)

Sensitivity: (Closed-loop) transfer function from r to e

S(s) =1

1 + L(s)


Classical methods for feedback control

Remember: the main objective is to assess/design the properties of theclosed-loop system by exploiting the knowledge of the open-loop system, andavoiding complex calculations. We have three main methods:Root Locus

Quick assessment of control design feasibility. The insights are correct andclear.Can only be used for finite-dimensional systems (e.g. systems with a finitenumber of poles/zeros)Difficult to do sophisticated design.Hard to represent uncertainty.

Bode plotsPotentially misleading results unless the system is open-loop stable andminimum-phase.Easy to represent uncertainty.Easy to draw, this is the tool of choice for sophisticated design.

Nyquist plotThe most authoritative closed-loop stability test. It can always be used (finiteor infinite-dimensional systems)Easy to represent uncertainty.Difficult to draw and to use for sophisticated design.



Evan’s Root Locus method

Invented in the late ’40s by Walter R. Evans.Useful to study how the roots of a polynomial (i.e., the poles of a system)change as a function of a scalar parameter, e.g., the “gain.”

k L(s)r y

−

Let us write the loop gain in the “root locus form”

kL(s) = kN(s)

D(s)= k

(s − z1)(s − z2) . . . (s − zm)

(s − p1)(s − p2) . . . (s − pn)

The sensitivity function is

S(s) =e(s)

r(s)=

1

1 + kL(s)=

D(s)

D(s) + kN(s)

The closed-loop poles are the values of s such that:

D(s) + kN(s) = 0.

Note: the closed loop poles depend on terms of the open loop system.


The root locus rules

How do the closed loop poles vary with k?

1 Since the degree of D(s) + kN(s) is the same as the degree of D(s), thenumber of closed-loop poles is the same as the number of open-loop poles.

2 For k → 0, D(s) + kN(s) ≈ D(s), and the closed-loop poles approach theopen-loop poles.

3 For k →∞,

and the degree of N(s) is the same as the degree of D(s), then1kD(s) + N(s) ≈ N(s), and the closed-loop poles approach the open-loop zeros.

If the degree of N(s) is smaller, then the “excess” closed-loop poles “go toinfinity” (we will look into this more).

4 The closed-loop poles need to be symmetric w.r.t. the real axis (i.e., eitherreal, or complex-conjugate pairs), because D(s) + kN(s) has real coefficients.


More rules: The angle and magnitude rules

Let us rewrite the closed-loop characteristic equation as

N(s)

D(s)= − 1

k

5 The angle rule — Take the argument on both sides:

∠(s − z1) + ∠(s − z2) + . . .+ ∠(s − zm)

− ∠(s − p1)− ∠(s − p2)− . . .− ∠(s − pn) =

{180◦(±q 360◦) if k > 00◦(±q 360◦) if k < 0

6 The magnitude rule — Take the magnitude on both sides:

|s − z1| · |s − z2| · . . . · |s − zm||s − p1| · |s − p2| · . . . · |s − pn|

=1

|k|


Graphical interpretation

All points on the complex plane that could potentially be a closed-loop pole(i.e., the root locus) have to satisfy the angle condition—which is essentiallyTHE rule for sketching the root locus.

Re

Im

s

∠s − p1

∠s − p2

∠s − z1

“The sum of the angles (counted from the real axis) from each zero to s,minus the sum of the angles from each pole to s must be equal to 180◦ (forpositive k) or 0◦ (for negative k), ± an integer multiple of 360◦.”


All points on the real axis are on the root locus

All points on the real axis to the left of an even number of poles/zeros (ornone) are on the negative k root locus

All points on the real axis to the right of an odd number of poles/zeros areon the positive k root locus1.

When two branches come together on the real axis, there will be ‘breakaway”or “break-in” points.

Practical rule: Start with a negative root locus (RL for k < 0) on the +∞ side ofthe real axis, and switch signs of the RL every time you meet an odd number ofpoles or zeros.

1Swarthmore College, Linear Physical Systems Analysis: tinyurl.com/yahgqtn2.J. Tani, E. Frazzoli (ETH) Lecture 8: Control Systems I 09/11/2018 10 / 40

https://tinyurl.com/yahgqtn2

Singular Points

Singular points are those points for which there are multiples RL branches fora single value of k (multiple solutions).

The singular points (s, k) are given by the solution to these equations:

n∏i=1

(s − pi ) + km∏i=1

(s − zi ) = 0

d

ds

n∏i=1

(s − pi ) + kd

ds

m∏i=1

(s − zi ) = 0

There are at most n + m − 1 singular points.

Branches of the RL converging in a singular point always have alternatingdirections.

Practical take-away: You can typically find singular points through intuition (exactlocation not necessary for qualitative analysis), not by solving the equations.


Asymptotes

So what happens when k →∞ and there are more open-loop poles than zeros? Wecan see, e.g., from the magnitude condition, that the “excess” closed-loop poleswill have to go to “to infinity” (s →∞).

Since this is the complex plane, we need to identify “in which direction” they gotowards infinity. This is were we use the angle rule again.

If we “zoom out” sufficiently far, the contributions from all the finite open-looppoles and zeros will all be approximately equal to ∠s, and the angle rule isapproximated by (m − n)∠s = −(∠− k ± q360◦), q ∈ N.

In other words, as k →∞, the excess poles will go to infinity along asymptotes atangles of

∠s =∠− k ± q360◦

n −m

These asymptotes meet in a “center of mass” lying on the real axis at

scom =

∑ni=1 pi −

∑mj=1 zj

n −m

(note that poles are “positive” unit masses, and zeros are “negative” unit masses inthis analogy)

A convenient summary of the rules: tinyurl.com/yand99yq.J. Tani, E. Frazzoli (ETH) Lecture 8: Control Systems I 09/11/2018 12 / 40

http://lpsa.swarthmore.edu/Root_Locus/RootLocusRules.pdf

Graphical Conventions

Poles are indicated by crosses.

Zeros are indicated by circles.

The positive root locus (for k > 0) is drawn with a continuous line.

The negative root locus (for k < 0) is drawn with a dashed line.

Each root locus branch has a direction shown by an arrow.

For the positive branches, the arrow indicates k growing from 0 to ∞,

For the negative branches, the arrow indicates k growing from −∞ to 0.


Examples: two distinct real poles

k1

(s−1)(s−2)r y

−

Re

Im


Examples: complex coniugate poles

k1

s2+s+1

r y

−

Re

Im


Examples: adding a zero

ks+1

s2+s+1

r y

−

Re

Im


Examples: higher multiplicity poles

ks+1

s2(s+3)r y

−

Re

Im


Examples: how to find k of intersection with Im(·)?

k1

s(s+1)(s+2)r y

−

Re

Im


Root locus summary (for now)

Great tool for back-of-the-envelope control design, quick check forclosed-loop stability.

Qualitative sketches are typically enough. There are many detailed rules fordrawing the root locus in a very precise way: if you really need to do that,just use a computer or other methods.

Closed-loop poles start from the open loop poles, and are “repelled” by them.

Closed-loop poles are “attracted” by zeros (or go to infinity). Here you seean obvious explanation why non-minimum-phase zeros are in general to beavoided.

Remember that the root locus must be symmetric w.r.t. the real axis.

If denominator factorization is non trivial, how to find immaginary axiscrossings (k that makes the closed loop system unstable)?


Stability and Routh-Hurwitz Criterion

Take a(ny) polynomial p(s) = ansn + an−1s

n−1 + . . .+ a1s + a0.

the Routh-Hurwitz criterion is used to evaluate the signs of the roots of p(s),without calculating them.

1 Build the Routh table (see next slides)2 Check the number of sign variations of the coefficients in the first coloumn3 Routh criterion: assuming the table can be completed (no zeros in the first

column), the roots of p(s) all have (strictly) negative real part if and only ifthe coefficients of the first coloumn all have the same sign. The number ofsign variations indicates the number of roots of p(s) with positive real part.

x = Ax + Bu is asymptotically stable if the Routh table of det(sI − A) canbe completed and there are no sign variations in the first coloumn.

If the table cannot be completed, all we can say is that the system is notasymptotically stable.

The values of k for which we get a 0 in the first column of the Routh table ofD(s) + kN(s) = 0 correspond to the crossing of the RL with the Im(·) axis(i.e., k for which the closed loop system goes unstable).


Routh-Hurwitz Table

Consider the characteristic polynomial

p(s) = ansn + an−1s

n−1 + an−2sn−2 + . . .+ a1s + a0.

sn an an−2 an−4 . . .sn−1 an−1 an−3 an−5 . . .sn−2 bn−1 bn−3 bn−5 . . .sn−3 cn−1 cn−3 cn−5 . . ....

......

......

s0 hn−1

bn−1 = − 1

an−1

∣∣∣∣ an an−2

an−1 an−3

∣∣∣∣ ,


Routh-Hurwitz Table



n−1 + an−2sn−2 + . . .+ a1s + a0.

sn an an−2 an−4 . . .sn−1 an−1 an−3 an−5 . . .sn−2

bn−1 bn−3 bn−5 . . .

sn−3

cn−1 cn−3 cn−5 . . ....

......

......

s0 hn−1

bn−1 = − 1

an−1

∣∣∣∣ an an−2

an−1 an−3

∣∣∣∣ ,


Routh-Hurwitz Table



n−1 + an−2sn−2 + . . .+ a1s + a0.


......

......

s0 hn−1

bn−1 = − 1

an−1

∣∣∣∣ an an−2

an−1 an−3

∣∣∣∣ ,J. Tani, E. Frazzoli (ETH) Lecture 8: Control Systems I 09/11/2018 21 / 40

Routh-Hurwitz Table



n−1 + an−2sn−2 + . . .+ a1s + a0.


......

......

s0 hn−1

bn−3 = − 1

an−1

∣∣∣∣ an an−4

an−1 an−5


Routh-Hurwitz Table



n−1 + an−2sn−2 + . . .+ a1s + a0.


......

......

s0 hn−1

cn−1 = − 1

bn−1

∣∣∣∣an−1 an−3

bn−1 bn−3


Routh-Hurwitz Table



n−1 + an−2sn−2 + . . .+ a1s + a0.


......

......

s0 hn−1

cn−3 = − 1

bn−1

∣∣∣∣an−1 an−5

bn−1 bn−5


Routh-Hurwitz Condition: Example

The Routh-Hurwitz criterion states that the number of roots of p(s) withpositive real parts is equal to the number of sign changes in the first columnof the Routh table.

Example (from root locus):

p(s) = s(s + 1)(s + 2) + k = s3 + 3s2 + 2s + k

s3 1 2s2 3 k

s1 6− k

30

s0 k


Routh-Hurwitz Summary

General method borrowed in CS I;

Very useful to evaluate asymptotic (in)stability for non trivial polynomials,e.g., of closed loop transfer functions;

Can be used as a tool to synthesize controllers with variations on the theme.


Frequency response

Remember the definition of the transfer function: the steady-state responseof a linear, time-invariant systems to a complex exponential input of the formu(t) = est is yss(t) = G (s)est .

In particular, if we choose s = jω (i.e., the real part of the input isRe[e jωt ] = cos(ωt)), and we assume that the system is stable, then thesteady-state output will be given by

yss(t) = G (jω)e jωt = |G (jω)|e jωt+∠G(jω).

The real part of the output is now

Re[yss(t)] = |G (jω)| cos(ωt + ∠G (jω)).


Frequency response

In other words, the steady-state response to a sinusoidal input of frequency ωis a sinusoidal output of the same frequency such that:

1 the amplitude of the output is |G(jω)| times the amplitude of the input;2 the phase of the output lags the phase of the input by ∠G(jω)|.

0 5 10 15 20 25 30-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1Linear Simulation Results

Time (seconds)

Ampl

itude


How can we display/visualize the frequency response?


Frequency response plots

The frequency response G (jω) ∈ C is a complex function of a single realargument ω ∈ R.

We basically have two options to plot the frequency response:

1 A parametric curve showing G(jω) in the complex plane, in which ω isimplicit. This leads to the polar plot and eventually to the Nyquist plot.

2 Two separate plots for, e.g., real and imaginary part of G(jω) or — better —the magnitude and phase of G(jω) as a function of ω. The latter choice leadsto the Bode plot.


The Bode Plot

The Bode plot is composed of two plots: the magnitude and the phase ofthe frequency response of the system.

On the horizontal axis of both plots, we report the frequency ω on alogarithmic scale (base 10).

On the vertical axis we report1 The logarithm of G(jω) (in base 10), or, equivalently, in dB (deciBels). Note

that we use the convention that

|G(jω)|[dB] = 20 log10 |G(jω)|.

Note that “one decade” = 20 dB.2 The phase ∠G(jω). Usually given in degrees (ok to use radians though).

Note: Since magnitudes multiply (i.e., their logs add) and phases add, thesechoices of vertical coordinates makes it possible to just add Bode plots ofserial connections.

Also, inverting the transfer function is equivalent to reflection about thehorizontal axis, in both Bode plots.


Bode plots from data

Assuming that we have a stable plant, we can choose several values of ω, andfor each measure amplitude and phase of the steady-state response to asinusoidal input sin(ωt).

No need for an analytical models—but one can derive an analytical modelfrom the experimental frequency response.

One can also maintain error bounds for the uncertainty in the frequencyresponse.


Bode plots from transfer function

1 Substitute s ← jω in a transfer function

2 Express the transfer function in Bode form, i.e., such that only the followingfactors appear:

constant or static gain: k,

monomial or integrator/differentiator: jω,

binomial or single, stable pole/zero: 1 + jωτ , where τ is the time constant,

trinomial or complex-coniugate, stable poles/zeros: 1 + 2ζj ωωn

+(j ωωn

)2

,

where ζ is thee damping and ωn the natural frequency.

3 Draw each factor independently

4 Add magnitudes (and phases) of numerator terms, subtract magnitudes (andphases) of denominator terms.


Bode plots — static gain

If G (s) = k , then:

|G (jω)|dB = 20log10|k |, ∠G (jω) =

{0 k > 0−180◦ k < 0

-40

-30

-20

-10

0

10

20

30

40

Mag

nitu

de (d

B)

10-2 10-1 100 101 102-180

-135

-90

-45

0

45

90

Phas

e (d

eg)

G(s) = 10

Frequency (rad/s)


Bode plots — Integrator

The frequency response of G (s) = 1/s is G (jω) = 1jω = −j 1

ω , hence

|G (jω)|dB = −20log10ω, ∠G (jω) = −90◦.


Asymptotic Bode plots — single real, stable pole

Consider G (s) = 1 + τs to start with.The magnitude is |G (jω)|dB = 20log10

√1 + ω2τ 2

The phase is ∠G (jω) = tan−1 ωτ . It is useful to construct approximationsfor:

ω << 1|τ | → |G(jω)|dB ' 20log101 = 0dB ,

∠G(jω) ' 0ω >> 1

|τ | → |G(jω)|dB ' 20log10|ωτ | = 20log10ω + 20log10τ ,

∠G(jω) '{

90◦ τ > 0−90◦ τ < 0

G (s) = 11+s →


Asymptotic Bode plots — complex-conjugate, stable poles

Consider G (s) = s2/ω2n + 2ζs/ωn + 1⇒ G (jω) = 1 + 2ζj

ω

ωn+

(jω

ωn

)2

|G (jω)|dB = 20log10

√(1− ω2

ω2n

)2

+ 4ζ2 ω2

ω2n

and ∠G (jω) = tan−1 2ζ ωωn

1−ω2

ω2n

For ω << ωn, G(jω) ≈ 1, i.e.,

|G(jω)| ' 1 = 0dB , ∠G(jω) ' 0.

For ω >> ωn, G(jω) ≈ ω2

−ω2n

, and

|G(jω)|dB ' 20log10

(ω

ωn

)2

= 40log10ω − 40log10ωn,

∠G(jω) '{

180◦ ζ > 0−180◦ ζ < 0


Asymptotic Bode plots — complex-conjugate, stable poles

Re

Im

jωjω


Putting it all together: Bode plots for complicated transfer functions


Example

Sketch the Bode plots of

G (s) =1

2

(s + 2)(s + 10)

(s2 + s + 1)(s + 5)

First thing: write the transfer function in the ”Bode” form:

G (s) = 2(s/2 + 1)(s/10 + 1)

(s2 + s + 1)(s/5 + 1)

Second: draw the Bode plot for each factor in the transfer function.

Third: add all of the above together to get the final Bode plot.


Example

G (s) = 2(s/2 + 1)(s/10 + 1)

(s2 + s + 1)(s/5 + 1)

-70

-60

-50

-40

-30

-20

-10

0

10

20M

agni

tude

(dB)

10-2 10-1 100 101 102 103-180

-135

-90

-45

0

45

90

Phas

e (d

eg)

Bode Diagram

Frequency (rad/s)


Example

G (s) = 2(s/2 + 1)(s/10 + 1)

(s2 + s + 1)(s/5 + 1)

-70

-60

-50

-40

-30

-20

-10

0

10

20

Mag

nitu

de (d

B)

10-2 10-1 100 101 102 103-180

-135

-90

-45

0

45

90

Phas

e (d

eg)

Bode Diagram

Frequency (rad/s)


Documents

Control Systems I - ETH Z · Bode plots Potentially misleading results unless the system is open-loop stable and minimum-phase. Easy to represent uncertainty. Easy to draw, this is