Feedback Handout (copyright)

EE 3CL4, §31 / 69

Tim Davidson

Transferfunctions

Open loopStability &Performance

Closed loopStability &Performance

Open loop vsclosed loop

Step resp.:2nd-orderG(s)A taste ofpole-placementdesign

Extensions

Steady-stateerror

Summary andplan

EE3CL4:Introduction to Linear Control Systems

Section 3: Fundamentals of Feedback

Tim Davidson

McMaster University

Winter 2014

EE 3CL4, §32 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Outline

1 Transfer Function (review)

2 Open loop controlStability & Performance

3 Closed loop controlStability & Performance

4 Open loop control vs closed loop control

5 Step resp.: 2nd-order G(s)A taste of pole-placement designExtensions

6 Steady-state error

7 Summary and plan

EE 3CL4, §34 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Transfer function

• Y (s) = G(s)U(s)

• Stability (more details later):

the output y(t) is bounded for all bounded inputs u(t)if and only ifthe poles of G(s) are in the open left half plane

• Note that in this setting u(t) is an arbitrary input, notnecessarily the unit step function

EE 3CL4, §36 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Open loop controlSomewhat of an oxymoron, but this is conventional terminology

• Denote input to process by U(s)

• Y (s) = G(s)U(s) = G(s)Gc(s)R(s)

• If G(s) = nG(s)dG(s) and Gc(s) = nC (s)

dC (s) ,

Y (s)

R(s)= Gc(s)G(s) =

nC(s)nG(s)

dC(s)dG(s)

• Key quantity, error: E(s) = R(s)− Y (s)

EE 3CL4, §37 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Stability & Performance

• E(s) =(

1−Gc(s)G(s))

R(s) + G(s)Td (s)

• For good tracking want Gc(s) ≈ 1G(s)

• One design strategy: with G(s) = nG(s)dG(s) , Gc(s) = nC(s)

dC(s)

• Put poles of Gc(s) near zeros of G(s), andput zeros of Gc(s) near poles of G(s)

• Known as “pole-zero cancellation”• That strategy very sensitive to knowledge of G(s)• Problematic if G(s) has zeros in the right half plane

EE 3CL4, §38 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan


• E(s) =(

1−Gc(s)G(s))

R(s) + G(s)Td (s)

• Controller cannot mitigate effect of Td (s)

• No capability for disturbance rejection

• If G(s) and Gc(s) are stable, so is the “open loop”.What if G(s) is not stable?

• In theory, controller can stabilize Y (s)R(s) by pole-zero

cancellation• but difficult to implement to sufficient accuracy• cannot stabilize Y (s)

Td (s)

EE 3CL4, §39 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan


• In practice, we rarely model G(s) accurately, and it maychange with age, temperature and many other effects

• What if G(s) not perfectly known?

• Let T (s) = Y (s)R(s) (when Td (s) = 0)

• Let actual process transfer function be G(s) + ∆G(s)

• What is the ratio of relative error in T (s) due to relativeerror in G(s) for small errors?

• That is, what is lim∆G(s)→0∆T (s)/T (s)∆G(s)/G(s)

• Can show (prove for yourself) that this “sensitivity” = 1• That means that for small errors the relative error in

T (s) is the same as the relative error in G(s);we can’t design Gc(s) to make it smaller

EE 3CL4, §311 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Closed loop control

• Error: E(s) = R(s)− Y (s)

• Measured error: Ea(s) = R(s)− H(s)(

Y (s) + N(s))

.

• In the general case, Ea(s) 6= E(s).

• When H(s) = 1 and N(s) = 0, Ea(s) = E(s).

EE 3CL4, §312 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

The output signal

What is the output Y (s)? (Calculate yourself for practice)

Y (s) =Gc(s)G(s)

1 + H(s)Gc(s)G(s)R(s)

+G(s)

1 + H(s)Gc(s)G(s)Td (s)

− H(s)Gc(s)G(s)

1 + H(s)Gc(s)G(s)N(s)

EE 3CL4, §313 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

The error signal, H(s) = 1

What is the error E(s) = R(s)− Y (s)?To simplify things, consider the case where H(s) = 1

E(s) =1

1 + Gc(s)G(s)R(s)

− G(s)

1 + Gc(s)G(s)Td (s)

+Gc(s)G(s)

1 + Gc(s)G(s)N(s)

Recall, Ea(s) = E(s) only if H(s) = 1 and N(s) = 0.

EE 3CL4, §314 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Loop gain, H(s) = 1

Define loop gain: L(s) = Gc(s)G(s)

E(s) =1

1 + L(s)R(s)− G(s)

1 + L(s)Td (s) +

L(s)

1 + L(s)N(s)

What do we want?

EE 3CL4, §315 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Stability, H(s) = 1

E(s) =1

1 + L(s)R(s)− G(s)

1 + L(s)Td (s) +

L(s)

1 + L(s)N(s)

• Stability: bounded inputs lead to bounded errors

• For simplicity, let Td (s) = 0, N(s) = 0

• G(s) = nG(s)dG(s) ; Gc(s) = nC (s)

dC (s) ; L(s) = nC (s)dC (s)

nG(s)dG(s)

• Hence,1

1 + L(s)=

dC(s)dG(s)

dC(s)dG(s) + nC(s)nG(s)

• =⇒ closed loop poles are roots of dC(s)dG(s) + nC(s)nG(s)

• These can be in left half plane even if G(s) is unstable,but they can also be in the right half plane if G(s) is stable

EE 3CL4, §316 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Performance: s-domain,H(s) = 1

E(s) =1

1 + L(s)R(s)− G(s)

1 + L(s)Td (s) +

L(s)

1 + L(s)N(s)

What else do we want, in addition to stability?• Good tracking: E(s) does depend only weakly on R(s)

=⇒ L(s) large where R(s) large

• Good disturbance rejection:=⇒ L(s) large where Td (s) large

• Good noise suppression:=⇒ L(s) small where N(s) large

EE 3CL4, §317 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

A taste of loop shaping,H(s) = 1

Possibly easier to understand in pure freq. domain, s = jω

Recall that L(s) = Gc(s)G(s),G(s): fixed; Gc(s): controller to be designed

• Good tracking: =⇒ L(s) large where R(s) large|L(jω)| large in the important frequency bands of r(t)

• Good dist. rejection: =⇒ L(s) large where Td (s) large|L(jω)| large in the important frequency bands of td (t)

• Good noise suppr.: =⇒ L(s) small where N(s) large|L(jω)| small in the important frequency bands of n(t)

Typically, L(jω) is a low-pass function. How big should |L(jω)| be?

Any constraints? Stability! Any others?

EE 3CL4, §318 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Inherent constraints, H(s) = 1

Define sensitivity: S(s) =1

1 + L(s)

Define complementary sensitivity: C(s) =L(s)

1 + L(s)

E(s) = S(s)R(s)− S(s)G(s)Td (s) + C(s)N(s)

Note that S(s) + C(s) = 1.Trading S(s) against C(s), with stability,

is a key part of the art of control design

EE 3CL4, §319 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Performance: time-domain

• Difficult for arbitrary inputs

• In classical control techniques, typically assessed via

• nature of transient component of step response• how fast does system respond?• how long does it take to settle to new operating point

• steady-state error for constant changes in position, orvelocity or acceleration; that is steady-state error for

• step input; ramp input, parabolic input

EE 3CL4, §320 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Trade-off exampleDisk drive system

Y (s) =5000Ka

s3 + 1020s2 + 20000s + 5000KaR(s)

+s + 1000

s3 + 1020s2 + 20000s + 5000KaTd (s)

Coarsely design Ka to balance properties of step responseand response to step disturbance

EE 3CL4, §321 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Responses for Ka = 10

Disturbance step response and step response

Low gain:• steady-state disturbance might not be negligible• slow transient response for step input

EE 3CL4, §322 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Responses for Ka = 10,100


Medium gain:• steady-state disturbance much reduced• faster transient response for step input,

but now some overshoot

EE 3CL4, §323 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Responses forKa = 10,100,1000


High gain:• steady-state disturbance almost completely rejected• fast transient response for step input,

but now significant overshoot• Actually can show by Routh Hurwitz technique (later)

that loop is unstable for Ka ≥ 4080

EE 3CL4, §324 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Robustness

• What if G(s) is not perfectly known?

• How does T (s) = Y (s)R(s) change as G(s) changes?

• As in open loop case, look at ratio of relative error in T (s)due to a relative error in G(s), for small errors,lim∆G(s)→0

∆T (s)/T (s)∆G(s)/G(s)

• For an open loop system this sensitivity is 1

• For the closed loop system, with H(s) = 1,this sensitivity is S(s) = 1

1+Gc(s)G(s)

• Now can design controller to manage sensitivity

EE 3CL4, §326 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Open loop vs closed loop

Error: E(s) = R(s)− Y (s); Loop gain: L(s) = Gc(s)G(s),

Open loop, with process input Gc(s)R(s) + Td (s)

Eol(s) =(

1− L(s))

R(s)−G(s)Td (s)

Closed loop, with H(s) = 1,

Ecl(s) =1

1 + L(s)R(s)− G(s)

1 + L(s)Td (s) +

L(s)

1 + L(s)N(s)

EE 3CL4, §327 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Advantages of feedback

• Can mitigate disturbances

• Can stabilize (most) unstable processes

• Can mitigate errors in the model of process

EE 3CL4, §328 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Price of feedback

• more components than open loop

• introduces noise, but can “shape” the response

• less gain: Gc(s)G(s)1+Gc(s)G(s) instead of Gc(s)G(s)

• potential for instability even when G(s) is stable

More complicated to analyze and design,but often advantages are worth the price

EE 3CL4, §330 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

A second-order systemNow examine, in detail, a particular class of second ordersystems

Y (s) =G(s)

1 + G(s)R(s) =

ω2n

s2 + 2ζωns + ω2n

R(s)

EE 3CL4, §331 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Step response• What is the step response?

• Set R(s) = 1/s; take inverse Laplace transform of Y (s)

• Y (s) =ω2

n

s(

s2+2ζωns+ω2n

)• For the case of 0 < ζ < 1,

y(t) = 1− 1β

e−ζωnt sin(ωnβt + θ)

where β =√

1− ζ2 and θ = cos−1 ζ.

• Recall pole positions of G(s)1+G(s) (ignore the zero):

EE 3CL4, §332 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Typical step responses, fixed ωn

EE 3CL4, §333 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Typical step responses, fixed ζ

EE 3CL4, §334 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Key parameters of(under-damped) step response

With β =√

1− ζ2 and θ = cos−1 ζ,

y(t) = 1− 1β


EE 3CL4, §335 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Peak time and peak value

y(t) = 1− 1β


• Peak time: first time dy(t)/dt = 0• Can show that this corresponds to ωnβTp = π

• Hence, Tp =π

ωn√

1− ζ2

• Hence, peak value, Mpt = 1 + e−(ζπ/√

1−ζ2)

EE 3CL4, §336 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Percentage overshoot

Let fv denote the final value of the step response.

Percentage overshoot defined as:

P.O. = 100Mpt − fv

fv

In our example, fv = 1, and hence

P.O. = 100 e−(ζπ/√

1−ζ2)

EE 3CL4, §337 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Overshoot vs Peak TimeThis is one of the classic trade-offs in control

EE 3CL4, §338 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Steady-state error, ess

In general this is not zero. (See “Steady-state error” section)

However, for our second-order system,

y(t) = 1− 1β


Hence ess = 0

EE 3CL4, §339 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Settling time

y(t) = 1− 1β


• How long does it take to get within ±x% of final value?• Approx. when e−ζωnt < x/100• When x = 2, that corresponds to ζωnTs ≈ 4;

that is, 4 time constants

• In that case, Ts =4ζωn

EE 3CL4, §340 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Rise time (under-damped)

y(t) = 1− 1β


• How long to get to the target (for first time)?• Tr , the smallest t such that y(t) = 1

EE 3CL4, §341 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

10%–90% Rise time

• What is Tr in over-damped case? ∞• Hence, typically use Tr1, the 10%–90% rise time

EE 3CL4, §342 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

10%–90% Rise time

• Difficult to get an accurate formula• Linear approx. for 0.3 ≤ ζ ≤ 0.8 (under-damped),

EE 3CL4, §343 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Design problem

For what values of K and p is• the settling time ≤ 4 secs, and• the percentage overshoot ≤ 4.3%?

T (s) =G(s)

1 + G(s)=

Ks2 + ps + K

=ω2

n

s2 + 2ζωns + ω2n,

where ωn =√

K and ζ = p/(2√

K )

EE 3CL4, §344 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Pole positions

Ts ≈4ζωn

P.O. = 100 e−(ζπ/√

1−ζ2)

• For Ts ≤ 4, ζωn ≥ 1• For P.O. ≤ 4.3%, ζ ≥ 1/

√2

Where should we put the poles of T (s)?

EE 3CL4, §345 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Pole positions

ζωn ≥ 1 ζ ≥ 1/√

2

s1, s2 = −ζωn ± jωn√

1− ζ2 = −ωn cos(θ)± jωn sin(θ)

where θ = cos−1(ζ).

EE 3CL4, §346 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Final design constraints

ζωn ≥ 1 ζ ≥ 1/√

2

p ≥ 2 p ≥√

2K

EE 3CL4, §347 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Caveat

• Our work on transient response has been for systemswith

T (s) =ω2

n

s2 + 2ζωns + ω2n

• What about other systems?

EE 3CL4, §348 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Poles, zeros and transientresponse

• Consider a general transfer function T (s) = Y (s)R(s)

• Step response: Y (s) = T (s)1s

• Consider case with DC gain = 1; no repeated poles• Partial fraction expansion

Y (s) =1s

+∑

i

Ai

s + σi+∑

k

Bks + Ck

s2 + 2αks + (α2k + ω2

k )

• Step response

y(t) = 1 +∑

i

Aie−σi t +∑

k

Dke−αk t sin(ωk t + θk )

EE 3CL4, §349 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

EE 3CL4, §351 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Steady-state error

E(s) = R(s)− Y (s) =1

1 + Gc(s)G(s)R(s)

If the the conditions are satisfied, the final value theoremgives steady-state tracking error:

ess = limt→∞

e(t) = lims→0

s1

1 + Gc(s)G(s)R(s)

One of the fundamental reasons for using feedback, despitethe cost of the extra components, is to reduce this error.

We will examine this error for the step, ramp and parabolicinputs

EE 3CL4, §352 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Step, ramp, parabolic

EE 3CL4, §353 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Step input

ess = limt→∞

e(t) = lims→0

s1

1 + Gc(s)G(s)R(s)

• Step input: R(s) = As

• ess = lims→0sA/s

1+Gc(s)G(s) = A1+lims→0 Gc(s)G(s)

• Now let’s examine Gc(s)G(s). Factorize num., den.

Gc(s)G(s) =K∏M

i=1(s + zi)

sN∏Q

k=1(s + pk )

where zi 6= 0 and pk 6= 0.• Limit as s → 0 depends strongly on N.• If N > 0, lims→0 Gc(s)G(s)→∞ and ess = 0• If N = 0,

ess =A

1 + Gc(0)G(0)

EE 3CL4, §354 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

System types

• Since N plays such a key role,it has been given a name

• It is called the type number

• Hence, for systems of type N ≥ 1,ess for a step input is zero

• For systems of type 0, ess = A1+Gc(0)G(0)

EE 3CL4, §355 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Position error constant

• For type-0 systems, ess = A1+Gc(0)G(0)

• Sometimes written as ess = A1+Kp

where Kp is the position error constant

• Recall Gc(s)G(s) =K

QMi=1(s+zi )

sNQQ

k=1(s+pk )

• Therefore, for a type-0 system

Kp = lims→0

Gc(s)G(s) =K∏M

i=1(zi)∏Qk=1(pk )

• Note that this can be computed from positions of thenon-zero poles and zeros

EE 3CL4, §356 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Ramp input

• The ramp input, which represents a step change invelocity is r(t) = At .

• Therefore R(s) = As2

• Assuming conditions of final value theorem aresatisfied,

ess = lims→0

s(A/s2)

1 + Gc(s)G(s)= lim

s→0

As + sGc(s)G(s)

= lims→0

AsGc(s)G(s)

• Again, type number will play a key role.

EE 3CL4, §357 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Velocity error constant

• For a ramp input ess = lims→0A

sGc(s)G(s)


QMi=1(s+zi )

sNQQ

k=1(s+pk )

• For type-0 systems, Gc(s)G(s) has no poles at origin.Hence, ess →∞

• For type-1 systems, Gc(s)G(s) has one pole at the origin.Hence, ess = A

Kv, where Kv =

KQ

i ziQk pk

• Note Kv can be computed from non-zero poles and zeros

• Suggests formal definition of velocity error constant

Kv = lims→0

sGc(s)G(s)

• For type-N systems with N ≥ 2, for a ramp input ess = 0

EE 3CL4, §358 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Parabolic input

• The parabolic input, which represents a step change inacceleration is r(t) = At2/2.

• Therefore R(s) = As3

• Assuming conditions of final value theorem aresatisfied,

ess = lims→0

s(A/s3)

1 + Gc(s)G(s)= lim

s→0

As2Gc(s)G(s)

• Again, type number will play a key role.

EE 3CL4, §359 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Acceleration error constant

• For a parabolic input ess = lims→0A

s2Gc(s)G(s)


QMi=1(s+zi )

sNQQ

k=1(s+pk )

• For type-0 and type-1 systems, Gc(s)G(s) has at most onepole at origin. Hence, ess →∞

• For type-2 systems, Gc(s)G(s) has two poles at the origin.Hence, ess = A

Ka, where Ka =

KQ

i ziQk pk

• Again, Ka can be computed from non-zero poles and zeros

• Suggests formal definition of acceleration error constant

Ka = lims→0

s2Gc(s)G(s)

• For type-N systems with N ≥ 3, for a parabolic input ess = 0

EE 3CL4, §360 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Summary of steady-state errors

EE 3CL4, §361 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Robot steering system,P control

Let’s examine a proportional controller:

Gc(s) = K1

• Gc(s)G(s) = K1K/(τs + 1)

• Hence, Gc(s)G(s) is a type-0 system.• Hence, for a step input,

ess =A

1 + Kp

where Kp = K1K .

EE 3CL4, §362 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Robot steering system,P control example

• Let G(s) = 1s+2 = 0.5

0.5s+1 .

• Proportional control, Gc(s) = K1. Choose K1 = 18.

• Since Gc(s)G(s) is type-0:• finite steady-state error for a step,• unbounded steady-state error for a ramp

• In this example, Kp = KK1 = 10

• The steady-state error for a step input will be1

1+Kp= 10% of the height of the step.

• For a unit step the steady-state error will be 0.1.

EE 3CL4, §363 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Robot steering system,P control example

EE 3CL4, §364 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Robot steering system,PI control

Let’s examine a proportional-plus-integral controller:

Gc(s) = K1 +K2

s=

K1s + K2

s

• When K2 6= 0, Gc(s)G(s) = K (K1s+K2)s(τs+1)

• Hence, Gc(s)G(s) is a type-1 system.• Hence, for a step input, ess = 0• For ramp input,

ess =AKv,

where Kv = lims→0 sGc(s)G(s) = KK2

EE 3CL4, §365 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Robot steering system,PI control example

• Same system: G(s) = 1s+2 = 0.5

0.5s+1 .

• Prop. + Int. control, Gc(s) = K1 + Kss = K1s+K2

s .Choose K1 = 18 and K2 = 20.

• Now since Gc(s)G(s) is type-1:• zero steady-state error for a step• finite-steady state error for a ramp

• In this example Kv = KK2 = 10

• The steady-state error for a ramp input will be1

Kv= 10% of the slope of the ramp.

• For a unit ramp the steady-state error will be 0.1.

EE 3CL4, §366 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Robot steering system,PI control example

EE 3CL4, §368 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Summary: Desirable properties

With H(s) = 1, E(s) = R(s)− Y (s), L(s) = Gc(s)G(s),

E(s) =1

1 + L(s)R(s)− G(s)

1 + L(s)Td (s) +

L(s)

1 + L(s)N(s)

• Stability

• Good tracking in the steady state

• Good tracking in the transient

• Good disturbance rejection (good regulation)

• Good noise suppression

• Robustness to model mismatch

EE 3CL4, §369 / 69

Tim Davidson

Transferfunctions





Extensions

Steady-stateerror

Summary andplan

Plan: Analysis and designtechniques

Rest of course: about developing analysis and designtechniques to address these goals

• Routh-Hurwitz:• Enables us to determine stability without having to find

the poles of the denominator of a transfer function

• Root locus• Enables us to show how the poles move as a single

design parameter (such as an amplifier gain) changes

• Bode diagrams• There is often enough information in the Bode diagram

of the plant/process to construct a highly effectivedesign technique

• Nyquist diagram• More advanced analysis of the frequency response that

enables stability to be assessed even for complicatedsystems

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