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EE 3CL4, §31 / 78
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 2018
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EE 3CL4, §32 / 78
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
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
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EE 3CL4, §34 / 78
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
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
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EE 3CL4, §36 / 78
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
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)
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EE 3CL4, §37 / 78
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
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
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EE 3CL4, §38 / 78
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
Stability & Performance
• 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)
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EE 3CL4, §39 / 78
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
Stability & Performance
• 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
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EE 3CL4, §311 / 78
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
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).
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EE 3CL4, §312 / 78
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
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)
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EE 3CL4, §313 / 78
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
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.
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EE 3CL4, §314 / 78
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
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?
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EE 3CL4, §315 / 78
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
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
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EE 3CL4, §316 / 78
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
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) depends 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
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EE 3CL4, §317 / 78
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
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?
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EE 3CL4, §318 / 78
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
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
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EE 3CL4, §319 / 78
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
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
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EE 3CL4, §320 / 78
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
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
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EE 3CL4, §321 / 78
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
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
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EE 3CL4, §322 / 78
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
Responses for Ka = 10,100
Disturbance step response and step response
Medium gain:• steady-state disturbance much reduced• faster transient response for step input,
but now some overshoot
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EE 3CL4, §323 / 78
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
Responses forKa = 10,100,1000
Disturbance step response and step response
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
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EE 3CL4, §324 / 78
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
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
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EE 3CL4, §326 / 78
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
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)
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EE 3CL4, §327 / 78
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
Advantages of feedback
• Can mitigate disturbances
• Can stabilize (most) unstable processes
• Can mitigate errors in the model of process
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EE 3CL4, §328 / 78
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
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
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EE 3CL4, §330 / 78
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
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)
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EE 3CL4, §331 / 78
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
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):
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EE 3CL4, §332 / 78
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
Typical step responses, fixed ωn
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EE 3CL4, §333 / 78
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
Typical step responses, fixed ζ
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EE 3CL4, §334 / 78
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
Key parameters of(under-damped) step response
With β =√
1− ζ2 and θ = cos−1 ζ,
y(t) = 1− 1β
e−ζωnt sin(ωnβt + θ)
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EE 3CL4, §335 / 78
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
Peak time and peak value
y(t) = 1− 1β
e−ζωnt sin(ωnβt + θ)
• 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)
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EE 3CL4, §336 / 78
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
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)
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EE 3CL4, §337 / 78
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
Overshoot vs Peak TimeThis is one of the classic trade-offs in control
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EE 3CL4, §338 / 78
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
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
e−ζωnt sin(ωnβt + θ)
Hence ess = 0
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EE 3CL4, §339 / 78
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
Settling time
y(t) = 1− 1β
e−ζωnt sin(ωnβt + θ)
• How long does it take to get (and stay) within ±x% offinal value?
• Tricky.• Instead, approximate by time constants of envelopes:
1± 1β
e−ζωnt
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EE 3CL4, §340 / 78
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
5% settling time
• After 3 time consts, env. error reduced by factor 1/e3;becomes ≈ 5% of envelope error at t = 0
• In figure, ζ = 0.5, ωn = 1, error bounds at ±0.05:Time constant, 1
ζωn= 2.
After t = 6, envelopes are almost within ±5%Response is within ±5%
• Hence, Ts,5 ≈ 3ζωn
; approx. good for ζ . 0.9
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EE 3CL4, §341 / 78
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
2% settling time
• After 4 time consts, env. error reduced by a factor 1/e4;becomes ≈ 2% of envelope error at t = 0
• In figure, ζ = 0.5, ωn = 1, error bounds at ±0.02:After t = 8, envelopes are almost within ±2%Response is also almost within ±2%
• Hence, Ts,2 ≈ 4ζωn
; approx. good for ζ . 0.9
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EE 3CL4, §342 / 78
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
Rise time (under-damped)
y(t) = 1− 1β
e−ζωnt sin(ωnβt + θ)
• How long to get to the target (for first time)?• Tr , the smallest t such that y(t) = 1
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EE 3CL4, §343 / 78
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
10%–90% Rise time
• What is Tr in over-damped case? ∞• Hence, typically use Tr1, the 10%–90% rise time
![Page 39: EE3CL4: Introduction to Linear Control Systemsdavidson/EE3CL4/slides/Feedback_handout.… · EE3CL4: Introduction to Linear Control Systems Section 3: Fundamentals of Feedback Tim](https://reader031.vdocuments.us/reader031/viewer/2022030909/5b56e1ec7f8b9a835c8cfe8b/html5/thumbnails/39.jpg)
EE 3CL4, §344 / 78
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
10%–90% Rise time
• Difficult to get an accurate formula• Linear approx. for 0.3 ≤ ζ ≤ 0.8 (under-damped),
![Page 40: EE3CL4: Introduction to Linear Control Systemsdavidson/EE3CL4/slides/Feedback_handout.… · EE3CL4: Introduction to Linear Control Systems Section 3: Fundamentals of Feedback Tim](https://reader031.vdocuments.us/reader031/viewer/2022030909/5b56e1ec7f8b9a835c8cfe8b/html5/thumbnails/40.jpg)
EE 3CL4, §345 / 78
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
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 )
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EE 3CL4, §346 / 78
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
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)?
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EE 3CL4, §347 / 78
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
Pole positions
ζωn ≥ 1 ζ ≥ 1/√
2
s1, s2 = −ζωn ± jωn√
1− ζ2 = −ωn cos(θ)± jωn sin(θ)
where θ = cos−1(ζ).
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EE 3CL4, §348 / 78
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
Final design constraints
ζωn ≥ 1 ζ ≥ 1/√
2
p ≥ 2 p ≥√
2K
![Page 44: EE3CL4: Introduction to Linear Control Systemsdavidson/EE3CL4/slides/Feedback_handout.… · EE3CL4: Introduction to Linear Control Systems Section 3: Fundamentals of Feedback Tim](https://reader031.vdocuments.us/reader031/viewer/2022030909/5b56e1ec7f8b9a835c8cfe8b/html5/thumbnails/44.jpg)
EE 3CL4, §349 / 78
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
Caveat
• Our work on transient response has been for systemswith
T (s) =ω2
n
s2 + 2ζωns + ω2n
• What about other systems?
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EE 3CL4, §350 / 78
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
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 )
![Page 46: EE3CL4: Introduction to Linear Control Systemsdavidson/EE3CL4/slides/Feedback_handout.… · EE3CL4: Introduction to Linear Control Systems Section 3: Fundamentals of Feedback Tim](https://reader031.vdocuments.us/reader031/viewer/2022030909/5b56e1ec7f8b9a835c8cfe8b/html5/thumbnails/46.jpg)
EE 3CL4, §351 / 78
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
![Page 47: EE3CL4: Introduction to Linear Control Systemsdavidson/EE3CL4/slides/Feedback_handout.… · EE3CL4: Introduction to Linear Control Systems Section 3: Fundamentals of Feedback Tim](https://reader031.vdocuments.us/reader031/viewer/2022030909/5b56e1ec7f8b9a835c8cfe8b/html5/thumbnails/47.jpg)
EE 3CL4, §353 / 78
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
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
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EE 3CL4, §354 / 78
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
Step, ramp, parabolic
![Page 49: EE3CL4: Introduction to Linear Control Systemsdavidson/EE3CL4/slides/Feedback_handout.… · EE3CL4: Introduction to Linear Control Systems Section 3: Fundamentals of Feedback Tim](https://reader031.vdocuments.us/reader031/viewer/2022030909/5b56e1ec7f8b9a835c8cfe8b/html5/thumbnails/49.jpg)
EE 3CL4, §355 / 78
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
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)
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EE 3CL4, §356 / 78
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
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)
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EE 3CL4, §357 / 78
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
Position error constant
• For type-0 systems, ess = A1+Gc(0)G(0)
• Sometimes written as ess = A1+Kposn
where Kposn is the position error constant
• Recall Gc(s)G(s) =K∏M
i=1(s+zi )
sN∏Q
k=1(s+pk )
• Therefore, for a type-0 system
Kposn = 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
![Page 52: EE3CL4: Introduction to Linear Control Systemsdavidson/EE3CL4/slides/Feedback_handout.… · EE3CL4: Introduction to Linear Control Systems Section 3: Fundamentals of Feedback Tim](https://reader031.vdocuments.us/reader031/viewer/2022030909/5b56e1ec7f8b9a835c8cfe8b/html5/thumbnails/52.jpg)
EE 3CL4, §358 / 78
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
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.
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EE 3CL4, §359 / 78
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
Velocity error constant
• For a ramp input ess = lims→0A
sGc(s)G(s)
• Recall Gc(s)G(s) =K
∏Mi=1(s+zi )
sN∏Q
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 =
K∏
i zi∏k 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
![Page 54: EE3CL4: Introduction to Linear Control Systemsdavidson/EE3CL4/slides/Feedback_handout.… · EE3CL4: Introduction to Linear Control Systems Section 3: Fundamentals of Feedback Tim](https://reader031.vdocuments.us/reader031/viewer/2022030909/5b56e1ec7f8b9a835c8cfe8b/html5/thumbnails/54.jpg)
EE 3CL4, §360 / 78
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
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.
![Page 55: EE3CL4: Introduction to Linear Control Systemsdavidson/EE3CL4/slides/Feedback_handout.… · EE3CL4: Introduction to Linear Control Systems Section 3: Fundamentals of Feedback Tim](https://reader031.vdocuments.us/reader031/viewer/2022030909/5b56e1ec7f8b9a835c8cfe8b/html5/thumbnails/55.jpg)
EE 3CL4, §361 / 78
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
Acceleration error constant
• For a parabolic input ess = lims→0A
s2Gc(s)G(s)
• Recall Gc(s)G(s) =K
∏Mi=1(s+zi )
sN∏Q
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 =
K∏
i zi∏k 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
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EE 3CL4, §362 / 78
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
Summary of steady-state errors
The Kp in this table corresponds to Kposn
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EE 3CL4, §363 / 78
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
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 + Kposn
where Kposn = K1K .
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EE 3CL4, §364 / 78
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
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 = 9
• 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.
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EE 3CL4, §365 / 78
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
Robot steering system,P control example
• Left: y(t) for unit step input, r(t) = u(t)• Right: y(t) for unit ramp input, r(t) = tu(t)
![Page 60: EE3CL4: Introduction to Linear Control Systemsdavidson/EE3CL4/slides/Feedback_handout.… · EE3CL4: Introduction to Linear Control Systems Section 3: Fundamentals of Feedback Tim](https://reader031.vdocuments.us/reader031/viewer/2022030909/5b56e1ec7f8b9a835c8cfe8b/html5/thumbnails/60.jpg)
EE 3CL4, §366 / 78
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
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
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EE 3CL4, §367 / 78
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
Robot steering system,PI control example
• Same system: G(s) = 1s+2 = 0.5
0.5s+1 .
• Prop. + Int. control, Gc(s) = K1 + K2s = 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.
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EE 3CL4, §368 / 78
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
Robot steering system,PI control example
• Left: y(t) for unit step input, r(t) = u(t)• Right: y(t) for unit ramp input, r(t) = tu(t)
![Page 63: EE3CL4: Introduction to Linear Control Systemsdavidson/EE3CL4/slides/Feedback_handout.… · EE3CL4: Introduction to Linear Control Systems Section 3: Fundamentals of Feedback Tim](https://reader031.vdocuments.us/reader031/viewer/2022030909/5b56e1ec7f8b9a835c8cfe8b/html5/thumbnails/63.jpg)
EE 3CL4, §369 / 78
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
Robot steering system,PI2I control
Let’s examine a PI plus double integral controller:
Gc(s) = K1 +K2
s+
K3
s2 =K1s2 + K2s + K3
s2
• When K3 6= 0, Gc(s)G(s) = K (K1s2+K2s+K3)s2(τs+1)
• Hence, Gc(s)G(s) is a type-2 system.• Hence, for a step input or a ramp input, ess = 0• For parabolic input,
ess =AKa,
where Ka = lims→0 s2Gc(s)G(s) = KK3
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EE 3CL4, §370 / 78
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
Robot steering system,PI2I control example
• Same system: G(s) = 1s+2 = 0.5
0.5s+1 .
• Prop. + Int. + double int. control, Gc(s) = K1 + K2s + K3
s2 .Choose K1 = 18, K2 = 20, K3 = 20.
• Now since Gc(s)G(s) is type-2:• zero steady-state error for a step or a ramp• finite-steady state error for a parabolic
• In this example Ka = KK3 = 10
• The steady-state error for a parabolic input would be1
Kv= 10% of the curvature of the parabola.
• For a unit parabola the steady-state error would be 0.1.
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EE 3CL4, §371 / 78
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
Robot steering system,PI2I control example
• Left: y(t) for unit step input, r(t) = u(t)• Right: y(t) for unit ramp input, r(t) = tu(t)
![Page 66: EE3CL4: Introduction to Linear Control Systemsdavidson/EE3CL4/slides/Feedback_handout.… · EE3CL4: Introduction to Linear Control Systems Section 3: Fundamentals of Feedback Tim](https://reader031.vdocuments.us/reader031/viewer/2022030909/5b56e1ec7f8b9a835c8cfe8b/html5/thumbnails/66.jpg)
EE 3CL4, §372 / 78
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
Robot steering system,PI2I control example
• y(t) for unit step input, r(t) = u(t), extended time scale
![Page 67: EE3CL4: Introduction to Linear Control Systemsdavidson/EE3CL4/slides/Feedback_handout.… · EE3CL4: Introduction to Linear Control Systems Section 3: Fundamentals of Feedback Tim](https://reader031.vdocuments.us/reader031/viewer/2022030909/5b56e1ec7f8b9a835c8cfe8b/html5/thumbnails/67.jpg)
EE 3CL4, §373 / 78
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
Robot steering system,PI2I control example
• y(t) for unit parabolic input, r(t) = t2u(t)• For this slide only, the gains have been reduced to
illustrate the effects, K1 = 1.8, K2 = 0.2, K3 = 0.02
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EE 3CL4, §374 / 78
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
Transient responses and polesShould we have been able to predict transient responses from pole (andzero) positions? Return to case of K1 = 18,K2 = K3 = 20
Closed loop transfer functions, T (s) = Y (s)R(s)
:
P one real pole, time const. = 1/20 = 0.05s
PI one real pole near the P one; plus another real pole (timeconst. ≈ 1s) that is close to a zero
PI2I one real pole near the P one; plus a conjugate pair withtime const. ≈ 2s, angle ≈ 60◦, but near zeros
![Page 69: EE3CL4: Introduction to Linear Control Systemsdavidson/EE3CL4/slides/Feedback_handout.… · EE3CL4: Introduction to Linear Control Systems Section 3: Fundamentals of Feedback Tim](https://reader031.vdocuments.us/reader031/viewer/2022030909/5b56e1ec7f8b9a835c8cfe8b/html5/thumbnails/69.jpg)
EE 3CL4, §375 / 78
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
Step responsesTo highlight the impacts of the different poles, we have done a partial fractionexpansion of the transfer function and used that to compute the step response
Control T (s) = Y (s)R(s)
Step Response, for t ≥ 0
P = 18s+20 = 0.9− 0.9e−20t
PI = 18s+20s2+20s+20
' 17.94s+18.94 + 0.0557
s+1.056 ' 1− 0.947e−18.94t − 0.053e−1.056t
PI2I = 18s2+20s+20s3+20s2+20s+20
' 17.89s+19.00 + 0.1106(s+0.5578)
s2+0.9971s+1.0525' 1− 0.942e−19.00t . . .
−0.108e−0.498t sin(0.897t + 2.57)
Notes:• 10% steady state error in the P case; it is zero in other cases• Second term for each system has a similar decay rate (similar pole
positions)• Third term in PI case decays much more slowly; third term in PI2I case even
slower (small real parts of these poles)• Terms related to poles that are near zeros have comparatively small
magnitudes
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EE 3CL4, §377 / 78
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
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
![Page 71: EE3CL4: Introduction to Linear Control Systemsdavidson/EE3CL4/slides/Feedback_handout.… · EE3CL4: Introduction to Linear Control Systems Section 3: Fundamentals of Feedback Tim](https://reader031.vdocuments.us/reader031/viewer/2022030909/5b56e1ec7f8b9a835c8cfe8b/html5/thumbnails/71.jpg)
EE 3CL4, §378 / 78
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
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