Control Systems
Lecture 9 Frequency Response
Classical Control – Prof. Eugenio Schuster – Lehigh University 1
Frequency Response
• We now know how to analyze and design systems via s-domain methods which yield dynamical information • The responses are described by the exponential modes – The modes are determined by the poles of the response Laplace Transform
• We next will look at describing system performance via frequency response methods • This guides us in specifying the system pole and zero positions
Classical Control – Prof. Eugenio Schuster – Lehigh University 2
1 Sinusoidal Steady-State Response Consider a stable transfer function with a sinusoidal input: Aω u(t) = Acos(ωt) ⇔ U(s) = s2 +ω 2 The Laplace Transform of the response has poles • Where the natural system modes lie – These are in the open left half plane Re(s)<0 • At the input modes s=+jω and s=-jω"
(s − z1)(s − z2 )(s − zm ) Aω Y(s) = G(s)U(s) = K 2 2 (s − p1)(s − p2 )(s − pn ) (s +ω ) Only the response due to the poles on the imaginary axis remains after a sufficiently long time This is the sinusoidal steady-state response Classical Control – Prof. Eugenio Schuster – Lehigh University 3
Sinusoidal Steady-State Response
• Input u(t) = Acos(ωt +φ) = Acosωtsinφ − Asinωt cosφ
s ω • Transform U (s) = −Acosφ + Asinφ s2 +ω 2 s2 +ω 2
• Response Transform k k* k k k Y(s) = G(s)U(s) = + + 1 + 2 ++ N s − jω s + jω s − p1 s − p2 s − pN • Response Signal forced response natural response j t * j t p t p t p t y(t) = ke ω + k e− ω + k e 1 + k e 2 ++ k e N 1 2N forced response natural response • Sinusoidal Steady State Response t → ∞ jωt * − jωt ySS (t) = ke + k e 0 Classical Control – Prof. Eugenio Schuster – Lehigh University 4
2 Sinusoidal Steady-State Response
• Calculating the SSS response to u( t ) = A cos(ω t + φ ) • Residue calculation k = lim [(s − jω)Y(s)]= lim [(s − jω)G(s)U (s)] s→ jω s→ jω * scosφ −ωsinφ ' * jω cosφ −ωsinφ ' = lim (G(s)(s − jω)A % = G( jω)A( % s→ jω ) (s − jω)(s + jω)& ) 2 jω & 1 1 = AG( jω) e jφ = AG( jω)e j(φ+∠G( jω)) 2 2 • Signal calculation
* −1' k k $ yss (t) = L & + # %s − jω s + jω " = k e j∠K e jωt + k e− j∠K e− jωt = 2 k cos(ωt + ∠K )
yss (t) = AG( jω) cos(ωt +φ + ∠G( jω))
Classical Control – Prof. Eugenio Schuster – Lehigh University 5
Sinusoidal Steady-State Response
• Response to u(t) = Acos(ωt +φ)
is yss = G( jω) Acos(ωt +φ + ∠G( jω))
– Output frequency = input frequency – Output amplitude = input amplitude × |G(jω)| – Output phase = input phase + ∠ G(jω)
• The Frequency Response of the transfer function G(s) is given by its evaluation as a function of a complex variable at s=jω • We speak of the amplitude response and of the phase response – They cannot independently be varied » Bode s relations of analytic function theory
Classical Control – Prof. Eugenio Schuster – Lehigh University 6
3 Frequency Response
• Find the steady state output for v1(t)=Acos(ωt+φ) + sL + R V (s) V1(s) _ 2 - • Compute the s-domain transfer function T(s) R – Voltage divider T(s) = sL + R • Compute the frequency response R 1&ωL# T( jω) = , ∠T( jω) = −tan− 2 2 $ ! R + (ωL) % R "
• Compute the steady state output AR 1 v (t) = cos ωt +φ − tan− (ωL / R) 2SS 2 2 [ ] R + (ωL)
Classical Control – Prof. Eugenio Schuster – Lehigh University 7
Bode Diagrams • Log-log plot of mag(T), log-linear plot of arg(T) versus ω" Bode Diagram 0
-5
-10
-15 Magnitude (dB)
-20 Magnitude (dB) Magnitude
-25
0
-45 Phase (deg) Phase (deg) Phase (deg)
-90 4 5 6 10 10 10 Frequency (rad/sec)
Classical Control – Prof. Eugenio Schuster – Lehigh University 8
4 Frequency Response
u(t) = Acos(ωt +φ) G(s) yss = G( jω) Acos(ωt +φ + ∠G( jω))
Stable Transfer Function
• After a transient, the output settles to a sinusoid with an amplitude magnified by G ( j ω ) and phase shifted by ∠ G ( j ω ) .
• Since all signals can be represented by sinusoids (Fourier series and transform), the quantities G ( j ω ) and ∠ G ( j ω ) are extremely important.
• Bode developed methods for quickly finding G ( j ω ) and ∠ G ( j ω ) for a given G ( s ) and for using them in control design.
Classical Control – Prof. Eugenio Schuster – Lehigh University 9
Bode Diagrams (s − z )(s − z )(s − z ) G(s) = K 1 2 m (s − p1)(s − p2 )(s − pn )
z z z K z z z p p p r1 r2 rm i[φ +(φ1 +φ2 ++φm )−(φ1 +φ2 ++φn )] G(s) = K p p p e r1 r2 rn The magnitude and phase of G(s) when s=jω is given by:
Nonlinear in the magnitudes z z z r1 r2 rm G( jω) = K p p p , r1 r2 rn K z z z p p p ∠G( jω) = φ + (φ1 +φ2 ++φm )− (φ1 +φ2 ++φn ) Linear in the phases
Classical Control – Prof. Eugenio Schuster – Lehigh University 10
5 Bode Diagrams Why do we express G ( jω ) in decibels?
G( jω) dB = 20log G( jω) z z z r1 r2 rm G( jω) = K p p p ⇒ G( jω) dB = ? r1 r2 rn By properties of the logarithm we can write:
z z z p p p 20logG(s) = 20log K + (20logr1 + 20logrm ++ 20logrm )− (20logr1 + 20logr2 ++ 20logrn ) The magnitude and phase of G(s) when s=jω is given by: Linear in the magnitudes (dB)
G(s) = K + r z + r z ++ r z − r p + r p ++ r p dB dB ( 1 dB 2 dB m dB ) ( 1 dB 2 dB n dB ) K z z z p p p ∠G(s) =φ + (φ1 +φ2 ++φm )− (φ1 +φ2 ++φn ) Linear in the phases Classical Control – Prof. Eugenio Schuster – Lehigh University 11
Bode Diagrams Why do we use a logarithmic scale? Let s have a look at our example: R R R 1 T (s) = ⇒ T ( jω) = ⇒ T ( jω) = = sL R j L R 2 2 2 + ω + R + (ωL) & ωL # 1+ $ ! % R " Expressing the magnitude in dB:
2 2 ,ωL ) & ,ωL ) # T( jω) dB = 20log1− 20log 1+ * ' = −10log$1+ * ' ! + R ( % + R ( " Asymptotic behavior:
ω → 0: T( jω) dB → 0 ( ω % R ω → ∞ : T( jω) dB → −20log& # = 20log(R / L)− 20logω = − 20logω ' R / L $ L dB LINEAR FUNCTION in log !!! We plot as a function of log . ω G( jω) dB ω
Classical Control – Prof. Eugenio Schuster – Lehigh University 12
6 Bode Diagrams Why do we use a logarithmic scale? Let s have a look at our example:
R R 1 T(s) = ⇒ T( jω) = = sL R j L R L + ω + jω +1 R Expressing the phase:
& ωL # −1& ωL # ∠T( jω) = ∠log1− ∠$1+ j ! = − tan $ ! % R " % R " Asymptotic behavior:
∠G( jω) """→0 ω→0 G( j ) R 45 ∠ ω ω= = − L ∠G( jω) "ω"→"∞→−90
LINEAR FUNCTION in logω!!! We plot ∠ G ( j ω ) as a function of logω.
Classical Control – Prof. Eugenio Schuster – Lehigh University 13
Bode Diagrams • Log-log plot of mag(T), log-linear plot of arg(T) versus ω" Bode Diagram 0
-5
-10 -20dB/dec
-15 Magnitude (dB)
-20 Magnitude (dB) Magnitude
-25
0
-45 Phase (deg) Phase (deg) Phase (deg)
-90 4 5 6 10 10 10 Frequency (rad/sec)
Classical Control – Prof. Eugenio Schuster – Lehigh University 14
7 Bode Diagrams
Decade: Any frequency range whose end points have a 10:1 ratio A cutoff frequency occurs when the gain is reduced from its maximum passband value by a factor 1 / 2 :
( 1 % 20log& T # = 20logT − 20log 2 ≈ 20logT − 3dB ' 2 MAX $ MAX MAX Bandwith: frequency range spanned by the gain passband Let s have a look at our example:
1 #ω = 0 T( jω) =1 T( j ) ω = 2 ⇒ " *ωL ' !ω = R / L T( jω) =1/ 2 1+ ( % ) R & This is a low-pass filter!!! Passband gain=1, Cutoff frequency=R/L The Bandwith is R/L!
Classical Control – Prof. Eugenio Schuster – Lehigh University 15
General Transfer Function (Bode Diagrams) 2 q m n &, jω ) jω # G( jω)= K ( jω) ( jωτ +1) $* ' + 2ζ +1! o * ω ' ω %$+ n ( n "! The magnitude (dB) (phase) is the sum of the magnitudes (dB) (phases) of each one of the terms. We learn how to plot each term, we learn how to plot the whole magnitude and phase Bode Plot.
Classes of terms:
1- G( jω)= K o 2- m G( jω)= ( jω) 3- n G j j 1 ( ω)= ( ωτ + ) 2 q 4- &, jω ) jω # G( jω) = $* ' + 2ζ +1! * ω ' ω %$+ n ( n "! Classical Control – Prof. Eugenio Schuster – Lehigh University 16
8 General Transfer Function: DC gain
G( jω)= Ko
Magnitude and Phase: G( jω)dB = 20log Ko
# 0 if Ko > 0 ∠G( jω) = " if K 0 G(s)= −10 !± π o < 40 200
180 35 160 30 140 25 120
20 100
Phase0(deg) 80 15 Magnitude0(dB ) 60 10 40 5 20
0 0 #1 0 1 2 #1 0 1 2 10 10 10 10 10 10 10 10 Frequency0(rad/sec) Frequency0(rad/sec) Classical Control – Prof. Eugenio Schuster – Lehigh University 17
General Transfer Function: Poles/zeros at origin G( jω)= ( jω)m
Magnitude and Phase: G( jω)dB = m⋅ 20logω 1 π m = −1,G(s) = ∠G( jω) = m s 2 20 0
10 #20 G(1)dB = 0
0 #40
#10 #60 Phase0(deg) Magnitude/(dB#20 ) dB #80 m⋅ 20 #30 dec #100
#40 #120 #1 0 1 2 #1 0 1 2 10 10 10 10 10 10 10 10 Frequency/(rad/sec) Frequency0(rad/sec)
Classical Control – Prof. Eugenio Schuster – Lehigh University 18
9 General Transfer Function: Real poles/zeros
G( jω)= ( jωτ +1)n
Magnitude and Phase:
2 2 G( jω)dB = n ⋅10log(ω τ +1) ∠G( jω) = n tan−1(ωτ )
Asymptotic behavior:
G( j ) 0 ∠G( jω) """→0 ω dB $ω$<<1$/τ → ω<<1/τ G( j ) n n 20log G( j ) n 90 ω dB $ω$>>1$/τ → ⋅τ dB + ⋅ ω ∠ ω "ω">>1"/τ → ⋅
Classical Control – Prof. Eugenio Schuster – Lehigh University 19
General Transfer Function: Real poles/zeros
10 n = −1,τ =1/10 5 0 1 dB G(s)= #5 n ⋅ 20 s n⋅3dB dec +1 #10 10 #15
#20 Magnitude0(dB ) #25 G( j0) dB = 0dB #30 G( j1/τ ) dB = n ⋅3dB #35 G(∞) = sgn(n)∞dB #40 dB #1 0 1 2 3 10 10 10 10 10 Frequency0(rad/sec)
Classical Control – Prof. Eugenio Schuster – Lehigh University 20
10 General Transfer Function: Real poles/zeros
10 n = −1,τ =1/10 0 #10 1 G(s)= #20 s +1 #30 10 #40 #50
Phase4(deg) #60 n⋅45 #70 ∠G( j0) = 0 #80 G( j1/ ) n 45 #90 ∠ τ = ⋅ #100 #1 0 1 2 3 ∠G( j∞) = n ⋅90 10 10 10 10 10 Frequency4(rad/sec)
Classical Control – Prof. Eugenio Schuster – Lehigh University 21
General Transfer Function: Complex poles/zeros 2 q &, jω ) jω # G( jω) = $* ' + 2ζ +1! * ω ' ω %$+ n ( n "! Magnitude and Phase: 2 2 /& ω 2 # & ω # , G( jω) = q ⋅10log-$1− ! + $2ζ ! * dB $ ω 2 ! $ ω ! .-% n " % n " +* & 2ζω /ω # G j q tan−1 n ∠ ( ω) = ⋅ $ 2 2 ! %1−ω /ωn " Asymptotic behavior:
G( jω) $$$→0 ∠G( jω) """→0 dB ω<<ωn ω<<ωn G( jω) $$$→−2q ⋅ωn + q ⋅40logω ∠G( jω) """→q ⋅180 dB ω>>ωn dB ω>>ωn
Classical Control – Prof. Eugenio Schuster – Lehigh University 22
11 General Transfer Function: Complex poles/zeros
20 q = −1,ωn =1,ζ = 0.05 q ⋅ 2ζ 0 dB 1 #20 dB G(s) = 2 q ⋅ 40 s + 0.1s +1 dec #40
#60 Mag nitude1(dB )
#80 G( j0) dB = 0dB
G( jωn ) = q ⋅ 3dB + ζ #100 dB ( dB )
G( j∞) dB = sgn(q)∞dB #120 #1 0 1 2 3 10 10 10 10 10 Frequency1(rad/sec) MAX ω G( j ) G( j ) q 2 1 2 n ω dB = ωr dB = ⋅( ζ −ζ )dB ⇔ ω = ωr = 1−ζ 2 Classical Control – Prof. Eugenio Schuster – Lehigh University 23
General Transfer Function: Complex poles/zeros
q 1, 1, 0.05 20 = − ωn = ζ = 0 #20 1 G(s) = 2 #40 s + 0.1s +1 #60 #80 #100 ∠G( j0) = 0 q 90 Phase0(deg) ⋅ #120 ∠G( j1/τ ) = q ⋅90 #140 ∠G( j∞) = q ⋅180 #160 #180 #200 #2 #1 0 1 2 10 10 10 10 10 Frequency0(rad/sec)
Classical Control – Prof. Eugenio Schuster – Lehigh University 24
12 Frequency Response: Poles/Zeros in the RHP
• Same G ( j ω ) .
• The effect on ∠ G ( jω ) is opposite than the stable case.
An unstable pole behaves like a stable zero An unstable zero behaves like a stable pole
1 Example: G(s) = s − 2
This frequency response cannot be found experimentally but can be computed and used for control design.
Classical Control – Prof. Eugenio Schuster – Lehigh University 25
Neutral Stability U (s) + KG(s) Y (s) - 1 G(s) = s(s +1)2
Root locus condition:
KG(s) =1, ∠G(s) = −180
At points of neutral stability RL condition hold for s=jω
KG( jω) =1, ∠G( jω) = −180
Stability: At ∠G( jω) = −180 KG( jω) <1 If ↑K leads to instability KG( jω) >1 If ↓K leads to instability
Classical Control – Prof. Eugenio Schuster – Lehigh University 26
13 Stability Margins
The GAIN MARGIN (GM) is the factor by which the gain can be raised before instability results.
UNSTABLE SYSTEM GM <1(GM dB < 0)⇒
GM is equal to at the frequency 1/ KG( jω) (− KG( jω) dB ) where ∠ G ( jω ) = − 180 .
The PHASE MARGIN (PM) is the value by which the phase can be raised before instability results.
PM < 0 ⇒ UNSTABLE SYSTEM PM is the amount by which the phase of G ( j ω ) exceeds
-180° when KG( jω) =1 (KG( jω) dB = 0)
Classical Control – Prof. Eugenio Schuster – Lehigh University 27
Stability Margins 1 G(s) = s(s +1)2 GM
PM
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14 Frequency Response
u(t) = Acos(ωt +φ) G(s) yss = G( jω) Acos(ωt +φ + ∠G( jω))
Stable Transfer Function
j G( j ) G( jω) = G( jω)e ∠ ω BODE plots
G( jω) = Re{G( jω)}+ j Im{G( jω)} NYQUIST plots
Classical Control – Prof. Eugenio Schuster – Lehigh University 29
Nyquist Diagrams G( jω) = Re{G( jω)}+ j Im{G( jω)}= G( jω)e j∠G( jω)
How are the Bode and Nyquist plots related? They are two way to represent the same information j Im{G( jω)}
G( jω) ∠G( jω)
Re{G( jω)}
Classical Control – Prof. Eugenio Schuster – Lehigh University 30
15 Nyquist Diagrams
Find the steady state output for v1(t)=Acos(ωt+φ) + sL + R V (s) V1(s) _ 2 - Compute the s-domain transfer function T(s) R Voltage divider T(s) = sL + R Compute the frequency response R −1&ωL# T( jω) = , ∠T( jω) = −tan $ ! 2 2 % R " R + (ωL) Compute the steady state output AR 1 v (t) = cos ωt +φ − tan− (ωL / R) 2SS 2 2 [ ] R + (ωL)
Classical Control – Prof. Eugenio Schuster – Lehigh University 31
Bode Diagrams • Log-log plot of mag(T), log-linear plot of arg(T) versus ω" Bode Diagram 0
-5 10 -20dB/dec G(s) = -10 s +10 -15
Magnitude (dB) R / L =10 -20 Magnitude (dB) Magnitude
-25
0
-45 Phase (deg) Phase (deg) Phase (deg)
-90 4 5 6 10 10 10 Frequency (rad/sec) [ω]= rad /sec, ω = 2πf , [f ]= Hz Classical Control – Prof. Eugenio Schuster – Lehigh University 32
16 Nyquist Diagrams R R R − jωL R2 − jωRL T ( jω) = = = R + jωL R + jωL R − jωL R2 + ω 2L2
R −1&ωL # T( jω) = , ∠T( jω) = −tan $ ! R2 + (ωL)2 % R " R2 ωRL Re{T( jω)}= , Im{T( jω)}= − R2 +ω 2L2 R2 +ω 2L2
1- ω → 0: T( jω) →1, ∠T( jω) → 0 T( jω) =1
R 2- ω → ∞: T( jω) →0, ∠T( jω) → −90 T( jω) !!!→− j !!!→0 ω→∞ ωL ω→∞
3- Re{T( jω)}= 0 ⇔ω = ∞
4- Im{T( jω)}= 0 ⇔ω = 0,ω = ∞
Classical Control – Prof. Eugenio Schuster – Lehigh University 33
Nyquist Diagrams Im{G( jω)} vs. Re{G( jω)}
10 G(s) = s +10 R / L =10
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17 Nyquist Diagrams
General procedure for sketching Nyquist Diagrams:
• Find G(j0)
• Find G(j∞)
• Find ω* such that Re{G(jω*)}=0; Im{G(jω*)} is the intersection with the imaginary axis.
• Find ω* such that Im{G(jω*)}=0; Re{G(jω*)} is the intersection with the real axis.
• Connect the points
Classical Control – Prof. Eugenio Schuster – Lehigh University 35
Nyquist Diagrams 1 Example: G(s) = s(s +1)2
1 1 (− jω)(1− jω)2 − 2ω + j(ω 2 −1) G( jω) = 2 = 2 2 = 2 jω( jω +1) jω( jω +1) (− jω)(1− jω) ω(ω 2 +1)
1- ω → 0:G( jω) = −2 − j∞
1 2- ω → ∞ :G( jω) !!!→ j !!!→0 ω→∞ ω3 ω→∞
3- Re{G( jω)}= 0 ⇔ω = ∞
1 Im{G( jω)}= 0 ⇔ω =1,ω = ∞ Re{G( j1)}= − 4- 2
Classical Control – Prof. Eugenio Schuster – Lehigh University 36
18 Nyquist Diagrams 1 Example: G(s) = s(s +1)2
Classical Control – Prof. Eugenio Schuster – Lehigh University 37
Nyquist Diagrams from Bode Diagrams 1 G(s) = s(s +1)2 dB − 20 dec dB − 60 dec
−90 −180
− 270 = 90
Classical Control – Prof. Eugenio Schuster – Lehigh University 38
19 Neutral Stability U (s) + KG(s) Y (s) - 1 G(s) = s(s +1)2
Root locus condition:
KG(s) =1, ∠G(s) = −180
At points of neutral stability RL condition hold for s=jω
KG( jω) =1, ∠G( jω) = −180
Stability: At ∠G( jω) = −180 KG( jω) <1 If ↑K leads to instability KG( jω) >1 If ↓K leads to instability
Classical Control – Prof. Eugenio Schuster – Lehigh University 39
Stability Margins
The GAIN MARGIN (GM) is the factor by which the gain can be raised before instability results.
UNSTABLE SYSTEM GM <1(GM dB < 0)⇒
GM is equal to at the frequency 1/ KG( jω) (− KG( jω) dB ) where ∠ G ( jω ) = − 180 .
The PHASE MARGIN (PM) is the value by which the phase can be raised before instability results.
PM < 0 ⇒ UNSTABLE SYSTEM PM is the amount by which the phase of G ( j ω ) exceeds
-180° when KG( jω) =1 (KG( jω) dB = 0)
Classical Control – Prof. Eugenio Schuster – Lehigh University 40
20 Stability Margins 1 G(s) = s(s +1)2 GM
PM
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Stability Margins 1 G(s) = s(s +1)2 1/GM
PM
Classical Control – Prof. Eugenio Schuster – Lehigh University 42
21 Nyquist Stability Criterion
Case 1: No pole/zero within contour
Case 2: Pole/zero within contour
Argument Principle: A contour map of a complex function will encircle the origin Z-P times, where Z is the number of zeros and P is the number of poles of the function inside the contour.
Classical Control – Prof. Eugenio Schuster – Lehigh University 43
Nyquist Stability Criterion
Let us consider this contour and closed-loop system
U (s) Y (s) + KG(s) -
The closed-loop poles are the solutions (roots) of:
1+ KG(s) = 0 The evaluation of H(s) will encircle the origin only if H(s) has a RHP zero or pole
Classical Control – Prof. Eugenio Schuster – Lehigh University 44
22 Nyquist Stability Criterion
Let us apply the argument principle to the function H(s) =1+KG(s).
If the plot of 1+KG(s) encircles the origin, the plot of KG(s) encircles -1 on the real axis.
Classical Control – Prof. Eugenio Schuster – Lehigh University 45
Nyquist Stability Criterion
By writing b(s) a(s) + Kb(s) 1+ KG(s) =1+ K = a(s) a(s) we can conclude that the poles of 1+KG(s) are also the poles of G(s). Assuming no pole of G(s) in the RHP, an encirclement of the point -1 by KG(s) indicates a zero of 1+KG(s) in the RHP, and thus an unstable pole of the closed-loop system.
A clockwise contour of C1 enclosing a zero of 1+KG(s) will result in KG(s) encircling the -1 point in the clockwise direction. A clockwise contour of C1 enclosing a pole of 1+KG(s) will result in KG(s) encircling the -1 point in the counterclockwise direction.
The net number of clockwise encirclements of the point -1, N, equals the number of zeros (closed-loop poles) in the RHP, Z, minus the number of poles (open-loop poles) in the RHP, P: N = Z − P Classical Control – Prof. Eugenio Schuster – Lehigh University 46
23 Nyquist Stability Criterion
U (s) + G(s) Y (s) -
When is this transfer function Stable?
NYQUIST: The closed loop is asymptotically stable if the number of counterclockwise encirclements (N negative) of the point (-1+j0) by the Nyquist curve of G(jω) is equal to the number of poles of G(s) with positive real parts (unstable poles) (P).
Corollary: If the open-loop system G(s) is stable (P=0), then the closed-loop system is also stable provided G(s) makes no encirclement of the point (-1+j0) (N=0) .
Classical Control – Prof. Eugenio Schuster – Lehigh University 47
Nyquist Stability Criterion 1 1 G(s) = G(s) = s4 + 2s3 + 3s2 + 3s +1 s4 + 5s3 + 3s2 + 3s +1
Classical Control – Prof. Eugenio Schuster – Lehigh University 48
24 Specifications in the Frequency Domain
1. The crossover frequency ωc, which determines bandwith ωBW, rise time tr and settling time ts.
2. The phase margin PM, which determines the
damping coefficient ζ and the overshoot Mp.
3. The low-frequency gain, which determines the steady-state error characteristics.
Classical Control – Prof. Eugenio Schuster – Lehigh University 49
Specifications in the Frequency Domain
The phase and the magnitude are NOT independent!
Bode s Gain-Phase relationship:
1 +∞dM ∠G( jωo )= W (u)du π ∫−∞ du
M = ln G( jω)
u = ln(ω /ωo ) W (u) = ln(coth u / 2)
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25 Specifications in the Frequency Domain
The crossover frequency: ωc ≤ ωBW ≤ 2ωc
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Specifications in the Frequency Domain
The Phase Margin: PM vs. Mp
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26 Specifications in the Frequency Domain
PM The Phase Margin: PM vs. ζ ζ ≅ 100
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Frequency Response – Phase Lead Compensators
Ts +1 D(s) = , α <1 αTs +1
1− sinφ α = MAX 1+ sinφMAX 1 1 1 log &log, ) log, )# ωMAX = $ * ' + * '! 2 % +T ( +αT ("
It is a high-pass filter and approximates PD control. It is used whenever substantial improvement in damping is needed. It tends to increase the speed of response of a system for a fixed low-frequency gain.
Classical Control – Prof. Eugenio Schuster – Lehigh University 54
27 Frequency Response – Phase Lead Compensators
1. Determine the open-loop gain K to satisfy error or bandwidth requirements: - To meet error requirement, pick K to satisfy error
constants (Kp, Kv, Ka) so that ess specification is met. - To meet bandwidth requirement, pick K so that the open-loop crossover frequency is a factor of two below the desired closed-loop bandwidth.
2. Determine the needed phase lead → α based on the PM specification. 1− sinφMAX α = 1+ sinφMAX
3. Pick ωMAX to be at the crossover frequency.
4. Determine the zero and pole of the compensator. 1/2 -1/2 z=1/T= ωMAX α p=1/ α T= ωMAX α 5. Draw the compensated frequency response and check PM.
6. Iterate on the design. Add additional compensator if needed.
Classical Control – Prof. Eugenio Schuster – Lehigh University 55
Frequency Response – Phase Lag Compensators
Ts +1 D(s) = α , α >1 αTs +1
It is a low-pass filter and approximates PI control. It is used to increase the low frequency gain of the system and improve steady state response for fixed bandwidth. For a fixed low-frequency gain, it will decrease the speed of response of the system.
Classical Control – Prof. Eugenio Schuster – Lehigh University 56
28 Frequency Response – Phase Lag Compensators
1. Determine the open-loop gain K that will meet the PM requirement without compensation.
2. Draw the Bode plot of the uncompensated system with crossover frequency from step 1 and evaluate the low- frequency gain.
3. Determine α to meet the low frequency gain error requirement.
4. Choose the corner frequency ω=1/T (the zero of the compensator) to be one decade below the new crossover
frequency ωc.
5. The other corner frequency (the pole of the compensator) is then ω=1/ α T.
6. Iterate on the design
Classical Control – Prof. Eugenio Schuster – Lehigh University 57
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