frequency response time-domain vs frequency-domain ? revision 8.5 of july 17, 2015 adapted from ch.9...

Frequency Response

Time-domain vs

Frequency-domain ?

Revision 8.5 of July 17, 2015

Adapted from Ch.9 of Romagnoli & Palazoglu’s book

see also Ch.17 and 18 - Stephanopoulos, “Chemical process control: an Introduction to theory and practice”

Frequency Response: definition

2

In simplest terms, if a sine wave is inputed to a system at a given frequency, a linear system will respond with another oscillating wave at that same frequency with a certain magnitude and a certain phase angle with respect to the input.

A curve representing the output-to-input relationship of a dynamic system as a function of frequency (e.g., with the input being the sine function)

System Transfer Function Response to the input Asin(ωt)

1st order

purely capacitive 1st order

2nd order

dead time

PID controller

Summary of Linear Dynamic Systems

3

2

1

222

)(1

2tan

)sin()2()(1

)(

for

tAK

ty

t

P

1s

K)s(G

p

p

s

KsG p

'

)(

Let’s calculate Re[G(j and Im[G(j, then mod G(j)and arg (G(j))

Introduction to Process Control Romagnoli & Palazoglu

Frequency Response:First-Order Process

4

By comparison to AR and from the response to A sin(ωt):

Angle of a complex number in Matlab

>> phi=angle(p)

Frequency Response: a lucky circumstance !

6

for a 1st order system:•AR=B/A=mod[G1st(jω)]•Φ=arg G1st(jω)

This result is GENERAL !

1.retake G(s)2.place s = jω3.determine mod G(jω) and arg G(jω)4.AR=B/A=mod[G(jω)]5.Φ=arg G(jω) see ch.17

Bode-Nyquist Diagrams

The response of a linear constant coefficient system to a sinusoidal input signal is an output sinusoidal signal at the same frequency as the input.

The frequency response of a system is characterized by its Amplitude Ratio AR phase angle Φ

However, the magnitude and phase of the output signal differ from those of the input sinusoidal signal, and the amount of difference is a function of the input frequency.

frequency ω is a parameter


Bode Diagrams(Logarithmic Plot)

Nyquist Diagram(Polar Plot)

7

10-2

10-1

100

101

102

-100

-50

0


Bode Diagrams consist of a pair of plots showing:How AR varies with frequencyHowvaries with frequency

10-2

10-1

100

101

102

0

1

2AR

Bode Diagrams

8

sometimes a normalized AR/K is plotted and a normalized ω is used

10-2

10-1

100

101

10-2

10-1

100

101

10-2

10-1

100

101

-100

-80

-60

-40

-20

0

c=0.33

AR


Bode Diagram

Bode DiagramsExample 1 - First-Order Process

9

Corner Fequency: ωc = 1/P

asymptotes

slope -1


Bode DiagramsExample 1 - First-Order Process

10

Corner frequency:inflection point at ϕ = 45°

ωc = 1/P

Low frequency asymptote

High frequency asymptote

0)(lim

)(lim

0

0

jG

KjG P

90)(lim

0)(lim

jG

jG

Example 1

ωc = 1/3 rad/s

1st order system

Bode DiagramsFirst-Order Purely Capacitive Process

j

KjG p

'

)(

Corner Fequency:none



90)(lim

)(lim

0

0

jG

jG

90)(lim

0)(lim

jG

jG

slope -1

11

Bode DiagramsSecond-Order Overdamped Process

Corner Fequency:ωc = 1/

Low frequency asymptotes

High frequency asymptotes

0)(lim

)(lim

0

0

jG

KjG p

slope -2

12

Bode DiagramsEx. of 2nd-Order Process: Corner Fequency

13

14.1

12

ss

sG

ωc = 1/ =1rad/s

G(s)=N(s)/D(s)

N(s)=1D(s)=s2+1.4s+1

D(s)= s2+2ζωns+ωn2

ωn=1rad/sζ=0.7

-80

-60

-40

-20

0

20

Mag

nitu

de (

dB)

10-2

10-1

100

101

102

-180

-135

-90

-45

0

Pha

se (

deg)

Bode Diagram

Frequency (rad/sec)

asymptotes

slope -2

Example 2

Bode Diagrams2nd Order Process: Phase diagram

180)(lim

90)(lim

0)(lim

1

0

jG

jG

jG

14

2

1

)(1

2tan

jG

0)(1 2

0)(1 2

Bode DiagramsSecond-Order Critically Damped Process




0)(lim

)(lim

0

0

jG

KjG p

180)(lim

0)(lim

jG

jG

15

Try construction of Bode Diagramsin Matlab and/or Sisotool !

Bode DiagramsSecond-Order Underdamped Process




0)(lim

)(lim

0

0

jG

KjG p

slope -2

16

5.0 with

12)(

22

ss

KsG p

Bode Diagrams2nd-Order Underdamped Process: resonance frequency

r n 1 2 2

0.707

Mp G r 1

2 1 2

0.707

adapted form Pribeiro - Calvin College 17

The abscissa of the max:

The existence condition:

The ordinate of the max:

Bode Plot 2nd-Order Underdamped

adapted form Pribeiro - Calvin College18

Bode Plot 2nd-Order Underdamped

adapted form Pribeiro - Calvin College 19

slope -2

!

!

Bode DiagramsSecond-Order Undamped Process



0)(lim

)(lim

0

0

jG

KjG p

20

1 with 1s

K)s(G

22

p

)j(Glimn

At the Resonance Fequency ωn = 1/ Bode diagrams generated by SisoTool

(with Kp = = 1)

Bode DiagramsSummary of Second-Order Systems

21

Bode DiagramsDead Time

Corner Fequency:none



)(lim

1)(lim

jG

jG

22

Bode diagrams generated by Matlab

Bode DiagramsPI controller

23

Try with Sisotool !

Bode DiagramsPD controller

Corner Fequency:ωc = 1/D



24

slope +1

Bode DiagramsPID controller

1 with

11

ID

IDcPID s

sG

Corner Fequency:ωc = 1/D = 1/I



slope +1slope -1

25 Bode diagrams generated by SisoTool


Corner Fequency:ωc1 = 1/D

ωc2 = 1/I



26

see risposta_in_frequenza_PID.PDF




27

Corner Fequency:ωc1 = 1/D

ωc2 = 1/I

see risposta_in_frequenza_PID.PDF

The plots in Ch.17 - Stephanopoulos, “Chemical process control: an Introduction to theory and

practice”

are WRONG !

Bode DiagramsSystems in Series

the presence of a constant >0 in the overall transfer function will move the entire AR curve vertically by a constant amount, with no effect on the phase shift

see pag. 330-331, Ch.17 - Stephanopoulos, “Chemical process control: an Introduction to theory and practice”

28

Asymptotic Bode Diagrams

29

If the transfer function is rational, the Bode plots can be replaced by straight line approximations that take the name of straight line Bode plots or uncorrected Bode plots or Asymptotic Bode Diagrams. The Asymptotic Bode Diagrams consider piecewise straight lines only. They are useful because they can be drawn by hand following a few simple rules. For example a First-Order System:

1s

K)s(G

P

P

http://lpsa.swarthmore.edu/Bode/BodeHow.html

Asymptotic Bode Diagramsin MatLab® (asbode.m)

30

The script asbode.m

is required in the folder Asymptotic Bode

http://www.diee.unica.it/giua/ANSIS/


31

; Examination problem of May 2, 2005 - Section 4 Part A; (Problema d’esame del 2.05.05 - Sez. 4 Parte A)

>> Gs=tf([10 0],[1 4 8]) Transfer function: 10 s-------------s^2 + 4 s + 8

>> num=[10 0]num = 10 0

>> den=[1 4 8]den = 1 4 8

>> asbode(num,den)

Guadagno (gain): K = 1.250, K_db = 2 db, phi = 0 degPoli in origine (poles at origin): nu = -1Poli complessi: p,p' = -2.000 +/- j 2.000, omega_n = 2.828, zeta = 0.71(complex poles) beta = 5.1, omega_s = 0.555, omega_d = 14.410 phi = da 0 a -180 deg, Delta M_db = -3 db


32

Examination test of May 2, 2005 – Section 4 Part A

overall system


33

Examination test of May 2, 2005 – Section 4 Part A

Line Element Value Corner freq.(omega_n)rad/s

Left freq. (omega_s)rad/s

Right freq. (omega_d)rad/s

AR ϕ

____ Gain 1.250 NA NA NA 1.250 0 deg

____ Zero at origin

0 NA NA NA from 0 to ∞

+90 deg

____ Coniugate poles

-2.000 +/- j 2.000

2.828 0.555 14.410 from 1 to 0

from 0 to −180 deg

____ Asymptotic Bode

2.828 0.555 14.410 from 0 to 0


- - - - Actual Bode

2.828 from 0 to 0


Frequency Response Methods:Bode Stability Criterion

Bode Stability Criterion: The closed-loop process is stable if the Amplitude Ratio (AR) of the corresponding open-loop transfer function is smaller than 1 (< 0 dB) at the crossover frequency ωco, i.e. the frequency at which the Phase Shift becomes −1800.

Introduction to Process Control Romagnoli & Palazoglu34

see Ch.18 - Stephanopoulos, “Chemical process control: an Introduction to theory and practice”

The gain margin is defined as the change in open loop gain required to make the system closed-loop unstable. The gain margin is the difference between the magnitude curve and 0dB at the point corresponding to the frequency that gives us a phase of -180 deg (the phase crossover frequency, ωco).

• Systems with greater gain margins can withstand greater changes in system parameters before becoming unstable in closed loop.

• Keep in mind that unity gain in magnitude is equal to a gain of zero in dB.The phase margin is defined as the change in open loop phase shift required to make a closed loop system unstable. The phase margin is the difference in phase between the phase curve and -180 deg at the point corresponding to the frequency that gives us a gain of 0dB (the gain cross over frequency, ωgc).

Bode DiagramsGain and Phase Margin

We have the following system: where K is a variable (constant) gain and G(s) is the plant under consideration.


-180

Gain and Phase Margin


dB

ϕ

-180

Bode Stability CriterionExample: 2nd order overdamped & PI

10-2

10-1

100

101

102

10-5

100

105

10-2

10-1

100

101

102

-200

-180

-160

-140

-120

-100

-80

AR

crossover

frequency ωco

k 0.5, I 0.5

This closed-loop system is stable as the AR is less than 1 at the crossover frequency.

The distance to instability can be quantified as gain and phase margins.

Gain margin is defined as GM=1/ARc, where ARc is the

amplitude ratio at the crossover frequency.

Phase margin is defined as PM=|-1800 - c|=1800-|c|,

where c is the phase shift

corresponding to an AR of 1.

Introduction to Process Control Romagnoli & Palazoglu37

Limitations to the Bode Stability Criterion

The Bode stability criterion is not straightforward when there are multiple crossover points or at least one of the Bode plots is non-monotone.

When the magnitude does not decrease monotonously (see this example!), we need to assess the stability situation at higher frequencies, or in other words, crossover at phase angles larger than –180°.

Theoretically, we need to look at all multiples: (–180° k360) with kN

Process Control P.C. Chau © 2002 38

Limitations to the Bode Stability Criterion

The Bode stability criterion is not applicable in a direct way, that is just according to its simple statement, when:

•the gain is negative

•the transfer function is not open-loop stable (poles with positive real part) •the transfer function has zeroes with positive real part•the phase plot exhibits more than one crossover frequency (at ϕ = –180°)•the AR plot presents more than one crossover (at AR=1).

39

Sisotool reports a statement about closed loop system stability or instability even in these cases !

Bode Stability CriterionExample: crossover frequency NOT located at –180°

When no limitations hold for the application of the Bode Stability Criterion, the crossover frequency is to be located at a multiple of –180°(–180° k360°with kN)

40

Bode diagrams generated by Sisotool

The Nyquist diagram contains the same information as the Bode diagrams for the same

system !!!


Nyquist Diagram

41

The Nyquist diagram maps the positive imaginary axis from the s-plane (s=jω) into a curve on the G-plane (G(jω)) !!!


Nyquist Diagram

42

s=jω, ω≥0 G(jω)

A Nyquist diagram (polar plot) is an alternative way to represent the frequency response.

A specific value of frequency defines a point on this plot,

e.g., for the point A.

Im G (j), ordinateRe G (j), abscissa


Nyquist Diagram

0

0

AR

Im

Re

A

43

First-Order Process


bja1

k)(j

1

k =

1

)1j(k

)1j(

)1j(

1j

k)j(g

2222

22

bja1

k)(j

1

k =

1

)1j(k

)1j(

)1j(

1j

k)j(g

2222

22

g(s) k

s 1 g(j)

k

j 1

1

k)()j(GImb ;

1

k)j(GRea

2222

1

k)()j(GImb ;

1

k)j(GRea

2222

44


Nyquist Diagram (for 0 ≤ < +)

Example - First-Order Process

1

)()(Im

1)(Re

22

22

kjG

kjG

1

)()(Im

1)(Re

22

22

kjG

kjG

45

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-1

-0.9

-0.8

-0.7

-0.6

-0.5

-0.4

-0.3

-0.2

-0.1

0

Real

Imag

inar

y

ω

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-1

-0.9

-0.8

-0.7

-0.6

-0.5

-0.4

-0.3

-0.2

-0.1

0

RealIm

agin

ary


Bode – Nyquist comparisonExample - First-Order Process

1

)()(Im

1)(Re

22

22

kjG

kjG

1

)()(Im

1)(Re

22

22

kjG

kjG

10-2

10-1

100

101

10-2

10-1

100

101

10-2

10-1

100

101

-100

-80

-60

-40

-20

0

c=0.33

AR

arg(g(j)) tan 1( 3)

Bode Diagram

AR g(j) 2

92 1

46



))j(Garg(

)j(GAR

Nyquist – Pure Integrator

G(s) 1

s

47



))j(Garg(

)j(GAR

Nyquist - Second-Order Process

G(s) Kp

2s2 2s 1

overdamped

48



))j(Garg(

)j(GAR


G(s) Kp

2s2 2s 1

critically damped

49



))j(Garg(

)j(GAR


50


unitcircle


))j(Garg(

)j(GAR


51



)j(Glim

)j(Glim

0)j(Glim

K)j(Glim

1

1

p0

52

1s

K)s(G

22

p

Resonance Fequency: ωn = 1/

Undamped ProcessNyquist Diagram (for 0 ≤ < +)

! real always

1

K

1j

K

)j(G

22

p

22

p


53

1s

K)s(G

22

p

Actually plotted by SisoTool

(with Kp = = 1)

Undamped ProcessNyquist Diagram (for 0 ≤ < +)

!

!

! real always

1

K

1j

K

)j(G

22

p

22

p


))j(Garg(

)j(GAR

Nyquist – Dead Time

54



Nyquist Diagram (for 0 < < +)

))j(Garg(

)j(GAR

Nyquist – PI controller

G(s) Kc 11

Is

Kc=1

55


))j(Garg(

)j(GAR

Nyquist – PD controller

G(s) Kc 1 Ds

Kc=1

56



))j(Garg(

)j(GAR

Nyquist – PID controller

G(s) Kc 1 Ds 1

Is

Kc=1

57


Some Nyquist plots (e.g., PID controllers) are wrong !

Nyquist – PID controllers

§ 17.4 - Stephanopoulos, “Chemical process control: an Introduction to theory and practice”, Prentice Hall, 1984

58

10-2

10-1

100

101

10-2

10-1

100

101

10-2

10-1

100

101

-400

-300

-200

-100

0

AR

)5.0()3(tan))(arg(

19

2)(

1

2

jg

jgAR

With the addition of a delay term, AR remains the same, but the Phase Shift is significantly affected.


13

2)(

5.0

s

esg

sBode Diagram

Example - First-Order Process with Delay

59

-0.5 0 0.5 1 1.5 2-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

Real

Imag

inar

y

g(s) 2 e 0.5s

3s 1

The net effect of the delay is to alter the phase characteristics of the process, which results in the circling of the origin at high frequencies with a decreasing radius.


Nyquist Diagram

Example - First-Order Process with Delay

60


Nyquist Diagram:Gain and Phase Margins

61

PM

GM-1

unit circle

PM = |-1800 - c| =

= 1800 - |c|

Background of the Nyquist Criterion:The Contour

62

A contour is a complicated mathematical construct, but luckily we only need to worry ourselves with a few points about them. We will denote contours with the Greek letter Γ (gamma).

Contours are lines, drawn on a graph on the complex plane, that follow certain rules:

1.The contour must close (it must form a complete loop)2.The contour may not cross directly through a pole of

the system.3.Contours must have a direction (clockwise, generally).

A contour is called "simple" if it has no self-intersections. We only consider simple contours here.

Background of the Nyquist Criterion:The Cauchy's argument principle

63

If we have a contour, Γ, drawn in one plane (say the complex Laplace plane), we can map that contour into another plane, the F(s) plane, by transforming the contour with the function F(s).

The resultant contour will circle the origin point of the F(s) plane N times, where N is equal to the difference between Z and P (the number of zeros and poles of the function F(s), respectively).

Cauchy’s theorem thus tells us that there is a relationship between the value of a contour integral, and the poles that reside within the contour.

Background of the Nyquist Criterion:The Nyquist Contour

64

Since we’re interested in whether there are any right half-plane roots of the closed loop tranfer function, we choose the Nyquist contour encircling the entire unstable region, that is the right half of the complex s plane.

The Nyquist contour is an infinite semi-circle that encircles the entire right-half of the s plane:1.The semicircle travels up the imaginary axis from negative infinity to positive infinity. 2.From positive infinity, the contour breaks away from the imaginary axis, in the clockwise direction.3.Finally, it forms a giant semicircle.

s=jω

s=-jω

s-plane


Background of the Nyquist Criterion:The extended Nyquist Diagram

65

G(jω)s=jω

s=-jω

s-plane

the resulting map is symmetric about the real axis

The extended Nyquist plot is the “mapping” of the Nyquist contour from the s-plane into a curve on the G-plane using a transfer function, G(jω), as the mapping function !!!


Nyquist Diagram extension to - < < + for a 1st-Order Process

Extended Nyquist Diagram. 1st case: the Nyquist plot is closed

Mirror image with respect to the real axis

- = 0

Obtain a unique contour (polar plot) on the complex

plane

Follow the G(s) contour for - < < +

66


Nyquist Diagram (for − < < +)

1s3

2)s(G

Extended Nyquist Diagram. Example 1: a First-Order Process

1

)()(Im

1)(Re

22

22

kjG

kjG

1

)()(Im

1)(Re

22

22

kjG

kjG

67

Nyquist Diagram extension to - < < + for the following transfer function:

Extended Nyquist Diagram. 2nd case: the Nyquist plot is not closed

68

s + 2 G(s) = -----

s^2

-

= 0+

+

= 0-

The Nyquist diagram must be closed. Connect manually the “ω=0-” to “ω=0+”. This should be done clockwise. In this example’s case the clockwise path is not the shortest.

Nyquist Diagram extension to - < < +

Extended Nyquist Diagram. 2nd case: the Nyquist plot is not closed

69

In order to obtain a closed polar plot, we introduce closure at infinity, which consists in rotating clockwise of π angle with an infinite radius for every pole with Re(.)=0

see pag. 14-20prof. Lanari

“Stability – Nyquist”

The (-1, 0) point is so important in the Nyquist plot.

The reason can be deduced from the characteristic equation

1 + GOL(s) = 0

This equation can also be written as GOL(s) = -1, which implies that at the (-1, 0) point:AROL = 1 and ϕOL = -180°

The (-1, 0) point is referred to as the critical point.

Background of the Nyquist Criterion:The critical point

70

Important property of the Nyquist plot

Extended Nyquist Diagram. 3rd case: the Nyquist contour passes through the critical point

71

The closed loop TF GCL(s) has poles with a null real part if and only if the Nyquist plot of the open loop TF GOL(s) passes through the critical point (-1, 0)

marginal stability

orBIBO instability

depending on the multiplicity of poles with a null real part

see pag. 6prof. Lanari

“Stability – Nyquist”

The Extended Nyquist Diagramin MatLab®

72

unit circle

>> nyquist(G)

Right mouse click >>> Characteristics >>> Minimum Stability margins

the Matlab nyquist command does not take poles or zeros on the jw axis into account and therefore produces an incorrect plot.

Stephanopoulos’ formulation:

Most process control problems are open-loop stable. Consequently,

the closed-loop system is unstable if the extended Nyquist plot for GOL(jω) encircles the -1 (critical point), one or more times.

Nyquist Stability Criterion:simplified form I: open-loop stable systems

see:§ 18.4 - Stephanopoulos, “Chemical process control: an Introduction to theory and practice”, Prentice Hall, 1984

73

Nyquist Stability Criterion:simplified form I

Real Axis

Imag

inar

y A

xis

0

0

2

2-2

-2

(-1,0).

unstable

stable

74

The Nyquist diagr. here is without the “mirrored” part!

The unstable one derives from the same transfer function of the stable one after multiplication by a constant K>1

Nyquist Stability Criterion:simplified form I

see:§ 18.4 - Stephanopoulos, “Chemical process control: an Introduction to theory and practice”, Prentice Hall, 1984

75

Stephanopoulos’ practical procedure:1.Have your Nyquist plot on a piece of paper2.

Nyquist Stability Criterion: Given the closed-loop characteristic equation 1 + GOL(s) = 0, if N is the number of times that the extended polar plot of GOL(jω) encircles the (–1,0) point in the clockwise direction as ω is varied from –∞ to +∞, and if P≥0 is the number of the poles of GOL(s) in the right half plane (RHP),then Z = N + P is the number of unstable roots of the closed-loop characteristic equation.

Notes:N Z (NB: a counterclockwise encirclement has a negative sign)

GOL(jω) must not intercept the (–1,0) critical point

Nyquist Stability Criterion:general form

Chemical Process Control - A First Course with MATLAB Chau p.161 76

Wikipedia’s practical procedure:An easy way to remember how to count N in the clockwise direction in the equation:

Z = N + P

draw a 2-color oblique half-straight line from the critical point crossing RED gives +1 crossing BLUE gives -1

Nyquist Stability Criterion

Wikipedia http://commons.wikimedia.org/wiki/File:Nyquist_Criteria.svg

completeNyquist

plot

77

1. It provides a necessary and sufficient condition for closed-loop stability based on the open-loop transfer function.

2. The Nyquist stability criterion allows stability to be determined without computing the closed-loop poles.

3. A negative value of N indicates that the -1 point is encircled in the opposite direction (counter-clockwise). This situation implies that each countercurrent encirclement can stabilize one unstable pole of the open-loop system.

4. Unlike the Bode stability criterion, the Nyquist stability criterion is applicable to open-loop unstable processes.

5. Unlike the Bode stability criterion, the Nyquist stability criterion can be applied when multiple values of c or gc occur.

Nyquist stability criterion:Important features

78

Example 14.6

Evaluate the stability of the closed-loop system in Fig. 14.1 (see below) for:

4( )5 1

s

pesGs

(the time constants and delay have units of minutes)

Gv = 2, Gm = 0.25, Gc = Kc

1.Obtain ωc and Kcu from a Bode plot. 2.Let Kc =1.5Kcu and draw the Nyquist plot for the resulting open-loop system.

79

Example 14.6

Figure 14.1 Block diagram with a disturbance D and measurement noise N.

80

Figure 14.7 Bode plot for Example 14.6, Kc = 1.

10

The Bode plot for GOL and Kc = 1 is shown in Figure 14.7.

For ωc = 1.69 rad/min, OL = -180° and AROL = 0.235. For Kc = 1, AROL = ARG and Kcu

can be calculated from Eq. 14-10. Thus, Kcu = 1/0.235 = 4.25. Setting Kc = 1.5Kcu gives Kc = 6.38.

SolutionExample 14.6

81

Figure 14.8 Nyquist plot for Example 14.6, Kc = 1.5Kcu = 6.38.

SolutionExample 14.6

82

The Nyquist diagr. here is without the “mirrored” part!

Example 2a 4th order Transfer Function

83

1s3s2s5.0s

K)s(G 2c1

with: initial value:

Kc = 1P = 0

08.0jGlim)0(G0

1

Nyquist plot initial point ( = 0):

0jGlim)(G1

Nyquist plot final point ( ):

Extended Nyquist plot initial point (negative frequency ):

0jGlim)(G1

frequency response time-domain vs frequency-domain ? revision 8.5 of july 17, 2015 adapted from ch.9...

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