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Ch. 13 Frequency analysisPowerPoint Presentation

Ch. 13 Frequency analysis

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Ch. 13 Frequency analysis

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Ch. 13 Frequency analysis

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Force linear system with input x(t) = A sin t .

Here is the output y(t):

Chapter 14

u(t) = A sin(!t)

u

y

As t!1:

y(t) = AR¢ A sin(!t + Á)

General (VERY SIMPLE):

AR = |G(j!)|

Á = Å G(j!)

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SINUSOIDAL RESPONSE OF FIRST-ORDER SYSTEM

k = 1, ¿ = 1 s

y(t) = AR¢ sin(!t + Á)

u(t) = sin(!t)

Plots: VARY ! from 0.1 to 30 rad/s

w=0.3; tau=1; t = linspace(0,20,1000);

u = sin(w*t);

AR = 1/sqrt((w*tau)^2+1)

phi = - atan(w*tau), phig=phi*180/pi, dt=-phi/w

y = AR*sin(w*t+phi);

plot(t,y,t,u)

Chapter 14

Mathematics. Complex numbers, j2=-1

Im(G)

G(j!)=R+jI

I

Á=Å G

R

Re(G)

Chapter 14

Polar form

Multiply complex numbers:

Multiply magnitudes and add phases

Chapter 14

Simple method to find sinusoidal response of system G(s)

- Input signal to linear system: u = u0 sin(! t)
- Steady-state (“persistent”, t!1) output signal: y = y0 sin(! t + Á)
- What is AR = y0/u0 and Á?
- Solution (extremely simple!)
- Find system transfer function, G(s)
- Let s=j! (imaginary number, j2=-1) and evaluate G(j!) = R + jI (complex number)
- Then (“believe it or not!”)
- AR = |G(j!)| (magnitude of the complex number)
- Á = Å G(j!) (phase of the complex number)

Im(G)

G(j!)=R+jI

I

R

Re(G)

|G|

Å G

Example 13.1:

1.

2.

Chapter 14

R I

3.

Gain and phase shift

of sinusoidal response!

SIMPLER: Polar form; see board

Chapter 14

Chapter 14

-1 rad = -57o at !µ = 1

=-!µ

Figure 14.4 Bode diagram for a time delay, e-qs.

Peak goes to infinity when ³! 0

Phase increases for LHP zero

Oops! Phase drops for RHP zero

Bode Diagram

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30

Magnitude (dB)

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90

45

Phase (deg)

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-45

-90

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-1

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Frequency (rad/s)

- |G| (dB) = 20 log10|G|

s=tf('s')

g = 10*(100*s+1)/[(10*s+1)*(s+1)]

bode(g) % gives AR in dB

*

*To change magnitude from dB to abs: Right click + properties + units

Other way: |G| = 10|G|(dB)/20

GM=2 is same as GM = 6dB

ASYMPTOTES

Frequency response of term (Ts+1): set s=j!.

Asymptotes:

(j!T + 1) ¼ 1 for !T ¿ 1

(j!T+ 1) ¼j!Tfor !T À1

Note: Integrator (1/s) has one “asymptote”:

=1/(j!)= -j!-1 at all frequencies

Rule for asymptotic Bode-plot, L = k(Ts+1)/(¿s+1)….. :

Start with low-frequency asymptote

(a) If constant (L=k): Gain=k and Phase=0o

(b) If integrator (L=k/s):

Phase: -90o.

Gain: Has slope -1 (on log-log plot) and gain=1 at !=k

2. Break frequencies:

Chapter 14

Figure 14.6 Bode plots of ideal parallel PID controller and series PID controller with derivative filter (α = 0.1).

Ideal parallel:

Series with Derivative Filter:

Chapter 14

CLOSED-LOOP STABILITY

- L = gcgm = loop transfer function with negative feedback
- Bode’s stability condition: |L(!180)|<1|
- Limitations
- Open-loop stable (L(s) stable)
- Phase of L crosses -180o only once

- Limitations
- More general: Nyquist stability condition:
Locus of L(j!) should encircle the

(-1)-point P times in the anti-clockwise

direction (where P = no. of unstable

poles in L).

Stable plant (P=0): Closed-loop stable if L has no encirclements of -1

(=Bode’s stability condition)

|L(j!)| =

c

180

Chapter 14

Å L(j!)=

c

180

Sigurd’s preferred notation in red

Time delay margin, ¢µ = PM[rad]/!c

EXAMPLE

Slope=-1

-2

-1

GM=1/0 = 1

-2

Slope

Help lines

-2

-1

!=0.5

!=0.01

!=0.05

-90o

-90o

PM=57o

-180o

-180o

L(s): SIMC PI-control with ¿c=4 for g(s) = 1/(100s+1)(2s+1)

!c = 0.19 rad/s

!180 = 1

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Slope=-1

With added delay, e-µs with µ=2

No change in gain

-2

-1

GM=1/0,4=2.5

-2

!=0.5

!=0.01

!=0.05

-90o

-90o

With added delay, e-µs with µ=2.

Contribution to phase is:

-5.7o at !=0.1/µ = 0.05

-57o at !=1/µ = 0.5

PM=35o

= 0.61 rad

-180o

-180o

!180 = 0.4

!c = 0.19

- g(s) =k’e-µs/s
- Derive: Pu = 4µ and Ku = (¼/2)/(k’µ)

- PI-controller, c(s) = Kc (1+1/¿Is)

Task: Compare Bode-plot (L=gc), robustness and simulations (use k’=1, µ=1).

Bode Diagram

Bode Diagram

Gm = 2.96 (at 1.49 rad/s) , Pm = 46.9 deg (at 0.515 rad/s)

Gm = 1.87 (at 1.35 rad/s) , Pm = 24.9 deg (at 0.76 rad/s)

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Magnitude (abs)

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Phase (deg)

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-720

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Frequency (rad/s)

SIMC-PI

Ziegler-Nichols PI

GM

GM

PM

PM

SIMC is a lot more robust:

¢µ = PM[rad]/!c

ZN: ¢µ = 24.9*(3.14/180)/0.76 = 0.572s

SIMC: ¢µ = 46.9*(3.14/180)/0.515 = 1.882s

s=tf('s')

g = exp(-s)/s

Kc=0.707, taui=3.33

c = Kc*(1+1/(taui*s))

L1 = g*c

figure(1), margin(L1) % Bode-plot with margins

% To change magnitude from dB to abs: Right click + properties + units

Kc=0.5, taui=8

c = Kc*(1+1/(taui*s))

L2 = g*c

figure(2), margin(L2)

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SIMC

OUTPUT, y

ZN

Simulink file: tunepid4

s=tf('s')

g = exp(-s)/s

Kc=0.707, taui=3.33, taud=0 % ZN

sim('tunepid4')

plot(Tid,y,'red',Tid,u,'red')

Kc=0.5, taui=8, taud=0 % SIMC

sim('tunepid4')

plot(Tid,y,'blue',Tid,u,'blue')

INPUT, u

t=0: setpoint change, t=20: input disturbance

- Conclusion: Ziegler-Nichols (ZN) responds faster to the input disturbance,
- but is much less robust.
- ZN goes unstable if we increase delay from 1s to 1.57s.
- SIMC goes unstable if we increase delay from 1s to 2.88s.

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SIMC

OUTPUT, y

ZN

INPUT, u

t=0: setpoint change, t=20: input disturbance

ZN is almost unstable when the delay is increased from 1s to 1.5s.

SIMC does not change very much

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e

SIMC: Ms=1.70

ZN: Ms = 2.93

SIMC

ZN

w = logspace(-2,1,1000);

[mag1,phase]=bode(1/(1+L1),w);

[mag2,phase]=bode(1/(1+L2),w);

figure(1), loglog(w,mag1(:),'red',w,mag2(:),'blue',w,1,'-.')

axis([0.01,10,0.001,10])

NO EFFECT

BAD

Control: GOOD