control systems - dis.uniroma1.itlanari/controlsystems/cs - lectures/2014_lec... · electric guitar...

Control Systems

System as a filterL. Lanari

Monday, November 3, 2014

Lanari: CS - Filter 2

Outline

• Mass-Spring-Damper

• 7-mass system

• first order system

• second order system

• microphone

• quarter-car system



Mass - Spring - Damper

k

µ

p

uF (s) =

1

ms2 + µs+ km

0 < µ < 2√km p1,2 =

− µm ± j

�4�

km

�−� µm

�2

2

µ = 2√km

transfer function

• over-damped

• critically-damped

• under-damped

µ > 2√km

p1,2 = − µ

2m

p1,2 =− µ

m ±�� µ

m

�2 − 4�

km

�

2

complex conjugate poles

real coincident poles

real distinct poles

asymptotically stable system for µ > 0




0 < µ < 2√km

µ = 2√km

• over-damped

• critically-damped

• under-damped

µ > 2√km

trinomial factor

two coincident binomial factors

two distinct binomial factors

ωn =

�k

mζ =

µ

2√km

1

τ= −p1,2 =

µ

2m=

�k

m

1

τ1= −p1 >

�k

m

1

τ2= −p2 <

�k

m

gain = 1/k



10!1

100

101

102

!75

!70

!65

!60

!55

!50

!45

!40

!35

!30

!25

frequency (rad/s)

Magnitude (dB)

MSD ! Magnitude with µ variable

Real coincident

Real distinct

Complex

Single binomials


m = 1 kgk = 100 N/m

magnitude in terms of the damping factor µ

the two single binomials are also shown (over-damped case) in order to put in evidence the dominant pole (corresponding to the ral pole closest to the origin)



input force on mass 1

7 - MSDsystem

(with damping)

100

101

102

103

!200

!150

!100

!50

Magnit

ude

100

101

102

103

!800

!600

!400

!200

0

Phase

7-mass

output position on mass 3

resonance peaks

anti-resonancepeak

natural frequenciesvery close to each other

119.6311 107.5098 78.7957 73.3411 42.2283 19.0596 11.0478

natural frequencies



10!2

100

102

!20

!15

!10

!5

0

5

frequency (rad/s)

Magnitude (dB)

F1

F2

F3

F4

F1(s) =1

1 + 10s, F2(s) =

1

1 + s, F3(s) =

1

1 + 0.1s, F4(s) =

1

1 + 0.01s

4 different systems(all first order and with unit gain)

example 1



0 2 4 6 83

2

1

0

1

2

3T

Input

100

101

102

!0.5

0

0.5

frequency (rad/s)

Frequency spectrum of the input

u(t) = −0.6 sin(f1t)−0.4 sin(f2t)+0.5 sin(f3t)+0.5 sin(f4t)−0.3 sin(f5t)−0.2 sin(f6t)+0.2 sin(f7t)−0.2 sin(f8t)

f1 = 2π0.75, f2 = 2π1.25, f3 = 2π1.5, f4 = 2π3, f5 = 2π5, f6 = 2π6, f7 = 2π8, f8 = 2π11

!2

0

2

!2

0

2

!2

0

2

!2

0

2

!2

0

2

!2

0

2

!2

0

2

0 0.5 1 1.5 2 2.5 3 3.5 4!2

0

2

adding one contribution at a time

sameinformationin the frequencydomain time

time

input signal



10!2

10!1

100

101

102

!0.5

0

0.5

10!2

10!1

100

101

102

!0.1

0

0.1

10!2

10!1

100

101

102

0

0.5

1

Magnitude

System 2

magnitudenot in dB

we need to multiply

input

system

output

1

s+ 1

0 2 4 6 83

2

1

0

1

2

3T

Input

frequency time

?



0 2 4 6 8

!2

!1

0

1

2

System 1

time (s)

Input

Output

10!2

100

102

0

0.2

0.4

0.6

0.8

1

frequency (rad/s)

Magnitude

System 1

10!2

100

102

0

0.2

0.4

0.6

0.8

1

frequency (rad/s)

Magnitude

System 2

0 2 4 6 8

!2

!1

0

1

2

System 2

time (s)

Input

Output

these are plotted for unit magnitude sinusoids



0 2 4 6 8

!2

!1

0

1

2

System 3

time (s)

Input

Output

0 2 4 6 8

!2

!1

0

1

2

System 4

time (s)

Input

Output

10!2

100

102

0

0.2

0.4

0.6

0.8

1

frequency (rad/s)

Magnitude

System 4

10!2

100

102

0

0.2

0.4

0.6

0.8

1

System 3

frequency (rad/s)

Magnitude

it just shows at whichfrequencies the input

has contributions



2 systems with same magnitude but different phase

10!2

100

102

0

0.2

0.4

0.6

0.8

1

frequency (rad/s)

Magnitude

10!2

100

102

!700

!600

!500

!400

!300

!200

!100

0

100

Phase (deg)

frequency (rad/s)

FA

FB

0 10 20 30 40 50!2.5

!2

!1.5

!1

!0.5

0

0.5

1

1.5

2

2.5System A

time (s)

InputOutput

0 10 20 30 40 50!2.5

!2

!1.5

!1

!0.5

0

0.5

1

1.5

2

2.5System B

time (s)

InputOutput

FA(s) =1

1 + s/10

FB(s) =(s2 − 0.2s+ 1)(1− 10s)

(s2 + 0.2s+ 1)(1 + 10s)(1 + s/10)

differences in phase can make the difference



10!2

100

102

!10

!5

0

5

10

frequency (rad/s)

Magnitude (dB)

F1

F2

F3

F4

F (s) =1

(1 + 2ζs/ωn + s2/ω2n)

³ = 0.2

!n = {1, 8, 10, 100}

4 different systems(all second order with same damping and unit gain)

example 2



0 2 4 6 8!2.5

!2

!1.5

!1

!0.5

0

0.5

1

1.5

2

2.5System 1

time (s)

InputOutput

0 2 4 6 8!2.5

!2

!1.5

!1

!0.5

0

0.5

1

1.5

2

2.5System 2

time (s)

InputOutput

10!2

100

102

0

0.5

1

1.5

2

2.5

frequency (rad/s)

Magnitude

System 1

10!2

100

102

0

0.5

1

1.5

2

2.5

frequency (rad/s)

Magnitude

System 2



10!2

100

102

0

0.5

1

1.5

2

2.5

System 3

frequency (rad/s)

Magnitude

0 2 4 6 8!2.5

!2

!1.5

!1

!0.5

0

0.5

1

1.5

2

2.5System 3

time (s)

InputOutput

0 2 4 6 8!2.5

!2

!1.5

!1

!0.5

0

0.5

1

1.5

2

2.5System 4

time (s)

InputOutput

10!2

100

102

0

0.5

1

1.5

2

2.5

frequency (rad/s)

Magnitude

System 4


Lanari: CS - Filter

0 1 2 3 4!3

!2

!1

0

1

2

3System 3

time (s)

yySS

16

0 1 2 3 4!1

!0.5

0

0.5

1System 3 ! transient

time (s)

forced response and steady-state transient as difference

transient

transient, if the system admits steady-state, also exists for any other input. We decided to characterize transient with specific quantities on the step input.

example: transient for a sinusoidal input



dynamic bass microphone(tailored for kick drum, works well with any low frequency instrument,

low frequency peak at 100 Hz)

a 10.000 € voice microphone ...

electric guitar microphone

microphones



Quarter-car suspension model

wheel

body

road

mb

mw

xb

xw

r

ks bs

kt

fs

Model of a quarter part of a car with its wheel and tire

The body with mass mb represents the car chassis connected to the wheel by a passive spring (ks), and a shock absorber represented by a damper (bs).The spring (kt) models the compressibility of the tire pneumatic.

In an active suspension an hydraulic actuator (fs)between the chassis and wheel assembly may help in balancing conflicting objectives as passenger comfort, road handling and suspension deflection.

mbxb + ks(xb − xw) + bs(xb − xw) = fs

mwxw − ks(xb − xw) + bs(xb − xw) + kt(xw − r) = −fs

fs acts on both the body and the wheel assembly

r can be seen as an input affecting the evolution of the system through the tire (disturbance)



state vector

several outputs of interest

�xb xb xw xw

�T

C1 =�1 0 0 0

�

C2 =�−ks/mb −bs/mb ks/mb bs/mb

�

body (passenger) position

body (passenger) acceleration

suspension deflectionC3 =�1 0 −1 0

�

two inputs (one, fs, can be controlled, the other is the disturbance r)

by setting one of the two inputs to zero and choosing the output of interest, we have a SISO system with corresponding transfer function

Passenger comfort is associated to small passenger acceleration

Physical limitation of the actuator (limits on maximum displacements) defines a constraint

these are two of the possible outputs



150

100

50

0

50

100

To: a

b

100

101

102

200

150

100

50

0

50To

: sd

Magnitude from road disturbace and actuator force (fs) to body acceleration and suspension travel

Frequency (rad/sec)

Mag

nitu

de (d

B)

Input: road disturbance (r)Input: actuator force (fs)

actuator to body accelerationfrequency response magnitude

actuator to suspension deflectionfrequency response magnitude

road to body accelerationfrequency response magnitude

road to suspension deflectionfrequency response magnitude

rattlesnake frequency

tire-hop frequency



rattlesnake frequency: pure imaginary zeros in the transfer function from the actuator to the suspension deflection, anti-resonance at 22.97 rad/s

tire-hop frequency: pure imaginary zeros in the transfer function from the actuator to the body acceleration (also from actuator to body position), anti-resonance at 56.27 rad/s

at these frequencies it is difficult to counteract any effect of the road on acceleration or on the suspension deflection


control systems - dis.uniroma1.itlanari/controlsystems/cs - lectures/2014_lec... · electric guitar...

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