chapter 3fivedots.coe.psu.ac.th/software.coe/240-371/slide/controlch3.pdf240-371: modelling in the...

ผูเรียบเรียง ธเนศ เคารพาพงศ, อํานวย สิทธิเจรญิชัย แกไข 20 มีนาคม 2545 หนา 1240-371 : Modelling in the Frequency Domain

Chapter 3

Modelling in the Frequency Domain


Introduction

Signals and systems Transfer functionDeveloping a mathematical model by

apply the fundamental physical laws of science end engineering in frequency domain


Introduction to Signals and Systems

Signal Taxonomy

1 Deterministic signal x(t) = sin(t), x(t)=, x(t)=t 2 Random signalCausal system – physical signal generator is turned on at time t=0 then the produced causal signal y(t) satisfies.

y t ={x t for t≥00 for t06


Continuous-Time Signals


Discrete -Time Signals


Element signals

Unit step functions

u t ={0 for t01 for t≥06

u kT s=u k ={0 for k01 for k≥06


Unit impulse, Dirac impulse

t =du t

dtu t =∫−∞

td


1.∫−∞

∞d =1

2. limt 0

t =∞

3.t =0 for t≠04.t =−t

Properties of Dirac


∫−∞

∞x t t−t 0dt=∫−∞

∞x t 0t−t 0dt=x t 0

Sampling property

Kronecker delta function

kT s=k ={0 if k ≠01 for k =06


Property of discrete-time impulse

1. ∑k=−∞

∞

K[k ]=1

2.K[k ]=

K[−k ]

Sampling property of discrete-time impulse function

∑k=−∞

∞

x [k ]K[k−k 0]= ∑

k=−∞

∞

x [k 0]K [k−k 0]=x [k 0]


Rectangular Pulse

rect t={1 if ∣t ∣ / 20 otherwise 6

rect k

K ={1 if ∣k ∣K /20 otherwise 6

Continuous-time

Discrete-time


Triangular Pulse

Continuous-time

Discrete-time

tri t={1−∣ t∣ if ∣t ∣

0 otherwise6

tri k

K={1−∣ kK ∣ if ∣k ∣K

0 otherwise6


Ramp function

ramp t

T ={ t

Tif 0≤t T

0 otherwise6

ramp k

K={ k

Kif 0≤k K

0 otherwise6

Continuous-time

Discrete-time


Complexe causal exponentials

x t =Ae tu t

x k = A e k T

s u k

= j

ej t

=cos t j sin t


Linear system

Y(t) = (Sx)(t)


S a n x n ... a 1 x 1 t = S a n x n t ...Sa 1 x 1 t

= a n S x n t ... a 1 S x 1 t


Cascaded interconnections

Subsystem Subsystem Subsystemr t

Inputc t

Output

- Input (Reference input)- Output (Controlled variable)

r t

c t

SystemInput Output


Laplace Transform

For solving linear differential equations by convert to complex plane.

s = σ + jω s = complex variable σ = real part ω = imaginary part


Definition

Laplace Transform

Inverse Laplace Transform

L [f t ]=F s=∫0∞f t e−st

dt

L−1[F s ]=

12 j

∫− j ∞

− j ∞F s e st

ds


Inverse Laplace TransformPartial-function expansion

1 Roots of the Denominator of F(s) are Real and distinct

F s =N s

D s

=N s

a n sna n−1 s

n−1...a 0

=N s

sP 1 sP 2 ... sP m sP n

=K 1

s p 1

K 2

s p 2

...K m

s p m

K n

s p n


F s =N s

s p 1 s p 2 ... s p m s p n

=K 1

s p 1

K 2

s p 2

...K m

s p m

...K n

s p n

s p m F s = s p m K 1

s p 1

s p m K 2

s p 2

K m s p m K n

s p n

K m=N p m

− p m p 1 − p m p 2 ... − p m p n


F s =1

s1 s 2 s4

=K 1

s 1

K 2

s 2

K 3

s 4

K 1=1

−1 2 −1 4 =

13

K 2=1

− 2 1 − 2 4 =

12

K 3=1

−4 1 −4 42=

16


F s =1 /3

s1

1 /2 s 2

1 /6

s4

f t =13e

− t

12e

−2t

16e

− 4t

F s =N s

D s

=N s

s p 1r s p 2 ... s p n

2. Root of the Denominator of F(s) are real and repeated


=K 1

s p 1 r

K 2

s p 1 r − 1

...K r

s p 1

K r 1

s p 2 ...

K n

s p n

F s =N s

s p 1r s p 2 ... s p n

= K 1 s p 1 K 2 s p 12 K 3 ... s p 1

r −1 K r

s p 1r K r 1

s p 2 ...

s p 1r K n

s p n


K i=[ 1 i −1 !

di−1F s

dsi−1 ]

s p1

F s =2

s 1 s 22

F s =K 1

s 2 2

K 2

s 2

K 3

s 1

s22คูณดวย2

s 1= K 1 s 2 K 2

s 2 2 K 3

s 1

Example

i = 1, 2, ..., r 0! = 1

K 1=−2

........ (1)


− 2 s 1 2

= K 2− [ s 1 2 s 2 K 3]− s 2 2 K 3

s 1 2

=K 2 s2K 3[2 s1− s2]

s12

− 2 s 12

=K 2s s 2 K 3

s 12

K2=−2

First derivative of (1)


F s =−2

s 22

−2s 2

2

s1

2 s 22

= s 1 K 1

S 22

s 1 K 2

s 2K 3

K 3=2

f t =−2te−2t

− 2 e −2t 2 e −t


F s =N s

D s

=N s

s p 1 s2 asb ...

=K1

s p 1

K 2 s K 3

s2 asb

...

s2 asb

2. Root of the Denominator of F(s) are complex

- complex or imaginary


F s =3

s s22s5

3s s

22s5

=K 1

sK 2 sK 3

s22s5

K 1=35

s s22s5 คูณดวย

3= s2 s 5 K 1 K 2 s

2 K 3 s

= K 1 K 2 s2 2K 1 K 3 s 5K 1

Example

และจัดสมการใหม


3=K 235 s

2K 3

65 s 3

K 235=0 K 3

65=0

K 2=−35

K 3=−65

F s =3

s s22s5

=3 /55

−35

s2s

22s5

แทนคา K1


L [Ae− t cos t ]= A s a

s a 22

L [Be−at sin t ]= B

s a 22

L [Ae−at cos t Be−at sin t ]= A sa B

sa 22

F s =3 /55

−35

s11 / 2 2 s12 22

f t =35−

35e

− tcos 2t 1

2sin 2t


Advantages : rapidly provide stability and transient response information.

Disadvantages : Limited applicability to linear to linear, time invariant system or system that can be approximated as such.

Advantages and Disadvantages


Transfer function

a n

dnc t

d tn

a n − 1d

n − 1c t

d tn − 1

⋯ a 0 c t

= b m

dmr t

dtm

b m − 1d

m − 1r t

dtm − 1

⋯ b 0 r t

an snC s...a0C s=bms

mR s...b0R s


Transfer function

C s

R s =G s =

b m smb m−1 s

m−1⋯b 0

a n sn a n−1 s

n−1⋯ a 0


Electric Network Transfer Functions


Block Diagram of Series RLC Network


Block Diagram of Series RLC Network

Cs

CLs2 RCs 1

V(s) I(s)


1. Replace passive element values with their impedances.2. Replace all sources and time variables with their Laplace Transform3. Assume a transform current and a current direction in each mesh.4. Write Kirchhoff’s voltage law around the output.5. Solve the simultaneous equations for the output.6. Form the Transfer function.

Complex Circuits via Mesh Analysis


Complex Circuits via Mesh Analysis


Translational Mechanical System Transfer Functions


Translational Mechanical System Transfer Functions

a. Mass, spring, and damper systemb. block diagram


Rotational Mechanical System Transfer Functions


a. Physical systemb. schematicc. block diagram


Transfer Functions for Systems with Gears


Transfer Functions for Systems with Gears

a. angular displacement inlossless gears.b. torque in lossless gears.


Electromechanical System Transfer Functions

DC motor : a. schematic; b. block diagram

chapter 3fivedots.coe.psu.ac.th/software.coe/240-371/slide/controlch3.pdf240-371: modelling in the...

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