chapter 3fivedots.coe.psu.ac.th/software.coe/240-209/slides/controlch3.pdf240-209: modelling in the...

ผ��เร�ยบเร�ยง ธเนศ เคารพาพงศ� แก�ไข 30 ต�ลาคม 2547 หน�า 1240-209 : Modelling in the Frequency Domain

Chapter 3

Modelling in the Frequency Domain


Outline

Signals and systems Transfer function Developing 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 t0}


Continuous-Time Signals


Discrete-Time Signals


Element signals

Unit step functions

u t ={0 for t01 for t≥0}

u kT s=u k ={0 for k01 for k≥0}


Unit impulse, Dirac impulse

t =du t dt

u 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=0}


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∣/2

0 otherwise }rect kK ={1 if ∣k∣K /2

0 otherwise }

Continuous-time

Discrete-time


Triangular Pulse

Continuous-time

Discrete-time

tri t={1−∣ t∣ if ∣t∣

0 otherwise }tri kK ={1−∣kK∣ if ∣k∣K

0 otherwise }


Ramp function

ramp tT ={ tT if 0≤tT

0 otherwise }ramp k

K ={kK if 0≤kK

0 otherwise }Continuous-time

Discrete-time


Complexe causal exponentials

x t =Ae t u t

x k =Ae kT s u k

= j

e j t=cos t j sin t


Linear system

Y(t) = (Sx)(t)


S an x n...a1 x 1t =S an x n t ...Sa1 x 1t

=an S x n t ...a1S x 1t

Superposition


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 est 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

an snan−1 s n−1...a0

=N s

sP 1sP 2 ...sP msP n

=K 1

sp 1

K 2

sp 2...

K m

spm

K n

sp n


F s = N s sp1sp 2 ...spmsp n

=K 1

sp 1

K 2

sp 2...

K m

spm...

K n

sp n

sp mF s =spmK 1

sp 1spm

K 2

sp 2

K mspmK n

sp n

K m=N pm

−pmp 1−pmp 2 ... −pmp n


F s = 1s1s2s4

=K 1

s1K 2

s2K 3

s4

K 1=1

−12−14=13

K 2=1

−21−24= 1

2

K 3=1

−41−42=1

6


F s = 1 /3s1

1/2s2

1/6s4

f t =13 e−t1

2 e−2t16 e−4t

F s =N s D s

=N s

sp 1r sp 2 ...sp n

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


=K 1

sp1r

K 2

sp1r−1...

K r

sp1

K r1

sp 2...

K n

sp n

F s = N s sp1

r sp 2 ...sp n

=K 1sp1K 2sp 12K 3...sp1

r−1K r

sp1r K r1

sp 2...

sp 1r K n

sp n


K i=[ 1 i−1!

d i−1 F s ds i−1 ]

s p 1

F s = 2s1s22

F s =K 1

s22

K 2

s2K 3

s1

s22ค�ณด�วย

2s1=K 1s2K 2

s22K 3

s1

Example

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

K 1=−2

........ (1)


−2s12

=K 2−[s12s2K 3 ]−s22K 3

s12

=K 2s2K 3[2 s1−s2]

s12

−2s12

=K 2s s2K 3

s12

K 2=−2

First derivative of (1)


F s = −2s22

−2s2

2s1

2s22

=s1K 1

S22s1K 2

s2 K 3

K 3=2

f t =−2te−2t−2 e−2t2 e−t


F s =N s D s

=N s

sp 1s2asb ...

= K1sp 1

K 2 sK 3

s 2asb...

s2asb

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

- complex or imaginary


F s = 3s s 22s5

3s s22s5

=K 1

s K 2 sK 3

s 22s5

K 1=35

s s 22s5ค�ณด�วย

3=s22s5K 1K 2 s2K 3 s

=K 1K 2s22K1K 3s5K1

Example

และจ�ดสมการใหม�


3=K 235s2K 3

65s3

K 235=0 K 3

65=0

K 2=−35 K 3=

−65

F s = 3s s 22s5

= 3/5s − 3

5s2

s22s5

แทนค�า K1


L [Ae−t cos t ]= Asasa22

L [Be−at sin t ]= Bsa22

L [Ae−at cost Be−at sin t ]=AsaBsa22

F s = 3/5s − 3

5s11/22s1222

f t =35−

35 e−t cos2t1

2 sin 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

and n c t dt n

an−1d n−1c t dt n−1 ⋯a0c t

=bmd m r t dt m

bm−1d m−1 r t dt m−1 ⋯b0 r t

an snC s ...a0C s =bm s

m R s ...b0R s


Transfer function

C sR s

=G s =bm s

mbm−1 sm−1⋯b 0

an sna n−1 s

n−1⋯a 0


Electric Network Transfer Functions



When v(t) is input and i(t) is output.


By Kirchhoff’s voltage law, we have

v t =


Transfer Function of RL circuit

I s V s

=

I(s) V(s)


By Kirchhoff’s voltage law, we have

When we define : v(t) is input and vc(t) is output.

v t =


Block Diagram of Series RLC Network

1LCs2RCs1

V(s) Vc(s)

When v(t) is input and vc(t) is output.


Block Diagram of Series RLC Network

CsLCs 2RCs1

V(s) I(s)

When v(t) is input and i(t) is output.


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

Example Find the transfer function of the circuit below when input is v(t) and output is i(t), vc(t)


Replace passive element values with their impedances and Replace all sources and time variables with their Laplace Transform


Transfer function of the circuit


Translational Mechanical System Transfer Functions


Translational Mechanical System Transfer Functions

Example Find Transfer function of the translational system below when input is f(t) and output is x(t)


Transfer function of the translational system


Rotational Mechanical System Transfer Functions


Example Find Transfer function of the rotational system below when input is T(t) and output is 2(t)


Schematic of the system


Equation on mass 1 (M1)

T t =J 1d 21t

d t

Equation on mass 2 (M2)

0=J 2d 22 t

d t


Transfer function of the system

2 s T s K J 1 s

2D 1 sK J 2 s2D 2 sK K 2


Transfer Functions for Systems with Gears


Transfer Functions for Systems with Gears

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


Electromechanical System


Electromechanical System Transfer Functions

Schematic of DC motor


ea(t) - Applied voltageia(t) - Armature currentvb(t) - Back emfTm(t) - Motor torquem(t) - Rotor displacementm(t)- Rotor angular velocity

Motor Variables


Motor parameters

La - Armature inductanceRa - Armature resistanceJm - Rotor inertiaDm - Viscous-friction coefficientkT - Torque constantkb - Back-emf constant


Electromechanical System Transfer Functions

Position control transfer function

Input is ea(t) and output is m(t)

Electrical equation

e a t =L ad i a t dt Ra i a t v b t


Mechanical Equation

Relation between Electrical and Mechanical

T m t =J md 2m t

dt 2 D md m t

dt

v b t =k bd m t

dt

T m t =k m i at


Position control transfer function

m s E as

=k m

L a sRa J m s2Dm s km k b s

k m

L a sRa J m s2D m s k m k b s

E as m s

chapter 3fivedots.coe.psu.ac.th/software.coe/240-209/slides/controlch3.pdf240-209: modelling in the...

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