Chapter 2 Modeling in the frequency domain Youngjoon, Han young@ssu.ac.kr

2.1 Introduction

Definition and examples Laplace Transform Definition and examples Unit Step Function u(t)

Laplace transform thorems-Properties

Laplace transform Pairs

Test During analyzing a circuit, input as a sinusoidal function must be Laplace transformed. It is to say that f(t) = e-atcoswtu(t). What is F(s)? F(s)=2/s2 , what is f(t)?

Partial Fraction

Partial Fraction

Partial fraction; repeated factor

Partial fraction; repeated factor

Roots of the Denominator of F(s) are Complex or Imaginary

The Transfer Function C(s)=G(s)R(s) Output=Transfer Function * Input

Example 2.4 dc(t)/dt+2c(t)=r(t) What is transfer function or gain? sC(s)+2C(s)=R(s) for zero initial condition (s+2)C(s)=R(s) G(s)=C(s)/R(s)=1/(s+2) Example 2.5 If r(t)=1, R(s)=1/s C(s)=1/[s(s+2)]=A/s+B/(s+2)

Solution to the example 2.5 1/[s(s+2)]=A/s+B/(s+2) A=1/2, B=-1/2 C(t)=½-½e-2t

Electric Network Transfer Function

The Analogy of Physical variable Type Rate Effort Electrical I (current) Voltage V Mech. (Trans) V (velocity) Force f Mech. (Rotational) w(ang. Vel.) Torque T Fluid q (flow rate) Pressure p Thermal Q (heat flow rate) Temp T

Analogy Quantity= (rate) dt Type Rate Quantity Electrical I (current) Charge Q Mech. (Trans) V (velocity) Displ x Mech. (Rotational) w(ang. Vel.) Angular disp q Fluid q (flow rate) Mass m Thermal Q (heat flow rate) Heat energy Q Quantity= (rate) dt

Transfer Function for RLC Circuit (1)

Transfer Function for RLC Circuit(1)

Transfer Function for RLC Circuit (1)

Transfer Function for RLC Circuit (2) (Ls+R+1/Cs)I(s)=V(s) Vc(s)=(1/Cs)I(s)=V(s)/[LCs2+RCS+1] G(s)=Vc(S)/V(s)=(1/LC)/[s2+(R/L)S+1/LC]

Complex Circuits Using Mesh Analysis Replace passive element values with their impedances Laplace transform Assume current in each mesh Kirchhoff’s voltage law around each mesh Solve the simultaneous equations Transfer function

Mesh Analysis method Mesh 1 Mesh 2 Impedance Laplace form Assume currents

Mesh Analysis method Write equations around the meshes Sum of impedance around mesh 1 Sum of applied voltages around the mesh Sum of impedance common to two meshes Sum of impedance around mesh 2

Mesh Analysis method Determinant

Mesh Analysis method

Complex Circuits Nodal Analysis Number of nodes equal to the simultaneous equations Use Kirchhoff current law Sum currents flowing from each node Conductance=1/resistance Admittance Y(s) Y(s)=1/impedance=1/Z(s) Y(s)=I(s)/V(s)

Nodal Analysis Method i1 + i2 +i3=0 i3 + i4 =0 i3 i1 i2 i4 Kirchhoff current law at these two nodes i2 i4 i1 + i2 +i3=0 i3 + i4 =0

Nodal Analysis Method Kirchhoff current law conductance

Nodal Analysis Method

Nodal Analysis Method Real part of admittance is conductance, imaginary part is susceptance; 1/R =G has only real part

Operational Amplifier High input impedancedoesn’t load, Zi= ∞ (ideal) The Currents into both terminals are zero The voltage across the input terminals is zero Low output impedance, Z0=0 (ideal) High constant gain amplification A= ∞

Operational Amplifier and Inverting OP-Amp

Non-inverting OP-Amp Noninverting Op-amp

Mechanical System Variables for translation movement

Mass, spring, and damper system

Free-body diagram of mass, spring, and damper system impedances

Transfer Function of mass, spring, and damper system

Impedance for mechanical components  [ Sum of impedance] X(s) = [ Sum of applied forces]

Two-degrees-of-freedom translational mechanical system

Two-degrees-of-freedom translational mechanical system

Two-degrees-of-freedom translational mechanical system

Two-degrees-of-freedom translational mechanical system

Three-degrees-of-freedom translational mechanical system

Three-degrees-of-freedom translational mechanical system

Skill-Assessment Exercise 2.8

Mechanical System variables for rotational movement

Transfer function – two equation of motion

Transfer function – two equation of motion

Transfer function – two equation of motion

Transfer function – two equation of motion 

Three-degrees-of-freedom rotational system

Skill-Assessment Exercise 2.9

Transfer Function For Gear System F, V must be compatible Energy=T1 Ө1= T2 Ө2 T2/T1=Ө1/ Ө 2= r2/r1= N2/N1

Transfer functions for angular displacement and torque

Rotational system driven by gears Mechanical Impedance of destination axis = ([#of destination teeth]/ [# of source teeth])2 * Mechanical Impedance of source axis (Js2+Ds+K) (N1/N2)q1=(N2/N1)T1 [J (N1/N2)2 s2+D (N1/N2)2 s+K (N1/N2)2 ] q1=T1 T2=(N2/N1)T1  q2=(N1/N2)q1

Example 2.21

Gear train

Transfer function –Gear with loss Target gear Source gear

Armature controlled DC servomotor Amature F=BlI

Armature controlled DC servomotor Stationary permanent magnets Voltage e=Blv; v is velocity Armature, rotor

Electromechanical System Voltage is proportional to velocity Electromotive force (emf) vb(t) vb(t)=Kbdqm(t)/dt Kb; back emf constant dqm(t)/dt; wm(t) angular velocity Vb(s)=Kbsqm(s) Around the armature loop RaIa(s)+ LasIa(s)+ Vb(s)= Ea(s)

Typical equivalent mechanical loading on a motor Tm(s)= (Jms2+ Dms) qm(s)

Electromechanical System The torque is proportional to the armature current Tm(s)=KtIa(s)  Ia(s)=Tm(s)/Kt Applying Ia(s)=Tm(s)/Kt to RaIa(s)+ LasIa(s)+ Kbsqm(s)= Ea(s) (Ra+Las) Tm(s)/Kt+ Kbsqm(s)= Ea(s) Applying Tm(s)= (Jms2+ Dms) qm(s) (Ra+Las) (Jms2+ Dms) qm(s) /Kt+ Kbsqm(s) =Ea(s)

Electromechanical System For small La, (La<<R) [(Ra /Kt)(Jms+Dm) +Kb]sqm(s) = Ea(s) G(s)= qm(s)/Ea(s)=K/[s(s+a)] = {Kt /(Ra Jm)}/[s{s+(Dm+ KtKb /Ra )/ Jm}]

DC motor driving a rotational mechanical load Jm=Ja+ JL(N1/N2)2 Dm=Da+ DL(N1/N2)2

Electromechanical System From the previous equation, put La=0 (Ra+Las) Tm(s)/Kt+ Kbsqm(s)= Ea(s) RaTm(t)/Kt+ Kbwm(t)= ea(s) Tm(t)=-(Kt Kb /Ra) wm+ (Kt/Ra)ea(t) Tm(t) Torque-speed curve wm

Torque-speed curves with an armature voltage, ea, as a parameter (Kt/Ra)ea(s) Kt/Ra=Tstall/ea Kb=ea(s)/wno-load ea(s)/ Kb Tm(t)=-(Kt Kb /Ra) wm+ (Kt/Ra)ea(s)

DC motor and load

DC motor and load From the graph Gear ratio

Nonlinearities A linear System possesses two properties: superposition and homogeneity

Nonlinearities

Linearization about a point A f(x) – f(x0) ≈ ma(x-x0)  δf(x) ≈ maδx Equilibrium point A

Example 2.26 Linearization for f(x)=5cosx at x= p/2 Using Taylor series expansion Linearization for f(x)=5cosx at x= p/2 f(x) – f(p/2 ) = df(x)/dx|x=p/2 (x - p/2 )  f(x)=-5(x-p/2)

Example 2.27 X”+2X+cos X=0 at X=p/4 Solution; X= dx+ p/4 cos(dx+ p/4)=cos(p/4)-sin(p/4) dx+ … dx” +2 dx+[cos(p/4)-sin(p/4) dx]=0 dx” +2 dx’-20.5/2 dx= -20.5/2

Example 2.28(Nonlinear electrical network) Ldi/dt+10 ln(ir/2)-20=v(t) ir=2e0.1vr vr=10 ln(ir/2)

Example 2.28(Nonlinear electrical network) Taylor series i0=14.78, L=1

Antenna Azimuth Position Control System- Schematic

Case Studies – Antenna Control: Transfer Function Subsystems of the antenna azimuth position control System

Case Studies – Antenna Control: Transfer Function Input Potentiometer; Output Potentiometer Preamplifier Power Amplifier

Case Studies – Antenna Control: Transfer Function Motor and Load

Case Studies – Antenna Control: Transfer Function