c2: basic of mechanical and electrical control system
TRANSCRIPT
Lecture 2
• Basic of mechanical and electrical control system
• By
• Oladokun Sulaiman
Note
Revise :
• Differential equation
• Partial fraction
2.0 Objective
• Free body and block diagram• Block diagram• Obtain the differential equation• Obtain Laplace transform of the differential
equations• Solve the resulting algebraic transform• Mathematical modeling of physical system
RECAP
implified block diagram of closed-loop control system
• Process or plant:– System to be controlled including actuators and power amplifiers
• Sensor:– Instrumentation that measures output and converts it to a signal
• Compensator or controller:– System added to enhance performance of control loop– Output or controlled variable is y(t) is the variable we wish to control– Input is a measure of (but not always equal to) the desired system
output u(t)– Error is desired output minus actual output: e(t)= y(t) - u(t)
Gc(s) Gp(s)Output
H(s)
Input
Y(s)U(s)
Compensatoror controller
Processor plant
Sensor
test waveform
•Refer to control engineering system page 19
Types of input
0 t1 t2 t3t4
Time, second
50
100
Forc
e,
N
0 t1 t2 t3t4
Time, second
50
100
Forc
e,
N
0 t1 t2 t3t4
Time, second
50
100
Forc
e,
N
0 t1 t2 t3
Time, second
50
100
Forc
e,
N
Two step input response
Ramp Input Response
Time Constant
• It is defined as time taken by a control plant to achieve output responseequal to 63% of its desired value.
Control System Response
• Time response
• Frequency response
• Steady state response
• Transient response
• Undershoot
• Overshoot
• Settling time
Control system stability
• Damping factor
• Damping ratio
• Routh’s stability criteria
• Nyquist stability criteria
fig_01_11
Control System Design Process
•Control engineering system page 16
2.1 Introduction : Mathematical modeling
• In order to analyze and design a control system knowledge of its behavior through mathematical terms is essential
• The mathematical equations are derived from law of physics i.e Newton
• Analysis can be done in two operating condition: Steady and Transient
• Change as a result of input or disturbance• Output will depend on the system variables and how they
interact• Description of dynamic system is obtain from differential
equation’• models of the physical system• Solutions of these mathematical equations simulate the
response of the physical system which they represent
2.2 Differential equation• Input and output relationship of a linear measurement
system - ordinary differential equation (ODE):
• u = input, y = output; u and y varies with t• n > m and a, b = constant coefficients
• DE contain variable and rate of change of or derivative of the variable in control system
• Ordinary differential equation (ODE) are main concern in control system , they contain singe dependent and independent variable which is usually time
• The order DE relate to the index of the highest derivative
yadt
dya
dt
da
dt
yda
n
n
nn
n
n 011
1
1
Example 2.1 : Spring mass balance damper
xo
xi
mk
c
=
Fs
FD
Fm
2
2
)()(dt
xdm
dt
dx
dt
dxcxxk ooi
oi
ii
ooo x
m
k
dt
dx
m
cx
m
k
dt
dx
m
c
dt
xd
2
2
2.3 Physical system modeling
• To obtain linear approximation of physical system Time response solution is obtain:
• Obtain the differential equation
• Obtain Laplace transform of the differential equations
• Solve the resulting algebraic transform
2.3a. Laplace transformDefinition of Laplace transformation of
f(t):
where s = + j = a complex variable– Inverse Laplace transformation– f(t) = L-1[F(s)]– L[Af(t)] = AL[f(t)]– L[f1(t) + f2(t)] = L[f1(t)] + L[f2(t)]
For step input f(t) = 0 t < 0– = A t > 0 – Laplace transform:
0 0
)()]([)()]([ dtetftfdtesFtfL stst
A
f(t)
t
dtAeALtfLsF st
0
][)()(
s
Ae
s
A st
0
Example 2• Find the time response xo(t) for this system if step input xi(t)=1
and initial condition xo(0)=0
• Differential equation: • For k =1,c = 1;
• Laplace transform
• Partial fraction
• Inverse Laplace transform:
0)( 00
dt
dxcxxk i
xi
k
c
xo
k = 1c = 1 dt
dxc 0
k(xi-x0)
ioo kxkx
dt
dxc
ioo xx
dt
dx
ssXssX
1)()( 00
)1(
11)(
sssX o
tetx 1)(0
Transient response Steady-state response
xo(t)
t
1
table_02_01
•Control engineering system page 33
table_02_02•Control engineering system page 34
2.3b. Transfer Functions
• Defined as the ratio of the Laplace Transform of the output to the Laplace Transform of the input to the system
• G(s) = Y(s)/X(s)
• X(s) Y(s)G(s)
Transfer Function
• An assembly of linked components within a boundary.
• The motor car is a good example; mechanical, electrical, control and suspension sub-systems within a body-chassis boundary.
• A system may have one input and a related output dependent on the effect of that system (transfer function G).
•0 = G I
•The boundary, represented as a "black box", may include a complex system which need not be analysed if G is provided. •More complex systems have interconnecting links to related systems. •A system must have input, process, output, and in most systems a source of power and a means of control.
Transfer Function Expression
Characteristics Equation
0012
2 aSaSa
•Denominator of the transfer function equated to zero is the characteristics equation of the system
•Characteristics equation of the system determines the response of the control system
012
2
01)(aSaSa
bSbsG
Characteristics Equation
Order of control systems
• Zero Order System
• First Order System
• 2nd Order System
)( 2121
01
asass
bsb
)( 2120
01
asass
bsb
)( 2122
01
asass
bsb
(Source: Instrumentation and Control Systems by Leslie Jackson)
Poles and Zeros
• Roots of the Characteristics equation are called poles of the system
Roots of numerator of the TF are called zeros of the system
;3,2,1;5;1)2(
;2,1,2)1(
6116
56)2
23
42)1
23
2
2
poleszeros
poleszerosAns
sss
ss
ss
s
Example
2.4 System modeling
Step for drawing block diagram:
• Step 1: Free body diagram
• Step 2: Mathematical equations
• Step 3: Block diagram
2.4a Mechanical system: Spring
• Spring• Where k = stiffness, x= displacemnet
• Fs= Fx or
• Transfer function = k or 1/k• Output variable = transfer function input variable• Spring with free at both ends
• Block diagram Fs = k(x1 – x2)
Fs xx
Fsk
kx1 x2
kx1(t)
x2(t)
+-
x1- x2 Fs(t)
4. Mechanical system: Mass
• Fm = ma
• Force F acting on mass m
• Use D-operator where: D = d/dt and D2 =d2/dt2
• Fm = mD2x
m
Fm(t) x(t)
2
1
mD
2
2
dx
xdmFm
Fm(t)x(t)
Spring mass system
•Equations:
• Fs = k(xi – x0)• Since Fm = ma and Fs = Fm• mD2x0(t) = Fs = k(xi – x0)kxi(t)xi(t)-xo(t)-+
• Block diagram for spring-mass system
m
xi(t)xo(t)
kxi(t) xi(t)-xo(t)
-+
Damper
• c = damping coeff.
• x = displacement
• dx(t)/dt = velocity
• FD(t) = cdx(t)/dt
• FD(t) = cDx
x(t) FD(t)
cD
Spring-damper system
• Force on spring: Fs = k(xi – xo)• Force on damper: FD = cDxo
xi(t)
xo(t)
kxi(t) Fs=FD
-+
Block diagram for spring-damper system
1/cD
Spring-mass-damper system
F = ma• Fs1 - FD = Fm
Fs1 FD
Fm
kxi(t) Fs
-+
FD
-+
xo(t)Fm
cD
2
1
mD
Spring Mass System
)(2
2
2
2
ForceDisturbingFkydt
dyf
dt
ydm
kyForceSpringdt
dyfForceDamping
dt
ydmForcengAccelerati
ksfmssF
sYsGTF
sFsYkfssm
2
2
1
)(
)()(
)()()(
fig_02_15
Example 2
•Draw free body diagram
•Determine forces and direction i.e applied force to the right and impeding forces to the left spring, viscous damper, acceleration
•Write the differential equation
Contd
2 ( ) ( ) ( ) ( )vMS s f sX s KX s F s
2( )vMS f s K
•Take la place transform
•Or
•Solve for transfer function
•G(s)=X(s)/F(s) = 1/
2( ) ( ) ( )vMS f s K X s F s
•Control engineering system page 60
fig_02_17
•Refer to control engineering system page 63
fig_02_18
fig_02_20
•Refer to control engineering system page 63
fig_02_21
fig_02_22
table_02_05
2.4b. Electrical system modeling Electrical components: resistance R, capacitance C and inductance LVariables: voltage V and current i
• Resistor -> VR = iR
• Inductance -> VL = Ldi/dt = LDi•
• Capacitance->i = CdV/dt = CDV
Circuit theory• Series
V = V1 + V2• Parallel• V = V1 = V2
– i = i1 + i2
+ -VR
+ -
VL
+ -
VCi
+ V
V1 V2
V1 V2
+
-
V
Example 3• Vi = VL + VC + VR; VR = Vo • VL = Ldi/dt = LDi
• CdVC/dt = i; VC = i/CD• Vo = iR
VL
Vo
+
-
Vi
VC
L1
Vo
+
-
Vi
VA
L2
C
+
-
R
i1
i2 i3
Equations:i1 = i2 + i3Vi – VA = L1Di1VA = L2Di2i3 = (VA – Vo)CDVo = i3R
Block diagram
•Use D-operator where D = d/dt and D2 =d2/dt2
Vi i1
-
+
-
+
VA
DL1
1L2D
+R
-
VA
i3
i2 CDi3
Vo
Vo
Example
• Subtitute i(t)=dq(t)
• Subtitute capacitor voltage charge relationship Q(T)=CvC(t)
• Take la place transform, rearrange terms and simplify
• Solve TF Vc(s)/V(s)
Summing the voltage around the loop ->assuming zero initial condition
2( 1) ( )CLCs RCs v V s
2 1
( ) 1/
( )C
CLC
v s LCRv s s SL
R-L-C Circuit
)()(11
)()(11
)()(
1
1
0
0
sEsIsC
sEsIsC
sIRsIsL
givestransformLaplaceTaking
eidtC
eidtC
Ridt
diL
i
i
•Combining two equations from previous slide gives
1
1)(
2
sRCsCLsGTF
TF of spring mass system and RLC circuit are mathematically similar and will give identical response
•Solution for electrical can also be done through KVL, KCL,
• voltage divider, current divider
•Refer to control engineering system page 45
Steps for electrical modeling
• Replace passive element value wit their impedance
• Replace all source and time value with their Laplace transform
• Assume transform current and current direction is each loop
• Write KVL around each loop• Solve simultaneous equation for the output• Form the transfer function
fig_02_05
fig_02_11
2.3 Block diagram manipulation
• Block diagram – can be simplified to fewer blocks
• Block diagram transformation and reduction –refer to table
• Output input relationship – transfer function
Block diagram manipulation
2 = F1(D)1 ; 3 = F2(D)2 3 = F1(D) F2(D)1 • Transfer function = 3 / 1 = F1(D) F2(D)
F1(D)1
F2(D)3
2
F1(D)F2(D)1
3
F1(s)
F2(s)
++
3
41 2
F1(D)+F2(D)1
4
Series block diagram reduction
Parallel block diagram reduction
4 = 2 + 3 4 = F1(D)1 + F2(D)1 4 = [F1(D) + F2(D)]1
Transfer function = 4 / 1 = F1(D) + F2(D)
Closed loop block diagram
• The negative feedback:• E(s) = U(s) – B(s) = U(s) – H(s)Y(s)• Y(s) = G(s)E(s) = G(s)[U(s) - H(s)Y(s)]• Y(s)[1 + G(s)H(s)] = G(s)U(s)
G(s)
H(s)
E(s)
-+
Input
Output
B(s)
)()(1
)(
)(
)(
sHsG
sG
sU
sY
U
(s)
Y(s)
)()(1
)(
sHsG
sG
Close-loop transfer function = (Forward transfer function)/(1+ Open-loop transfer function)
Example
• Derive transfer function for spring-mass system
200 mD
kxxx i
-+
xixo
2mD
k
kmD
k
x
x
i
o
2
x
i
x
o
kmD
k
2
Multiple-loop feedback control systemTo eliminate G3(s)G4(s)H(s), move H2 behind G4(s)Rule 4:
G4
H2 /G4
G4
H2
G3G4H1 = positive feedback control systemRule 6 – eliminate feedback loops
•Eliminate inner loop containing H2/G4 and G2 and G3G4/(1-G3G4H1)
Reduce the loop containing H3
Revise your digital last semester digital electronics
Tutorial Exercises • Define transfer function of a control
system.
• Define what is time constant.
• Describe the advantages of control and automation
Home work
Chapter 2 : NISE Control ->• Answer all short questions• Problem 7,9,10,16,17,20,28
• Due date January 7
Summary
• Free body and block diagram
• Block diagram
• Obtain the differential equation
• Obtain Laplace transform of the differential equations
• Solve the resulting algebraic transform
• Mathematical modeling of physical system