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Theory of Machines and Automatic ControlWinter 2016/2017
Lecturer: Sebastian Korczak, PhD, Eng.
Warsaw University of TechnologyThe Faculty of Automotive
and Construction Machinery EngineeringInstitute of Machine Design Fundamentals
Department of Mechanicshttp://www.ipbm.simr.pw.edu.pl/
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Lecture 14
Material repeat.Informations about the exam.
Questionnaires.
Materials license: only for educational purposes of Warsaw University of Technology students.
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Degrees of freedom
material point (2D)
material point (3D)
rigid body (2D)
rigid body (3D)
2 DoF
3 DoF
3 DoF
6 DoF
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Kinematic pairs (3D)
Class V
rotary
= 6 - 1
translatory screw-type
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Kinematic pairs (3D)
Class IV
cylindrical
= 6 - 2
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Kinematic pairs (3D)
Class III = 6 - 3
spherical
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Kinematic pairs (3D)
Class II = 6 - 4
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Kinematic pairs (3D)
Class I = 6 – 5
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Kinematic pairs (2D)
Class V
rotary
= 6 - 1
translatory
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Kinematic pairs (2D)
Class IV = 6 - 2
cam joint
cam follower (tapper)
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Kinematic pairs
lower kinematic pair – surface contact
higher kinematic pair – line or point contact
closed pair (self-closed pair) – contact because of shape
open pair (force-closed pair) – force required for constant contact
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Kinematic chain - examples
Four-bar chain
a
d
b
c
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Kinematic chain - examples
connecting rod
piston
crank
crank pin
gudgeon pin (wrist pin)
Reciprocating motion (reciprocation)
Crank-slider mechanism
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Kinematic chain - examples
rcrank
yoke
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Kinematic chain mobility
F > 1 – movableF = 1 – constrainedF < 1 – locked or overconstrained
(3 D chain) F=6 N−p1−2 p2−3 p3−4 p4−5 p5
(2 D chain) F=3 N−p4−2 p5
N−number of moving bodies
pi−number of i−type classes
kinematic chain mobility – structural formula
(the Chebychev–Grübler–Kutzbach criterion)
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Kinematic chain - examples
F = 0 Locked? No!
overconstrained
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Classification of kinematic chains
Simple kinematic chain – every member has maximum two kinematic pairs.
Complex kinematic chain – at least one member has three kinematic pairs.
Open kinematic chain – at least one member has only one kinematic pair.
Closed kinematic chain – every member has minimum two kinematic pairs.
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Kinematics of mechanisms
Kinematic analysis of a mechanism – determination of velocities and accelerations of selected mechanism members' points at considered configuration. Mechanism structure must be given (geometry of members, kinematic pairs) and drive method must be known.
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Methods of velocities and accelerationdetermination
Graphical methods - velocity projection method, - instantaneous center of rotation method, - instantaneous center of acceleration method, - method of rotated velocities, - velocity decomposition method, - acceleration decomposition method, - velocity scheme method, - accelerations scheme method.
Analytical method
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Velocity projection method
A
B
v A
vB
Projections of velocities of two rigid body's points onto common line are equal.
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Instantaneous center of rotation method
A
Bv A
vB
α
S
α
center of instantaneous rotation(center of velocities)
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AB
+AB
AB =
vB= v A+ vBA
absolute velocity of point B velocity of a linear motion
Angular velocity of point B in rotation around point A.
vBA=ω× AB
Velocity decomposition method2nd example
ω
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Velocity scheme method
Velocity scheme of a rigid body – geometry created by the ends of it's velocity vectors moved to the common starting point (pole).
Velocity scheme is similar to the corresponding rigid body: it is
scaled and rotated by an 90o angle in the direction of body's
angular velocity.
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Velocities in relative motion
A1
A2
v A 2=v A 1+ v A 2 A 1
absolute velocity of
point A2
transportation velocity
relative velocity
A
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Instantaneous center of acceleration
A
BaA
aB
α
P
α
center of acceleration
ψψ
ψ=atan εω2
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AB
AB
ω+A
B
=
aB=aA+ aBA=aA+ aBAn + aBA
t
absolute acceleration of point B
Angular acceleration of point B in rotation around point A.
Acceleration decomposition methodExample
AB
ε+
absolute acceleration of point A Centripetal acceleration
(normal)
Rotary acceleration (tangential)
aBA=ω×(ω× AB )=−ω2 AB
aBA=ε× AB
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Acceleration scheme (diagram)
Acceleration scheme of a rigid body – geometry created by the ends of it's acceleration vectors moved to the common starting point (acceleration scheme's pole).
Acceleration scheme is similar to the corresponding rigid body:
it is scaled and rotated by (180o-ψ) angle in the direction of
body's angular velocity if sgnω=sgnε (or opposite direction if sgnω≠sgnε).
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Accelerations in relative motion
B1
B2
B
aB 2=aB 1u + aB 2 B1
w + ac
absolute acceleration of point B2
Transportation acceleration (absolute acceleration of point B1)
Relativeacceleration
Coriolisacceleration
ac=2 ωu×vB 2B 1
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Procedure of analytical determination of velocities and accelerations in planar mechanisms.
1. Set up Cartesian coordinate system OXY
.
2. Substitute the mechanism's members with set of vectors. All vectors can move with mechanism's elements, change their size, location and orientation.3. Vectors must to create closed polygons.4. Define “directed angles” for all vectors defined in the same manner. Assume that this angles are created by the positive x axis counter- clockwise rotation.5. Fore each of polygon write down sum of vectors, e.g.:
∑i=1
i=n
l i=0
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Procedure of analytical determination of velocities and accelerations in planar mechanisms.
6a. Write down projections of each polygon onto coordinate system's axes:
x: ∑i=1
i=n
li cosφ i=0 y: ∑i=1
i=n
li sin φ i=0
(we do not need to analyze signs because of „directed angles” setup procedure)6b. Define which vectors' lengths and angles are given and/or constant (related to geometry), and which are variable in time and unknown.(for a proper defined system number of unknown variables is equal to the number of equations)7. Solve the equations. The resulting functions describes motion of the mechanism.
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Procedure of analytical determination of velocities and accelerations in planar mechanisms.
8. Differentiate functions achieved in p.7 to obtain velocities. Differentiate once again to obtain accelerations.
9. If desired informations was not obtained in p.8, differentiate equations from p.6. Sometimes rotation of the coordinate system is useful here.
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Cam-follower
Cam-follower mechanism – mechanism build of a cam and a follower (tappet) connected as a IV class kinematic pair.
Cam is rotating (sometimes is translating)
follower is reciprocating (sometimes is swinging/oscillating)
advantages• simple to construction,• simple to create,• any dimensions,• simple to create advanced motions.
disadvantages• small strength with hight loads,• no adaptation possible.
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Cam-follower
Classification
flat / spatial
with in-line (central) follower / with offset (eccentric) follower
closed with geometry / closed with force
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Cam-follower
Followers
rollerknife-edge
spherical-faced
source: T. Kołacin, „Podstawy teorii maszyn i automatyki”, OW PW
flat-faced
flat-faced
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Cam-follower
Followers
swingingeccentric follower
flat frame
source: T. Kołacin, „Podstawy teorii maszyn i automatyki”, OW PW
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Cam-follower
examples
cylindrical cam
globoidal
translating cam
source: T. Kołacin, „Podstawy teorii maszyn i automatyki”, OW PW
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Analysis and synthesis of cam-follower mechanism
Analysis – calculation of displacement, velocity and acceleration functions for a follower motion with respect to a cam's rotations angle for arbitrary given geometry.
Synthesis – calculation of a cam geometry needed to obtain given displacement/velocity/acceleration functions. Limitations must be included, i.e. some maximum values, geometry limitations and jerk values (third derivative).
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Analysis and synthesis of cam-follower mechanisms
Analysis Syntesis
● substitution of IV. class kinematic pair with V. class kinematic pairs + graphical method (velocity and acceleration scheme)
● graphical determination of a follower movement and graphical differentiation
● analytical method (substitution with polygones of vectors)
● graphical determination of cam outline by a base circle rotation with follower movement
● analytical designing with a function description
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Analytical method
Synthesis of cam-follower mechanisms
For a given function of velocity or acceleration, function of a follower displacement can be found by integration.
Follower displacement as a function of cam angle could be used to obtain cam outline directly (or after change of coordinates).
For a knife-edge follower we will obtain exact real displacement.For a roller-ended follower some errors are possible.
Roller-ended follower give us limitation of a maximum velocity (there is a relation between roller radius and cam size).
Usually we are designing symmetric and smooth cams.
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Minimal cam size
Cam-follower mechanisms
1st condition: minimum cam radius in case of material strength and wear resistance.
2nd condition: maximum angle of contact in case of follower bending resistanceand pressure inside the socket.
3rd condition: maximum distance of contact in case of follower's stem bending (for a flat-faced followers).
Follower offset towards direction opposite to direction of rotation decrease angle of contact
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Overview
Dynamics of planar mechanisms
Members description as rigid bodies and and material points.
Graphical determination of inertial forces and torques.
Reaction forces in kinematic pairs.
Driving and operating forces/torques.
Inverse and direct dynamics problems.
Graphical, analytical and graphical-analytical method.
Friction in kinematic pairs.
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Members description
Dynamics of planar mechanisms
For a planar mechanism member represented by a rigid body:
● mass
● location of a center of a mass
● mass moment of inertia wrt the axis perpendicular to the motion
plan in center of a mass
● location of connection points
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Members description
Dynamics of planar mechanisms
a set of material points
Material points method
● same masses● same center of mass
● same inertia
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Inverse dynamics problem – calculation of forces and torques that cause given motion of a mechanism.
Direct dynamics problem – calculation of mechanism's motion caused by external forces and torques.
Dynamics of planar mechanisms
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Inverse dynamics problem
Dynamics of planar mechanisms
Calculation of forces and torques that cause given motion of a mechanism
0. Mechanism and it's geometry, driving and operating forces/torques, displacement, velocity and acceleration functions are given.
1. Calculation of inertia forces and torques acting moving members of the mechanism.
2. Decomposition of the mechanism with reaction disclosure.
3. Write down vector sums of external forces, reactions and inertia forces (d'Alembert equations).
4. Solve the equations with graphical and/or analytical method.
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Overview
Machine dynamics
time
angu
lar
velo
city
starting steady-state motion
stopping
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Kinetic energy
Reduction of masses
mr(t)F r(t )
xr( t)
I r(t)
M r(t)
φ r (t)
Total kinetic energy
T = 12
I r ωr2
reduced moment of inertia
T = 12
mr v r2
reduced mass
or
v r=dxr (t )
dt
ωr=d φ r (t)
dt
T=∑i=1
n
(12
mi v i2+1
2I iωi
2)
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System power
Reduction of forces
Total system's power
P=M rωr
reduced torque
P (F i , M i ,ωi , v i , ...)
P=Fr vr
reduced force
or
mr(t)F r(t )
xr( t)
I r(t)
M r(t)
φ r (t)
v r=dxr (t )
dt
ωr=d φ r (t)
dt
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Linear motion
Machine equation of motion
dT=dW
d ( 12
m(t) v (t)2)=F ( t)dx
12
dm(t )v (t)2+m( t)v ( t)dv (t)=F (t)dx
12
dm( t)v ( t)2+m(t )dx (t )
dtdv (t )=F (t )dx
dm(t)dx
v (t)2
2+m
dv (t)dt
=F (t )
dm(t )dt
v (t )2
+mdv (t )
dt=F (t)
if m=const . ⇒ mdv (t)
dt=P(t) o r m x ( t )=F (t )
m(t)F (t)
v (t)
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Angular motion
Machine equation of motion
dT=dW
d (I ω(t)2
2 )=M ( t)d φ
...
...dI (t)d φ
ω(t)2
2+ I (t)
dω(t )dt
=M (t )
dI ( t)dt
ω( t)2
+ I ( t)dω(t)
dt=M ( t)
if I=const . ⇒ Idω( t)
dt=M (t ) o r I φ (t)=M (t )
I (t)
M (t)
φ ( t)
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engine machine
φ ( t) I R
φ ( t)
t
ωmax
ωmin
T max=12
I Rωmax2 T min=
12
I Rωmin2
W=T max−Tmin=δ I Rωmean2
δ=ωmax−ωmin
ωmeanωmean=
ωmax+ωmin
2
Non-uniformity of machine motion
Non-uniformity of machine motionSteady-state motion
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ω( t)
φ ( t )
engine machine
φ ( t) I R
MD MP
φ ( t)
MD MP
π 2π
W
W=∫φmin
φmax
(M D−M P)d φ
δ=W
I Rωmean2
Non-uniformity of machine motionSteady-state motion
Example
ωmax
ωminW=T max−Tmin=δ I Rωmean2
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engine machine
φ ( t) I RI FW
I FW=( δ1
δ2−1) I R
W=δ1 I Rωmean2
assume I R≈const .
Flywheel
engine machine
φ ( t) I R
φ ( t )
t
ωmax
ωmin
Steady-state motion
φ ( t)
t
ωmax
ωmin
W=δ2( I R+ I FW )ωmean2
δ1 I Rωmean2 =δ2(I R+ I FW )ωmean
2
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Automatic control
“Automatic control in engineering and technology is a wide generic term covering the application of mechanisms to the operation and regulation of processes without continuous direct human intervention.” - wikipedia
Control theory – branch of mathematics and cybernetics that deals with analysis and mathematical modeling of objects and processes threated as dynamical systems with feedback.
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Automatic control
Classical control theorymodern control theory
(1950-now)
single input, single output (SISO) multiple input, multiple output (MIMO)
usually linear systems often nonlinear systems
time independent systems time dependent systems
description by a transfer functions description by a state equations
time and frequency domain analysis time domain analysis
system response is the most important system state is the most important
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Linear time-invariant (LTI) system
Linear system
x (t ) - input, y (t )=h(x (t )) - output
h(α x (t ))=αh(x (t ))=α y (t ) scaling
h(x1(t)+x2(t))=h(x1(t))+h(x2(t )) superposition
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Linear time-invariant (LTI) system
Time-invariant system
output does not depend explicitly on time
if y (t)=h(x (t )) then y (t−τ)=h(x (t−τ))
Time-varying system
if y (t)=h(x (t )) then y (t−τ)≠h(x (t−τ))
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Closed loop control
SYSTEMu(t)=x (t ) y (t )
CONTROLLER
yd (t )
desiredoutput control
function
system output
system input
+
-
e(t)
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Laplace transform
Assumption: x (t ) - signal such that for t<0 x (t )=0
X (s)=L{x (t )}=∫0
∞
x (t)e−st dt
where: s∈ℂ , s=σ+ jω , j=√−1
A necessary condition for existence of the integral is that x(t) must be locally integrable on t in <0, ∞).
Laplace transform of x(t):
Inverse Laplacetransform of x(t): x (t )=L−1{X (s)}= 1
2π jlimω→∞
∫γ− jω
γ+ jω
X (s)est ds
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Transfer function
dn y (t )dtn +a1
dn−1 y (t )dt n−1 +...+an−1
dy (t)dt
+an y (t )=dm x (t )
dtm +b1
dm−1 x (t )dtm−1 +...+bm−1
dx (t)dt
+bm x (t)
Linear time-invariant SISO system for continuous-time input signal x(t) and output y(t) in a form
after Laplace transformation with zero initial conditions
sn Y (s)+a1 sn−1 Y (s)+...+an−1 s Y (s)+an Y (s)=sm X (s)+b1 sm−1 X (s)+...+bm−1 s X (s)+bm X (s)
(sn+a1 sn−1+...+an−1 s+an)Y (s)=(sm+b1 sm−1+...+bm−1 s+bm)X (s)
H (s)=Y (s)X (s)
=sm+b1 sm−1+...+bm−1 s+bm
sn+a1 sn−1+...+an−1 s+an
Transferfunction
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Transfer function
H (s)=Y (s)X (s)
=sm+b1 sm−1+...+bm−1 s+bm
sn+a1 sn−1+...+an−1 s+an
H (s)=Y (s)X (s)
=(s−z1)(s−z2)...(s−zm)(s−p1)(s−p2)...(s−pn)
z1, z2 , ... , zm - zeroes
p1, p2 , ... , pn - poles
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Input and output
x (t ) y (t )=h(t)∗x (t )h(t)
X (s) Y (s)=H (s)X (s)H (s)
time domain
complex domain
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H (s) H ( jω)=P (ω)+ j Q(ω)
A (ω)=|H ( jω)|=√P2(ω)+Q2(ω)
φ (ω)=Arg H ( jω)=arctanQP
P(ω)
Q(ω)
ω=0ω=∞
y (t)=A sin (ω t+φ )
Transfer function – frequency response
input: x (t)=sin (ω t) output:H (s)transfer function:
Nyquist plot
A (ω)
φ (ω)
s= jω
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φ(ω
) [r
ad]
gain (magnitude) plot
Bode Plot
y (t)=A sin (ω t+φ )input: x (t )=sin (ω t ) output:H (s)transfer function:
phase (phase shift) plot
Transfer function – frequency response
ω [rad/s]
L(ω
) [d
B]
ω [rad/s]
L(ω)=20 log A (ω)
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Transfer function – frequency responseRC circuit example
A (gain) 20logA
1000 60
100 40
10 20
1 0
0.1 -20
0.01 -40
0.001 -60
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Classification of basic automatic systems
Element name Equation Transfer function
proportional k
first order (inertial)
integrator or
y (t )=ku (t )
Tdy (t )dt
+y (t )=ku (t )
y ( t )=k∫0
t
u (t )dt
dy (t )dt
=ku (t )
kTs+1
ks
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Classification of basic automatic systems
Element name EquationTransfer function
derivative
derivative with inertia
y (t )=kdu (t )
dt
Tdy (t )dt
+y (t )=kdu (t )
dt
ks
ksTs+1
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Classification of basic automatic systems
Element name EquationTransfer function
delay
second order (oscillator)
y (t )=u (t−τ )
T 12 d 2 y (t )
dt 2 +T 2
dy ( t )dt
+
+y (t )=ku (t )
e−τ s
k
T 12 s2 +T 2 s+1
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Proportional element
1. Element equation: y (t )=ku (t ) u(t ) - input, y (t ) - output
2. Static characteristic (steady state): y=ku for dydt=0∧ du
dt=0
3. Transfer function: H (s)=k
4. Step response: y (t)=k u0 1(t )
u
y
for u(t)=u0 1(t)
t
u0
u(t)k u0
y (t )
t
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Proportional element
5. Frequency response: H ( jω)=k P (ω)=k , Q (ω)=0
6. Nyquist plot:
P(ω)
Q(ω)
7. Bode plot:
φ(ω
) [r
ad]
ω [rad/s]
L(ω
) [d
B]
ω [rad/s]
L(ω)=20 log A (ω)A(ω)=√P2+Q2=|k| φ (ω)=arctan
QP={0 , dla k≥0
π , dla k<0 }
20 log|k|
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Proportional elementExamples
1
GEARBOX:input – angular velocity ω
1(t)
output – angular velocity ω2(t)
GEARBOX:input – rotation angle φ
1(t)
output – rotation angle φ2(t)
ω1(t)
ω2(t)
2φ
1(t)
φ2(t)
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Proportional elementExamples
4BEAM in steady state:input – force F
1
output – force F2
F1 F
2
3 OPERATIONAL AMPLIFIER:input – voltage v
1(t)
output – voltage v2(t)
Vsupply
0Vv
2(t)
v1(t)
R2
R1
v2 ( t )=v1 (t )(1+ R2
R1)
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Proportional elementExamples
5HYDRAULIC LEVER:input – displacement x
1(t)
output – displacement x2(t)
x1(t) x
2(t)
6 PRESSURE ACTUATOR:input – pressure p
1(t)
output – displacement x(t)
x(t)
p(t)
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First-order inertial element
1. Element equation: u(t) - inputy (t ) - output
2. Static characteristic (steady state): y=ku for dydt=0∧ du
dt=0
3. Transfer function: H (s)= kTs+1
u
y
Tdy (t )dt
+y (t )=ku (t )
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First-order inertial element
4. Step response:input: u (t)=u0 1(t)
u0
u(t)
Laplace of input: U ( s)=u01s
Laplace of output: Y (s)=H ( s)U (s)=k u0
s (Ts+1)
output: y (t)=L−1{Y (s)}=k u0(1−e−t /T )
k u0
y (t )
tT 2T 3T
0.950 k u0
0.865 k u0
0.632 k u0
t
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First-order inertial element
5. Frequency response: H ( jω)= kTjω+1
6. Nyquist plot:
P (ω)= kT 2ω2+1
, Q (ω)= −k T ωT 2ω2+1
P(ω)
Q(ω)ω=0ω=∞
k /2 k0
−k /2ω=1/T
for k>0
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First-order inertial element
7. Bode plot:
L(ω)=20 log A (ω)=20 log|k|−20 log√T 2ω2+1
A(ω)=√P2+Q2=|k|/√T 2ω2+1
φ (ω)=arctanQP=arctan (−T ω )
L(ω
) [d
B] ω [rad/s]1
10 T1T
20 log|k|−3
10 /T
20 log|k|−20φ(ω
) [r
ad]
−π2
−π4
1T
10T
ω [rad/s]
100T
110T
1100T
20 log|k|
20 log|k|−40
for k>0
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First-order inertial elementExamples
1LINEAR MOTION OF A MATERIAL POINT WITH LINEAR DAMPING:input – force F(t)output – velocity v(t)
F(t)
v(t)
example: car driving on a flat surface with air resistance proportional to velocity, described using machine equation of motion, with assumption of constant reduced mass.
2ANGULAR MOTION OF A RIGID BODY WITH LINEAR DAMPING:input – torque M(t)output – angular velocity ω(t)
M(t)
ω(t)
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First-order inertial elementExamples
3p
1(t)
p2(t) AIR CONTAINER:
input – pressure p1(t)
output – pressure p2(t)
4 HEATED OBJECT WITH SMALL INERTIA:input – heater power h(t)output – object temperature T
i(t)
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Integrator
1. Element equation:u(t) - inputy (t ) - output
2. Static characteristic (steady state): for dydt=0∧ du
dt=0
3. Transfer function: H (s)= ks
dy (t)dt
=k u(t )
u=0
u
y
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Integrator
4. Step response:
input: u (t)=u0 1(t)
u0
u(t)
u0
y (t )
t
Laplace of input: U ( s)=u01s
Laplace of output: Y (s)=H ( s)U (s)=k u0
s2
output: y (t)=L−1{Y (s)}=k u0 t
1/kt
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Integrator
5. Frequency response: H ( jω)= kjω
6. Nyquist plot:
P (ω)=0 , Q(ω)=− kω
P(ω)
Q(ω)
ω=∞0
for k>0
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Integrator
7. Bode plot:
L(ω)=20 log A (ω)=20 log| kω|
A(ω)=√P2+Q2=| kω|
φ (ω)=arctanQP=arctan(−∞)
φ(ω
) [r
ad]
−π2
ω [rad/s]
for k>0
L(ω
) [d
B] ω [rad/s]
k /10 k
−20dB/dek
10k0
20
40
100k
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IntegratorExamples
1PRISM LIQUID TANK:input – liquid inflow f(t)output – liquid level h(t)
h(t)
f(t)
2 OPERATIONAL AMPLIFIER:input – voltage v
1(t)
output – voltage v2(t)V
supply
0Vv
2(t)
v1(t)
CR
v2(t )=1
RC∫0t
v1(t)dt
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IntegratorExamples
3GEARBOX:input – angular velocity ω(t)output – rotation angle φ(t)
ω(t)
φ(t)
4 HYDRAULIC CYLINDER:input – volume inflow f(t)output – displacement x(t)
x(t)
f(t)
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Differentiator
1. Element equation:u(t) - inputy (t ) - output
2. Static characteristic (steady state): y=0 for dydt=0∧ du
dt=0
3. Transfer function: H (s)=k s
u
y
y (t)=kdu(t)
dt
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Differentiator
4. Step response:
input: u (t)=u0 1(t)
u0
u(t) y (t )
t
Laplace of input: U ( s)=u01s
Laplace of output: Y (s)=H ( s)U (s)=k u0
output: y (t)=L−1{Y (s)}=k u0δ(t )
t
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Differentiator
5. Frequency response: H ( jω)= j k ω
6. Nyquist plot:
P (ω)=0 , Q(ω)=k ω
P(ω)
Q(ω)
ω=00
for k>0
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Differentiator
7. Bode plot:
L(ω)=20 log A (ω)=20 log|kω|
A(ω)=√P2+Q2=|k ω|
φ (ω)=arctanQP=arctan(∞)
φ(ω
) [r
ad]
π2
ω [rad/s]
for k>0
L(ω
) [d
B]
ω [rad/s]k /10
k
+20dB/dek
10k0
20
40
−20
−40
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DifferentiatorExamples
1GEARBOX:input – rotation angle φ(t)output – angular velocity ω(t)
ω(t)
φ(t)
2 OPERATIONAL AMPLIFIER:input – voltage v
1(t)
output – voltage v2(t)
v2(t )=−RCdv1(t )
dt
Vsupply
0Vv
2(t)
v1(t)
C R
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Real differentiator (derivative+1st order)
1. Element equation: u(t) - inputy (t ) - output
2. Static characteristic (steady state): for dydt=0∧ du
dt=0
3. Transfer function: H (s)= k sTs+1
Tdy (t)
dt+ y (t )=k
du(t )dt
y=0
u
y
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Real differentiator (derivative+1st order)
4. Step response:input: u (t)=u0 1(t)Laplace of input: U ( s)=u0
1s
Laplace of output: Y (s)=H ( s)U (s)=k u0
Ts+1
output: y (t)=L−1{Y (s)}=k u0 e−t /T
k u0
y (t )
tT 2T 3T0.050 k u0
0.135 k u0
0.368 k u0u0
u(t)
t
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Real differentiator (derivative+1st order)
5. Frequency response: H ( jω)= k jωTjω+1
6. Nyquist plot:
P (ω)= k T ω2
T 2ω2+1, Q(ω)= k ω
T 2ω2+1
P(ω)
Q(ω)
ω=0 ω=∞k
2TkT
0
−k /2ω=1/Tfor k>0
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Real differentiator (derivative+1st order)
7. Bode plot:
L(ω)=20 log A (ω)=20 log|kω|−20 log√T 2ω2+1
A(ω)=√P2+Q2=|k ω|/√T 2ω2+1
φ (ω)=arctanQP=arctan ( 1
T ω )
φ(ω
) [r
ad]
π2
π4
1T
10T
ω [rad/s]
100T
110T
1100T
for k>0
L(ω
) [d
B]
ω [rad/s]110T 1
T
20 log|k /T|−3
10 /T
20 log|k /T|−20
20 log|k /T|
20 log|k /T|−40
0
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Real differentiator (derivative+1st order)Examples
1 RC CIRCUIT:input – voltage u
1(t)
output – voltage u2(t)
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Delay
1. Element equation:u(t) - inputy (t ) - output
2. Static characteristic (steady state): y=u for dydt=0∧ du
dt=0
3. Transfer function: H (s)=e−τ s
u
y
y (t)=u(t−τ)
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Delay
4. Step response:input: u (t)=u0 1(t)Laplace of input: U ( s)=u0
1s
Laplace of output: Y (s)=H ( s)U (s)=u0
se−τ s
output: y (t)=L−1{Y (s)}=u01(t−τ)
u0
u(t)
t
u0
y (t )
t
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Delay
5. Frequency response: H ( jω)=e−τ jω
6. Nyquist plot:
P (ω)=cos(τω), Q (ω)=−sin (τ ω)
P(ω)
Q(ω)
ω=00
ω= π2 τ
ω=πτ
ω=3π2 τ
1
1
−1
−1
for k>0
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Delay
7. Bode plot:
L(ω)=20 log A (ω)=20 log 1=0A(ω)=√P2+Q2=1
φ (ω)=arctanQP=arctan (−tan( τω))=−τω
φ(ω
) [r
ad]
−π
πτ
ω [rad/s]
10πτ
L(ω
) [d
B]
ω [rad/s]
110 T
1T
10T
0
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DelayExamples
1 WIRELESS TRANSMISSION:input – sent dataoutput – received data
data pocket data pocket
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Second-order inertial element
1. Element equation:
2. Static characteristic (steady state): y=ku for dydt=0∧ du
dt=0
3. Transfer function: H (s)= k
T 12 s2+T 2 s+1
u
y
T 12 d 2 y (t)
dt 2 +T 2
dy (t )dt
+ y (t )=k u(t)
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Second-order inertial element
4. Step response:
u0
u(t)
t
k u0
y (t )
t
h<ω0
h=ω0
h>ω0
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Second-order inertial element
5. Frequency response: H ( jω)= k
−T 12ω2+T 2 jω+1
6. Nyquist plot:
P (ω)=k (1−T 1
2ω2)
(1−T 12ω2)2+T 2
2ω2 , Q (ω)=−k T 2ω
(1−T 12ω2)2+T 2
2ω2
P(ω)
Q(ω)ω=0ω=∞
k0
ω=1/T
for k>0
for h<ω0
for h=ω0
for h>ω0
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Second-order inertial element
7. Bode plot:
L(ω)=20 log A (ω)
A(ω)=√P2+Q2
φ (ω)=arctanQP
L(ω
) [d
B]
ω [rad/s]
110 T 1
1T 1
20 log|k|−20
20 log|k|
20 log|k|−40
for k>0
10T 1
for h<ω0
for h=ω0
for h>ω0
φ(ω
) [r
ad]
π
π2
1T
10T
ω [rad/s]
100T
110 T
1100 T
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Second-order inertial elementExamples
material point of mass m
linear spring with stiffness k
linear damper with damping c
1 VIBRATING SYSTEM:input – force F(t)output – displacement y(t)
y(t)
F(t)
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Second-order inertial elementExamples
LINEAR MOTION OF A MATERIAL POINT WITH LINEAR DAMPING:input – force F(t)output – displacement x(t)
F(t)
x(t)
example: car driving on a flat surface with air resistance proportional to velocity, described using machine equation of motion, with assumption of constant reduced mass.
2
3ANGULAR MOTION OF A RIGID BODY WITH LINEAR DAMPING:input – torque M(t)output – angle φ(t)
M(t)
φ(t)
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Second-order inertial elementExamples
4 HEATED OBJECT WITH HIGH INERTIA:input – heater power h(t)output – object temperature T
i(t)
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Classification of basic automatic systems
Element name Transfer function
proportional k
first order (inertial)
integrator
differentiator
differentiator with inertia
delay
second order (oscillator)
kTs+1
ksksks
Ts+1
e−τ s
k
T 12 s2 +T 2 s+1
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BLOCK DIAGRAM ALGEBRAinformation node
X(s)
one input,a few outputs,
X(s) X(s)
X(s)
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BLOCK DIAGRAM ALGEBRAsum node
A(s) +
B(s)
+
A(s)+B(s)-C(s)–
C(s)
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BLOCK DIAGRAM ALGEBRAserial connection
G1(s)
x(s) y(s) x(s) y(s)G
R(s)G
2(s)
GR(s)=G
1(s) G
2(s)
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BLOCK DIAGRAM ALGEBRAparallel connection
G1(s)x(s) y(s) x(s) y(s)
GR(s)
G2(s)
GR(s)= - G
1(s) + G
2(s) + G
3(s)
-
+
G3(s)
+
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BLOCK DIAGRAM ALGEBRAfeedback
G2(s)
+
–
x(s) y(s) x(s) y(s)G
R(s)G
1(s)
GR=G1
1+G1G2
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Closed loop control
SYSTEMu(t)=x (t ) y (t )
CONTROLLER
yd (t )
desiredoutput control
function
system output
system input
+
-
e(t)
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Types of controllers
● ON/OFF● three state
● proportional● integrator
● differentiator● proportional-inegral-derivative
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RELAY / ON-OFF / TWO STATE / BANG-BANG CONTROLLER
input
output
input
output
input
output
real(with hysteresis)
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THREE STATE CONTROLLER
input
output
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Types of controllers
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SATURATION
input
output
Symmetric hard limiting saturation
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DEAD ZONE
input
output
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PID CONTROLLERstep responses
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P - CONTROLLER
time
input
output
G(s)=K p
K p x0
x0
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I - CONTROLLER
time
input
output
T i
G(s)= 1T i s
x0
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PI - CONTROLLER
time
input
output
T i
G(s)=K p( 1+ 1T i s )
K p x0
x0
2 K p x0
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D - CONTROLLER
time
input
output
G(s)=T d s+∞
x0
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D - CONTROLLER
time
input
output
G(s)=T d s
T s+1
x0
T
x0
T d
T
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PD - CONTROLLER
time
input
output
G(s)=K p( 1+T d s
T s+1)
x0
T
K p x0
K P x0( 1+T d
T )
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PID - CONTROLLER
time
input
output
G(s)=K p(1+ 1T i s
+T d s
T s+1 )
x0
K p x0
K P x0(1+T d
T )2 K p x0
T i
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PID CONTROLLERimportant notes
Proportional term – necessary part of the controller, creates a main part of control signal that bring output of the system closer to desired value; higher K
P coefficient gives lover errors;
control signal is based on present error;
Integral term – this part of the controller accumulates error; for nonzero error control signal increases that helps to achieve zero error; control signal is based on past error values; “integral windup” problem;
Derivative term – this part of the controller reacts on error changes; for constant error control signal is zero; control signal is based on the trend of feature error; this term is very sensitive to noise;
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PID CONTROLLERintegral windup problem
After a large change in a setpoint the integral term can produce very large control signal (higher than maximum possible) – system input is very hight until accumulated error goes back close to zero.
Possible solution: disabling and zeroing integral term outside the small region around the setpoint.
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PID CONTROLLERtuning methods
Analytical With a simulation Experimental
1st step: calculation of the system's reduced
transfer function2nd step: calculation of
the system's step response
3rd step: tuning of the Kp, Ki and Kd
coefficients to obtain desired shape of step
response
1st step: calculation of the system's reduced
transfer function2nd step: numerical
implementation of the system's reduced transfer function
3rd step: tuning of the Kp, Ki and Kd
coefficients to obtain desired shape of the system's simulated
outputs
Manual tuning
or
Ziegler-Nichols method
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PID CONTROLLERZiegler-Nichols tuning method (PID in standard form)
1. Disable integral and derivative terms of the controller. Set proportional gain to small value.2. Observe a step response of the output of control loop. Go to point 3, if you observe stable and consistent oscillations. If not, increase proportional gain and repeat step 2.
3. For the ultimate gain Ku
from step 2 and oscillation period Tu
calculate
parameters of the controller according to the table:
kp
Ti
Td
classic Z-N 0.6 Ku
0.5 Tu
0.125 Tu
Pessen 0.7 Ku
0.4 Tu
0.15 Tu
no overshoot 0.2 Ku
0.5 Tu
0.333 Tu
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General stability criterion
Re p1<0 ∧ Re p2<0 ∧...∧ Re pn<0
G (s)=(s−z1)(s− z2)...( s−zm)( s− p1)(s− p2)...( s− pn)
LTI SISO system is assymptotically stable if real part of every pole of the system's transfer function is less than zero.
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Hurwitz criterion
LTI SISO system with a transfer function
H (s)=bm sm+bm−1 sm−1+...+b1 s+b0
an sn+an−1 sn−1+...+a1 s+a0
=(s−z1)(s−z2) ...(s−zm)(s−p1)(s−p2)...(s−pn)
an>0 , an−1>0 , ... , a1>0 , a0>01
2
M n=[ an−1 an 0 0 0 0 an−3 an−2 an−1 an 0 0 an−5 an−4 an−3 . . . . . . . . . 0 0 0 a0 a1 a2
0 0 0 0 0 a0
]Δ2 Δ3 Δn−1
detΔ2>0detΔ3>0...detΔn−1>0
is stable if:
- leading principal minor of order i
Δi
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Nyquist stability criterion
G z (s)=y( s)x (s)
=G1(s)
1+G1(s)G2(s)
G1(s)
G2(s)
+
–
x(s) y(s)
G1G2 =−1Unstable if:
G1(s)
G2(s)
Gopen(s)=a(s)x (s)
=G1(s)G2(s)
a(s)
y(s)x(s)
a(s)
ω=0
ReGopen
Im G open
ω→−∞ω→+∞
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Nyquist stability criterion (particular)
The closed-loop system is stable if:1) open-loop transfer function is stable AND
2) open-loop transfer function not enclosing the point (-1,j0).
ReGopen
Im G open
ReGopen
Im G open
-1 -1
stable unstable
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TRANSFER FUNCTION
P(ω)
Q(ω)
Δ M
-1
gain margin
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TRANSFER FUNCTION
P(ω)
Q(ω)
-1
Δφ
phase margin
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Theory of Machines and Automatic ControlWinter 2016/2017
Field of studies: Electric and Hybrid Vehicle Engineering (full-time)
form of studies: 30 hrs lecture, 15 hrs projects
ECTS: 4
24.01.2017 TM&AC, Lecture 14, Sebastian Korczak, only for educational purposes of WUT students. 142
EXAM – TERM 1
3rd February 2017 (Friday)
13:25 – lecture hall (room 2.5) openning
13:25-13:30 – preparation
13:30-14:30 – exam
6th February 2017 (Monday)
to 12:00 – publication of exam effects on the website
myinventions.pl/dydaktyka/ and USOSweb
12:00-14:00 – filling of indexes and erasmus papers
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EXAM – TERM 2
10th February 2017 (Friday)
13:25 – lecture hall (room 2.5) openning
13:25-13:30 – preparation
13:30-14:30 – exam
12th February 2017 (Sunday) – last session day
do 23:59 – publication of exam effects on the website
myinventions.pl/dydaktyka/ and USOSweb
13th February 2017 (Monday)
12:00 – 14:00 – filling of indexes and erasmus papers
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EXAM – IMPORTANT NOTES
● You have to pass the project class to attend the exam.● Index, student card or erasmus paper is needed on the exam.● Please write the exam clearly on the A4 paper.● Everyone must to return the exam.● You can not use any electronic devices during the exam
(mobile phones, smart watches, calculators).● Table of Laplce transform will be displayed on the screen.● Additional persons are delegated to help during the exam.● Any cheating behaviors will cause exam failure.● Topics will be distributed in printed form.
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EXAM – IMPORTANT NOTES
● Your answers will be rated with points.● Exam mark will be based on the total number of points
achieved with the rules: < 50% - mark 2 (exam failed) 51%-60% - mark 3,0 61%-70% - mark 3,5 71%-80% - mark 4,0 81%-90% - mark 4,5 >90% - mark 5,0
● Final_mark = 0.5 * project_mark + 0.5 * exam_mark
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EXAM – IMPORTANT NOTES
MAIN GROUPS OF TOPICS
1. Mechanisms – kinematic pairs, movability, velocities and accelerations, dynamics.2. Machine dynamics – system reduction, equation of machine motion, flywheel.3. Laplace transform. Transfer function.4. Clasification of basic automatic systems and their characteristics (step responses, Nyquist and Bode plots).5. Block diagram algebra (information and sum nodes, serial, parallel and feedback connections).6. Controllers (on/off, PID).7. Stability criterions.
Please prepare carfully for the exam.