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Theory of Machines and Automatic ControlWinter 2017/2018

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/

24.01.2017 TM&AC, Lecture 14, Sebastian Korczak, only for educational purposes of WUT students. 2

Lecture 14

Material repeat.Informations about the exam.

WUT questionnaires.

Materials license: only for educational purposes of Warsaw University of Technology students.


Lecture 1

kinematic pairs, mechanisms, mobility

Materials license: only for education purposes of Warsaw University of Technology students.


Degrees of freedom

material point (2D)

material point (3D)

rigid body (2D)

rigid body (3D)

2 DoF

3 DoF

3 DoF

6 DoF


Kinematic pairs & chains

A kinematic pair is a movable coupling of two rigid members that imposes restraints on the relative motion of the members by the conditions of linkage.

A kinematic chain is an assembly of kinematic pairs.

A base is a fixed (motionless) member of mechanism.


Kinematic pairs (3D)

Class V

rotary

= 6 - 1

translatory screw-type



Class IV

cylindrical

= 6 - 2



Class III = 6 - 3

spherical



Class II = 6 - 4



Class I = 6 – 5



Class V

rotary

= 6 - 1

translatory



Class IV = 6 - 2

cam joint

cam follower (tapper)


Kinematic pairs

lower kinematic pair – surface contact

higher kinematic pair – line or point contact


Kinematic pairs

closed pair – contact because of shape

open pair – force required for constant contact


Kinematic chain mobility

F > 1 – movableF = 1 – constrainedF < 1 – locked or overconstrained

(3D chain) F=6N − p1 − 2 p2 − 3 p3 − 4 p4 − 5 p5

(2D chain) F=3N − p4 − 2 p5

N− numberof moving bodies

pi − number of i − type classes

kinematic chain mobility – structural formula

(the Chebychev–Grübler–Kutzbach criterion)


Lecture 2

Structural classification,velocities in planar mechanisms.

Materials license: only for education purposes of Warsaw University of Technology students.


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.


Structural classification of mechanisms

Structural group – the simplest part of mechanism that has zero

mobility.

Planar mechanism with only 5th class pairs: F=3n − 2 p5=0

p5

n=

3

2=

6

4=

9

6=.. .

n=2 p5=3

IInd structural

group

IIIrd structural

group

n=4 p5=6 n=6 p5=9

IVth structural group


Structural classification of mechanisms

crank drive

Ist structural group – drive

n=1 p5=1 + drive

linear drive rotary drive


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.


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


Methods of velocities and accelerationdetermination

Graphical methods Analytical method

advantage

better understanding of mechanism motion,

analysis of very complicated mechanisms,

computers not needed,

functions of configuration as a solution,

analysis of very complicated mechanisms,

disadvantage

great workload, needs to repeat graphs

for every configuration, graphical errors.

computer needed for complicated mechanisms,

complicated systems of equations to solve,

solution interpretation may be complicated.


Velocity projection method

A

B

vA

vB

Projections of velocities of two rigid body's points onto

common line are equal.


Instantaneous center of rotation method

A

BvA

vB

S

center of instantaneous rotation

(center of velocities)


AB

+AB

AB =

vB= v A+ v BA

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 method

2nd example

ω


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.


Velocity scheme method

A

B

vA

vB

C

vC

vB

vA

Ov

a

b

c

90o

velocity scale: the samegeometry scale: new!

Example

Graph in scale! e.g.:

geometry scale: 1cm → 10cmvelocity scale: 1cm → 1m/s


Velocities in relative motion

A1

A2

v A2=v A1+ v A2 A1

absolute

velocity of

point A2

transportation

velocity

relative

velocity

A


Instantaneous center of acceleration

A

BaA

aB

P

center of acceleration

ψ=atan εω2


AB

A

Bω

+AB

=

aB=aA+ aBA=aA+ aBAn+ aBA

t

absolute acceleration

of point B

Angular acceleration of point B

in rotation around point A.

Acceleration decomposition method

Example

AB

ε+


of point A Centripetal acceleration

(normal)

Rotary acceleration

(tangential)


AB

A

Bω

+AB

=

aB=aA+ aBA=aA+ aBAn+ aBA

t

aBA=ω×(ω× AB )= − ω2AB

Acceleration decomposition method

Example

AB

ε+

Centripetal acceleration

(normal)

Rotary acceleration

(tangential)

aBA=ε× AB


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-) angle in the direction of

body's angular velocity if sgnω=sgnε (or opposite direction

if sgnω≠sgnε).


Acceleration scheme method

A

B

aA

aB

C

Oa

a

b

acceleration scale, e.g.: 1cm → 1m/sgeometry scale wrt. original dimensions

Example

Given: aA and aB + geometry

Searched: aC

c

aA

aB

aC


Accelerations in relative motion

B1

B2

B

aB2=aB1u+ aB2B1

w+ a

c


of point B2

Transportation acceleration

(absolute acceleration of

point B1)

Relative

acceleration

Coriolis

acceleration

ac=2 ωu×v B2B1


Lecture 4

Analytical method. Cam mechanisms.



Procedure of analytical determination of velocities and accelerations in planar mechanisms.

1. Set up Cartesian coordinate system OXY.

2. Subsitiute 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



6a. Write down projections of each polygon onto coordinate system's axes:

x: ∑i=1

i=n

licosφ 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.



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.


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.


Cam-follower

Classification

flat / spatial

with in-line (central) follower / with offset (eccentric) follower

closed with geometry / closed with force


Analysis and synthesis of cam-follower mechanism

Analysis – calculation of displacement, velocity and accelerationfunctions for a follower motion with respect to a cam's rotations anglefor arbitrary given geometry.

Synthesis – calculation of a cam geometry needed to obtain givendisplacement/velocity/acceleration functions. Limitations must beincluded, i.e. some maximum values, geometry limitations and jerkvalues (third derivative).


Lecture 5

Cam-follower mechanisms cont.Dynamics of planar mechanisms.



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


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.


Members description


Material points method - example

Given:Geometry, mass m,center of a mass location (pt. C) and mass moment of inertia IC

m1 m2 m3

m

C

m1+m2+m3=m

− am1+bm3

m1+m2+m3

=0a b

m1a2+m3b

2=IC

x

y

x


Inertia forces and torques


C aC

ε

BC= − maC

Inertia torque

MC= − IC ε

MC

BC

inertia force


Inverse dynamics problem – calculation of forces and torques thatcause given motion of a mechanism.

Direct dynamics problem – calculation of mechanism's motion causedby external forces and torques.



Inverse dynamics problem


Calculation of forces and torques that cause given motion of a mechanism(kinetostatics)

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 themechanism.

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.


Lecture 6

Machine dynamics.Reduction of masses and forces.

Machine equation of motion.



Kinetic energy

Reduction of masses

mr(t)Fr(t )

xr( t)

I r(t)

Mr(t)

φ r (t)

Total kinetic energy

T =1

2I r ωr

2

reduced moment of inertia

T =1

2mr vr

2

reduced mass

or

vr=dxr (t )

dt

ωr=d φ r (t)

dt

T=∑i=1

n

(12 mi vi2+

1

2I iωi

2)


System power

Reduction of forces

Total system's power

P=Mrωr

reduced torque

P (Fi , Mi ,ωi , vi , ...)

P=Fr vr

reduced force

or

mr(t)Fr(t )

xr( t)

I r(t)

Mr(t)

φ r (t)

vr=dxr (t )

dt

ωr=d φ r (t)

dt


Kinetic energy

Reduction of masses

T= ∑i=1

n1

2mi vi

2+ ∑

j=1

k1

2I jω j

2

n – translating elementsk – rotating elements

1

2mr vr

2= ∑

i=1

n1

2mi vi

2+ ∑

j=1

k1

2I jω j

2

mr= ∑i=1

n

mi

vi2

vr2+ ∑

j=1

k

I jω j

2

vr2

1

2I rωr

2= ∑

i=1

n1

2mi vi

2+ ∑

j=1

k1

2I jω j

2

I r= ∑i=1

n

mi

vi2

ωr2+ ∑

j=1

k

I jω j

2

ωr2

– arbitrary chosen velocitiesvr , ωr


Work

Reduction of forces

dW= ∑i=1

n

Pi dsi cosαi+ ∑j=1

k

M j d φ j

n – translating elementsk – rotating elements

Pr dsr=∑i=1

n

Pi dsicosα i+∑j=1

k

M j d φ j Mr dφ r=∑i=1

n

Pi dsicosαi+∑j=1

k

M j d φ j

Pr=∑i=1

n

Pi

dsidsr

cosα i+∑j=1

k

M j

d φ j

dsr

Pr=∑i=1

n

Pi

vi dt

vr dtcosαi+∑

j=1

k

M j

ω j dt

vr dt

Pr=∑i=1

n

Pi

vivr

cosαi+∑j=1

k

M j

ω j

vr

Mr=∑i=1

n

Pi

dsid φ r

cosαi+∑j=1

k

M j

d φ j

d φ r

Mr= ∑i=1

n

Pi

vi dt

ωr dtcosα i+ ∑

j=1

k

M j

ω j dt

ωr dt

Mr= ∑i=1

n

Pi

viωr

cosαi+ ∑j=1

k

M j

ω j

ωr


Linear motion

Machine equation of motion

dT=dW

d(1

2m(t) v (t)2)=F ( t)dx

1

2dm (t)v (t)

2+m( t)v ( t)dv (t)=F (t)dx

1

2dm ( 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+m

dv (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)


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)


Lecture 7

Non-uniformity of machine motion.Introduction to automatic control.



engine machine

φ ( t) I R

φ ( t)

t

ωmax

ωmin

T max=1

2IRωmax

2 Tmin=1

2I Rωmin

2

W=Tmax − Tmin=δ I Rωmean2

δ=ωmax− ωmin

ωmeanωmean=

ωmax+ωmin

2

Non-uniformity of machine motion

Non-uniformity of machine motionSteady-state motion

pumps combustion engines generators

δ=1/5÷1/30 δ=1/50÷1/150 δ=1/200÷1/300


x (t)

t

vmax

v min

δ=vmax − vminvmean

vmean=vmax+vmin

2

Non-uniformity of machine motion

machine

mR

FR( t) x (t)


W=Tmax − Tmin=δmRvmean2


ω( t)

φ ( t)

engine machine

φ ( t) I R

MD MP

φ ( t)

MD MP

π 2π

W

W=∫φmin

φmax

(MD − MP)d φ

δ=W

IRωmean2


Example

ωmax

ωminW=Tmax − Tmin=δ IRωmean

2


engine machine

φ ( t) I RI FW

I FW=(δ1

δ2

− 1)I R

W=δ1 I Rωmean

2assume

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 (if velocity of a flywheel same as analyzed velocity)


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.


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


Single Input Single Output (SISO) system

SYSTEMx (t) y (t )


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


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))


Open loop control

SYSTEMu(t)=x (t ) y (t )

CONTROLLERyd(t )

desiredoutput

controlfunction

system output

system input


Closed loop control

SYSTEMu(t)=x (t ) y (t )

CONTROLLER

yd(t )

desiredoutput control

function

system output

system input

+

-

e(t)


Lecture 8

Laplace transform.Transfer function.

Inputs and outputs in time domain.



Laplace transform

Assumption: x (t ) - signal such that for t<0 x (t )=0

X (s)=L{x (t )}= ∫0

∞

x (t)e − stdt

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)}=

12π j

limω → ∞

∫γ − jω

γ+ jω

X (s)est ds


Transfer function

dny (t )

dtn +a1

dn − 1

y (t )

dtn − 1 +...+an − 1

dy(t )

dt+an y (t )=

dmx (t )

dtm +b1

dm − 1

x (t )

dtm − 1 +...+bm − 1

dx(t)

dt+bm x (t)

Linear time-invariant SISO system for continous-time input signal x(t) and output y(t) in a form

after Laplace transformation with zero initial conditions

snY (s)+a1 s

n − 1Y (s)+...+an − 1sY (s)+anY (s)=s

mX (s)+b1 s

m − 1X (s)+...+bm − 1s X (s)+bm X (s)

(sn+a1s

n − 1+...+an − 1 s+an)Y (s)=(s

m+b1 s

m − 1+...+bm − 1 s+bm)X (s)

H (s)=Y (s)

X (s)=sm+b1 s

m − 1+...+bm − 1 s+bm

sn+a1 sn − 1+...+an − 1 s+an

Transferfunction


Transfer function

H (s)=Y (s)

X (s)=sm+b1 s

m − 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


Input and output

H (s)=Y (s)

X (s)Transfer function:

Y (s)=H (s)X (s)Laplace transform of output:

Output in time domain: y(t )=L − 1{Y (s)}

y(t )=L − 1{H (s)X (s)}=L − 1 {H (s)}∗ L − 1{X (s)}=h(t)∗ x(t)

Convolution h(t) ∗ g(t )= ∫0

∞

h(τ) x (t − τ)d τ

h(t) - system impulse response ( y(t) when x (t )=δ(t))


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

L L L-1


Exemplary input signals

No input: x (t )=0

Unit impulse (Dirac delta pseudofunction): δ(t)={0 , t<0 ∞ , t=00 , t>0

Unit step function (Heviside step function): 1(t)={0 , t<01 , t>0

H (t ) or 1+ (t)

Ramp function: x (t )={0 , t<0t , t>0

Harmonic function: x (t )=a sin (ω t)


Input and output

h(t)impulse responsey (t ) for x (t )=δ(t )

a(t)step responsey (t ) for x (t )=1(t)

d a(t )

dt=h(t)

t

ax (t)

a(t)

t

x (t)

h(t )


Lecture 9

Frequency response.Classification of basic automatic systems.



Transfer function

Linear time-invariant SISO system for continous-time input signal x(t) and output y(t) in a form

H (s)=Y (s)

X (s)Transferfunction

Y (s) - Laplace'a transform of an output

X (s) - Laplace'a transform of an input

30.11.2017 TMiPA, Wykład 9, Sebastian Korczak, tylko do użytku edukacyjnego studentów PW 80

Transfer function – frequency response

H (s)

Transfer function(Laplace domain)

H ( jω)

s= jω

Full system description(for every possible input)

Description of a system in steady state with harmonic input

Frequency response(Fourier domain)


H (s) H ( jω)=P (ω)+ jQ(ω)

A (ω)=|H ( jω)|= √P2(ω)+Q

2(ω)

φ (ω)=Arg H ( jω)=arctanQ

P

P(ω)

Q(ω)

ω=0ω= ∞

y (t)=A sin (ω t+φ)


input: x (t)=sin (ω t) output:H (s)transfer function:

Nyquist plot

A (ω)

φ (ω)

s= jω

DELAY

GAIN


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


ω [rad/s]

L(ω

) [d

B]

ω [rad/s]

L(ω)=20 log A (ω)



A (gain) 20logA [dB]

1000 60

100 40

10 20

1 0

0.1 -20

0.01 -40

0.001 -60


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 )

k

Ts+1

ks



Element name EquationTransfer function

derivative

derivative with inertia

y (t )=kdu (t )

dt

Tdy (t )

dt+y (t )=k

du (t )

dt

ks

ks

Ts+1



Element name EquationTransfer function

delay

second order (oscillator)

y (t )=u (t− τ )

T 1

2 d 2 y (t )

dt2

+T 2

dy (t )

dt+

+y (t )=ku (t )

e − τ s

k

T 12s

2+T 2 s+1


Lecture 10

Classification of basic automatic systemswith examples.



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 ∧dudt

=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

for k>0


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+Q

2=|k| φ (ω)=arctan

QP={0 , dla k≥ 0

π , dla k<0}

20 log|k|

for k>0


Proportional elementExamples

1

GEARBOX:input – angular velocity ω1(t)output – angular velocity ω2(t)

GEARBOX:input – rotation angle φ1(t)output – rotation angle φ2(t)

ω1(t)

ω2(t)

2

φ1(t)

φ2(t)



4BEAM in steady state:input – force F1output – force F2

F1 F2

3 OPERATIONAL AMPLIFIER:input – voltage v1(t)output – voltage v2(t)

Vsupply

0Vv2(t)

v1(t)

R2R1

v2 (t )=v1 (t )(1+R2

R1)



5

HYDRAULIC LEVER:input – displacement x1(t)output – displacement x2(t)

x1(t)

x2(t)

6 PRESSURE ACTUATOR:input – pressure p1(t)output – displacement x(t)

x(t)

p(t)


First-order inertial element

1. Element equation: u(t) - inputy (t ) - output


=0 ∧dudt

=0


Ts+1

u

y

Tdy (t )

dt+y (t )=ku (t )

for k>0



4. Step response:

input: u (t)=u01(t)

u0

u(t)

Laplace of input: U ( s)=u0

1

s

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



5. Frequency response: H ( jω)=k

Tjω+1

6. Nyquist plot:

P (ω)=k

T 2ω2+1, Q (ω)=

− k T ω

T 2ω2+1

P(ω)

Q(ω)

ω=0ω= ∞

k /2 k0

− k /2ω=1/T

for k>0



7. Bode plot:

L(ω)=20 log A (ω)=20 log|k| − 20 log √T 2ω2+1

A(ω)= √P2+Q

2=|k|/ √T

2ω

2+1

φ (ω)=arctanQ

P=arctan ( − T ω)

L(ω

) [d

B] ω [rad/s]1

10T1

T

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


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 is driving on a flat surface with air resistance proportionalTo its velocity, described using machine equation of motion, with assumptionof 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)


First-order inertial elementExamples

3p1(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 Ti(t)


Integrator

1. Element equation:u(t) - inputy (t ) - output

2. Static characteristic (steady state): for dydt

=0 ∧dudt

=0

3. Transfer function: H (s)=ks

dy (t)

dt=k u(t )

u=0

u

y


Integrator

4. Step response:

input: u (t)=u01(t)

u0

u(t)

u0

y (t )

t


1

s


s2

output: y (t)=L − 1{Y (s)}=k u0 t

1/kt

for k>0


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


Integrator

7. Bode plot:

L(ω)=20 log A (ω)=20 log|kω|

A(ω)= √P 2+Q2=|kω|

φ (ω)=arctanQ

P=arctan ( − ∞)

φ(ω

) [r

ad]

− π2

ω [rad/s]

L(ω

) [d

B] ω [rad/s]

k /10 k

− 20 dB/dek

10k0

20

40

100k

for k>0


IntegratorExamples

1PRISM LIQUID TANK:input – liquid inflow f(t)output – liquid level h(t)

h(t)

f(t)

2 OPERATIONAL AMPLIFIER:input – voltage v1(t)output – voltage v2(t)Vsupply

0Vv2(t)

v1(t)

CR

v2(t )=1

RC ∫0

t

v1(t)dt


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)


Differentiator


2. Static characteristic (steady state): y=0 for dydt

=0 ∧dudt

=0

3. Transfer function: H (s)=k s

u

y

y (t)=kdu(t)

dt


Differentiator

4. Step response:

input: u (t)=u01(t)

u0

u(t) y (t )

t


1

s


output: y (t)=L − 1{Y (s)}=k u0δ(t )

t


Differentiator

5. Frequency response: H ( jω)= j k ω

6. Nyquist plot:

P (ω)=0 , Q (ω)=k ω

P(ω)

Q(ω)

ω=0

0

for k>0


Differentiator

7. Bode plot:

L(ω)=20 log A (ω)=20 log|kω|

A(ω)= √P2+Q

2=|k ω|

φ (ω)=arctanQ

P=arctan ( ∞)

φ(ω

) [r

ad]

π2

ω [rad/s]

for k>0

L(ω

) [d

B]

ω [rad/s]k /10

k

+20 dB/dek

10k0

20

40

− 20

− 40


DifferentiatorExamples

1GEARBOX:input – rotation angle φ(t)output – angular velocity ω(t)

ω(t)

φ(t)

2 OPERATIONAL AMPLIFIER:input – voltage v1(t)output – voltage v2(t)

v2(t )= − RCdv1(t )

dt

Vsupply

0Vv2(t)

v1(t)

C R


Real differentiator (derivative+1st order)

1. Element equation: u(t) - inputy (t ) - output

2. Static characteristic (steady state): for dydt

=0 ∧dudt

=0

3. Transfer function: H (s)=k s

Ts+1

Tdy (t)

dt+ y (t )=k

du(t )

dt

y=0

u

y



4. Step response:

input: u (t)=u01(t)


1

s


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

for k>0



5. Frequency response: H ( jω)=k jω

Tjω+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= ∞

k2T

k

T

0

− k /2ω=1/T

for k>0



7. Bode plot:

L(ω)=20 log A (ω)=20 log|kω| − 20 log √T 2ω2+1

A(ω)= √P2+Q

2=|k ω|/ √T

2ω

2+1

φ (ω)=arctanQ

P=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

1T

20 log|k /T| − 3

10 /T

20 log|k /T| − 20

20 log|k /T|

20 log|k /T| − 40

0


Real differentiator (derivative+1st order)Examples

1 RC CIRCUIT:input – voltage u1(t)output – voltage u2(t)


Delay


2. Static characteristic (steady state): y=u for dydt

=0 ∧dudt

=0

3. Transfer function: H (s)=e− τ s

u

y

y (t)=u(t− τ)


Delay

4. Step response:

input: u (t)=u01(t)


1

s

Laplace of output: Y (s)=H ( s)U (s)=u0

se − τ s

output: y (t)=L − 1{Y (s)}=u0 1(t − τ)

u0

u(t)

t

u0

y (t )

t

τ


Delay

5. Frequency response: H ( jω)=e − τ jω

6. Nyquist plot:

P (ω)=cos(τω), Q (ω)= − sin (τω)

P(ω)

Q(ω)

ω=00

ω= π2 τ

ω=πτ

ω=3π2 τ

1

1

− 1

− 1

for k>0

e − x=cos x − j sin x


Delay

7. Bode plot:

L(ω)=20 log A (ω)=20 log 1=0

A(ω)= √P2+Q

2=1

φ (ω)=arctanQ

P=arctan ( − tan ( τω))= − τω

φ(ω

) [r

ad]

− π

πτ

ω [rad/s]

10πτ

L(ω

) [d

B]

ω [rad/s]

110T

1T

10T

0


DelayExamples

1 WIRELESS TRANSMISSION:input – sent dataoutput – received data

data pocket data pocket


Second-order inertial element

1. Element equation:


=0 ∧dudt

=0


T 12s

2+T 2 s+1

u

y

T 12 d

2y (t)

dt 2+T 2

dy (t )

dt+ y (t )=k u(t)



4. Step response:

input: u (t)=u01(t)


1

s


s(T 12 s2+T 2 s+1)

output: y (t )=L− 1 {Y (s )}=

={k u0

T 1

2(1− e

− ht (cosω t+hω sinω t )), for h≤ ω0

k u0

T 12 (1+e

− ht((h+w

2w− 1)e− wt

−h+w

2 we

wt)), for h≥ ω0

where: h=T 2

2T 12, ω0=

1T 1

, ω= √ω02 − h2 , w= √h2 − ω0

2



4. Step response:

u0

u(t)

t

k u0

y (t )

t

h<ω0

h=ω0

h>ω0



5. Frequency response: H ( jω)=k

− T 12ω

2+T 2 jω+1

6. Nyquist plot:

P (ω)=k (1 − T 1

2ω

2)

(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



7. Bode plot:

L(ω)=20 log A (ω)

A(ω)= √P2+Q

2

φ (ω)=arctanQ

P

L(ω

) [d

B]

ω [rad/s]

110T 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

110T

1100T


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)



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

3

ANGULAR MOTION OF A RIGID BODY WITH LINEAR DAMPING:input – torque M(t)output – angle φ(t)

M(t)

φ(t)



4 HEATED OBJECT WITH HIGH INERTIA:input – heater power h(t)output – object temperature Ti(t)



Element name Transfer function

proportional k

first order (inertial)

integrator

differentiator

differentiator with inertia

delay

second order (oscillator)

k

Ts+1

ks

ksks

Ts+1

e − τ s

k

T 12s

2+T 2 s+1


Lecture 11

Block diagram algebra.Control and controllers.



BLOCK DIAGRAM ALGEBRATransfer function

X(s) Y(s)G(s)


BLOCK DIAGRAM ALGEBRAinformation node

X(s)

one input,a few outputs,

X(s) X(s)

X(s)


BLOCK DIAGRAM ALGEBRAsum node

A(s) +

B(s)

+

A(s)+B(s)-C(s)–

C(s)


BLOCK DIAGRAM ALGEBRAserial connection

G1(s)x(s) y(s) x(s) y(s)

GR(s)G2(s)

GR(s)=G1(s) G2(s)


BLOCK DIAGRAM ALGEBRAparallel connection

G1(s)x(s) y(s) x(s) y(s)

GR(s)

G2(s)GR(s)= - G1(s) + G2(s) + G3(s)

-

+

G3(s)

+


BLOCK DIAGRAM ALGEBRAfeedback

G2(s)

+

–

x(s) y(s) x(s) y(s)GR(s)G1(s)

GR=G1

1+G1 G2


BLOCK DIAGRAM ALGEBRAchange of information points order

X(s) X(s)


BLOCK DIAGRAM ALGEBRAorder change of sum node and block

G(s)+

+

X(s) Y(s)

A(s)

G(s)

G(s)+

+

X(s) Y(s)

A(s)

Y=(X+A)G Y=XG+AG


-

+

A(s)

C(s)B(s)

+

- +

+

A(s)

B(s)C(s)

-

+

BLOCK DIAGRAM ALGEBRAorder change of sum nodes

attention to signs!


BLOCK DIAGRAM ALGEBRAorder change of block and information node

G(s) G(s)

G(s)


Closed loop control

SYSTEM/PROCESS/

PLANT

u(t)=x (t) y (t)CONTROLLER

yd(t )

desiredoutput(setpoint) control

function

system output(measured)

system input

+

-

e (t )

error


Closed loop control

u(t)CONTROLLER

controlfunction

e(t)

error


Types of controllers

ON/OFF

three state

Proportional (P)

Integrator (I)

Differentiator (D)

Proportional-inegral-derivative (PID)


RELAY / ON-OFF / TWO STATE / BANG-BANG CONTROLLER

u(t)

Control signal

e(t)

erroru(t)

e(t)

u(t)={umax , if e>e0

umin , if e< − e0

no change, in other situations}

umax

umin

e0 − e0

eo - mechanical or programmed hysteresis


NONLINEARITIES

input

output

Symmetric hard limiting saturation


input

output

NONLINEARITIESDead zone


EXAMPLE 1

mdv (t )

dt= f (t ) − d (t )

Car on a flat surfacem – mass,f(t) – driving force,d(t)=c*v(t) – air resistance,v(t) – velocity H (s)=

V (s)

F (s)=

1

ms+c

Speed control (cruise control, autocruise, tempomat)

CARf (t ) v (t )

PROPORTIONALCONTROLLER

vd(t )

desiredvelocity

+

-

e(t)

realvelocity

driving force


EXAMPLE 2Water level control

x1(t)[m3/ s ] - inflow of a liquid (controlled)

x2(t )[m3 /s ] - outflow of a liquid (not controlled)

w (t )[m] - level of a liquid in a tank

A [m2] - constant surface area

w(t)

x2(t)

x1(t)


Lecture 12

PID controller.Stability.



PID transfer functions

kP

1

Ti s

Td s

T d s

T s+1

Controller Transfer function

Proportional (P)

Integral (I)

Ideal derivative (D)

Real derivative (D)



kP(1+ 1

Ti s+Td s)

kP+ki1

s+kd s


Proportional-integral-derivative (PID)

in standard formwith ideal derivative


in parallel formwith ideal derivative



kP(1+ 1

Ti s+Td s

Ts+1)

kP+ki1

s+kd

s

Ts+1



in standard formwith real derivative


in parallel formwith real derivative


PID CONTROLLERstandard form with ideal derivative

1

+

+

1Ti s

kP

T d s

+


PID CONTROLLERparallel form with ideal derivative

+

+

ki1s

kP

kd s

+


PID CONTROLLERstep responses

u(t)CONTROLLER

controlsignal

e(t)

error

?


P - CONTROLLER

time

input

output

G(s)=K p

K p x0

x0


I - CONTROLLER

time

input

output

Ti

G(s)=1

Ti s

x0


PI - CONTROLLER

time

input

output

Ti

G(s)=Kp(1+1

Ti s)

K p x0

x0

2K p x0


D - CONTROLLER

time

input

output

G(s)=T d s

+ ∞

x0


D - CONTROLLER

time

input

output

G(s)=T d s

x0

discrete time

x0

Δ t

Δ t


PD - CONTROLLER

time

input

output

G(s)=KP (1+T d s)+ ∞

x0

K p x0


czas

wejście

wyjście

G(s)=Kp(1+1Ti s

+T d s)

x0

K p x0

2K p x0

Ti

+ ∞

PID – CONTROLLERstandard form, ideal derivative


D - CONTROLLER

time

input

output

G(s)=Td s

T s+1

x0

T

0,368umax

T T

0,135umax

0,05umax

umax=x0

Td

T


PD - CONTROLLER

time

input

output

G(s)=Kp(1+T d s

T s+1)

x0

T

K p x0

KP x0(1+Td

T )


time

input

output

G(s)=K p(1+ 1Ti s

+T d s

T s+1 )

x0

K p x0

KP x0(1+ T d

T )2K p x0

T i

PID – CONTROLLERstandard form, real derivative


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 KP 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;


PID CONTROLLERInfluence of errors onto control signal

Control signal

P

Control error

I

D

0,00 0,10 0,20 0,30 0,40 0,50 0,60 0,70 0,80 0,90 1,00

-1-0,8-0,6-0,4-0,20

0,20,40,60,81

0,00 0,10 0,20 0,30 0,40 0,50 0,60 0,70 0,80 0,90 1,00

-1,00-0,80-0,60-0,40-0,200,000,200,400,600,801,00

0,00 0,10 0,20 0,30 0,40 0,50 0,60 0,70 0,80 0,90 1,00

-1,00

-0,60

-0,20

0,20

0,60

1,00

0,00 0,10 0,20 0,30 0,40 0,50 0,60 0,70 0,80 0,90 1,00

-1,00-0,80-0,60-0,40-0,200,000,200,400,600,801,00


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.


Quality of the control process

Overshoot: w=e1

e0

100%

Risetime

steady-stateerror

Settlingtime

czas

setpoint

System output

e1

e2

eST

TR ±5%e0 region

TS

e0

Damping: d=e2

e1

100 %


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

response3rd 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

methods: Ziegler-Nichols Pessen Cohen-Coon Åström–Hägglund


PID CONTROLLERinteractive simulation and tuning

Download spreadsheet file from the website


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 Tucalculate parameters of the controller according to the table:

kp Ti Td

ClassicZiegler-Nichols

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


PID CONTROLLERprogramming

dt = 0.1p_error = 0.sum = 0.Kp = 2.Ki = 0.5Kd = 0.01start:

setpoint = …measurement = …error = setpoint – measurementsum = sum + error * dtderivative = (error – p_error) / dtoutput = Kp*error + Ki*sum + Kd*derivativep_error = errorwait(dt)goto start

SYSTEMPIDsetpoint

controlfunction

system output+

– measurement

error


PID CONTROLLERinteractive simulation

PID for a car position control – real-time simulation


BIBO stability

Bounded Input, Bounded Output stability(in signal processing and control theory)

a LTI SISO system is called BIBO stable if its outputwill stay bounded for any bounded input.

∀t ⩾ 0

if |x (t)|⩽ A , then |y (t )|⩽ B ∃0<B< ∞

∃0<A< ∞

x (t ) - input

y (t) - output


General stability criterion

Hurwitz criterion

Nyquist stability criterion

STABILITY CRITERIA


ℜ ( p1)=a1 , ℑ ( p1)=b1 ℜ y (t )=e

a1 tcos b1 tG (s)=

Y (s)

X (s)=

1s − p1

ℜ ( p1)<0asymptotically stable


General stability criterion

G (s) is stable if: Re p1<0 ∧ Re p2<0 ∧ ... ∧ Re pn<0

LTI SISO system is asymptotically stable if real part of every pole of the system's transfer function is less than zero.

G (s)=(s − z1)(s − z2)...(s − zm)

(s − p1)(s − p2) ...(s − pn)


Lecture 13

Stability criteria.Gain margin and phase margin.

System correction.



Hurwitz criterion

mathematics

a necessary and sufficient condition whether all

the roots of the polynomial are in the left half of the complex plane

control theory

a necessary and sufficient condition whether all the poles of transfer function of a linear

system have negative real parts


Hurwitz criterion

LTI SISO system with a transfer function

H (s)=bm s

m+bm − 1 s

m − 1+...+b1 s+b0

an sn+an − 1 s

n − 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>0

detΔ3>0

...detΔn − 1>0

is stable if:

- leading principal minor of order i

Δi


Nyquist stability criterion

Gz (s)=y( s)

x (s)=

G1( s)

1+G1(s)G2( s)

G1(s)

G2(s)

+

–

x(s) y(s)

G1 G2 = − 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

Re Gopen

Im G open

ω → − ∞

ω → + ∞


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

Re Gopen

Im G open

-1 -1

stable unstable


Gain margin

P(ω)

Q(ω)

ΔM

-1

L(ω

) [d

B]

ω [rad/s]

φ(ω

) [r

ad] ω [rad/s]

− π

ΔM0

Closed-loop system will loose its stability if we add additional gain (in serial) greater or equals to gain margin.


Phase margin

P(ω)

Q(ω)

-1

Δφ

L(ω

) [d

B]

ω [rad/s]

φ(ω

) [r

ad] ω [rad/s]

− πΔφ

0

Closed-loop system will loose its stability if we add additional

delay (in serial) greater or equals to phase margin.


Summing of Bode plots – example

G (s) = 10

s2+s = 10 ⋅

1

s+1 ⋅

1

s

L(ω

) [d

B] φ(ω

) [r

ad]

ω [rad/s]

− π2

0

201 100,1

40

60

− π

− 20

− 40

− 60

ω [rad/s]

1 100,1


Lecture 14

Material repeat.Informations about the exam.

WUT questionnaires.



Lecture 15 – modern control theory overview,

experiment with control system,

consultations


Theory of Machines and Automatic ControlWinter 2017/2018

Field of studies: Electric and Hybrid Vehicle Engineering (full-time)

form of studies: 30 hrs lecture, 15 hrs project class

ECTS: 4

Lecture: Tuesdays at 8:15 (room 3.3)

Projects: Wednesdays at 9:15 (room 3.8)1st meeting on 8th November

Lecturer: Sebastian Korczak, PhD, Eng.

room: 2.8b

e-mails: [email protected], [email protected]

consultations: Tuesdays at 10:00-11:00 and Thursdays at 12:00-13:00

website with materials and marks: http://myinventions.pl/lectures/


Assessment method

Exam: written examination on skills and knowledge after completing and successful attestation of project classes.

2 terms in the winter examination session (1.02, 8.02)

1 term in the autumn examination session (3.09 – 19.09)

Negativ mark: 2,0Positiv marks: 3,0; 3,5; 4,0; 4,5; 5,0


EXAM – IMPORTANT NOTES

You have to pass the project class to attend the exam.

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 or displayed.



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

If marks from project class and exam are positive, then

Final_mark = 0.5 * project_mark + 0.5 * exam_mark



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.


EXAM – TERM 1

1st February 2018 (Thursday)

12:00 – 13:15 exam for lecture in polish

13:25 – lecture hall (room 3.4) openning

13:25-13:30 – preparation

13:30-14:30 – exam

5th February 2018 (Monday)

to 12:00 – publication of exam effects on the website

http://myinventions.pl/lectures/ and USOSweb

12:00-14:00 – filling of indexes and erasmus papers


EXAM – TERM 2

8th February 2018 (Thursday)

12:00 – 13:15 exam for lecture in polish

13:25 – lecture hall (room 3.4) openning

13:25-13:30 – preparation

13:30-14:30 – exam

11th February 2017 (Sunday) – last session day

to 23:59 – publication of exam effects on the website

http://myinventions.pl/lectures/ and USOSweb

13th February 2017 (Monday)

12:00 – 14:00 – filling of indexes and erasmus papers

warsawuniversity of technologymyinventions.pl/dydaktykasimr/tmac_17-18-lecture_14_notes.pdf ·...

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