TLTE.3120 Computer Simulation in Communication
and Systems (5 ECTS)
Lecture 8
28.10.2015
Timo Mantere
http://www.uwasa.fi/~timan/TLTE3120/
UNIVERSITY of VAASA
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Matlab features
 We cannot go through all the possibilities of Matlab and
Simulink in this course, so most important is to know where
to get help when you start doing real things with Matlab
 Basically most of the things you need to know is available
in Mathworks homepage
 http://se.mathworks.com/
 Use find there with proper keywords and you probably find
what you are looking for, lot of examples and example
codes
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Matlab features
 Matlab tutorial videos available
 Data analysis and visualization
 http://se.mathworks.com/products/matlab/description3.html
 Programming and algorithm development
 http://se.mathworks.com/products/matlab/description4.html
 Application development
 http://se.mathworks.com/products/matlab/description5.html
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Matlab features
 Simulink tutorial videos available
 Simulink key features






Building the model
Simulating the model
Analyzing simulation results
Managing projects
Connecting to hardware
http://se.mathworks.com/products/simulink/features.html#key_features
 Videos
 http://se.mathworks.com/products/simulink/videos.html
 Examples
 http://se.mathworks.com/examples/product-group/matlab-signalprocessing-and-communications
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Matlab features
 Features important to our students
 Digital signal processing toolbox
 http://se.mathworks.com/products/dsp-system/
 Control system toolbox
 http://se.mathworks.com/products/control/
 Communications system toolbox
 http://se.mathworks.com/products/communications/
 Also:
 Design and simulate radio systems
 http://se.mathworks.com/products/simrf/
 Radio frequency toolbox
 http://se.mathworks.com/products/rftoolbox/
 LTE toolbox
 http://se.mathworks.com/products/lte-system/
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Matlab features
 Features important to our students
 Digital filter design
 http://se.mathworks.com/discovery/filter-design.html
 Filterbuilder





http://se.mathworks.com/help/dsp/gs/design-a-filter-using-filterbuilder.html
Type filterbuilder
Select filter type
Select features
View filter response
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Matlab features
 Filter design with Matlab, example, copy paste parts
s1=[20, 30, -10, 40, 40, 30, 20]
%first signal
s2=[20, 30, 40, 0, -1, 0, 40, 30, 20] %second signal
b1=fir1(3,0.5)
%create FIR filter with fir1
[b2, a]=butter(3,0.5) %create IIR filter
freqz(b2,a,100,1000) %frequency response of IIR filter
%hold
%freqz(b1,1,100,1000) %frequency response of FIR filter
%filter 1. signal with FIR and IIR filters
F1=filter(b1,1,s1) %FIR filter a=1
I1=filter(b2,a,s1)
%filter 2. signal with FIR and IIR filters
F2=filter(b1,1,s2)
I2=filter(b2,a,s2)
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Matlab features
 Filter design with Matlab, example 2
N=50;
%fir degree
%create rectangular window
window=ones(N+1,1);
%frequncies and responses for drawing
f=[0 1];
m=[0 1];
%fir filter creation with fir1 or fir2 function
b=fir2(N, f, m, window)
%b=fir2(N, f, m, window);
%calclate fir attenuation
[h,w]=freqz(b,1,1000);
%draw frequncy response, wanted and obtained
plot(f,m,w/pi,abs(h))
legend('Ideal','fir2 Designed');
title('Comparison of Frequency Response Magnitudes');
%freqz(b,1,512)
%attenuation
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Matlab features
 Filter design with Matlab, example 3
% Define bandbass
wn=[0.2 0.6];
N=50;
%FIR degree
window=ones(N+1,1);
b=fir1(N, wn, window) %Fir
%attenuation
[h,w]=freqz(b,1,512)
%frequncies for calulating responses
f=[0 0.2 0.2 0.6 0.6 1];
m=[0 0 1 1 0 0];
%draw wanted and obtained response
plot(f,m,w/pi,abs(h))
legend('Ideal','fir1 Designed');
title('Comparison of Frequency Response Magnitudes');
%freqz(b,1,512)
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Matlab features
 Also available other filter models e.g.
 Butterfield
 % Generate low pass filter eg. len=10, Fn=0.5
 [b,a]=butter(len,Fn);
 Chebyshev
 Cheby1 [b,a] = cheby1(N,R,Wp)
 Chepy2 [b,a] = cheby2(N,R,Wst)
 Elliptic
 [b,a] = ellip(N,Rp,Rs,Wp)
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Matlab features




Matlab C code generation
C conversion can be done with codegen command
Help codegen
Examples:
 http://se.mathworks.com/examples/matlab-coder/1866-c-code-generationfor-a-matlab-kalman-filtering-algorithm#1
 http://se.mathworks.com/examples/search?utf8=%E2%9C%93&q=C+Code
+Generation+for+a+MATLAB
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Simulation of communication systems
 Communication systems have become more and more
complex over the years
 Systems requires more analyses during the design phase
 Analyses must be done fast and cost-effectively
 These things have lead to the computer based simulations
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Simulation of communication systems
 Communication systems can be analyzed and
evaluated e.g. by




Based on formulas and their calculations (usually simplified models)
Waveform-level simulations
Hardware prototyping and measurements (quite costly)
Combinations of previous
 Usually simulation based evaluation is most flexible, one can
combine mathematics with models or measured data
 Communication system could be whatever, satellite-based, terrestrial
wireless, optic cables, Ethernet, etc.
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Simulation of communication systems
 Different simulation techniques are used to
evaluate performance, e.g.
 At network level flow of packets and messages over the
network is simulated, calculating throughput, packet loss,
response time, resource utilization etc
 Communication links are measured in terms of error
characteristics, bit error rate etc.
Link could contain different parts that are simulated
separately; modulators, encoders, filters, amplifiers,
decoders, demodulators, etc.
 Also the hardware parts can be simulated by using hardware
description language like VHDL
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Simulation of communication systems
 Quite often the simulation results are presented as
 Scope Y-T plot
 X-Y plot
 Frequency curve
 Frequency response curve (filter, channel)
 Eye diagram
 Error rate plot (SNR)
 I-Q plot (real-imaginary axis)
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A Refreshment of System Analysis
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Introduction: What is a Signal?
 Signals describe quantities that change. This change can
be with time, frequency, location, ..etc.
 Figure (1) in the next slide shows signal x(t) changes with
time, this signal represent voice signal for word “Finland”.
 Figure (2) shows signal y(f) which represents certain signal
in frequency domain
 Figure (3) shows signal z(n) which represents the grades of
one of one of telecommunication courses in the University
of Vaasa
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Introduction: What is a Signal?
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y(f)
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Figure (1)
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time [2]
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Figure (2)
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frequency [Hz]
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x 10
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Number of students
x(t)
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Figure (3)
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Group1
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Grade
4
5
Introduction: What is a Signal?
 A signal is a function or sequence of values that represent
information.
 The signals can be classified according to various criteria
as shown in the table:
Continuous (in time)
discrete (in time)
Amplitude-continuous
Amplitude-discrete
analogue
digital
Real-valued
Complex-valued
Uni-dimensional
Multi-dimensional
Deterministic
stochastic
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Introduction: What is a Signal?
 E.g. signal processing
s = importdata('test.wav')
soundsc(s.data,s.fs)
plot(s.data)
plot(s.data(:,1))
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% load sound sample
% play sample
% plot sound sample
% plot first channel
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x 10
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-0.8
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x 10
Introduction: What is a Signal?
 E.g. spectral density
DF=fft(s.data(:,1)); % Fourier transform of 1. channel
N=length(s.data); % Length
FF= DF.*conj(DF)/N;
% spectral density
f = s.fs/N*(0:(N-1)/2);
% s.fs sample rate
plot(f,FF(1:length(f)));
3.5
3
See signal processing examples
2.5
http://se.mathworks.com/products/signal/index.html
2
1.5
1
0.5
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0
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x 10
Introduction: What is a System?
 A system is the abstraction of a process or object that puts a
number of signals into some relationship.
 Usually, we can classify system’s signals as input and output
signals.
 Input signals exist independently of the system and are not affected
by the system; instead the system reacts to these signals.
 Output signals carry information generated by the system, often as
a response to input signals.
x
System
y
Single Input, Single Output
(SISO)
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x1
x2
xM
System
y1
y2
yN
Multiple Input - Multiple Output (MIMO)
Derivative Notation
 Following notations of derivative are equivalent:
dx(t )
 x '(t )  x(t )
dt
dx 2 (t )
 x ''(t )  x(t )
2
dt
dx 3 (t )
 x '''(t )  x (t )
3
dt
 Usually the last form (x with dots) is favored in system analysis
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Example: Static and Dynamic Models
 A system is called dynamic, if its state is a
function of its previous states.
 For example: The effect of external force F to
object location x can be derived from the
force balance equation (m is a mass, k is a
spring constant and B is a damping
coefficient)
F(t)
x(t)
m
B
k
mx(t )  Bx (t )  kx (t )  F (t )
 A system is called static, if its current state
does not depend on previous state
 For example: the effect of temperature T to
pressure p in closed and isolated tank can be
derived from the ideal gases law (n is the
molar amount, V volume and R gas constant)
p(t )V  nRT (t )
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p(t)
T(t)
n
V
Example: Static and Dynamic Models
 In the simulations plotted on the right
side, the external force in the mechanic
system and the temperature in the gas
tank model are changed step-vise
 In a dynamic system, the response will
change long after the input has already
stabilized. One can not define the
respose just by knowing the input at the
same time moment.
 In a static system the input and the
response are changing only on same
time monents, and the system response
can be defined just by knowing the input
at the same time moment
Mass location
External force
Pressure
Temperature
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Linear and Nonlinear Systems
 A system is linear, if it satisfies the following two conditions:
1) If input u1 generates response y1, then input Ku1 generates response Ky1 (K is
a constant).
2) If input u1 generates response y1 and input u2 generates response y2, then
input (u1+u2) generates response (y1+y2).
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Linear and Nonlinear Systems
 In general, a differential equation is linear, if each term in it has a form
constant(variable or variable’s derivative)
 For example:

y (t )  3 y (t )  5 y (t )  2u1 (t )  u1 (t )  3u2 (t )  2u2 (t )
 The real systems are usually nonlinear, but one can often approximate
them by using linearization
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Continuous Time and Discrete Time Models
 Dynamic, continuous time models are described by using differential
equations and differential equation groups
 For example, the damped oscillator differential equation
mx(t )  Bx (t )  kx (t )  F (t )
 Dynamic, discrete time models are described by using difference
equations and difference equation groups
 For example, a rate formula:
y (t k  1)  107
.  y (t k )  u(t k  1)
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Time Variant and Time Invariant Systems
 In time variant systems the model parameters are changing as a function
of time
 Further analysis of time variant systems is out of the scope of this course
m(t ) 
x (t )  Bx (t )  kx (t )  F (t )
 In time invariant systems the model parameters are assumed to be
constants over time
 Most of the real-world systems are time variant (worning, dirt and dust,
changing environmental conditions etc.), but in many cases the effect of
time invariance is so small that it can be ignored in the model
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State-Space Representation
 There always exist a first order differential equation group which is
equivalent to higher order differential equation
 The solution is not unique; usually there are several first order groups equivalent to
higher order differential equation
 State-space representation is a compact way to present higher order
differential equations
 System state in a certain time moment is a complete system
description. If we know the initial state as well as all input values after
the initial state, we can define system state and output values in every
time moment. That makes state space representation well suitable for
modeling.
 State space representation is also applied in the context of
multivariable systems
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State-Space Representation
 In state-space representation, a higher order differential equation (or a
group of higher order differential equations) is represented as a group
of first order differential equations
 The number of first order equations will be equal to the order of the higher
order differential equation
 There are several ways to select the state variables  state-space
representation is not unique
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State-Space Representation
 The general form of the state-space representation is
x(t)=f(x(t),u(t)
y(t)=g(x(t),u(t))
 x(t) contains the state variables, u(t) control (input) variables and y(t)
output variables
 These variables can be either matrices or scalars
 f(x(t),u(t)) is the system equation which describes the system dynamics
 g(x(t),u(t)) is the output equation which describes how system output
depends on states and on control input
 If u(t) has a scaler value u(t) and y(t) has a scalar value y(t), the system
is SISO. The dimension of state matrix x(t) can still be higher.
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Example: A Fluid Flow System (1/5)
 In a flow process a liquid one which have C 1 F1(t)
a chemical consentration C1 is mixed with
liquid two which have a chemical
concentration C2.
 We are targeting to have a production
(flow F) of such a product, which
concentration is C by controlling the flows
F1 and F2.
 There is a free flow from the mixing tank
to the air pressure  the outflow is
propotional to the square root of the tank
surface level:
F (t )  k h(t )
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C 2 F2 (t)
h(t) C(t)
A
C(t)
F(t)
Example: A Fluid Flow System (2/5)
 Form a differential equations for mass balance (will be simplified to
volume balance) and for concentration balance:
dh(t )
 dV (t )

A
 F1 (t )  F2 (t )  F (t )
 dt
dt

 dC (t )V (t )  C F (t )  C F (t )  C (t ) F (t )
1 1
2 2

dt
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Linear State-Space Representation
 If the system is linear, variables and parameters can be separated. In
that case the standard form of the state-space representation is
X(t)
x (t )  f (x(t ), u(t ))  Ax(t )  Bu(t )
R
S
y(t )  g(x(t ), u(t ))  Cx(t )  Du(t )
Y(t)
T
 A is system matrix, B control matrix, C output matrix and D direct effect
matrix.
 It is often so that there is no direct effect. In that case D = 0 (so-called
strictly proper system).
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Linear State-Space Representation
A differential equation group
 x1  a11 x1 
x  a x 
 2
21 1


 xn  an1 x1 
 a1n xn  b11u1 
 b1mum
 a2 n xn  b21u1 
 b2 mum
 ann xn  bn1u1 
 bnmum
Can be represented as a matrix equation
 x1   a11
 
  
 xn   an1
a1n   x1   b11
 
  
ann   xn  bn1
 x  Ax  Bu
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b1m   u1 
 
 
bnm  um 
Example: a Circuit Board (1/4)

A model for the circuit board
illustrated on the left
 Input is v0(t), outputs are
voltages v1(t) ja v2(t).
 Equations for the capacitors and
currents are
C1
dv1 (t )
dv (t )
 i3 (t ), C2 2  i2 (t )
dt
dt
 Kirchoff I
 Kirchoff II
vR1 (t )  R1i1 (t ), vR 2 (t )  R2i2 (t )
i1 (t )  i2 (t )  i3 (t )
v0 (t )  v1 (t )

i
(
t
)

1
R1
v0 (t )  vR1 (t )  v1 (t )
v0 (t )  R1i1 (t )  v1 (t )

 
 

v
(
t
)

v
(
t
)

v
(
t
)
v
(
t
)

R
i
(
t
)

v
(
t
)
1
R2
2
1
2 2
2
i (t )  v1 (t )  v2 (t )
 2
R2
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Example: a Circuit Board (2/4)
 Since the system input and system responses are all voltages, it is
reasonable to eliminate the currents from the equations.
v0 (t )  v1 (t ) v1 (t )  v2 (t )
1
 dv1 (t ) 1

i
(
t
)

i
(
t
)

i
(
t
)


1

3
2
 dt
C1
C1
R1C1
R2C1


 dv2 (t )  1 i (t )  v1 (t )  v2 (t )
2
 dt
C2
R2C2
 dv1 (t )
 1
 1 
 1 
1 



v
(
t
)

v
(
t
)



 1

 2

 v0 (t )
 dt
 R1C1 R2C1 
 R2C1 
 R1C1 
 
 dv2 (t )   1  v (t )   1  v (t )

 1

 2
 dt
R
C
R
C
 2 2
 2 2

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Example: a Circuit Board (3/4)
 We got first order differential equation group
 dv1 (t )
 1
 1 
 1 
1 
 


 v1 (t )  
 v2 (t )  
 v0 (t )
 dt
 R1C1 R2C1 
 R2C1 
 R1C1 

 dv2 (t )   1  v (t )   1  v (t )

 1

 2
 dt
R
C
R
C
 2 2
 2 2

 Capasitor voltages are natural choise to be the state variables. Select
the output to be only the voltage of the latter capasitor v2:
v1(t)
v (t ) O
L
,
x(t )  M P
, u(t )  v (t ),
v ((t)
t )Q
N
1
0
22
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y ( t )  v2 ( t )
Example: a Circuit Board (4/4)
By applying these selections we will get

  1

1 
1
1

x2 (t ) 
u (t ) 

 
 x1 (t ) 
R2C1
R1C1
x(t )    R1C1 R2C1 
  f (x(t ), u (t ))



1
1

x1 (t ) 
x2 (t )



R2C2
R2C2



 y (t )  x2 (t )  g (x(t ), u (t ))
and finally

  1
1 
1 

 1 

 


 x(t )    R1C1 R2C1  R2C1  x(t )   R C  u (t )  Ax(t )  Bu (t )
1 1





1
1

 0 




R2C2
R2C2 


UNIVERSITY
of VAASA

y
(
t
)
  0 Engineering
1 x(t )  0Group
u (t )  Cx(t )  Du (t )
Telecommunication

Finding the State Variables
 How to select the state variables?
 Make such a selection, which is physically reasonable (like in the previous
examples)
 Use derivative operator p (so-called p-technique)
 Use chanonical forms
 Physically reasonable state variable selection is usually the easiest
way
 Applying chanonical forms is just applying of a straightforward formula,
but it might be difficult to see the physical meaning of the state
variables
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Telecommunication Engineering Group
Method 3: Use Chanonical Forms (2/2)
 Apply chanonical forms
mx(t )  Bx(t )  kx(t )  F (t )  x(t )  mB x(t )  mk x(t )  m1 F (t )
 A result by using controllable chanonical form is similar with the one
achieved by using derivative operator
y(t )  mB y(t )  mk y(t )  0u(t )  m1 u(t )  y(t )  a1 y(t )  a2 y(t )  b1u (t )  b2u (t )


  mB 1 
0
 a1 1 
 b1 
x(t )   1  u (t )
x(t )    u(t )
x(t )   k
 x (t )  


 
 a2 0 
b2 

m
 m 0
 y (t )  1 0 x(t )
 y (t )  1 0 x(t )
 




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The Dimensions of the Linear State-Space Representation
 The system order is the same as the sum of the order of the differential
equations which describe the system. It is also the dimension of the
system matrix A.




The number of inputs (control signals) is nu
The number of outputs is ny
System order is nS
The dimensions of the linear state-space representation are as follows:
x(t )  A x(t )  B u(t )
 nS 1 nS nS nS 1 nS nu nu 1

y (t )  C x(t )  D u(t )
 ny 1 ny nS nS 1 ny nu nu 1
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Solutions in Time and Laplace Domains
Time domain
problem
 y(t )  2 y(t )  et

 y(0)  1
Laplace-domain
problem

 1 
sY ( s)  y (0)  2Y ( s)  

 s  1

 y (0)  1

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Telecommunication Engineering Group
Time domain
solution
y (t )  et
Laplace-domain
solution
1
Y ( s) 
s 1
Laplace-transform
 F(s) is a Laplace transform of time domain function f(t). The Laplacetransform and its inverse transform are defined as follows:
F (s)  L  f (t ) 


f (t )e st dt
f (t )  L1 F ( s) 
0
1
b  j
2 j b j
F ( s)e st ds
 The following two theorems can be applied, if the respective limits
exist:
 Final value theorem
lim f (t )  lim sF ( s)
 Initial value theorem
lim f (t )  lim sF ( s)
t 
t 0
s 0
s 
 The most usual Laplace-transforms and inverse transforms are
tabulated
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Laplace-transforms (1/5)
Laplace-transform Inverse Laplace transform
F (s)
f (t )
T1
C1 F1 ( s )  C2 F2 ( s )
C1 f1 (t )  C2 f 2 (t )
T2
F ( s  a)
e  at f (t )
T3
ta
0,

 f (t  a ), t  a
T4
f (at )
T5
f (t )t
T6
e
 as
F (s)
1 s
F 
a a
d
 F ( s)
ds
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Laplace-transforms (2/5)
Laplace-transform
Time-domain function

 F ( )d
f (t )
s
t
 f ( ) f
F1 ( s ) F2 ( s )
1
2
1
t
(t   )d
T7
T8
0
sF ( s )  f (0)

s 2 F ( s )  sf (0)  f (0)

s n F ( s )  s n 1 f (0)  s n  2 f (0)
t

1
1
F ( s )    f ( )d 
s
s0

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f (t )
T9
f (t )
T10
f ( n ) (t )
T11
 f ( )d
T12

 f ( n 1) (0)

t
t 0
0
Laplace-transforms (3/5)
Laplace-transform
Laplace-transform Time domain
1
1
s
1
s2
1
s n 1
1
sa
1
( s  a)2
1
( s  a ) n 1
1
s(s  a)
 (t )
M1
1
M2
t
M3
tn
n!
M4
e  at
M5
te
 at
M6
n  at
t e
n!
1
1  e  at 

a
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M7
M8
Telecommunication Engineering Group
1
( s  a )( s  b)
1
s ( s  a )( s  b)
a
s2  a2
s
s2  a2
a
( s  b) 2  a 2
sb
( s  b) 2  a 2
sa
sb
Time domain
1
e  bt  e  at 
M9

a b
1
1

ae  bt  be  at  M10

ab ab(b  a )
sin( at )
M11
cos(at )
M12
e  bt sin(at )
M13
e  bt cos(at )
M14
 (t )  (a  b)e  bt
M15
Impulssi
Askel
Penger
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Diracin delta
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Deterministic Test Functions
 Following signals u(t) are often used as system inputs
 Unit impulse function (Dirac delta function)
; t  0
u (t )   (t )  
0; t  0
 Unit step function
U(t) = 0; t <0
= 1; t >0
U ( s)  ( s)  1
U s ( s) 
1
s
 Unit ramp function
U(t) = 0; t <0
= t; t >0
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U r ( s) 
1
s2
Example: Damped Oscillator
The differential equation of the damped oscillator: F(t)
mx(t )  Bx (t )  kx (t )  F (t )
x(t)
m
Solve the system response (output), when
 K=5
 x(0)  1
k

 x(0)  1
 B=2
 F (t )  2 (t ) (impulse)

 M=1
In time domain the model is x(t )  2 x (t )  5x(t )  2 (t )
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B
Example: Damped Oscillator
Solve location X(s) from the Laplace-transformed expression:
X ( s) 
s3
s  1 2
s 1
2



s2  2 s  5 ( s  1) 2  2 2 ( s  1) 2  2 2 ( s  1) 2  2 2
Then solve x(t) by computing the inverse Laplace transform
back to the time domain:
x(t )  L1  X (s)  et cos(2t )  et sin(2t )  et sin(2t )  cos(2t ) 
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Zero-State and Zero-Control Responses
 System response can be divided to the part y0(t) (zerocontrol response) caused by system initial values and to the
part yu(t) caused by the external input (zero-state response).
Linear system response is a sum of these two parts:
y ( t )  yu ( t )  y0 ( t )
 System response is equal to zero-control response y0(t) if
the external inputs ui(t) = 0.
 System response is equal to zero-state response yu(t), if all
system initial values y(n)(0) and ui(n)(0) are zeros.
 Term ”response” refers often to zero-state response, in
other words system response to some external input
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Example: Damped Oscillator
 Zero-control response starts from the system initial state
(location 1, velocity -1)
 Zero-state response starts from zero once the external force
starts to effect
 System response is a sum of these two components
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Transfer Function
 One of the crucial topics in system analysis is to analyze that how
external signals and disturbances affect the system response
 System transfer function is system response divided by system
external input in Laplace-domain:
Y ( s)
G ( s) 
U ( s)
 It follows from the definition that if system initial values are zero,
system output in Laplace-domain is given by
Y ( s)  G( s) U ( s)
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U(s)
G(s)
Y(s)
h e rsTransfer
äi ti fcner vResponse
t ao s f t
Input
Transfer Function
 A differential equation (or equation group) characterizing the
system is first Laplace-transformed and then Y(s) / U(s) is
solved.
 If we have several input and output variables (MIMO-model),
we will get a transfer function matrix
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Transfer Function
General form of the linear differential equation is
y ( n ) (t )  a1 y ( n1) (t ) an1 y (1) (t )  an y(t )  b1u( n1) (t ) bn1u(1) (t )  bn u(t )
Laplace-transform gives (initial values are assumed to be zero)
s
n
 a1s n 1 
 an 1s  an  Y (s )   b1s n 1 
Then it is easy to solve the transfer function
b1s n1   bn 1s  bn
Y ( s)
G(s) 
 n
U ( s) s  a1s n1   an 1s  an
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 bn 1s  bn U (s )
Static Gain
System static gain quantifies how much signal is strengthtened of faded
once it goes through the system
 For unit step respons, the static gain describes in which level the
system response will set (asymptotic stable system)
 For unit ramp response, the static gain gives the response angular
coefficient in continuous state (asymptotic stable system)
Static gain can be solved as a limit of system transfer function. It can be
solved also for nonstabile systems, but in that case there is no physical
connection to system response end value.
System static gain is given by
For stabile system
k  limG(s)  lim  y(t )
s 0
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t 
Telecommunication Engineering Group
l q
k  lim G( s)
s0
From State-Space Representation to Transfer Function
A transfer function can be counted directly from the state-space
representation such that
G(s)  C( sI  A)1 B  D
If the direct effect term D = 0, the formula is simplified to
G(s)  C( sI  A)1 B
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Example: Damped Oscillator
 Solve unit step response by using a transfer function
1
 s 1   0 
G ( s )  C( sI  A ) B  1 0 
 1 
5
s

2

  
 s  2 1 0 
1 0 
 1 

5
s
1





 2
2
s  2s  5
s  2s  5
1
 The response is


1
y(t )  L Y ( s)  L G (s)U (s )  L  2

s
(
s

2
s

5)


1
1

 15 1   et cos(2t )  12 et sin(2t ) 
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1

1
U ( s) 
s