Lavi Shpigelman, Dynamic Systems and control – 76929 –
Linear Time Invariant systems




definitions,
Laplace transform,
solutions,
stability
1
Lavi Shpigelman, Dynamic Systems and control – 76929 –
Lumpedness and causality
 Definition: a system is lumped if it can be
described by a state vector of finite
dimension. Otherwise it is called distributed.
Examples:
• distributed system: y(t)=u(t- t)
• lumped system (mass and spring with friction)
 Definition: a system is causal if its current
state is not a function of future events (all
‘real’ physical systems are causal)
3
Linearity and Impulse Response
description of linear systems
Lavi Shpigelman, Dynamic Systems and control – 76929 –
 Definition: a function f(x) is linear if
(this is known as the superposition property)
Impulse response:
 Suppose we have a SISO (Single Input Single Output) system
system as follows:
where:
 y(t) is the system’s response (i.e. the observed output) to the
control signal, u(t) .
 The system is linear in x(t) (the system’s state) and in u(t)
4
Lavi Shpigelman, Dynamic Systems and control – 76929 –
Linearity and Impulse Response
description of linear systems
 Define the system’s impulse response, g(t,), to be the
response, y(t) of the system at time t, to a delta function
control signal at time  (i.e. u(t)=t,) given that the
system state at time  is zero (i.e. x()=0 )
 Then the system response to any u(t) can be found by
solving:
Thus, the impulse response contains all the information
on the linear system
5
Lavi Shpigelman, Dynamic Systems and control – 76929 –
Time Invariance
 A system is said to be time invariant if its response to an
initial state x(t0) and a control signal u is independent of
the value of t0.
So g(t,) can be simply described as g(t)=g(t,0)
 A linear time invariant system is said to be causal if
 A system is said to be relaxed at time 0 if x(0) =0
 A linear, causal, time invariant (SISO) system that is
relaxed at time 0 can be described by
causal
relaxed
Time invariant
Convolution
6
LTI - State-Space Description
Lavi Shpigelman, Dynamic Systems and control – 76929 –
Fact: (instead of using the impulse response representation..)
 Every (lumped, noise free) linear, time invariant
(LTI) system can be described by a set of
equations of the form:
Linear, 1st order ODEs
Linear algebraic equations
Dynamic
Process
Observation
Process
Controllable
inputs u
u
D
B
+
1/s
x
Observations
y
+
C
A
State x
Disturbance
(noise) w
Measurement
Error (noise) n
Plant
7
What About
nth Order Linear ODEs?
Lavi Shpigelman, Dynamic Systems and control – 76929 –

Can be transformed into n 1st order ODEs
1. Define new variable:
2. Then:
Dx/dt
=
y = [I 0 0  0] x
A
x
+
B
u
8
Lavi Shpigelman, Dynamic Systems and control – 76929 –
Using Laplace Transform to
Solve ODEs
 The Laplace transform is a very useful tool in the
solution of linear ODEs (i.e. LTI systems).
 Definition: the Laplace transform of f(t)
 It exists for any function that can be bounded by
aet (and s>a ) and it is unique
 The inverse exists as well
Laplace transform pairs are known for many
useful functions (in the form of tables and Matlab
functions)
 Will be useful in solving differential equations!
9
Some Laplace Transform
Properties
Lavi Shpigelman, Dynamic Systems and control – 76929 –
 Linearity (superposition):
 Differentiation
10
Lavi Shpigelman, Dynamic Systems and control – 76929 –
 Remember integration by parts:
 Using that and the transform definition:
11
Some Laplace Transform
Properties
Lavi Shpigelman, Dynamic Systems and control – 76929 –
 Linearity (superposition):
 Differentiation
 Convolution
12
Lavi Shpigelman, Dynamic Systems and control – 76929 –
 Using definitions
 Integration over triangle 0 <  < t
 Define  = t-, then d = dt and region is  > 0, t > 0
13
Some Laplace Transform
Properties
Lavi Shpigelman, Dynamic Systems and control – 76929 –
 Linearity (superposition):
 Differentiation
 Convolution
 Integration
14
Lavi Shpigelman, Dynamic Systems and control – 76929 –
 By definition:
 Switch integration order
 Plug  = t-
15
Some specific Laplace
Transforms (good to know)
Lavi Shpigelman, Dynamic Systems and control – 76929 –
 Constant (or unit step)
 Impulse
 Exponential
 Time scaling
16
Homogenous (aka Autonomous / no input)
1st order linear ODE
Lavi Shpigelman, Dynamic Systems and control – 76929 –
 Solve:
 Do the Laplace transform
 Do simple algebra
 Take inverse transform
Known as zero
input response
17
1st order linear ODE
with input (non-homogenous)
Lavi Shpigelman, Dynamic Systems and control – 76929 –
 Solve:
 Do the Laplace transform
 Do simple algebra
 Take inverse transform
Known as the zero state response
18
Lavi Shpigelman, Dynamic Systems and control – 76929 –
Example: a 2nd order system
 Solve:
 Do the Laplace transform
 Do simple algebra
 Take inverse transform
19
Using Laplace Transform to
Analyze a 2nd Order system
Lavi Shpigelman, Dynamic Systems and control – 76929 –
 Consider the autonomous (homogenous) 2nd order system
 To find y(t), take the Laplace transform
(to get an algebraic equation in s)
 Do some algebra
characteristic polynomial
determined by Initial condition
 Find y(t) by taking the inverse transform
20
2nd Order system Inverse Laplace
Lavi Shpigelman, Dynamic Systems and control – 76929 –
 Solution of inverse transform depends on nature of the
roots 1,2 of the characteristic polynomial p(s)=as2+bs+c:
• real & distinct, b2>4ac
• real & equal, b2=4ac
• complex conjugates b2<4ac
 In shock absorber example:
a=m, b=damping coeff., c=spring coeff.
 We will see:
Re{} exponential effect
Im{} Oscillatory effect
21
Real & Distinct roots (b2>4ac)
Lavi Shpigelman, Dynamic Systems and control – 76929 –
 Some algebra helps fit the polynomial to Laplace
tables.
 Use linearity, and a table entry
To conclude:
p(s)=s2+3s+1
y(0)=1,y’(0)=0
1=-2.62
2=-0.38
• Sign{}  growth or decay
• ||  rate of growth/decay
y(t)=-0.17e-2.62t+1.17e-0.38t
22
Real & Equal roots (b2=4ac)
Lavi Shpigelman, Dynamic Systems and control – 76929 –
 Some algebra helps fit the polynomial to Laplace
tables.
 Use linearity, and a some table entries to
conclude:
p(s)=s2+2s+1
y(0)=1,y’(0)=0
1=-1
• Sign{}  growth or decay
• ||  rate of growth/decay
y(t)=-e-t+te-t
23
Complex conjugate roots (b2<4ac)
Lavi Shpigelman, Dynamic Systems and control – 76929 –
 Some algebra helps fit the polynomial to Laplace
tables.
 Use table entries (as before) to conclude:
 Reformulate y(t) in terms of  and 
Where:
24
Complex conjugate roots (b2<4ac)
 E.g. p(s)=s2+0.35s+1 and initial condition y(0)=1 , y’(0)=0
Lavi Shpigelman, Dynamic Systems and control – 76929 –
 Roots are =+i=-0.175±i0.9846
 Solution has form:
with constants
A=||=1.0157
r=0.5-i0.0889
=arctan(Im(r)/Re(r))
=-0.17591
 Solution is an
exponentially
decaying oscillation
 Decay governed by 
oscillation by .
25
The “Roots” of a Response
Lavi Shpigelman, Dynamic Systems and control – 76929 –
Im(s)
Marginally
Stable
Stable
Unstable
Re(s)
26
(Optional) Reading List
 LTI systems:
Lavi Shpigelman, Dynamic Systems and control – 76929 –
• Chen, 2.1-2.3
 Laplace:
• http://www.cs.huji.ac.il/~control/handouts/laplace_Boyd.pdf
• Also, Chen, 2.3
 2nd order LTI system analysis:
• http://www.cs.huji.ac.il/~control/handouts/2nd_order_Boyd.pdf
 Linear algebra (matrix identities and
eigenstuff)
• Chen, chp. 3
• Stengel, 2.1,2.2
27