Transcript LAPLACE TRANSFORMS INTRODUCTION Definition  Transforms -- a mathematical conversion from one way of thinking to another to make a problem easier to solve problem in original way.

LAPLACE TRANSFORMS
INTRODUCTION
Definition

Transforms -- a mathematical conversion
from one way of thinking to another to
make a problem easier to solve
problem
in original
way of
thinking
transform
solution
in transform
way of
thinking
2. Transforms
solution
in original
way of
thinking
inverse
transform
problem
in time
domain
Laplace
transform
solution
in
s domain
• Other transforms
• Fourier
• z-transform
• wavelets
2. Transforms
inverse
Laplace
transform
solution
in time
domain
All those signals……….
Time
Amplitude
111
continuous
110
100
discrete
011
010
Ts
001
time
1
2
3
4
5
continuous
indexing
discrete
000
n=0
continuous
sampling
101
w(t)
Amplitude
discrete
Signal
continuous-time
analog signal
w(t)
discrete-time
analog signal
w(nTs)
continuous
discrete-time
sequence
w[n]
sampling
discrete
discrete-time
digital signal
Cn
…..and all those transforms
Sample in time,
period = Ts
Continuous-time
analog signal
C
w(t)
Laplace
Transform
W(s) C

 w(t) e
 st
0
-  s  

dt
Continuous
Fourier
Transform
C
W(f)
 w(t) e
- j 2 p ft

dt
z-Transform
W(z) C

 w [n] z
n = 
  z
-  f  
z=
s = j
=2pf
C
Continuous-variable
D
Discrete-variable
Discrete-time
analog sequence
D
w [n]
ejW
-n
Discrete-Time
Fourier
Transform
W(W) C

 w [n] e
- jW n
Discrete
Fourier
Transform D
W(k)
N 1
-j
 w [n] e
n = -
n=0
0  W  2p
0  k  N 1
=2pf
W =  Ts,
scale
amplitude
by 1/Ts
2 p nk
Sample in
frequency,
W = 2pn/N,
N = Length
of sequence
N
Laplace transformation
time domain
linear
differential
equation
time
domain
solution
Laplace transform
inverse Laplace
transform
Laplace
transformed
equation
algebra
Laplace
solution
Laplace domain or
complex frequency domain
4. Laplace transforms
Basic Tool For Continuous Time:
Laplace Transform
L [ f ( t )] = F ( s ) =


f (t ) e
 st
dt
0

Convert time-domain functions and operations into
frequency-domain




f(t)  F(s) (tR, sC)
Linear differential equations (LDE)  algebraic expression
in Complex plane
Graphical solution for key LDE characteristics
Discrete systems use the analogous z-transform
The Complex Plane (review)
Imaginary axis (j)
u = x  jy
y
 u   = tan
r


y
x
x
Real axis
| u | r  | u |=
r
 y
1
u = x  jy
(complex) conjugate
x  y
2
2
Laplace Transforms of Common
Functions
Name
Impulse
f(t)
F(s)
t =0
1
f (t ) = 
0
Step
f (t ) = 1
Ramp
f (t ) = t
Exponential
f (t ) = e
Sine
1
t 0
1
s
1
s
1
at
f ( t ) = sin(  t )
2
sa
1
 s
2
2
Laplace Transform Properties
Addition/S caling
L [ af 1 ( t )  bf 2 ( t )] = aF 1 ( s )  bF 2 ( s )
Differenti ation
d

L
f ( t )  = sF ( s )  f ( 0  )
 dt

Integratio n
L
 f ( t ) dt  = F s( s )  1s  f ( t ) dt 
t=0
t
Convolutio n

f 1(t  τ)f 2 (τ ) dτ = F1 ( s ) F 2 ( s )
0
Initial - value theorem
f ( 0  ) = lim sF ( s )
Final - value theorem
lim f ( t ) = lim sF ( s )
s 
t 
s 0
LAPLACE TRANSFORMS
SIMPLE TRANSFORMATIONS
Transforms (1 of 11)

Impulse --  (to)

e-st  (to) dt
F(s) =
0
= e-sto
f(t)
 (to)
t
4. Laplace transforms
Transforms (2 of 11)

Step -- u (to)

e-st u (to) dt
F(s) =
0
= e-sto/s
f(t)
1
u (to)
t
4. Laplace transforms
Transforms (3 of 11)

e-at

e-st e-at dt
F(s) =
0
= 1/(s+a)
4. Laplace transforms
Transforms (4 of 11)
Linearity
f1(t)  f2(t)
F1(s) ± F2(s)
Constant multiplication
a f(t)
a F(s)
Complex shift
eat f(t)
F(s-a)
Real shift
f(t - T)
eTs F(as)
Scaling
f(t/a)
a F(as)
4. Laplace transforms
Transforms (5 of 11)

Most mathematical handbooks have tables
of Laplace transforms
4. Laplace transforms
Table of Laplace
Transforms
Definition of Laplace transform
L { f ( t )} =


e
 st
f ( t ) dt
0
1
f ( t ) = L { F ( s )} F ( s ) = L { f ( t )}
1
1
s
t
e
n!
n
s
at
sin  t
cos  t
n 1
1
sa

2
2
s 
s
2
2
s 
Translation and Derivative Table for
Laplace Transforms
First translation and derivative theorems
f (t ) = L
e
at
1
F  s )
F  s ) = L  f (t ) 
F (s  a)
f (t )
n
(  1)
t f (t )
d
ds
n
n
F (s)
sF ( s )  f ( 0 )
f ' (t )
s F ( s )  sf ( 0 )  f ' ( 0 )
2
f ' ' (t )
f ' ' ' (t )
n
s F ( s )  s f ( 0 )  sf ' ( 0 )  f ' ' ( 0 )
3
2
Unit step and Dirac delta function
Unit step and Dirac delta function
f (t ) = L
1
F  s )
F  s ) = L  f (t ) 
u (t  a )
f (t )u (t  a )
f (t  a )u (t  a )
 (t  a )
e
 as
s
e
 as
e
L{ f ( t  a )}
 as
e
F (s)
 as
Convolution theorem
Convolution theorem
f (t ) = L
1
F  s )
F  s ) = L  f (t ) 
t
f (t )  g (t ) =
 f ( ) g (t   ) d 
0
F ( s )G ( s )
LAPLACE TRANSFORMS
SOLUTION PROCESS
Solution process (1 of 8)

Any nonhomogeneous linear differential
equation with constant coefficients can be
solved with the following procedure, which
reduces the solution to algebra
4. Laplace transforms
Solution process (2 of 8)

Step 1: Put differential equation into
standard form



D2 y + 2D y + 2y = cos t
y(0) = 1
D y(0) = 0
Solution process (3 of 8)

Step 2: Take the Laplace transform of both
sides

L{D2 y} + L{2D y} + L{2y} = L{cos t}
Solution process (4 of 8)

Step 3: Use table of transforms to express
equation in s-domain

L{D2 y} + L{2D y} + L{2y} = L{cos  t}





L{D2 y} = s2 Y(s) - sy(0) - D y(0)
L{2D y} = 2[ s Y(s) - y(0)]
L{2y} = 2 Y(s)
L{cos t} = s/(s2 + 1)
s2 Y(s) - s + 2s Y(s) - 2 + 2 Y(s) = s /(s2 + 1)
Solution process (5 of 8)

Step 4: Solve for Y(s)

s2 Y(s) - s + 2s Y(s) - 2 + 2 Y(s) = s/(s2 + 1)


(s2 + 2s + 2) Y(s) = s/(s2 + 1) + s + 2
Y(s) = [s/(s2 + 1) + s + 2]/ (s2 + 2s + 2)

= (s3 + 2 s2 + 2s + 2)/[(s2 + 1) (s2 + 2s + 2)]
Solution process (6 of 8)

Step 5: Expand equation into format covered by
table




Y(s) = (s3 + 2 s2 + 2s + 2)/[(s2 + 1) (s2 + 2s + 2)]
= (As + B)/ (s2 + 1) + (Cs + E)/ (s2 + 2s + 2)
(A+C)s3 + (2A + B + E) s2 + (2A + 2B + C)s + (2B
+E)
Equate similar terms





1=A+C
2 = 2A + B + E
2 = 2A + 2B + C
2 = 2B + E
A = 0.2, B = 0.4, C = 0.8, E = 1.2
Solution process (7 of 8)

(0.2s + 0.4)/ (s2 + 1)



= 0.2 s/ (s2 + 1) + 0.4 / (s2 + 1)
(0.8s + 1.2)/ (s2 + 2s + 2)
= 0.8 (s+1)/[(s+1)2 + 1] + 0.4/ [(s+1)2 + 1]
Solution process (8 of 8)

Step 6: Use table to convert s-domain to
time domain





0.2 s/ (s2 + 1) becomes 0.2 cos t
0.4 / (s2 + 1) becomes 0.4 sin t
0.8 (s+1)/[(s+1)2 + 1] becomes 0.8 e-t cos t
0.4/ [(s+1)2 + 1] becomes 0.4 e-t sin t
y(t) = 0.2 cos t + 0.4 sin t + 0.8 e-t cos t + 0.4
e-t sin t
LAPLACE TRANSFORMS
PARTIAL FRACTION EXPANSION
Definition


Definition -- Partial fractions are several
fractions whose sum equals a given fraction
Example -



(11x - 1)/(x2 - 1) = 6/(x+1) + 5/(x-1)
= [6(x-1) +5(x+1)]/[(x+1)(x-1))]
=(11x - 1)/(x2 - 1)
Purpose -- Working with transforms requires
breaking complex fractions into simpler
fractions to allow use of tables of transforms
Partial Fraction Expansions
s 1
( s  2 ) ( s  3)
s 1
( s  2 ) ( s  3)
=
=
( s  2 ) ( s  3)
s2

B
s3
A ( s  3)  B s  2 )
( s  2 ) ( s  3)
A  B =1
s 1
A
3A  2B =1
=
1
s2

2
s3




Expand into a term for
each factor in the
denominator.
Recombine RHS
Equate terms in s and
constant terms. Solve.
Each term is in a form so
that inverse Laplace
transforms can be
applied.
Example of Solution of an ODE
2
d y
dt
2
 6
dy
y (0) = y ' (0) = 0 
 8y = 2
ODE w/initial conditions
dt
s Y (s)  6s Y (s)  8 Y (s) = 2 / s
2
2
Y (s) =
Y (s) =
1
4s


s (s  2) (s  4)
1
2 (s  2)
y (t ) =

1
4

e

1
4 (s  4)
2 t

2

e
4 t
4

Apply Laplace transform
to each term
Solve for Y(s)
Apply partial fraction
expansion
Apply inverse Laplace
transform to each term
Different terms of 1st degree


To separate a fraction into partial fractions
when its denominator can be divided into
different terms of first degree, assume an
unknown numerator for each fraction
Example -




(11x-1)/(X2 - 1) = A/(x+1) + B/(x-1)
= [A(x-1) +B(x+1)]/[(x+1)(x-1))]
A+B=11
-A+B=-1
A=6, B=5
Repeated terms of 1st degree (1 of 2)

When the factors of the denominator are of
the first degree but some are repeated,
assume unknown numerators for each
factor


If a term is present twice, make the fractions
the corresponding term and its second power
If a term is present three times, make the
fractions the term and its second and third
powers
3. Partial fractions
Repeated terms of 1st degree (2 of 2)

Example -






(x2+3x+4)/(x+1)3= A/(x+1) + B/(x+1)2 +
C/(x+1)3
x2+3x+4 = A(x+1)2 + B(x+1) + C
= Ax2 + (2A+B)x + (A+B+C)
A=1
2A+B = 3
A+B+C = 4
A=1, B=1, C=2
3. Partial fractions
Different quadratic terms


When there is a quadratic term, assume a
numerator of the form Ax + B
Example -






1/[(x+1) (x2 + x + 2)] = A/(x+1) + (Bx +C)/ (x2 +
x + 2)
1 = A (x2 + x + 2) + Bx(x+1) + C(x+1)
1 = (A+B) x2 + (A+B+C)x +(2A+C)
A+B=0
A+B+C=0
2A+C=1
A=0.5, B=-0.5, C=0
3. Partial fractions
Repeated quadratic terms

Example -







1/[(x+1) (x2 + x + 2)2] = A/(x+1) + (Bx +C)/ (x2 +
x + 2) + (Dx +E)/ (x2 + x + 2)2
1 = A(x2 + x + 2)2 + Bx(x+1) (x2 + x + 2) +
C(x+1) (x2 + x + 2) + Dx(x+1) + E(x+1)
A+B=0
2A+2B+C=0
5A+3B+2C+D=0
4A+2B+3C+D+E=0
4A+2C+E=1
A=0.25, B=-0.25, C=0, D=-0.5, E=0
3. Partial fractions
Apply Initial- and Final-Value
Theorems to this Example
Y (s) =
lim
t
lim
 f (t )  =
t0
 f (t )  =
2

Laplace
transform of the
function.

Apply final-value
theorem

Apply initialvalue theorem
s (s  2) (s  4)
2 (0)
(0) (0  2) (0  4)
2 ( )
( ) (  2) (  4)
=
1
4
= 0
LAPLACE TRANSFORMS
TRANSFER FUNCTIONS
Introduction

Definition -- a transfer function is an
expression that relates the output to the
input in the s-domain
r(t)
r(s)
differential
equation
transfer
function
5. Transfer functions
y(t)
y(s)
Transfer Function

Definition




H(s) = Y(s) / X(s)
X(s)
H(s)
Y(s)
Relates the output of a linear system (or
component) to its input
Describes how a linear system responds to
an impulse
All linear operations allowed

Scaling, addition, multiplication
Block Diagrams



Pictorially expresses flows and relationships
between elements in system
Blocks may recursively be systems
Rules



Cascaded (non-loading) elements: convolution
Summation and difference elements
Can simplify
Typical block diagram
reference input, R(s)
plant inputs, U(s)
error, E(s)
pre-filter
G1(s)
control
Gc(s)
feedback
H(s)
feedback, H(s)Y(s)
5. Transfer functions
output, Y(s)
plant
Gp(s)
post-filter
G2(s)
Example
R
L
v(t)
C
v(t) = R I(t) + 1/C
I(t) dt + L di(t)/dt
V(s) = [R I(s) + 1/(C s) I(s) + s L I(s)]
Note: Ignore initial conditions
5. Transfer functions
Block diagram and transfer function

V(s)



= (R + 1/(C s) + s L ) I(s)
= (C L s2 + C R s + 1 )/(C s) I(s)
I(s)/V(s) = C s / (C L s2 + C R s + 1 )
V(s)
I(s)
C s / (C L s2 + C R s + 1 )
5. Transfer functions
Block
diagram
reduction
rules
Series
U
G1
Parallel
U
G2
+
Y
U
G1 G 2
Y
Y
G1
U
+
G1 + G2
Y
G2
Feedback
U
+
G1
Y
U
G2
5. Transfer functions
G1 /(1+G1 G2)
Y
Rational Laplace Transforms
F (s) =
A(s)
B (s)
A ( s ) = a n s  ...  a 1 s  a 0
n
B (s) = bm s
m
 ...  b1 s  b 0
Poles : s *  B ( s *) = 0 (So, F ( s *) =  )
Zeroes : s *  A ( s *) = 0 (So, F ( s *) = 0 )
Poles and zeroes are complex
Order of system = # poles = m
First Order System
Y (s)
R (s)
=
K
1  K  sT

K
1  sT
Reference
R (s)
S
E (s)
B (s)
K
U (s)
1
1  sT
1
Y (s)
First Order System
Impulse
response
Exponential
K
1  sT
Step response
K
-
s
Step,
exponential
K
s 1/T
Ramp response
K
s
2
-
KT
s
-
KT
s 1/T
No oscillations (as seen by poles)
Ramp, step,
exponential
Second Order System
Impulse response :
Y (s)
R (s)
Oscillates
Js
2
=
 Bs  K
s  2 N s   N
if poles have non - zero imaginary
Damping ratio :  =
B
Bc
Undamped
=
N
2
K
where B c = 2 JK
natural frequency
: N =
K
J
2
2
part (ie, B  4 JK  0 )
2
Second Order System: Parameters
Interpreta tion of damping ratio
 = 0 : Undamped
oscillatio n (Re = 0, Im  0)
0    1 : Underdampe
1   : Overdamped
d (Re  0  Im)
(Re  0, Im = 0)
Interpreta tion of undamped
 N gives the frequency
natural frequency
of the oscillatio n
Transient Response Characteristics
2
1.75
mumixam =
toohsrevo
1.5
p
M
1.25
1
0.75
0.5
0.25
0.5
d
t
r
t
1
p
t
1.5
s
2
2.5
3
t
t d : Delay until reach 50% of steady
state value
t r : Rise time = delay until first reach steady
state value
t p : Time at which peak value is reached
t s : Settling
time = stays within specified
% of steady
state
Transient Response


Estimates the shape of the curve based on
the foregoing points on the x and y axis
Typically applied to the following inputs




Impulse
Step
Ramp
Quadratic (Parabola)
Effect of pole locations
Oscillations
(higher-freq)
Im(s)
Faster Decay
(e-at)
Re(s)
Faster Blowup
(eat)
Basic Control Actions: u(t)
Proportion al control :
u (t ) = K p e (t )
t
Integral control :
u ( t ) = K i  e ( t )dt
0
Differenti al control :
u (t ) = K d
d
dt
e (t )
U (s)
= K
E (s)
U (s)
=
p
Ki
E (s)
s
U (s)
= Kds
E (s)
Effect of Control Actions

Proportional Action


Integral Action



Adjustable gain (amplifier)
Eliminates bias (steady-state error)
Can cause oscillations
Derivative Action (“rate control”)



Effective in transient periods
Provides faster response (higher sensitivity)
Never used alone
Basic Controllers


Proportional control is often used by itself
Integral and differential control are typically
used in combination with at least proportional
control

eg, Proportional Integral (PI) controller:
G (s) =
U (s)
E (s)
= K

1 


= K p  1 

s
T
s
i 

KI
p
Summary of Basic Control




Proportional control
 Multiply e(t) by a constant
PI control
 Multiply e(t) and its integral by separate constants
 Avoids bias for step
PD control
 Multiply e(t) and its derivative by separate constants
 Adjust more rapidly to changes
PID control
 Multiply e(t), its derivative and its integral by separate constants
 Reduce bias and react quickly
Root-locus Analysis


Based on characteristic eqn of closed-loop transfer
function
Plot location of roots of this eqn



Multiple parameters are ok



Same as poles of closed-loop transfer function
Parameter (gain) varied from 0 to 
Vary one-by-one
Plot a root “contour” (usually for 2-3 params)
Quickly get approximate results

Range of parameters that gives desired response
LAPLACE TRANSFORMS
LAPLACE APPLICATIONS
Initial value

In the initial value of f(t) as t approaches 0
is given by
f(0 ) = Lim s F(s)

s
Example
f(t) = e -t
F(s) = 1/(s+1)
f(0 ) = Lim s /(s+1) = 1

s
6. Laplace applications
Final value

In the final value of f(t) as t approaches 
is given by
f(0 ) = Lim s F(s)
s
0
Example
f(t) = e -t
F(s) = 1/(s+1)
f(0 ) = Lim s /(s+1) = 0
s
0
6. Laplace applications
Apply Initial- and Final-Value
Theorems to this Example
Y (s) =
lim
t
lim
 f (t )  =
t0
 f (t )  =
2

Laplace
transform of the
function.

Apply final-value
theorem

Apply initialvalue theorem
s (s  2) (s  4)
2 (0)
(0) (0  2) (0  4)
2 ( )
( ) (  2) (  4)
=
1
4
= 0
Poles




The poles of a Laplace function are the
values of s that make the Laplace function
evaluate to infinity. They are therefore the
roots of the denominator polynomial
10 (s + 2)/[(s + 1)(s + 3)] has a pole at s =
-1 and a pole at s = -3
Complex poles always appear in complexconjugate pairs
The transient response of system is
determined by the location of poles
6. Laplace applications
Zeros



The zeros of a Laplace function are the
values of s that make the Laplace function
evaluate to zero. They are therefore the
zeros of the numerator polynomial
10 (s + 2)/[(s + 1)(s + 3)] has a zero at s =
-2
Complex zeros always appear in complexconjugate pairs
6. Laplace applications
Stability


A system is stable if bounded inputs
produce bounded outputs
The complex s-plane is divided into two
regions: the stable region, which is the left
half of the plane, and the unstable region,
which is the right half of the s-plane
s-plane
j
x
x
x
x
x
x
stable
x
unstable

LAPLACE TRANSFORMS
FREQUENCY RESPONSE
Introduction



Many problems can be thought of in the
time domain, and solutions can be
developed accordingly.
Other problems are more easily thought of
in the frequency domain.
A technique for thinking in the frequency
domain is to express the system in terms
of a frequency response
7. Frequency response
Definition


The response of the system to a sinusoidal
signal. The output of the system at each
frequency is the result of driving the system
with a sinusoid of unit amplitude at that
frequency.
The frequency response has both amplitude
and phase
7. Frequency response
Process

The frequency response is computed by
replacing s with j  in the transfer function
Example
f(t) = e -t
magnitude in dB
F(s) = 1/(s+1)
F(j ) = 1/(j  +1)
Magnitude = 1/SQRT(1 + 2)
Magnitude in dB = 20 log10 (magnitude)
Phase = argument = ATAN2(- , 1)
7. Frequency response

Graphical methods



Frequency response is a graphical method
Polar plot -- difficult to construct
Corner plot -- easy to construct
7. Frequency response
Constant K
60 dB
40 dB
20 dB
0 dB
-20 dB
-40 dB
-60 dB
magnitude
20 log10 K
+180o
+90o arg K
0o
-90o
-180o
-270o
0.1
phase
1
10
, radians/sec
7. Frequency response
100
Simple pole or zero at origin, 1/ (j)n
magnitude
60 dB
40 dB
20 dB
0 dB
-20 dB
-40 dB
-60 dB
+180o
+90o
0o
-90o
-180o
-270o
0.1
1/ 
phase
1/ 3
1/ 2
1/ 
1/ 2
1/ 3
1
10
, radians/sec
G(s) = n2/(s2 + 2 ns +  n2)
100
Simple pole or zero,
1/(1+j)
magnitude
60 dB
40 dB
20 dB
0 dB
-20 dB
-40 dB
-60 dB
+180o
+90o
0o
-90o
-180o
-270o
0.1
phase
1
T
7. Frequency response
10
100
Error in asymptotic approximation
T
dB
arg (deg)
0.01
0
0.5
0.1
0.043
5.7
0.5
1
26.6
0.76
2
37.4
1.0
3
45.0
1.31
4.3
52.7
1.73
6.0
60.0
2.0
7.0
63.4
5.0
14.2
78.7
10.0
20.3
84.3
7. Frequency response
Quadratic pole or zero
magnitude
60 dB
40 dB
20 dB
0 dB
-20 dB
-40 dB
-60 dB
+180o
+90o
0o
-90o
-180o
-270o
0.1
phase
1
T
7. Frequency response
10
100
Transfer Functions




Defined as G(s) = Y(s)/U(s)
Represents a normalized model of a process,
i.e., can be used with any input.
Y(s) and U(s) are both written in deviation
variable form.
The form of the transfer function indicates the
dynamic behavior of the process.
Derivation of a Transfer Function
M
dT
dt
= F1 T1  F 2 T 2  ( F1  F 2 ) T
 T = T  T0
M
dT
dt
 T1 = T1  T0

 T 2 = T 2  T0 
= F1  T1  F 2  T 2  ( F1  F 2 )  T

Dynamic model of
CST thermal
mixer
Apply deviation
variables
Equation in terms
of deviation
variables.
Derivation of a Transfer Function
T (s) =
G (s) =
T (s)
T1 ( s )
F1 T1 ( s )  F 2 T 2 ( s )
M
=
s  F1  F 2
M


F1
s  F1  F 2



Apply Laplace transform
to each term considering
that only inlet and outlet
temperatures change.
Determine the transfer
function for the effect of
inlet temperature
changes on the outlet
temperature.
Note that the response
is first order.
Poles of the Transfer Function
Indicate the Dynamic Response
G (s) =
Y (s) =
1
( s  a ) ( s  bs  c ) ( s  d )
2
A
(s  a)
y (t ) = A e

 at

B
( s  bs  c )
2
 B e
pt

C
(s  d )
sin(  t )  C  e
dt
For a, b, c, and d positive constants, transfer
function indicates exponential decay, oscillatory
response, and exponential growth, respectively.
Poles on a Complex Plane
Im
Re
Exponential Decay
Re
y
Im
Time
Damped Sinusoidal
Re
y
Im
Time
Exponentially Growing Sinusoidal
Behavior (Unstable)
Re
y
Im
Time
What Kind of Dynamic Behavior?
Im
Re
Unstable Behavior



If the output of a process grows without
bound for a bounded input, the process is
referred to a unstable.
If the real portion of any pole of a transfer
function is positive, the process
corresponding to the transfer function is
unstable.
If any pole is located in the right half plane,
the process is unstable.