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Transcript
```Laplace Transform
Specific objectives for today:



Introduce Laplace transform
Investigate Laplace transform of exponential signals
Derive region of convergence of Laplace transform
Introduction to the Laplace Transform

The Laplace Transform can be applied to the
analysis of un-stable system (signals with
infinite energy) and play a role in the analysis
of system stability.

The Laplace transform is a very useful tool in
the solution of linear Ordinary Differential
Equation (ODEs).
The Laplace Transform





The response of an LTI system with impulse response h(t) to a
st
complex exponential input, x(t)=est, is y (t )  H ( s)e
where s is a complex number and H ( s) 



h(t )est dt
when s is complex, this is the Laplace transform of h(t), H(s)
The Laplace transform of a general signal x(t) is:

X (s)   x(t )est dt

and is usually expressed as:
L
x(t )  X ( s)
Laplace Transforms
L f (t )    f (t ) e

 st
dt  F ( s )

Useful for solving linear differential equations.

Approach is to apply Laplace transform to differential
equation. Then algebraically solve for Y(s). Finally, apply
inverse Laplace transform to directly determine y(t).

Tables of Laplace transforms are available.
Notation


L f (t )  F (s)
means that the Laplace transform
of f (t ) is F (s ) .
L1F (s)  f (t )
means that f (t ) is the function
whose Laplace transform is F (s )
(called the inverse of the Laplace
transform)
Example 1: Laplace Transform
x(t )  e at u (t )

Consider the signal

The Laplace transform is:

X ( s)   e u (t )e
at
 st
  ( s  a )t
e
dt
0
dt  
  (  a )t  jt
e
e
dt
0

L
1
e u (t )  X ( s ) 
, Re{s}  a
sa
at
Example 2: Laplace Transform


Consider the signal x(t )  e at u (t )
The Laplace transform is:

X ( s )     e at e st u (t )dt
    e  ( s  a )t dt
0
1

sa



Convergence requires that Re{s-a}<0 or Re{s}<a.
The Laplace transform expression is identical to Example 1
(similar but different signals), however the regions of
convergence of s are mutually exclusive (non-intersecting).
For a Laplace transform, we need both the expression and the
Region Of Convergence (ROC).
Region of Convergence

The Region Of Convergence (ROC) of the Laplace transform
is the set of values for s (   j ) for which the Fourier
transform of x(t)e-st converges (exists).
The ROC is generally displayed by drawing separating
line/curve in the complex plane, as illustrated below for
Examples 1 and 2, respectively.
Re{s}  a
Re{s}  a

The shaded regions denote the ROC for the Laplace transform

Example 3: sin(t)u(t)

The Laplace transform of the signal x(t) = sin(ωt)u(t) is:
X ( s)  

1
2
 j

1
2j


0
e
e
j t

 e  jt u (t )e  st dt
 ( s  j ) t
dt 
1
2j


0
e ( s  j ) t dt
 e  ( s  j ) t  e  ( s  j ) t  

 21j 

  ( s  j )
( s  j ) 0 
0




1
1

 

 ( s  j ) ( s  j ) 
 2 j 
 21j  2
2 
 s  
1
2j

 2
s 2
Re{ s}  0
Re{ s}  0
Example 4: Laplace Transform

Consider a signal that is the sum of two real exponentials:

The Laplace transform is then:

X ( s )   3e  2t u (t )  2e t u (t ) e  st dt
x(t )  3e 2t u (t )  2e t u (t )





 3 e u (t )e dt  2 e t u (t )e  st dt

 2t
 st


Using Example 1, each expression can be evaluated as:
3
2
X ( s) 

s  2 s 1

The ROC associated with these terms are Re{s}>-1 and
Re{s}>-2. Therefore, both will converge for Re{s}>-1, and the
s 1
Laplace transform: X ( s )  2
s 3s  2
Laplace transforms of simple functions
 Constants


 1 st 
 C C
LC   e Cdt   e C   0    
 s
0
 s s
0
 st
C
LC 
s
Laplace transforms of simple functions
 Powers of t



1 st
 1 st 
L
t   e tdt   e t    e dt
 s
0 s 0
0
 st

1  1 st 
1
 L
t  0  0   e   2
s s
0 s
1
Lt  2
s
Laplace transforms of simple functions
 Powers of t



1 st n1
 1 st n 
L{t }   e t dt   e t    e nt dt
 s
0 0 s
0
n
 st n

n st n 1
 L{t }  0  0   e nt dt
s0
n
n
 L{t }  L{t n 1}
s
n
Laplace transforms of simple
functions
 Similarly
n  1 n2
L{t } 
L{t }
s
n2
n2
L{t } 
L{t n3 }
s

n 1
1
0
L{t }  L{t }
s
n!
n
L{t }  n1
s
1
Laplace transforms of simple functions

L{e }   e e dt   e
 Exponentials
at
 st at
0

 L{e }   e
at
( s  a ) t
dt
0

( s  a ) t
0

 L{e }  0  

at

1 ( s  a ) t 

 
e

s

a

0
1 0 
1
e 
sa
 sa
1
L{e } 
sa
at
very
important!
Laplace transforms of simple functions
 Sine and Cosine
 1 it 1 -it 
Lcos t  L  e  e 
2
2

1 1
1 1
 Lcost 

2 s  i 2 s  i
1  s  i
s  i 
s
 Lcos t   2
 2
 2
2
2
2 s 
s   s 2
s
L{cos t}  2
s 2
Laplace transforms of simple functions
 Similarly

L{sin t}  2
2
s 
Laplace transform of derivatives
Given the Laplace transform:

Y ( s)  Ly(t )   e  st y(t )dt
0
Now find the Laplace transform of the
derivative

 dy 
 st dy
L    e
dt
dt
 dt  0
Laplace transform of derivatives
We use integration by parts with f=e-st, g=y.
Then the transform of the derivative is:


 st

dy
dy
de
 
 st
 st
L    e
dt  e y 0  
ydt
dt
dt
 dt  0
0






 dy 
 st
 st
 L   e y 0    se ydt
 dt 
0




 dy 
 st
 L    e y 0  s  e st ydt
 dt 
0
Laplace transform of derivatives
We assume that y(t) grows more slowly than
eat (for some a) as t   , and therefore e-st
dominates (we can choose s >a). Therefore
 st
e y (t )  0 as t  

e
 st
y


0
 0  e y(0)   y(0 )
0
Laplace transform of derivatives
Now we can simplify the formula above




 dy 
 st
 st
L    e y 0  s  e ydt
 dt 
0
 dy 
 L     y (0)  sY ( s )
 dt 
Laplace transform of derivatives
From above the transform of the derivative
is:
 dy 
L    sY ( s )  y (0  )
 dt 
Or
 dy 
L    sLy (t ) y (0  )
 dt 
Laplace transform of derivatives
Similarly we can calculate formulae for 2nd
and higher derivatives
 d 2 y  2
dy
L  2   s Y ( s )  (0)  sy (0  )
dt
 dt 
 d n y  n
d n1 y
L  n   s Y ( s )  n1 (0)  
 dt 
dt
n  2 dy
s
(0)  s n1 y (0  )
dt
Example of Laplace Transforms
+
+
The Unilateral Laplace Transform and Properties.

The Unilateral Laplace Transform of a signal x(t) is defined by

X s  

xt e  st dt
0


The lower limit of 0- implies that we do not include discontinuities and
impulses that occur at t = 0 in the integral. H(s) depends on x(t)for t
>= 0.
The relationship between X(s) and x(t) as
xt   
 X s 
Lu

The unilateral and bilateral Laplace transforms are equivalent for
signals that are zero for time t<0.
Cont’d…
Properties of Unilateral Laplace Transform.
 Scaling
Lu
 s 
1

xat 

X


a
a

Linearity,

Time Shift

for a>0
axt   by t 
 aX s   bY s 
Lu
xt  t 
 e  st X s 
Lu
for all t such that x(t - t)u(t) = x(t - t)u(t - t)
A shift in t in time correspond to multiplication of the Laplace
transform by the complex exponential e-st.
Cont’d…

s-Domain Shift

L
s
t
u  X s  s
e o xt 

o
Multiplication by a complex exponential in time introduces a
shift in complex frequency s into the Laplace transform.
Time shifts for which the unilateral Laplace transform time-shift property does
not apply.
(a) A nonzero portion of x(t) that occurs at times t  0 is shifted to times t < 0.
(b) A nonzero portion x(t) that occurs at times t < 0 is shifted to times t  0.
Cont’d…

Convolution.
xt * y t 
 X s Y s .
Lu


Convolution in time corresponds to multiplication of Laplace
transform. This property apply when x(t)=0 and y(t) = 0 for
t < 0.
Differentiation in the s-Domain.
d
 tx t 

X s .
ds
Lu

Differentiation in the s-domain corresponds to multiplication by t in the time domain.
Cont’d…

Differentiation in the Time Domain.
 
d
Lu

xt   sX s   x 0
dt

Initial and Final Value Theorem.
 
Lim sX s   x 0 .


s 


The initial value theorem allow us to determine the initial value,
x(0+), and the final value, x(infinity, of x(t) directly from X(s).
The initial value theorem does not apply to rational functions X(s)
in which the order of the numerator polynomial is greater than or
equal to hat of the denominator polynomial.
Cont’d…
Example 1: Applying Properties.
Find the unilateral Laplace Transform of x(t)=(-e3tu(t))*(tu(t)).
Solution:
Find the Unilateral Laplace Transform.
1
 e u (t ) 
( s  3)
3t
And
Apply s-domain differentiation property,
Use the convolution property,
Lu
1
u (t ) 
s
Lu
1
tu(t )  2
s
Lu
Lu
x (t ) 1 (e 3t u (t ) * (tu(t )) 
X ( s) 
.
1
s 2 ( s  3)
Inverse Laplace Transform
Example 1
Find the inverse Laplace Transform of:
X ( s) 
3s  4
( s  1)( s  2)
2
Use partial fraction expansion:
A3
A1
A2
X ( s) 


s  1 s  2 (s  2) 2
Solving for A1 , A2 and A3
1
1
2
X ( s) 


s  1 s  2 (s  2) 2
Taking Inverse Laplace Transform for each term :
L
1
e u (t ) 
s 1
t
L
1
 e u (t )  
s2
L
2
 2t
2te u (t ) 
( s  2) 2
 2t
So,
t
 x(t )  e u (t )  e
 2t
u (t )  2te
 2t
u (t )
Partial Fraction Expansions
s 1
A
B


( s  2) ( s  3) s  2 s  3
s 1
A( s  3)  Bs  2

( s  2) ( s  3)
( s  2) ( s  3)
A  B 1
3 A  2 B 1
s 1
1
2


( s  2) ( s  3) s  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 2
Find the inverse Laplace transform of:
X ( s) 
2s  9s  4s  10
3
2
s  3s  4
2
Use Long Division method!
By using the long division method to express X(s):
X ( s )  2s  3 
3s  2
s  3s  4
2
Use partial fraction expansion to obtain:
1
2
X ( s)  2s  3 

s 1 s  4
Term by term inversion of X(s) yields:
t
x(t )  2 t  3 (t )  e u(t )  2e u(t )
(1)
4t
Method for Solving Linear ODE’s
using Laplace Transforms
sY(s) - y(0) =
F(s,Y)
Y(s) = H(s)
Laplace Domain
Time Domain
dy/dt = f(t,y)
y(t) = h(t)
Solution of ODE’s
Perform Laplace transform.
Obtain polynomial equation for Y(s).
Solve for Y(s) - very simple.
‘Invert’ to obtain the solution for y(t) - difficult.
Example 1
Use Laplace Transform to find y(t) in response
to x(t)=(3/5)e-2tu(t) , with initial condition
y(0-) = -2 and RC = 0.2s for:
d
1
1
y (t ) 
y (t ) 
x(t )
dt
RC
RC
Solution:
Letting RC = 0.2s
d
y (t )  5 y (t )  5 x(t )
dt
Take Laplace transform for each term and use
differentiation property:

sY ( s )  y (0 )  5Y ( s )  5 X ( s )

sY ( s )  5Y ( s )  5 X ( s )  y (0 )
1

Y (s) 
[5 X ( s )  y (0 )
s5
Next, we use X ( s )  3 / 5 and the initial condition y(0-) = -2
s2
Obtaining:
3
2
Y ( s) 

( s  2)( s  5) s  5
Then, use partial fraction:
1
1
2
Y ( s) 


s2 s5 s5
1
3


s2 s5
Taking the inverse LT for each term:
y(t )  e
2t
u(t )  3e
5t
u(t )
Example 2 – Simple ODE
 Use the Laplace transform method to solve
d2y
y0
2
dt
dy
y (0 )  1, (0)  0
dt

 Step 1. Define your transform

Y ( s)  Ly(t )   e
0
 st
y(t )dt
 Step 2. Transform the ODE
d 2 y

L 2  y   L0
 dt



d y

 st
  e  2  y dt   e 0dt
 dt

0
0

 st
 e
0
 st
2
2

d y
 st
dt

e
ydt  0
2

dt
0
 Use the formula for the transform of a 2nd derivative:
dy
 2
 
 s Y ( s )  dt (0)  sy (0 )  Y ( s )  0
 Tidy up:


dy
s  1 Y (s) 
(0)  sy (0  )
dt
2
 Step 3. Use the initial conditions and solve for Y(s).
Remember:
dy
y (0 )  1, (0)  0
dt



 s  1 Y ( s)  0  s
2
s
 Y ( s)  2
s 1
 Step 4. Invert (using tables and useful rules). Look up
tables and find:
s
s
Lcos(t )  2
 Lcos t  2
2
s 
s 1
 Therefore
 s 
y (t )  L Y ( s )  L  2
  cos t
 s  1
1
1
y (t )  cos t
Example 3
d2y
dy

6
 8 y  2 y(0)  y ' (0)  0  ODE with initial conditions
2
dt
dt
s 2 Y ( s )  6s Y ( s )  8 Y ( s )  2 / s
2
Y (s) 
s ( s  2) ( s  4)
1
1
1
Y (s) 


4 s 2 ( s  2) 4 ( s  4)
1 e 2t
e 4t
y (t )  

4
2
4

Apply Laplace transform to
each term

Solve for Y(s)

Apply partial fraction
expansion
Apply inverse Laplace
transform to each term

Example 4
Solve
y  2y  y  e t
Create the
subsidiary eq. by
taking the
Laplace
Transform the
diff eq
y( 0 )  1 y( 0 )  1
s 2Y  sy (0)  y(0)  2[ sY  y (0)]  Y 
Substitute in
initial
conditions
1
s 1
s 2 Y  s (  1 )  1  2 [ sY  (  1 )]  Y 
Group like
terms, factor,
and set for Y
Y ( s 2  2s  1 )   s  1 
Solve for Y
Now take the
Inverse Laplace
Transform
Y 
1
s1
1
1

s  1 ( s  1) 3
1

y( t )   t 2  1 e t
2

1
s1
LT Methods in Circuit Analysis.


We can use the differentiation and integration properties to transform
circuits involving capacitive and inductive elements so that we can
solve using Laplace transforms.
Replace resistive, capacitive, and inductive elements by the Laplace
transform equivalent.
Laplace transform circuit models for use with Kirchhoff ’s voltage law. (a) Resistor. (b)
Inductor with initial current iL(0–). (c) Capacitor with initial voltage vc(0–).
Cont’d…
(a) Resistor
v



t   Ri t 
R
R

s   RI s 
R
R
d
v t L
i t 
L
dt L

Transforming the above equation
V

V
Transforming the above equation
Representation of the transform resistor element.
(b) Inductor.


 



sLI
s

Li
 0 .
L s
L
L


Representation of the transform inductor element.
Cont’d…
(c) Capacitor.
t
vc t  
1
C

 
ic t dt  vc  0  .
 
0

Transforming the above equation


vc  0  .
1


Vc s  
I c s  
sC
s

Representation of the transform capacitor element.
Cont’d…


When apply Kirchhoff ’s voltage law to solve a circuit, use equation
(6.29),(6.30) and (6.31).
When apply Kirchhoff ’s current law to solve a circuit, rewrite equation
(6.29),(6.30) and (6.31) to express current as a function of voltage.
Laplace transform circuit models for use with Kirchoff ’s current law. (a) Resistor. (b)
Inductor with initial current iL(0–). (c) Capacitor with initial voltage vc(0–).
Cont’d…
Example 1: Solving the Second Order Circuit.
Use the Laplace transform circuit model to determine the voltage y(t) in the
circuit below for the applied voltage x(t)=3e-10tu(t) V. the voltage across the
capacitor at time t=0- is 5V.
Solution:
Electrical circuit (a) Original circuit.
Step 1: Transform the circuit (a) to (b).
I1(s) and I2(s) are the currents through the branch.
Electrical circuit (a) Original circuit. (b) Transformed circuit.
Cont’d…
Step 2: Derive equation from the Transformed circuit
using KVL:
Y ( s )  1000( I1 ( s )  I 2 ( s ))
1
5
 X ( s)  Y ( s) 
I ( s)   0
4 1
s
s (10 )
1
5
X ( s)  Y ( s) 
I
(
s
)

1
s
s (10 4 )
X ( s )  Y ( s )  1000 I 2 ( s )
Step 3: Eliminate I1(s) and I2(s) to get Y(s).
s  10
5
Y (s)  [ X (s)
]
s  20 s  20
Step 4: Using X ( s ) 
3
s  10
we obtain:
2
Y (s) 
s  20
And thus conclude that:
y(t )  2e
20t
u(t ) V
Example 2
Use Laplace Transform circuit to determine the circuit below
assuming R = 1Ω and L = 0.5H. The current through the
inductor at time t = 0 - is 2A and x(t)=e-t u(t)
Solution:
Transform the circuit:
Find the KVL for that circuit:

X (s)  LiL (0 )  ( R  Ls)Y (s)
Rearrange for Y(s) and put in the value of R, L and iL(0-):
1 X (s)
i L (0  )
Y (s) 

L sR/L sR/L
2 X (s)
2


s2 s2
L
Using x(t )  e t u (t )  X ( s)  1
into equation above:
s 1
2
2
Y ( s) 

( s  1)( s  2) s  2
2
2
2
2




s 1 s  2 s  2 s 1
 y (t )  2e t u (t )
Transfer Function.


The transfer function of an LTI system is defined as the Laplace
transform of the impulse response.
Take the bilateral Laplace transform of both sides of the equation and
use the convolution properties result in,
Y ( s)  H ( s) X ( s)

Rearrange the above equation result in the ratio of Laplace transform
of the output signal to the Laplace transform of the input signal.
(X(s) is nonzero)
Y ( s)
H ( s) 
X ( s)
Transfer Function and Differential-Equation
System Description.
Given a differential equation.
M
dk
dk
ak k y(t )   bk k x(t )

dt
dt
k 0
k 0
N
Step 1: Substitute x(t) = est and y(t) = estH(s) into the equation.
y(t) = estH(s), substitute to the above equation result in,
M
 N
d k st 
d k st
  ak k e  H s    bk k e
dt
k 0
 k 0 dt

 
Step 2: Solve for H(s).
H(s) is a ratio of polynomial and
s is termed a rational transfer function.
 
M
H s  
k
b
s
 k
k 0
N
k
a
s
 k
k 0
Example 1: Find the Transfer Function.
Find the transfer function of the LTI system described by the differential equation
below
d2
d
d
yt   3 yt   2 yt   2 xt   3x(t )
2
dt
dt
dt
Solution:
Step 1: Substitute x(t) = est and y(t) = estH(s) into the equation.
 
 
 
   
d 2 st
d st
d st
st
st
e
H
(
s
)

3
e
H
(
s
)

2
e
H
(
s
)

2
e

3
e
dt 2
dt
dt
Step 2: Solve for H(s).
 
d k st
k st
e

s
e
k
dt
 d 2 st
d st
H ( s) 2 e  3 e  2 e st
dt
 dt
 

d st
  2 e  3 e st
dt

   
   
Cont’d… example
   
d st
2
e  3 e st
dt
H ( s) 
 d 2 st
d
 2 e  3 e st  2 e st
dt
 dt
 
   
 
The transfer function is,



d k st
k st
e

s
e
k
dt
2s  3
H ( s)  2
s  3s  2
System Stability




System stability is very important
A continuous-time LTI system is stable if its
impulse response is absolutely integrable
This translates into the frequency domain as
the requirement that all the poles of the
system transfer function must lie in the open
left half-plane of the s plane (pp. 675-676)
“Open left half-plane” means not including the
w axis
The relationship between the locations of poles and the impulse response in a causal system. (a) A
pole in the left half of the s-plane corresponds to an exponentially decaying impulse response. (b)
A pole in the right half of the s-plane corresponds to an exponentially increasing impulse response.
The system is unstable in this case (b).
The relationship between the locations of poles and the impulse response in a stable system. (a)
A pole in the left half of the s-plane corresponds to a right-sided impulse response. (b) A pole in
the right half of the s-plane corresponds to an left-sided impulse response. In this case, the
system is noncausal.
A system that is both stable and causal must have a transfer function
with all of its poles in the left half of the s-plane, as shown here.
THANK
YOU!
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