Chapter 7 Laplace Transforms
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Transcript Chapter 7 Laplace Transforms
Chapter 7
Laplace Transforms
Applications of Laplace Transform
• Easier than solving differential equations
– Used to describe system behavior
– We assume LTI systems
– Uses S-domain instead of frequency domain
• Applications of Laplace Transforms/
– Circuit analysis
• Easier than solving differential equations
• Provides the general solution to any arbitrary wave (not just LRC)
– Transient
– Sinusoidal steady-state-response (Phasors)
– Signal processing
– Communications
• Definitely useful for Interviews!
notes
Building the Case…
http://web.cecs.pdx.edu/~ece2xx/ECE222/Slides/LaplaceTransformx4.pdf
Laplace Transform
Laplace Transform
• We use the following notations for Laplace Transform pairs
– Refer to the table!
Laplace Transform Convergence
• The Laplace transform does not converge to a finite value for all signals
and all values of s
• The values of s for which Laplace transform converges is called the
Region Of Convergence (ROC)
• Always include ROC in your solution!
• Example:
f (t ) e at u (t );
0+ indicates greater than
zero values
F ( s ) f (t )e st dt
1
e at e st dt
e( s a )t
0
sa
1 ( j a )t
e
sa
0
1
; Re( s ) a
sa
; note : s j
0
1 ( a )t jt
e
e
sa
Re( s a) 0
0
Remember: e^jw is
sinusoidal; Thus, only the
real part is important!
Example of Bilateral Version
f (t ) e at u (t );
Find F(s):
ROC
F (s)
S-plane
0
e at e st dt
Re(s)<a
a
Find F(s):
f (t )e st dt
1
e( s a )t
sa
0
1
; Re( s a ) 0
sa
1
1
; Re( s ) a
; Re( s ) a
sa
sa
f (t ) e at u (t );
F ( s ) f (t )e st dt
Remember
These!
0
e at e st dt
1
e ( s a )t
sa
0
1
; Re( s a ) 0
sa
1
1
; Re( s ) a
; Re( s ) a
sa
sa
Note that Laplace can also be found for periodic functions
Example – RCO may not always exist!
f (t ) e 2t u (t ) e 3t u (t )
F ( s ) f (t )e st dt
1
e u (t )
; Re( s ) 2
s2
1
3t
e u (t )
; Re( s ) 3
s3
1
1
F (s)
;
s2 s3
2t
Note that there is no common ROC Laplace Transform can not be applied!
Example – Unilateral Version
• Find F(s):
f (t ) e at u (t ); a 0
• Find F(s):
F ( s) f (t )e st dt
F ( s) f (t )e dt
st
0
0
e st (t t 0 )dt e st0
e at e st dt
0
0
• Find F(s):
1
; Re( s a) 0 Re( s) a
sa
f (t ) e at ; a 0
• Find F(s):
F ( s ) f (t )e st dt
0
e e dt
at st
0
1
[0 1]
sa
1
; Re( s a) 0 Re( s ) a
sa
f (t ) (t t 0 )
f (t ) (t )
F ( s) 1; s
f (t ) u (t )
F ( s ) f (t )e st dt
0
1
e st u (t )dt [lim t e st 1]
0
s
1
[lim t e ( j ) t 1]
s
1
; Re( s ) 0
s
Example
f (t ) cos(bt )
f t 1 / 2e
jbt
1 / 2e
jbt
F ( s ) f (t )e st dt
0
1
e at
; Re( s ) a
sa
1/ 2
1/ 2
1 / 2e jbt
;1 / 2e jbt
; Re( s ) 0
s jb
s jb
1/ 2
1/ 2
s
F ( s)
2
; Re( s ) 0
s jb s jb s b 2
f (t ) sin( bt )
f t 1 / 2 je jbt 1 / 2 je jbt
F ( s ) f (t )e st dt
0
1
; Re( s ) a
sa
1/ 2 j
1/ 2 j
1 / 2 je jbt
;1 / 2e jbt
; Re( s ) 0
s jb
s jb
1/ 2 j 1/ 2 j
b
F ( s)
2
; Re( s ) 0
2
s jb s jb s b
e at
Example
f (t ) e at cos(bt )
f t 1 / 2e jbte at 1 / 2e jbte at
F ( s ) f (t )e st dt
0
1
e
; Re( s ) a
sa
1
1
1
F (s)
2 s (a jb) s (a jb)
sa
; Re( s a ) 0
2
2
(s a) b
at
Properties
• The Laplace Transform has many difference properties
• Refer to the table for these properties
Linearity
Scaling & Time Translation
Scaling
Do the time translation first!
e sb / a
f (at b)u(at b)
F (s / a)
a
Time Translation
b=0
Shifting and Time Differentiation
Shifting in s-domain
Differentiation in t
Read the rest of
properties on your
own!
Examples
f (t ) 5e 0.3t
1
e at
; Re( s) a
sa
1
F (s) 5
; Re( s) 0.3
s 0.3
f (t ) 5e 0.3( t 2 )u(t 2)
1
; with _ time _ shift
sa
1
F ( s) 5
e 2 s ; Re( s ) 0.3
s 0.3
e at
Note the ROC did not change!
f (t ) 5e 0.3(t )u (t 2)
1
e at
; with _ time _ shift
sa
f (t ) 5e 0.3(t )u (t 2){e 2 ( 0.3) e 2( 0.3) }
5e 2 ( 0.3) {e 0.3(t ) e 2( 0.3)u (t 2)}
5e 2 ( 0.3) {e 0.3(t 2 )u (t 2)}
e 2( 0.3) 2 s 2.744 2 s
F ( s) 5
e
e ; Re( s ) 0.3
s 0.3
s 0.3
Example – Application of Differentiation
g (t ) tf (t )
{tf (t )} ?
G ( s ) tf (t )e st dt
0
f (t )e st dt
ds 0
G (s) F (s)
ds
g (t ) t cos(bt )
{t cos(bt )} ?
G ( s) F ( s)
ds
s
{ 2
}; Re( s ) 0
2
ds s b
s 2 b2
; Re( s ) 0
2
2 2
s b
Read Section 7.4
Read about Symbolic Mathematics:
http://www.math.duke.edu/education/ccp/materials/diffeq/mlabtutor/mlabtut7.html
And
http://www.mathworks.de/access/helpdesk/help/toolbox/symbolic/ilaplace.html
Matlab Code:
Example
e sb / a
f (at b)u(at b)
F (s / a)
a
• What is Laplace of t^3?
– From the table: 3!/s^4 Re(s)>0
Time transformation
• Find the Laplace Transform:
g (t ) sin( 12t / 2)u (4t / 6)
g (t ) sin{ 3(4t / 6)}u (4t / 6) sin( 3 )u ( ) 4t / 6
Time _ Translatio n :
a 4; b / 6
e s / 24
G( s)
F ( s / a);
4
f ( ) sin( 3 )u ( ) F ( s )
3
s2 9
Note that without u(.) there will
be no time translation and thus,
the result will be different:
e s / 24
3
G( s)
;
4 ( s / 4) 2 9
Assume t>0
Given Laplace find f(t)!
A little about Polynomials
• Consider a polynomial function:
• A rational function is the ratio of two polynomials:
Has roots and zeros; distinct roots, repeated
roots, complex roots, etc.
• A rational function can be expressed as partial fractions
• A rational function can be expressed using polynomials
presented in product-of-sums
Finding Partial Fraction Expansion
• Given a polynomial
• Find the POS
(product-of-sums) for the denominator: G ( s) N ( s)
D( s)
• Write the
partial fraction expression
for the polynomial
G( s)
N ( s)
( s p1 )( s p2 )( s p3 )...
k1
k2
k3
....
( s p1 ) ( s p2 ) ( s p3 )
k1 [( s p1 ) I ( s )]
• Find the constants
– If the rational polynomial has
distinct poles then we can use the
following to find the constants:
s p1
k2 [( s p2 ) I ( s )]
s p2
k2 [( s p3 ) I ( s )]
s p3
......
http://cnx.org/content/m2111/latest/
Application of Laplace
• Consider an RL circuit with R=4, L=1/2. Find i(t) if
v(t)=12u(t).
di (t )
Ri (t ) v (t )
dt
di (t )
0.5
4i (t ) v (t )
dt
U sin g _ Laplace :
(0.5s 4) I ( s ) V ( s )
L
1
I ( s ) H ( s ).V ( s )
0 .5 s 4
Given
v (t ) 12u (t ) V ( s ) 12 / s
Partial fraction expression
24
k
k
I ( s ) H ( s ).V ( s )
1 2
s( s 8) s s 8
H ( s) I ( s) / V ( s)
k1 [( s p1 ) I ( s )]
k2 [( s p2 ) I ( s )]
I ( s)
p1 0
3
p 2 8
3
3 3
i (t ) 3u (t ) 3e 8t ; t 0
s s8
Matlab
Code
Application of Laplace
• What are the initial [i(0)] and final
values:
i (0 ) lim s sI ( s ) lim s
24
0
s 8
lim s i (t ) lim s 0 sI ( s ) lim s 0
24
3
s 8
– Using initial-value property:
– Using the final-value property
Note: using Laplace Properties
Initial _ value : f (0 ) lim f (t ) lim sF ( s )
t 0
t
Final _ value : f () lim f (t ) lim sF ( s )
t
t 0
Note that
Initial Value: t=0, then, i(t) 3-3=0
Final Value: t INF then, i(t) 3
i(t ) 3u(t ) 3e8t ; t 0
Using Simulink
v(t
)
H(s)
i(t)
Actual Experimentation
• Note how the voltage looks like:
di (t )
Ri (t ) v(t )
dt
di (t )
0.5
4i (t ) v(t )
dt
i (t ) 3u (t ) 3e 8t ; t 0
L
v(t ) 0.5
di (t )
12e 8t ; t 0
dt
Output Voltage:
Input Voltage:
Partial Fraction Expansion
(no repeated Poles/Roots) – Example
• Using Matlab:
• Matlab code:
b=[8 3 -21];
a=[1 0 -7 -6];
[r,p,k]=residue(b,a)
We can also use
ilaplace (F); but the
result may not be
simplified!
Finding Poles and Zeros
• Express the rational function as
the ratio of two polynomials each
represented by product-of-sums
• Example:
F ( s)
4s 8
2( s 2)
2 s 2 8s 6
( s 1)( s 3)
Pole
S-plane
zero
H(s) Replacing the Impulse Response
x(t)
h(t)
convolution
y(t)
X(s)
H(s)
multiplication
Y(s)
H(s) Replacing the Impulse Response
x(t)
h(t)
y(t)
X(s)
multiplication
convolution
Example: Find the output
X(t)=u(t); h(t)
H(s)
Y(s)
h(t)
1
h(t ) u (t ) u (t 1)
0
1
y(t)
1
0
1
1 es
H ( s)
s s
1
X (s)
s
e^-sF(s)
1 es 1 es
Y ( s ) H ( s ). X ( s )
2 2
s2
s
s
y (t ) tu(t ) (t 1)u (t 1)
This is commonly used in D/A converters!
Dealing with Complex Poles
• Given a polynomial
• Find the POS (product-of-sums)
for the denominator:
• Write the partial fraction
expression for the polynomial
• Find the constants
– The pole will have a real and
imaginary part: P=|k|f
• When we have complex poles
{|k|f} then we can use the
following expression to find the
time domain expression:
http://cnx.org/content/m2111/latest/
a Re( P); b Im( P); angle _ of _ P f
f (t ) 2 | k | e at cos(bt f )
Laplace Transform Characteristics
• Assumptions: Linear Continuous Time
Invariant Systems
• Causality
– No future dependency
– If unilateral: No value for t<0; h(t)=0
• Stability
– System mode: stable or unstable
– We can tell by finding the system
characteristic equation (denominator)
We need to add control
mechanism to make the
overall system stable
H ( s)
1
1
s 2 4 ( s 2)( s 2)
h(t ) Ae2t u (t ) Be 2t u (t )
• Stable if all the poles are on the left
plane
– Bounded-input-bounded-output (BIBO)
• Invertability
– H(s).Hi(s)=1
• Frequency Response
– H(w)=H(s);sjw=H(s=jw)
3s 1
s 2 2s 5
3 j 1
H ( )
2 2 j 5
H (s)
Frequency Response – Matlab Code
3s 1
s 2 2s 5
3 j 1
H ( )
2 2 j 5
H (s)
Inverse Laplace Transform
Example of Inverse Laplace Transform
Bilateral Transforms
• Laplace Transform of two
different signals can be the
same, however, their ROC can
1
x(t ) e at u (t )
;a Re( s ) Right sided
be different:
sa
• Very important to know the
1
x(t ) e at u (t )
;a Re( s ) Left sided
ROC.
sa
• Signals can be
– Right-sided Use the bilateral
Laplace Transform Table
– Left-sides
– Have finite duration
• How to find the transform of
signals that are bilateral!
See notes
How to Find Bilateral Transforms
• If right-sided use the table for unilateral Laplace Transform
• Given f(t) left-sided; find F(s):
– Find the unilateral Laplace transform for f(-t) laplace{f(-t)}; Re(s)>a
– Then, find F(-s) with Re(-s)>a
• Given Fb(s) find f(t) left-sided :
– Find the unilateral Inverse Laplace transform for F(s)=fb(t)
– The result will be f(t)=–fb(t)u(-t)
• Example
x(t ) 2e 5t u (t ) e 4t u (t )
2
;5 Re( s ) Right sided
s5
To _ Find :e 4t u (t )
Assume : f (t )
2e 5t u (t )
1
;4 Re( s ) Right sided
s4
1
F ( s)
;4 Re( s ) 4 Re( s ) Left sided
s4
2
1
F (s)
s5 s4
e 4t u (t )
Examples of Bilateral Laplace Transform
x(t ) 2e 5t u (t ) e 4t u (t )
2
1
;5 Re( s ) 4
s5 s4
x(t ) 2e 5t u (t ) e 4t u (t )
2
1
;4 Re( s ) & 5 Re( s )
s5 s4
x(t ) 2e 5t u (t ) e 4t u (t )
2
1
;4 Re( s );5 Re( s )
s5 s4
Find the unilateral Laplace transform for f(-t) laplace{f(-t)}; Re(s)>a
Then find F(-s) with Re(-s)>a
Alternatively: Find the unilateral Laplace transform for f(t)u(-t)
(-1)laplace{f(t)}; then, change the inequality for ROC.
Feedback System
Find the system function for the
following feedback system:
X ( s) R( s) E ( s) Y ( s) / F ( s)
R( s) Y ( s).G ( s)
X(t)
+
Sum
e(t)
y(t)
F(s)
+
r(t)
X ( s) (Y ( s).G ( s)) Y ( s) / F ( s)
G(s)
F (s)
Y ( s) / X ( s) H ( s)
1 F ( s).G ( s)
Equivalent System
X(t)
Feedback Applet:
http://physioweb.uvm.edu/homeostasis/simple.htm
H(s)
y(t)
Practices Problems
• Schaum’s Outlines Chapter 3
–
–
–
–
3.1, 3.3, 3.5, 3.6, 3.7-3.16, For Quiz!
3.17-3.23
Read section 7.8
Read examples 7.15 and 7.16
Useful Applet: http://jhu.edu/signals/explore/index.html