3-laplace_transform
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Transcript 3-laplace_transform
Introduction to Laplace
Transforms
Definition of the Laplace Transform
One-sided Laplace transform:
F ( s) L f (t ) 0 f (t )e
st
l Some functions may not have Laplace
transforms but we do not use them in circuit
analysis.
Choose 0- as the lower limit (to capture
discontinuity in f(t) due to an event such as
closing a switch).
dt
The Step Function
?
Laplace Transform of the Step Function
Lu( t ) 0 u( t )e
st
1 st
- e
s
0
st
e
dt
0
dt
1
s
The Impulse Function
The sifting property:
f (t ) (t a )dt f (a )
The impulse function is a derivative of the step function:
du(t )
(t )
dt
t
( x )dx u(t )
?
Laplace transform of the impulse function:
L ( t a ) 0 ( t a )e
For a 0,
L ( t ) 1
st
dt e
as
Laplace Transform Features
1) Multiplication by a constant
L f ( t ) F ( s )
L Kf ( t ) KF ( s )
2) Addition (subtraction)
L f1 ( t ) F1 ( s ),
L f 2 ( t ) F2 ( s ),
L f1 ( t ) f 2 ( t ) F1 ( s ) F2 ( s )
3) Differentiation
df ( t )
L
sF ( s ) f (0 ),
dt
d 2 f ( t ) 2
L
s
F
(
s
)
sf
(0
)
f
(0
)
2
dt
d n f ( t ) n
n 1
n 2
L
s
F
(
s
)
s
f
(0
)
s
f
(0
)
n
dt
sf ( n 2) (0 ) f ( n 1) (0
Laplace Transform Features (cont.)
3) Differentiation
df ( t )
L
sF ( s ) f (0 ),
dt
d 2 f ( t ) 2
)
(0
f
)
(0
sf
)
s
(
F
s
L
2
dt
.
.
.
d n f ( t ) n
n 2
n 1
)
(0
f
s
)
(0
f
s
)
s
(
F
s
L
n
dt
sf ( n 2) (0 ) f ( n 1) (0 )
Laplace Transform Features (cont.)
4) Integration
L
t
0
F ( s)
f ( x )dx
s
5) Translation in the Time Domain
L f (t a )u(t a ) e
as
F ( s ),
a
0
Laplace Transform Features (cont.)
6) Translation in the Frequency Domain
L e
at
f (t ) F ( s a )
7) Scale Changing
1 s
L f (at ) F ,
a a
a
0
Laplace Transform of Cos and Sine
?
Laplace Transform of Cos and Sine
Lcos t .u( t ) 0 cos t .u( t )e
st
e
0
j t
dt 0 cos te st dt
e j t st
1 ( s j )t
( s j ) t
e dt [ 0 e
dt 0 e
dt ]
2
2
1 1
e ( s j )t
2 s j
0
1 1
e ( s j )t
2 s j
1 1
1 1
s
2
2 s j 2 s j s 2
Lsin t .u(t ) 2
2
s
0
Name
Time function f(t) Laplace Transform
Unit Impulse
(t)
u(t)
t
tn
e-at
t n e-at
sin(bt)
cos(bt)
e-at sin(bt)
e-at cos(bt)
t sin(bt)
t cos(bt)
Unit Step
Unit ramp
nth-Order ramp
Exponential
nth-Order exponential
Sine
Cosine
Damped sine
Damped cosine
Diverging sine
Diverging cosine
1
1/s
1/s2
n!/sn+1
1/(s+a)
n!/(s+a)n+1
b/(s2+b2)
s/(s2+b2)
b/((s+a)2+b2)
(s+a)/((s+a)2+b2)
2bs/(s2+b2)2
(s2-b2) /(s2+b2)2
Some
Laplace
Transform
Properties
(t)
F(s)
A (t)
A F(s)
1(t) ± 2(t)
F1(s) ± F2(s)
Property
Scaling
Linearity
1 s
F
a a
a0
Time Scaling
(a·t)
Time Shifting
(delay)
(t–t0) u(t–t0)
e–s·t0 F(s) t00
Frequency Shifting
e–a·t (t)
F(s+a)
Time Domain
Differentiation
d f (t )
dt
s F(s) – (0)
Frequency Domain
Differentiation
t (t)
t
0
Time Domain
Integration
t
Convolution
0
f ( ) d
f1 ( ) f 2 ( t ) d
d F ( s)
ds
1
F ( s)
s
F1(s) F2(s)
Inverse Laplace Transform
Partial Fraction Expansion
Step1: Expand F(s) as a sum of partial fractions.
Step 2: Compute the expansion constants (four
cases)
Step 3: Write the inverse transform
Example:
s6
2
s( s 3)( s 1)
K3
K1
K2
K4
s6
2
2
s
s
3
s1
s( s 3)( s 1)
( s 1)
1
s6
L
2
s( s 3)( s 1)
K1 K 2e 3 t K 3te t K 4e t u( t )
Distinct Real Roots
K3
K2
96( s 5)( s 12) K1
F ( s)
s( s 8)( s 6)
s
s8 s6
96( s 5)( s 12)
96(5)(12)
K1 sF ( s ) s 0 s
120
s( s 8)( s 6) s 0
8(6)
96( s 5)( s 12)
96( 3)(4)
K 2 ( s 8)F ( s ) s 8 ( s 8)
72
s( s 8)( s 6) s 8
8( 2)
K 3 ( s 6)F ( s ) s 6 ( s 6)
96( s 5)( s 12)
96( 1)(6)
48
s( s 8)( s 6) s 6
6(2)
120
48
72
F ( s)
s
s6 s8
f ( t ) (120 48e 6 t 72e 8 t )u( t )
Distinct Complex Roots
100( s 3)
100( s 3)
F ( s)
2
( s 6)( s 6 s 25) ( s 6)( s 3 j 4)( s 3 j 4)
K3
K1
K2
s 6 s 3 j4 s 3 j4
100( s 3)
100( 3)
K1 2
12
25
s 6 s 25 s 6
K2
100( s 3)
100( j 4)
( s 6)( s 3 j 4) s 3 j 4 (3 j 4)( j 8)
6 j 8 10e
j 53.130
100( s 3)
100( j 4)
K3
( s 6)( s 3 j 4) s 3 j 4 (3 j 4)( j 8)
6 j 8 10e
j 53.130
Distinct Complex Roots
(cont.)
12 10 53.130 10 53.130
F ( s)
s6
s 3 j4
s 3 j4
f ( t ) ( 12e
10e
6 t
10e
j 53.130 (3 j 4) t
10e
e
3 t
(e
j 53.130 (3 j 4) t
e
10e
j (4 t 53.130 )
e
10e
j 53.130 (3 j 4) t
j 53.130 (3 j 4) t
e
j (4 t 53.130 )
)
20e 3 t cos(4t 53.130 )
f ( t ) [12e 6 t 20e 3 t cos(4t 53.130 )]u( t )
e
)u( t )
Whenever F(s) contains distinct complex roots at the denominator
as (s+-j)(s++j), a pair of terms of the form
K
K*
s j s j
appears in the partial fraction.
Where K is a complex number in polar form K=|K|ej=|K| 0
and K* is the complex conjugate of K.
The inverse Laplace transform of the complex-conjugate pair is
*
K
K
1
L
s j s j
2K e
t
cos( t )
Repeated Real Roots
K3
K2
K4
100( s 25) K 1
F ( s)
3
3
2
s ( s 5)
s5
s( s 5)
( s 5)
100( s 25)
100(25)
K1
20
3
125
( s 5) s 0
100( s 25)
100(20)
K2
400
s
5
s 5
d
s ( s 25)
3
K 3 ( s 5) F ( s )
100
100
2
s 5
ds
s
s 5
1 d2
1
2 s(25)
3
K4
( s 5) F ( s )
100
20
2
4
s
5
2 ds
2
s
s 5
20
400
100
20
F ( s)
3
2
s ( s 5)
s5
( s 5)
f ( t ) [20 200t 2e 5 t 100te 5 t 20e 5 t ]u( t )
Repeated Complex Roots
768
768
F ( s) 2
2
( s 6 s 25)
( s 3 j 4)2 ( s 3 j 4)2
K1
K1*
K2
K 2*
2
2
( s 3 j 4) ( s 3 j 4)
( s 3 j 4)
( s 3 j 4)
768
K1
( s 3 j 4)2
768
12
2
( j 8)
s 3 j 4
d
768
2(768)
K2
2
ds ( s 3 j 4) s 3 j 4
( s 3 j 4)3
2(768)
0
j
3
3
-90
( j 8)3
s 3 j 4
12
12
F ( s)
2
2
( s 3 j 4)
( s 3 j 4)
3 -900
3 900
s 3 j4 s 3 j4
f ( t ) [ 24te
3 t
cos 4t 6e
3 t
cos(4t 90 )]u( t )
0
Improper Transfer Functions
An improper transfer function can always be expanded into a
polynomial plus a proper transfer function.
s 4 13 s 3 66 s 2 200 s 300
F ( s)
s 2 9 s 20
30 s 100
2
s 4 s 10 2
s 9 s 20
20
50
2
s 4 s 10
s4 s5
d (t )
d ( t )
4 t
5 t
f (t )
4
10
(
t
)
(
20
e
50
e
)u( t )
2
dt
dt
2
POLES AND ZEROS OF F(s)
The rational function F(s) may be expressed as
the ratio of two factored polynomials as
K ( s z1 )( s z2 ) ( s zn )
F ( s)
( s p1 )( s p2 ) ( s pm )
The roots of the denominator polynomial –p1, -p2,
..., -pm are called the poles of F(s). At these values
of s, F(s) becomes infinitely large. The roots of the
numerator polynomial -z1, -z2, ..., -zn are called the
zeros of F(s). At these values of s, F(s) becomes
zero.
10( s 5)( s 3 j 4)( s 3 j 4)
F ( s)
s( s 10)( s 6 j 8)( s 6 j 8)
The poles of F(s) are at
0, -10, -6+j8, and –6-j8.
The zeros of F(s)
are at –5, -3+j4, -3-j4
num=conv([1 5],[1 6 25]);
den=conv([1 10 0],[1 12 100]);
pzmap(num,den)
Initial-Value Theorem
The initial-value theorem enables us to determine the
value of f(t) at t=0 from F(s). This theorem assumes that
f(t) contains no impulse functions and poles of F(s),
except for a first-order pole at the origin, lie in the left
half of the s plane.
lim f (t ) lim sF ( s)
t 0
s
100( s 3)
F ( s)
2
( s 6)( s 6 s 25)
f ( t ) [12e 6 t 20e 3 t cos(4t 53.130 )]u( t )
Final-Value Theorem
The final-value theorem enables us to determine the behavior of
f(t) at infinity using F(s).
lim f ( t ) lim sF ( s )
t
s 0
The final-value theorem is useful only if f(∞) exists. This condition is
true only if all the poles of F(s), except for a simple pole at the
origin, lie in the left half of the s plane.
100( s 3)
lim F ( s ) lim s
0
2
s 0
s 0 ( s 6)( s 6 s 25)
lim f ( t ) lim[12e 6t 20e 3t cos(4t 53.130 )]u( t ) 0
t
t