introduction to laplace transforms
DESCRIPTION
Introduction to Laplace Transforms. Def inition of the Laplace Transform. One-sided Laplace transform :. 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 - PowerPoint PPT PresentationTRANSCRIPT
DefDefinition of the Laplace inition of the Laplace TransformTransform
Some functions may not have Laplacetransforms but we do not use them in circuit analysis. Choose 0- as the lower limit (to capturediscontinuity in f(t) due to an event such asclosing a switch).
0
( ) ( ) ( ) stF s L f t f t e dt
One-sided Laplace transform:
Laplace Transform of the Step Laplace Transform of the Step FunctionFunction
0 0
0
( ) ( )
1 1 -
s
st st
st
L u t u t e dt e dt
es
The sifting property:
The impulse function is a derivative of the step function:
( ) ( ) ( )f t t a dt f a
( )( ) ( ) ( )
tdu tt x dx u t
dt
Laplace Laplace TransformTransform Features Features1) Multiplication by a constant
( ) ( ) ( ) ( ) L f t F s L Kf t KF s 2) Addition (subtraction)
1 1 2 2
1 2 1 2
( ) ( ), ( ) ( ),
( ) ( ) ( ) ( )
L f t F s L f t F s
L f t f t F s F s
3) Differentiation
22
2
1 2 ( 2) ( 1)
( ) ( )( ) (0 ), ( ) (0 ) (0 )
( )( ) (0 ) (0 ) (0 ) (0 )
nn n n n n
n
df t d f tL sF s f L s F s sf f
dt dt
d f tL s F s s f s f sf f
dt
3) Differentiation
22
2
1 2 ( 2) ( 1)
( )( ) (0 ),
( )( ) (0 ) (0 )
.
.
.
( )( ) (0 ) (0 ) (0 ) (0 )
nn n n n n
n
df tL sF s f
dt
d f tL s F s sf f
dt
d f tL s F s s f s f sf f
dt
Laplace Laplace TransformTransform Features (cont.) Features (cont.)
Laplace Laplace TransformTransform Features Features (cont.)(cont.)
4) Integration
0
( )( )
t F sL f x dx
s
5) Translation in the Time Domain
( ) ( ) ( ), a 0asL f t a u t a e F s
Laplace Laplace TransformTransform Features Features (cont.)(cont.)
6) Translation in the Frequency Domain
( ) ( )atL e f t F s a 7) Scale Changing
1( ) , 0
sL f at F a
a a
Laplace Transform of Cos and Laplace Transform of Cos and SineSine
0 0
( ) ( )0 0 0
( ) ( )
0 0
2 2
cos . ( ) cos . ( ) cos
1 [ ]
2 2
1 1 1 1
2 2
1 1 1 1
2 2
st st
j t j tst s j t s j t
s j t s j t
L t u t t u t e dt te dt
e ee dt e dt e dt
e es j s j
s
s j s j s
2 2sin . ( )L t u t
s
NameName Time function Time function f(t)f(t)
Laplace Laplace TransformTransform
Unit ImpulseUnit Impulse (t)(t) 11Unit StepUnit Step u(t)u(t) 1/s1/sUnit rampUnit ramp tt 1/s1/s22
nth-Order rampnth-Order ramp t t nn n!/sn!/sn+1n+1
ExponentialExponential ee-at-at 1/(s+a)1/(s+a)nth-Order nth-Order exponentialexponential
t t nn e e-at-at n!/(s+a)n!/(s+a)n+1n+1
SineSine sin(sin(bt)bt) b/(sb/(s22+b+b22))CosineCosine cos(cos(btbt)) s/(ss/(s22+b+b22))Damped sineDamped sine ee-at -at sin(sin(bt)bt) b/((s+a)b/((s+a)22+b+b22))Damped cosineDamped cosine ee-at -at cos(cos(bt)bt) (s+a)/(s+a)/
((s+a)((s+a)22+b+b22))Diverging sineDiverging sine tt sin( sin(bt)bt) 2bs/(s2bs/(s22+b+b22))22
Diverging cosineDiverging cosine tt cos( cos(bt)bt) (s(s22-b-b22)) /(s/(s22+b+b22))22
Partial Fraction ExpansionPartial 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
31 2 42 2
6
3 1( 3)( 1) ( 1)
KK K Ks
s s ss s s s
2
6
( 3)( 1)
s
s s s
12
31 2 3 4
6
( 3)( 1)
( )t t t
sL
s s s
K K e K te K e u t
Example:Example:
Distinct Real RootsDistinct Real Roots
31 2
1 00
2 88
3 6
96( 5)( 12)( )
( 8)( 6) 8 6
96( 5)( 12) 96(5)(12)( ) 120
( 8)( 6) 8(6)
96( 5)( 12) 96( 3)(4)( 8) ( ) ( 8) 72
( 8)( 6) 8( 2)
96(( 6) ( ) ( 6)
ss
ss
s
KK Ks sF s
s s s s s s
s sK sF s s
s s s
s sK s F s s
s s s
sK s F s s
6
6 8
5)( 12) 96( 1)(6)48
( 8)( 6) 6(2)
120 48 72( )
6 8
( ) (120 48 72 ) ( )
s
t t
s
s s s
F ss s s
f t e e u t
Distinct Complex RootsDistinct Complex Roots
0
2
31 2
1 26
23 4
53.13
3
100( 3) 100( 3)( )
( 6)( 3 4)( 3 4)( 6)( 6 25)
K
s 6 3 4 3 4
100( 3) 100( 3)12
256 25
100( 3) 100( 4)
( 6)( 3 4) (3 4)( 8)
6 8 10
1
s
s j
j
s sF s
s s j s js s s
KK
s j s j
sK
s s
s jK
s s j j j
j e
K
0
3 4
53.13
00( 3) 100( 4)
( 6)( 3 4) (3 4)( 8)
6 8 10
s j
j
s j
s s j j j
j e
Distinct Complex RootsDistinct Complex Roots (cont.)(cont.)
0 0
0 0
0 0
0 0
6 53.13 (3 4) 53.13 (3 4)
53.13 (3 4) 53.13 (3 4)
3 (4 53.13 ) (4 53.13 )
3
12 10 53.13 10 53.13( )
6 3 4 3 4
( ) ( 12 10 10 ) ( )
10 10
10 ( )
20 cos(4 53.1
t j j t j j t
j j t j j t
t j t j t
t
F ss s j s j
f t e e e e e u t
e e e e
e e e
e t
0
6 3 0
3 )
( ) [ 12 20 cos(4 53.13 )] ( )t tf t e e t 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| 0and K* is the complex conjugate of K.
The inverse Laplace transform of the complex-conjugate pair is *
1
2 cos( )t
K KL
s j s j
K e t
Repeated Real RootsRepeated Real Roots31 2 4
3 3 2
1 30
25
33 25
5
23
4 2 5
100( 25)( )
5( 5) ( 5) ( 5)
100( 25) 100(25)20
125( 5)
100( 25) 100(20)400
5
( 25)( 5) ( ) 100 100
1 1( 5) ( )
2 2
s
s
ss
s
KK K KsF s
s ss s s s
sK
s
sK
s
d s sK s F s
ds s
dK s F s
ds
45
3 2
2 5 5 5
2 (25)100 20
20 400 100 20( )
5( 5) ( 5)
( ) [20 200 100 20 ] ( )
s
t t t
s
s
F ss ss s
f t t e te e u t
Repeated Complex RootsRepeated Complex Roots
2 2 2 2
* *1 1 2 2
2 2
1 2 23 4
2 2 33 4 3 4
3
768 768( )
( 6 25) ( 3 4) ( 3 4)
( 3 4) ( 3 4)( 3 4) ( 3 4)
768 76812
( 3 4) ( 8)
768 2(768)
( 3 4) ( 3 4)
2(768) 3
( 8)
s j
s j s j
F ss s s j s j
K K K K
s j s js j s j
Ks j j
dK
ds s j s j
jj
03 -90
2 2
0 0
3 3 0
12 12( )
( 3 4) ( 3 4)
3 -90 3 90
3 4 3 4
( ) [ 24 cos4 6 cos(4 90 )] ( )t t
F ss j s j
s j s j
f t te t e t u t
Improper Transfer FunctionsImproper Transfer FunctionsAn improper transfer function can always be expanded into a polynomial plus a proper transfer function.
4 3 2
2
22
2
24 5
2
13 66 200 300( )
9 2030 100
4 109 20
20 50 4 10
4 5
( ) ( )( ) 4 10 ( ) ( 20 50 ) ( )t t
s s s sF s
s ss
s ss s
s ss s
d t d tf t t e e u t
dtdt
POLES AND ZEROS OF F(s)POLES AND ZEROS OF F(s)
The rational function F(s) may be expressed as the ratio of two factored polynomials as
1 2
1 2
( )( ) ( )( )
( )( ) ( )n
m
K s z s z s zF s
s p s p s p
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( 5)( 3 4)( 3 4)( )
( 10)( 6 8)( 6 8)
s s j s jF s
s s s j s j
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 TheoremInitial-Value TheoremThe 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.
0lim ( ) lim ( )t s
f t sF s
2
6 3 0
100( 3)( )
( 6)( 6 25)
( ) [ 12 20 cos(4 53.13 )] ( )t t
sF s
s s s
f t e e t u t
Final-Value TheoremFinal-Value Theorem
0lim ( ) lim ( )t s
f t sF s
The final-value theorem enables us to determine the behavior of f(t) at infinity using F(s).
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.
20 0
6 3 0
100( 3)lim ( ) lim 0
( 6)( 6 25)
lim ( ) lim[ 12 20 cos(4 53.13 )] ( ) 0
s s
t t
t t
sF s s
s s s
f t e e t u t