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Laplace Transform. Applications of the Laplace transform solve differential equations (both ordinary and partial) application to RLC circuit analysis Laplace transform converts differential equations in the time domain to algebraic equations in the frequency domain, thus 3 important processes: - PowerPoint PPT Presentation
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Laplace Transform
• Applications of the Laplace transform– solve differential equations (both ordinary and partial)
– application to RLC circuit analysis
• Laplace transform converts differential equations in the time domain to algebraic equations in the frequency domain, thus 3 important processes:(1) transformation from the time to frequency domain
(2) manipulate the algebraic equations to form a solution
(3) inverse transformation from the frequency to time domain
Definition of Laplace Transform
• Definition of the unilateral (one-sided) Laplace transform
where s=+j is the complex frequency, and f(t)=0 for t<0
• The inverse Laplace transform requires a course in complex variables analysis (e.g., MAT 461)
0
dtetfstf stFL
Singularity Functions
• Singularity functions are either not finite or don't have finite derivatives everywhere
• The two singularity functions of interest here are
(1) unit step function, u(t)(2) delta or unit impulse function, (t)
Unit Step Function, u(t)
• The unit step function, u(t)– Mathematical definition
– Graphical illustration
01
00)(
t
ttu
1
t0
u(t)
Extensions of the Unit Step Function• A more general unit step function is u(t-a)
• The gate function can be constructed from u(t)– a rectangular pulse that starts at t= and ends at t= +T
– like an on/off switch
at
atatu
1
0)(
1
t0 a
1
t0 +T
u(t-) - u(t- -T)
Delta or Unit Impulse Function, (t)
• The delta or unit impulse function, (t)– Mathematical definition (non-pure version)
– Graphical illustration
0
00 1
0)(
tt
tttt
1
t0
(t)
t0
Transform Pairs
The Laplace transforms pairs in Table 13.1 are important, and the most important are repeated here.
(t) F (s )
δ (t) 1
u (t) {a co ns ta n t}
s
1
e -a t
as 1
t2
1
s
t e -a t
2
1
as
Laplace Transform PropertiesT h e o r e m P r o p e r t y ( t ) F ( s )
1 S c a l i n g A ( t ) A F ( s )
2 L i n e a r i t y 1 ( t ) ±
2 ( t ) F 1 ( s ) ± F 2 ( s )
3 T i m e S c a l i n g ( a · t ) 01
aa
s
aF
4 T i m e S h i f t i n g ( t - t 0 ) u ( t - t 0 ) e - s · t 0 F ( s ) t 0 0
6 F r e q u e n c y S h i f t i n g e - a · t ( t ) F ( s + a )
9 T i m e D o m a i nD i f f e r e n t i a t i o n dt
tfd )( s F ( s ) - ( 0 )
7 F r e q u e n c y D o m a i nD i f f e r e n t i a t i o n
t ( t )ds
sd )(F
1 0 T i m e D o m a i nI n t e g r a t i o n
tdf
0)( )(
1s
sF
1 1 C o n v o l u t i o n t
dtff0 21 )()( F 1 ( s ) F 2 ( s )
Block Diagram Reduction
Block Diagram Reduction
Block Diagram Reduction
Block Diagram Reduction
Reference
Y(s) = ___K*G(s)R(s) 1+K*G*H(s)Closed Loop
KSum
H(s)
Y(s)R(s)-
Gplant
Y is the 'ControlledOutput'
Forward Path
Y(s) = ___K*G(s)R(s) 1+K*G*H(s)Closed Loop
Characteristic Equation:Den(s) = 1+K*GH(s) = 0
Stability: The response y(t) reverts to
zero if input r = 0.All roots (poles of Y/R) must have Re(pi) < 0
Characteristic Equation:Den(s) = 1+K*GH(s) = 0
Closed Loop Poles are theroots of the Characteristic
Equation, i.e. 1+K*GH(s) = 0
Poles and Stability
Poles and Stability
Poles and Stability
Underdamped System (2nd Order)