1. The Laplace transform converts differential equations describing systems from the time domain to the frequency domain by replacing functions of time with functions of a complex variable dependent on frequency.
2. The inverse Laplace transform converts the solution back from the frequency domain to the time domain to obtain the solution in terms of the time variable.
3. Partial fraction expansion is often used to break solutions into simpler terms that can be inverted using Laplace transform tables to find the solution in the time domain.
This presentation gives the basic idea about the methods of solving ODEs
The methods like variation of parameters, undetermined coefficient method, 1/f(D) method, Particular integral and complimentary functions of an ODE
This presentation describes the Fourier Transform used in different mathematical and physical applications.
The presentation is at an Undergraduate in Science (math, physics, engineering) level.
Please send comments and suggestions to improvements to solo.hermelin@gmail.com.
More presentations can be found at my website http://www.solohermelin.com.
its ppt for the laplace transform which part of Advance maths engineering. its contains the main points and one example solved in it and have the application related the chemical engineering
How to handle Initial Value Problems using numerical techniques?
#WikiCourses
https://wikicourses.wikispaces.com/Topic+Initial+Value+Problems
https://eau-esa.wikispaces.com/Topic+Initial+Value+Problems
Z Transform And Inverse Z Transform - Signal And SystemsMr. RahüL YøGi
The z-transform is the most general concept for the transformation of discrete-time series.
The Laplace transform is the more general concept for the transformation of continuous time processes.
For example, the Laplace transform allows you to transform a differential equation, and its corresponding initial and boundary value problems, into a space in which the equation can be solved by ordinary algebra.
The switching of spaces to transform calculus problems into algebraic operations on transforms is called operational calculus. The Laplace and z transforms are the most important methods for this purpose.
This presentation gives the basic idea about the methods of solving ODEs
The methods like variation of parameters, undetermined coefficient method, 1/f(D) method, Particular integral and complimentary functions of an ODE
This presentation describes the Fourier Transform used in different mathematical and physical applications.
The presentation is at an Undergraduate in Science (math, physics, engineering) level.
Please send comments and suggestions to improvements to solo.hermelin@gmail.com.
More presentations can be found at my website http://www.solohermelin.com.
its ppt for the laplace transform which part of Advance maths engineering. its contains the main points and one example solved in it and have the application related the chemical engineering
How to handle Initial Value Problems using numerical techniques?
#WikiCourses
https://wikicourses.wikispaces.com/Topic+Initial+Value+Problems
https://eau-esa.wikispaces.com/Topic+Initial+Value+Problems
Z Transform And Inverse Z Transform - Signal And SystemsMr. RahüL YøGi
The z-transform is the most general concept for the transformation of discrete-time series.
The Laplace transform is the more general concept for the transformation of continuous time processes.
For example, the Laplace transform allows you to transform a differential equation, and its corresponding initial and boundary value problems, into a space in which the equation can be solved by ordinary algebra.
The switching of spaces to transform calculus problems into algebraic operations on transforms is called operational calculus. The Laplace and z transforms are the most important methods for this purpose.
Application of Residue Inversion Formula for Laplace Transform to Initial Val...iosrjce
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2. There are techniques for finding the system response of a system
described by a differential equation, based on the replacement of
functions of a real variable (usually time or distance) by certain
frequency-dependent representations, or by functions of a complex
variable dependent upon frequency. The equations are converted from
the time or space domain to the frequency domain through the use of
mathematical transforms.
4. Let f(t) be a real function of a real variable t (time) defined for t>0. Then
is called the Laplace transform of f(t). The Laplace transform is a
function of a complex variable s. Often s is separated into its real and
imaginary parts: s= σ +jω , where σ and ω are real variables.
5. After a solution of the transformed problem has been obtained in terms
of s, it is necessary to "invert" this transform to obtain the solution in
terms of the time variable, t. This transformation from the s-domain into
the t-domain is called the inverse Laplace transform.
6. Let F(s) be the Laplace transform of a function f(t), t>0. The contour integral
where c> σ 0 (σ 0 as above) is called the inverse Laplace transform of F(s).
7. It is seldom necessary to perform the integration in the Laplace transform
or the contour integration in the inverse Laplace transform. Most often,
Laplace transforms and inverse Laplace transforms are found using tables
of Laplace transform pairs.
8. Time Domain Frequency Domain
f(t), t> 0 F(s)
1. δ 1
2. K K/s
3. Kt K/s2
4. Ke-at K/(s+ a)
5. Kte-at K/(s+ a) 2
6. Ksinωt Kω/(s2+ ω2)
7. Kcosωt Ks/(s2+ ω2)
8. Ke-at sinωt Kω/((s+ a) 2+ ω2))
9. Ke-at cosωt K(s+ a)/ ((s+ a) 2+ ω2))
9. Time Domain Frequency Domain
f(t), t> 0 F(s)
10. t s
11. f(t) F(s)
12. L-1{ F(s)} = f(t) L{ f(t )} = F(s)
13. Af 1(t) + Bf 2(t) AF1(s)+ BF2(s)
14.
15.
10. The inverse Laplace transform is usually more
difficult than a simple table conversion.
8( s + 3)( s + 8)
X ( s) =
s( s + 2)( s + 4)
11. If we can break the right-hand side of the
equation into a sum of terms and each term is in a
table of Laplace transforms, we can get the
inverse transform of the equation (partial fraction
expansion).
8( s + 3)( s + 8) K1 K2 K3
X ( s) = = + +
s( s + 2)( s + 4) s s+2 s+4
12. In general, there will be a term on the right-hand
side for each root of the polynomial in the
denominator of the left-hand side. Multiple roots
for factors such as (s+2)n will have a term for each
power of the factor from 1 to n.
8( s + 1) K1 K2
Y ( s) = = +
( s + 2) 2
s + 2 ( s + 2) 2
13. Complex roots are common, and they always
occur in conjugate pairs. The two constants in
the numerator of the complex conjugate terms
are also complex conjugates.
5.2 K K*
Z ( s) = 2 = +
s + 2 s + 5 ( s + 1 − j 2) ( s + 1 + j 2)
where K* is the complex conjugate of K.
14. The solution of each distinct (non-multiple)
root, real or complex uses a two step process.
The first step in evaluating the constant is to
multiply both sides of the equation by the factor
in the denominator of the constant you wish to
find.
The second step is to replace s on both sides of
the equation by the root of the factor by which
you multiplied in step 1
15. 8( s + 3)( s + 8) K1 K2 K3
X ( s) = = + +
s( s + 2)( s + 4) s s+2 s+4
8( s + 3)( s + 8) 8(0 + 3)(0 + 8)
K1 = = = 24
( s + 2)( s + 4) s=0
(0 + 2)(0 + 4)
8( s + 3)( s + 8) 8( −2 + 3)( −2 + 8)
K2 = = = −12
s( s + 4 ) s =−2
−2( −2 + 4 )
16. 8( s + 3)( s + 8) 8( −4 + 3)( −4 + 8)
K3 = = = −4
s( s + 2 ) s =−4
−4( −4 + 4)
The partial fraction expansion is:
24 12 4
X ( s) = − −
s s+2 s+4
17. The inverse Laplace transform is found from
the functional table pairs to be:
−2 t −4 t
x (t ) = 24 − 12e − 4e
18. Any unrepeated roots are found as before.
The constants of the repeated roots (s-a)m are
found by first breaking the quotient into a
partial fraction expansion with descending
powers from m to 0:
Bm B2 B1
++ +
(s − a) m
(s − a) 2
(s − a)
19. The constants are found using one of the
following:
1 d m −i
P( s)
Bi = m
(m − i )! ds m −i Q( s ) /( s − a1 ) s = a1
P(a )
Bm =
Q( s) / ( s − a ) m
s=a
20. 8( s + 1) K1 K2
Y ( s) = = +
( s + 2) 2
s + 2 ( s + 2) 2
8( s + 1)( s + 2) 2
K2 = = 8( s + 1) s=−2 = −8
( s + 2) 2
s =−2
21. 1 d 8( s + 1)
Bi = ( s + 2) 2 /( s + 2) 2 =8
(2 − 1)! ds s = −2
The partial fraction expansion yields:
8 8
Y ( s) = −
s + 2 ( s + 2) 2
22. The inverse Laplace transform derived from the functional
table pairs yields:
−2 t −2 t
y (t ) = 8e − 8te
23. 8( s + 1) K1 K2
Y ( s) = = +
( s + 2) 2
s + 2 ( s + 2) 2
8( s + 1) = K1 ( s + 2) + K 2
8s + 8 = K1s + 2 K1 + K 2
Equating like terms:
8 = K1 and 8 = 2 K1 + K 2
24. 8 = K1 and 8 = 2 K1 + K 2
8 = 2 × 8 + K2
8 − 16 = −8 = K 2
Thus
8 8
Y (s) = −
s + 2 ( s + 2) 2
y (t ) = 8e −2t − 8te −2t
25. 8( s + 1) K1 K2
Y (s) = = +
( s + 2) 2
s + 2 ( s + 2) 2
As before, we can solve for K2 in the usual manner.
8( s + 1)( s + 2) 2
K2 = = 8( s + 1) s=−2 = −8
( s + 2) 2
s =−2
26. 8( s + 1) 2 K1 8
( s + 2)
2
= ( s + 2) − ( s + 2) 2
( s + 2) 2
s+2 ( s + 2) 2
d [ 8( s + 1)] d [ ( s + 2 ) K1 − 8]
=
ds ds
8 = K1
8( s + 1) 8 8
Y (s) = = −
( s + 2) 2
s + 2 ( s + 2) 2
y (t ) = 8e −2t − 8te −2t
27. Unrepeated complex roots are solved similar to
the process for unrepeated real roots. That is you
multiply by one of the denominator terms in the
partial fraction and solve for the appropriate
constant.
Once you have found one of the constants, the
other constant is simply the complex conjugate.
28. 5.2 K K*
Z ( s) = 2 = +
s + 2 s + 5 ( s + 1 − j 2) ( s + 1 + j 2)
5.2( s + 1 − j 2)
K= = − j13
.
( s + 1 − j 2)( s + 1 + j 2) s =−1+ j 2
K = j13
*
.
29. 5.2 − j1.3 j1.3
Z ( s) = 2 = +
s + 2 s + 5 ( s + 1 − j 2) ( s + 1 + j 2)
e − j1.3 e j 1. 3
Z ( s) = +
( s + 1 − j 2) ( s + 1 + j 2)