inverse laplace transforms

Inverse Laplace Transforms There is no integral definition for finding an inverse Laplace transform. Inverse Laplace transforms are found as follows:1) For simple functions: Use tables of Laplace transform pairs.2) For complex functions: Decompose the complex function into two or more

simple functions using Partial Fraction Expansion (PFE) and then find the inverse transform of each function from a table of Laplace transform pairs.

1Chapter 12 EGR 272 – Circuit Theory II

Read: Ch. 12 in Electric Circuits, 10th Edition by NilssonHandouts: • Laplace Transform Properties and Common Laplace Transforms• Partial Fraction Expansion (using various calculators)


Table of Laplace Transforms (will be provided on tests)

Table of Laplace Transform Properties (will be provided on tests)

1 Linearity L{af(t)} = aF(s)2 Superposition L {f1(t) + f2(t) } = F1(s) + F2(s)3 Modulation L {e-atf(t)} = F(s + a)4 Time-Shifting L {f(t - )u(t - )} = e-sF(s)5 Scaling 1 sf(at) F

a a

L6 Real Differentiation d f(t) sF(s) - f(0)

dt

L7 Real Integration

0

1f(t)dt F(s)s

tL

8 Complex Differentiation dtf(t) F(s)ds

L9 Complex Integration

s

f(t) F(s)dst

L

10 Convolution L {f(t) * g(t)} = F(s)·G(s)


Example: Find f(t) for F(s) = 16s/(s2 + 4s + 29)(refer to the table of Laplace transforms on slide 5)


Example: Find f(t) for F(s) = 16/(s+8)(refer to the table of Laplace transforms on slide 5)

Partial Fraction Expansion (or Partial Fraction Decomposition)Partial Fraction Expansion (PFE) is used for functions whose inverse Laplace transforms are not available in tables of Laplace transform.

PFE involves decomposing a given F(s) intoF(s) = A1F1(s) + A2F2(s) + … + ANFN(s)

Where F1(s), F2(s), … , FN(s) are the Laplace transforms of known functions.

Then by applying the linearity and superposition properties:f(t) = A1f1(t) + A2f2(t) + … + ANfN(t)

In most engineering applications,


N(s) numerator in the form of a polynomial in sF(s) D(s) denominator in the form of a polynomial in s

Finding roots of the polynomials yields:

where zi = zeros of F(s)

and pi are the poles of F(s)

Note that:

1 2 M

1 2 N

K(s - z )(s - z ) (s - z )F(s) (s - p )(s - p ) (s - p )

i

i

s = z

s = p

F(s) 0

F(s)


Poles and zeros in F(s)Poles and zeros are sometimes plotted on the s-plane. This is referred to as a pole-zero diagram and is used heavily in later courses such as Control Theory for investigating system stability and performance. Poles and zeros are represented on the pole-zero diagram as follows:x - represents a poleo - represents a zero

ExampleSketch the pole-zero diagram for the following function:

jw

s-plane

100(s + 2)(s + 5)F(s) (s + 4)(s + 1 - j2)(s + 1 + j2)


Surface plots used to illustrate |F(s)|The names “poles” and “zeros” come from the idea of using a surface plot to graph the magnitude of F(s). If the surface, which represents |F(s)|, is something like a circus tent, then the zeros of F(s) are like “tent stakes” where the height of the tent is zero and the poles of F(s) are like “tent poles” with infinite height.

Example A surface plot is shown to the right.

Note: Pole-zero diagrams and surface plots for |F(s)| are not key topics for this course and will not be covered on tests. They are mentioned here as a brief introduction to future topics in electrical engineering.


An important requirement for using Partial Fractions Expansion

Show that expressing F(s) as

leads to an important requirement for performing Partial Fractions Expansion:

If F(s) does not satisfy the condition above, use long division to place it (the remainder) in the proper form (to be demonstrated later).

order of N(s) < order of D(s)

N1 2

1 2 N

AN(s) A AF(s) = D(s) (s - p ) (s - p ) (s - p )


Methods of performing Partial Fractions Expansion:1) common denominator method2) residue method3) calculators, MATLAB, etc

Example: (Simple roots)Use PFE to decompose F(s) below and then find f(t). Perform PFE using:1) common denominator method

4sF(s) (s 1)(s 2)(s 3)


Example: (continued)2) residue method

4sF(s) (s 1)(s 2)(s 3)


Repeated rootsA term in the decomposition with a repeated root in the denominator could in general be represented as:

(Note that in general the order of the numerator should be 1 less than the order of the denominator).F(s) above is inconvenient, however, since it is not the transform of any easily recognizable function. An equivalent form for F(s) works better since each part is a known transform:

F(t) F(s)Ke-atu(t) K

s + aKte-atu(t)

2K

s + a?

1 2

2

Ks + Ks + a


p-sA

p-sA F(s) 2

21

2p-sB As F(s)

Example: (Repeated roots) Find f(t) for F(s) shown below.


3s2s2s F(s) 2

Example: (Repeated roots) Find f(t) for F(s) shown below.


3

2

3s1 2s F(s)

Complex rootsComplex roots always yield sine and/or cosine terms in the time domain. Complex roots may be handled in one of two ways:

F(t) F(s)Kcos(wt)u(t)

22 w+sK(s)

Ksin(wt)u(t)22 w+s

K(w)

Ke-atcos(wt)u(t) 22 w+a) (sa) K(s

Ke-atsin(wt)u(t) 22 w+a) (sK(w)

Also note that cosine and sine terms can be represented as a single cosine term with a phase angle using the identity shown below:

1) using quadratic factors – Leave the portion of F(s) with complex roots as a 2nd order term and manipulate this term into the form of the transform for sine and cosine functions (with or without exponential damping). Keep the transform pairs shown to the right in mind:

2 2 -1

Useful trigonometric identity:

BAcos(wt) + Bsin(wt) = A + B cos wt - tanA

(or using phasors, convert (A,-B) to polar form)


2) using complex roots – a complex term can be represented using complex linear roots as follows:

*

1 22 2

A s AF(s) (s 2 s w ) (s jw) (s + jw)

B B

where the two terms with complex roots will yield a single time-domain term that is represented in phasor form as 2 2B B

or in time-domain form as 2Betcos(wt + )*

Note that may be found as usual using the residue method. It is not necessary to fin d .B B

The two methods for handling complex roots are summarized in the table below.

F ( t ) F ( s )

K e - a tc o s (w t )u ( t ) 22 w+a) (sa) K(s

K e - a ts in ( w t)u ( t ) 22 w+a) (sK(w)

2 B e - a tc o s ( w t + )u ( t )

*B B +

s + - jw s + + jw

Quadratic factor method

Complex linear root method


Example: (Complex roots) Find f(t) for F(s) shown below. Use both methods described above and show that the results are equivalent.1) Quadratic factor method


102s s1s2s F(s) 2

Example: (continued) 2) Complex linear root method


102s s1s2s F(s) 2

Example: (Time-delayed function) Find f(t) for

Example: (Order of numerator too large) Find f(t) for


4s3s2e F(s)

-2s

Hint: Form a new function such that F(s) = F1(s)e-2s.Find f1(t). f(t) is simply a delayed version of f1(t).

4ss3 2s F(s)

2

inverse laplace transforms

Documents