Inverse Laplace Transforms

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Lecture #3 EGR 261 – Signals & Systems. Read : Ch. 12, Sect. 1-9 in Electric Circuits, 9 th Edition by Nilsson Ch. 4, Sect. 1-3 and Sect. B.5 in Linear Signals & Systems, 2 nd Ed. by Lathi Handout on Bb and webpage : Partial Fraction Expansion (using various calculators). - PowerPoint PPT Presentation

Text of 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.Example: Find f(t) for F(s) = 16/(s+8)Lecture #3 EGR 261 Signals & SystemsRead: Ch. 12, Sect. 1-9 in Electric Circuits, 9th Edition by NilssonCh. 4, Sect. 1-3 and Sect. B.5 in Linear Signals & Systems, 2nd Ed. by LathiHandout on Bb and webpage: Partial Fraction Expansion (using various calculators)

• *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 pairs.

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)Example: Find f(t) for F(s) = 16s/(s2 + 4s + 29)Lecture #3 EGR 261 Signals & Systems

• *In most engineering applications,

Finding roots of the polynomials yields:

where zi = zeros of F(s)and pi are the poles of F(s)

Note that:

Lecture #3 EGR 261 Signals & Systems

• *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 zeroExampleSketch the pole-zero diagram for the following function:Lecture #3 EGR 261 Signals & Systems

• *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.Lecture #3 EGR 261 Signals & Systems

• *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)Lecture #3 EGR 261 Signals & Systems

• *Methods of performing Partial Fractions Expansion:1)common denominator method2)residue method3)calculators/softwareExample: (Simple roots)Use PFE to decompose F(s) below and then find f(t). Perform PFE using:1)common denominator methodLecture #3 EGR 261 Signals & Systems

• *Example: (continued)2)residue method3)calculators (demonstrate with TI-86, TI-89, and MathCAD)Lecture #3 EGR 261 Signals & Systems

• *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:Lecture #3 EGR 261 Signals & Systems

F(t)

F(s)

Ke-atu(t)

Kte-atu(t)

?

_1128536282.unknown

_1128536283.unknown

_1128536123.unknown

• *Example: (Repeated roots) Find f(t) for F(s) shown below. Lecture #3 EGR 261 Signals & Systems

• *Example: (Repeated roots) Find f(t) for F(s) shown below. Lecture #3 EGR 261 Signals & Systems

• *Complex rootsComplex roots always yield sine and/or cosine terms in the time domain. Complex roots may be handled in one of two ways: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:Lecture #3 EGR 261 Signals & Systems

F(t)

F(s)

Kcos(wt)u(t)

Ksin(wt)u(t)

Ke-atcos(wt)u(t)

Ke-atsin(wt)u(t)

_950309100.unknown

_950309170.unknown

_950309206.unknown

_950309069.unknown

• *2)using complex roots a complex term can be represented using complex linear roots as follows:where the two terms with complex roots will yield a single time-domain term that is represented in phasor form asor in time-domain form as 2Betcos(wt + )The two methods for handling complex roots are summarized in the table below.Lecture #3 EGR 261 Signals & Systems

• *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 Lecture #3 EGR 261 Signals & Systems

• *Example: (continued) 2) Complex linear root method Lecture #3 EGR 261 Signals & Systems

• *Example: (Time-delayed function) Find f(t) for Example: (Order of numerator too large) Find f(t) for Lecture #3 EGR 261 Signals & Systems Laplace Transforms ¢â‚¬â€œ recap for Inverse Laplace Transforms ¢â‚¬â€œ the Bromwich Integral This is a contour
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