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**The Inverse Laplace Transform**

The University of Tennessee Electrical and Computer Engineering Department Knoxville, Tennessee wlg

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**Inverse Laplace Transforms**

Background: To find the inverse Laplace transform we use transform pairs along with partial fraction expansion: F(s) can be written as; Where P(s) & Q(s) are polynomials in the Laplace variable, s. We assume the order of Q(s) P(s), in order to be in proper form. If F(s) is not in proper form we use long division and divide Q(s) into P(s) until we get a remaining ratio of polynomials that are in proper form.

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**Inverse Laplace Transforms**

Background: There are three cases to consider in doing the partial fraction expansion of F(s). Case 1: F(s) has all non repeated simple roots. Case 2: F(s) has complex poles: (expanded) Case 3: F(s) has repeated poles. (expanded)

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**Inverse Laplace Transforms**

Case 1: Illustration: Given: Find A1, A2, A3 from Heavyside

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**Inverse Laplace Transforms**

Case 3: Repeated roots. When we have repeated roots we find the coefficients of the terms as follows:

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**Inverse Laplace Transforms**

Case 3: Repeated roots. Example ? ? ?

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**Inverse Laplace Transforms**

Case 2: Complex Roots: F(s) is of the form; K1 is given by,

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**Inverse Laplace Transforms**

Case 2: Complex Roots:

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**Inverse Laplace Transforms**

Case 2: Complex Roots: Therefore: You should put this in your memory:

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**Inverse Laplace Transforms**

Complex Roots: An Example. For the given F(s) find f(t)

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**Inverse Laplace Transforms**

Complex Roots: An Example. (continued) We then have; Recalling the form of the inverse for complex roots;

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**Inverse Laplace Transforms**

Convolution Integral: Consider that we have the following situation. System x(t) y(t) h(t) x(t) is the input to the system. h(t) is the impulse response of the system. y(t) is the output of the system. We will look at how the above is related in the time domain and in the Laplace transform.

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**Inverse Laplace Transforms**

Convolution Integral: In the time domain we can write the following: In this case x(t) and h(t) are said to be convolved and the integral on the right is called the convolution integral. Comment: The convolution integral is extremely important in signals and systems. It can be implemented with a PC in real time and can be very useful in that form. However, if we do the integration by hand, it is tedious, difficult at times. It is much easier to use: Y(s) = X(s)H(s) And the take the inverse to find y(t). Here we simply multiply X(s) and H(s) together, expand in partial fractions and take the inverse to get y(t). Most always this is easier than carrying out the integration. It can be shown that, This is very important * note

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**Inverse Laplace Transforms**

Convolution Integral: Through an example let us see how the convolution integral and the Laplace transform are related. We now think of the following situation: h(t) x(t) y(t) H(s) Y(s) X(s)

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**Inverse Laplace Transforms**

Convolution Integral: From the previous diagram we note the following: h(t) is called the system impulse response for the following reason. Eq A If the input x(t) is a unit impulse, (t), the L(x(t)) = X(s) = 1. Since x(t) is an impulse, we say that y(t) is the impulse response. From Eq A, if X(s) = 1, then Y(s) = H(s). Since,

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**Inverse Laplace Transforms**

Convolution Integral: A really important thing here is that anytime you are given a system diagram as follows, X(s) Y(s) H(s) the inverse Laplace transform of H(s) is the system’s impulse response. This is important !!

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**Inverse Laplace Transforms**

Convolution Integral: Example using the convolution integral. x(t) y(t) = ? e-4t

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**Inverse Laplace Transforms**

Convolution Integral: Same example but using Laplace. x(t) = u(t) h(t) = e-4tu(t)

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**Inverse Laplace Transforms**

Convolution Integral: Practice problems: 1.5[(t) – 2e-2tu(t)] [e-6t - 6te-6t]u(t) [(-1/32) + (1/16)t - (1/32)e-4t + (1/16)te-4t]u(t) Answers given on note page

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**Inverse Laplace Transforms**

Circuit theory problem: You are given the circuit shown below. Use Laplace transforms to find v(t) for t > 0.

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**Inverse Laplace Transforms**

Circuit theory problem: We see from the circuit,

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**Inverse Laplace Transforms**

Circuit theory problem: Take the Laplace transform of this equations including the initial conditions on vc(t)

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**Inverse Laplace Transforms**

Circuit theory problem:

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