15.6 Application to Integrodifferential Equations

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15.6 Application to Integrodifferential Equations

The Laplace transform is useful in solving linear inte grodifferential equations. Using the differentiation and integration properties of Laplace transforms, each term in the integrodifferential equation is transformed.

Figure 15.23 For Example 15.14.

Figure 15.24

For Example 15.14; excitation and response.

Initial conditions are automatically taken into account. We solve the resulting algebraic equation in the s-domain. We then con vert the solution back to the time domain by using the in verse transform. The following examples illustrate the process.

Example 15.15 Use the Laplace transform to solve the differential equation

$$\frac{d^2v(t)}{dt^2} + 6\frac{dv(t)}{dt} + 8v(t) = 2u(t)$$

subject to v(0) = 1, v′(0) = −2.

Solution:

We tak e the Laplace transform of each term in the gi ven dif ferential equation and obtain

$$[s^2V(s) - sv(0) - v'(0)] + 6[sV(s) - v(0)] + 8V(s) = \frac{2}{s}$$

Substituting v(0) = 1, v′(0) = −2,

$$s^2V(s) - s + 2 + 6sV(s) - 6 + 8V(s) = \frac{2}{s}$$

or

$$(s2 + 6s + 8)V(s) = s + 4 + \frac{2}{s} = \frac{s2 + 4s + 2}{s}$$

Hence,

$$V(s) = \frac{s^2 + 4s + 2}{s(s+2)(s+4)} = \frac{A}{s} + \frac{B}{s+2} + \frac{C}{s+4}$$

where

$$A = sV(s) \Big|_{s=0} = \frac{s^2 + 4s + 2}{(s+2)(s+4)} \Big|_{s=0} = \frac{2}{(2)(4)} = \frac{1}{4}$$ $$B = (s+2)V(s) \Big|_{s=-2} = \frac{s^2 + 4s + 2}{s(s+4)} \Big|_{s=-2} = \frac{-2}{(-2)(2)} = \frac{1}{2}$$ $$C = (s+4)V(s) \Big|_{s=-4} = \frac{s^2 + 4s + 2}{s(s+2)} \Big|_{s=-4} = \frac{2}{(-4)(-2)} = \frac{1}{4}$$

Hence,

$$V(s) = \frac{\frac{1}{4}}{s} + \frac{\frac{1}{2}}{s+2} + \frac{\frac{1}{4}}{s+4}$$

By the inverse Laplace transform,

$$v(t) = \frac{1}{4}(1 + 2e^{-2t} + e^{-4t})u(t)$$

Solve the follo wing differential equation using the Laplace transform Practice Problem 15.15 method.

$$\frac{d^2v(t)}{dt^2} + 4\frac{dv(t)}{dt} + 4v(t) = 7e^{-t}$$

if v(0) = v′(0) = 2.

Answer:

$$(7e^{-t} - 5e^{-2t} - te^{-2t})u(t)$$

.

Solve for the response y(t) in the following integrodifferential equation. Example 15.16

$$\frac{dy}{dt} + 5y(t) + 6 \int_0^t y(\tau)d\tau = u(t), \qquad y(0) = 2$$

Solution:

Taking the Laplace transform of each term, we get

$$[sY(s) - y(0)] + 5Y(s) + \frac{6}{s}Y(s) = \frac{1}{s}$$

Substituting y(0) = 2 and multiplying through by s,

$$Y(s)(s^2 + 5s + 6) = 1 + 2s$$

or

$$Y(s) = \frac{2s+1}{(s+2)(s+3)} = \frac{A}{s+2} + \frac{B}{s+3}$$

where

$$A = (s + 2)Y(s) \Big|_{s=-2} = \frac{2s+1}{s+3} \Big|_{s=-2} = \frac{-3}{1} = -3$$ $$B = (s+3)Y(s) \Big|_{s=-3} = \frac{2s+1}{s+2} \Big|_{s=-3} = \frac{-5}{-1} = 5$$

s + 2

−1

Thus,

$$Y(s) = \frac{-3}{s+2} + \frac{5}{s+3}$$

Its inverse transform is

$$y(t) = (-3e^{-2t} + 5e^{-3t})u(t)$$

Use the Laplace transform to solve the integrodifferential equation Practice Problem 15.16

$$\frac{dy}{dt} + 3y(t) + 2\int_0^t y(\tau)d\tau = 2e^{-3t}, \qquad y(0) = 0$$

Answer: (−e^−t + 4e⁻2*^t* − 3e⁻3*^t* )u(t).

15.7 Summary

1. The Laplace transform allows a signal represented by a function in the time domain to be analyzed in the s-domain (or comple x frequency domain). It is defined as

$$\mathcal{L}[f(t)] = F(s) = \int_0^\infty f(t)e^{-st} dt$$

• 2. Properties of the Laplace transform are listed in Table 15.1, while the Laplace transforms of basic common functions are listed in Table 15.2.
• 3. The inverse Laplace transform can be found using partial fraction expansions and using the Laplace transform pairs in Table 15.2 as a look-up table. Real poles lead to exponential functions and complex poles to damped sinusoids.
• 4. The convolution of tw o signals consists of time-re versing one of the signals, shifting it, multiplying it point by point with the second signal, and integrating the product. The convolution integral relates the convolution of two signals in the time domain to the in verse of the product of their Laplace transforms:

$$\mathcal{L}^{-1}[F_1(s)F_2(s)] = f_1(t) * f_2(t) = \int_0^t f_1(\lambda)f_2(t - \lambda) d\lambda$$

5. In the time domain, the output y(t) of the network is the convolution of the impulse response with the input x(t),

$$y(t) = h(t) * x(t)$$

Convolution may be re garded as the flip-shift-multiply-time-area method.

6. The Laplace transform can be used to solve a linear integrodifferential equation.

Review Questions

15.1 Every function f(t) has a Laplace transform.

(a) True (b) False

15.2 The variable s in the Laplace transform H(s) is called

(a) complex frequency	(b) transfer function
(c) zero	(d) pole

15.3 The Laplace transform of u(t − 2) is:

(a)

$$\frac{1}{s+2}$$

(b) $\frac{1}{s-2}$
(c) $\frac{e^{2s}}{s}$ (d) $\frac{e^{-2s}}{s}$

15.4 The zero of the function

of the function

$$F(s) = \frac{s+1}{(s+2)(s+3)(s+4)}$$

is at
(a) −4	(b) −3
(c) −2	(d) −1

15.5 The poles of the function

$$F(s) = \frac{s+1}{(s+2)(s+3)(s+4)}$$

are at

(a) −4	(b) −3
(c) −2	(d) −1

15.6 If F(s) = 1∕(s + 2), then f(t) is

(a) e^2t u(t) (b) e⁻2*^t u*(t) (c) u(t − 2) (d) u(t + 2)

Problems 707