By the end of this chapter, you will be able to: - confidently solve Laplace transforms using both initial and final value theorems, - accurately find inverse Laplace transforms by applying partial fraction decomposition, - correctly solve differential equations using Laplace transforms.
Mastering these skills will help you tackle real engineering problems with confidence and precision in your trade.
Laplace transforms are a fundamental mathematical tool for Kenyan industrial control and installation professionals, especially when working with systems that can be modeled by differential equations. Whether analyzing control systems for industrial machinery, troubleshooting automation panels, or designing relay circuits, Laplace transforms simplify the process by converting differential equations into algebraic equations. Mastery of this technique allows for more efficient analysis and solution of real-world engineering problems encountered in Kenyan factories, processing plants, and automated production lines.
In industrial control, many processes, such as current flow in relay coils or temperature rise in a heating element, are described by functions of time. The Laplace transform is a mathematical operation that converts these time-domain functions, \( f(t) \), into a new function, \( F(s) \), in the frequency (or complex) domain. This transformation makes it easier to solve differential equations, analyze system response, and design controllers.
The Laplace transform of a function \( f(t) \) is defined as the integral:$$ L\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) dt $$ where \( s \) is a complex parameter. Common notations include \( L\{f(t)\} \), \( F(s) \), or sometimes \( f(s) \). In industrial control, engineers use Laplace transforms to analyze the dynamic behavior of electrical and mechanical systems.
Example 1: Find the Laplace transform of \( f(t) = 2 \).
Given: \( f(t) = 2 \)
$$ L\{2\} = \int_{0}^{\infty} e^{-st} \cdot 2 \, dt $$$$ = 2 \int_{0}^{\infty} e^{-st} dt $$$$ = 2 [ \frac{-1}{s} e^{-st} ]_{0}^{\infty} $$ As \( t \to \infty \), \( e^{-st} \to 0 \) (for \( s > 0 \)), so:$$ = 2 ( 0 - ( \frac{-1}{s} \cdot 1 ) ) $$$$ = 2 ( \frac{1}{s} ) $$ Answer: \( \boxed{\frac{2}{s}} \)
Example 2: Find the Laplace transform of \( f(t) = t \).
Given: \( f(t) = t \)
$$
L\{t\} = \int_{0}^{\infty} e^{-st} t \, dt
$$
Using integration by parts:
Let \( u = t \), \( dv = e^{-st} dt \), so \( du = dt \), \( v = \frac{-1}{s} e^{-st} \)$$
= u v |_{0}^{\infty} - \int_{0}^{\infty} v du
$$$$
= . t \cdot \frac{-1}{s} e^{-st} |_{0}^{\infty} - \int_{0}^{\infty} \frac{-1}{s} e^{-st} dt
$$
First term: as \( t \to \infty \), \( t e^{-st} \to 0 \)
At \( t = 0 \), value is 0
Second term:$$
- \int_{0}^{\infty} \frac{-1}{s} e^{-st} dt = \frac{1}{s} \int_{0}^{\infty} e^{-st} dt = \frac{1}{s} [ \frac{-1}{s} e^{-st} ]_{0}^{\infty}
$$$$
= \frac{1}{s} (0 - ( -\frac{1}{s} ))
$$$$
= \frac{1}{s} \cdot \frac{1}{s} = \frac{1}{s^2}
$$
Answer: \( \boxed{\frac{1}{s^2}} \)
Example 3: Find the Laplace transform of \( f(t) = 5e^{3t} \).
Given: \( f(t) = 5e^{3t} \)
$$ L\{5e^{3t}\} = \int_{0}^{\infty} e^{-st} \cdot 5e^{3t} dt $$$$ = 5 \int_{0}^{\infty} e^{-st} e^{3t} dt $$$$ = 5 \int_{0}^{\infty} e^{(3-s)t} dt $$$$ = 5 [ \frac{1}{3-s} e^{(3-s)t} ]_{0}^{\infty} $$ For \( s > 3 \), as \( t \to \infty \), \( e^{(3-s)t} \to 0 \)$$ = 5 ( 0 - \frac{1}{3-s} \cdot 1 ) $$$$ = -5 \cdot \frac{1}{3-s} $$$$ = \frac{5}{s-3} $$ Answer: \( \boxed{\frac{5}{s-3}} \)
Example 4: Find the Laplace transform of \( f(t) = 4t^2 \).
Given: \( f(t) = 4t^2 \)
Recall: \( L\{t^n\} = \frac{n!}{s^{n+1}} \)
Here, \( n = 2 \), so \( 2! = 2 \)$$
L\{4t^2\} = 4 \cdot L\{t^2\} = 4 \cdot \frac{2}{s^{3}}
$$$$
= \frac{8}{s^{3}}
$$
Answer: \( \boxed{\frac{8}{s^{3}}} \)
Example 5 (Industrial Control Context): The voltage across a relay coil is given by \( v(t) = 12e^{-2t} \). Find its Laplace transform.
Given: \( v(t) = 12e^{-2t} \)
$$ L\{12e^{-2t}\} = 12 \cdot L\{e^{-2t}\} $$ For \( L\{e^{at}\} = \frac{1}{s-a} \), here \( a = -2 \)$$ = 12 \cdot \frac{1}{s - (-2)} $$$$ = 12 \cdot \frac{1}{s + 2} $$ Answer: \( \boxed{\frac{12}{s+2}} \)
Laplace transforms have several important properties that are frequently used in industrial control system analysis. The most basic is linearity, which allows you to transform sums and scalar multiples easily.
Linearity:$$ L\{a f(t) + b g(t)\} = a L\{f(t)\} + b L\{g(t)\} $$ where \( a \) and \( b \) are constants.
Example 1: Find the Laplace transform of \( f(t) = 3t + 4 \).
Given: \( f(t) = 3t + 4 \)
$$ L\{3t + 4\} = 3 L\{t\} + 4 L\{1\} $$ Recall: \( L\{t\} = \frac{1}{s^2} \), \( L\{1\} = \frac{1}{s} \)$$ = 3 \cdot \frac{1}{s^2} + 4 \cdot \frac{1}{s} $$$$ = \frac{3}{s^2} + \frac{4}{s} $$ Answer: \( \boxed{\frac{3}{s^2} + \frac{4}{s}} \)
Example 2: Find the Laplace transform of \( f(t) = 2e^{5t} - 7 \).
Given: \( f(t) = 2e^{5t} - 7 \)
$$ L\{2e^{5t} - 7\} = 2 L\{e^{5t}\} - 7 L\{1\} $$ Recall: \( L\{e^{at}\} = \frac{1}{s-a} \), so \( a = 5 \)$$ = 2 \cdot \frac{1}{s-5} - 7 \cdot \frac{1}{s} $$$$ = \frac{2}{s-5} - \frac{7}{s} $$ Answer: \( \boxed{\frac{2}{s-5} - \frac{7}{s}} \)
Example 3: Find the Laplace transform of \( f(t) = 6t^2 - 4t + 9 \).
Given: \( f(t) = 6t^2 - 4t + 9 \)
$$ L\{6t^2 - 4t + 9\} = 6 L\{t^2\} - 4 L\{t\} + 9 L\{1\} $$ Recall: \( L\{t^2\} = \frac{2}{s^3} \), \( L\{t\} = \frac{1}{s^2} \), \( L\{1\} = \frac{1}{s} \)$$ = 6 \cdot \frac{2}{s^3} - 4 \cdot \frac{1}{s^2} + 9 \cdot \frac{1}{s} $$$$ = \frac{12}{s^3} - \frac{4}{s^2} + \frac{9}{s} $$ Answer: \( \boxed{\frac{12}{s^3} - \frac{4}{s^2} + \frac{9}{s}} \)
Example 4 (Industrial Control Context): The current in a solenoid is given by \( i(t) = 0.5 t^2 + 7e^{-t} \). Find its Laplace transform.
Given: \( i(t) = 0.5 t^2 + 7e^{-t} \)
$$ L\{0.5 t^2 + 7e^{-t}\} = 0.5 L\{t^2\} + 7 L\{e^{-t}\} $$ Recall: \( L\{t^2\} = \frac{2}{s^3} \), \( L\{e^{-t}\} = \frac{1}{s + 1} \)$$ = 0.5 \cdot \frac{2}{s^3} + 7 \cdot \frac{1}{s+1} $$$$ = \frac{1}{s^3} + \frac{7}{s+1} $$ Answer: \( \boxed{\frac{1}{s^3} + \frac{7}{s+1}} \)
Example 5: Find the Laplace transform of \( f(t) = -2t + 8 \).
Given: \( f(t) = -2t + 8 \)
$$ L\{-2t + 8\} = -2 L\{t\} + 8 L\{1\} $$$$ = -2 \cdot \frac{1}{s^2} + 8 \cdot \frac{1}{s} $$$$ = -\frac{2}{s^2} + \frac{8}{s} $$ Answer: \( \boxed{-\frac{2}{s^2} + \frac{8}{s}} \)
Standard functions commonly encountered in industrial control systems have well-known Laplace transforms. These are used to quickly analyze signals and system responses.
Example 1: Find the Laplace transform of \( f(t) = 1 \).
Given: \( f(t) = 1 \)
$$ L\{1\} = \int_{0}^{\infty} e^{-st} dt $$$$ = [ \frac{-1}{s} e^{-st} ]_{0}^{\infty} $$$$ = 0 - ( -\frac{1}{s} ) $$$$ = \frac{1}{s} $$ Answer: \( \boxed{\frac{1}{s}} \)
Example 2: Find the Laplace transform of \( f(t) = t^3 \).
Given: \( f(t) = t^3 \)
Recall: \( L\{t^n\} = \frac{n!}{s^{n+1}} \), \( n = 3 \), \( 3! = 6 \)$$ L\{t^3\} = \frac{6}{s^4} $$ Answer: \( \boxed{\frac{6}{s^4}} \)
Example 3: Find the Laplace transform of \( f(t) = 7e^{2t} \).
Given: \( f(t) = 7e^{2t} \)
$$ L\{7e^{2t}\} = 7 L\{e^{2t}\} $$ Recall: \( L\{e^{at}\} = \frac{1}{s-a} \), \( a = 2 \)$$ = 7 \cdot \frac{1}{s-2} $$ Answer: \( \boxed{\frac{7}{s-2}} \)
Example 4 (Industrial Control Context): The temperature in a furnace increases as \( T(t) = 3t^2 \). Find its Laplace transform.
Given: \( T(t) = 3t^2 \)
Recall: \( L\{t^2\} = \frac{2}{s^3} \)$$ L\{3t^2\} = 3 \cdot \frac{2}{s^3} $$$$ = \frac{6}{s^3} $$ Answer: \( \boxed{\frac{6}{s^3}} \)
Example 5: Find the Laplace transform of \( f(t) = 2t^2 + 5t + 7 \).
Given: \( f(t) = 2t^2 + 5t + 7 \)
$$ L\{2t^2 + 5t + 7\} = 2 L\{t^2\} + 5 L\{t\} + 7 L\{1\} $$$$ = 2 \cdot \frac{2}{s^3} + 5 \cdot \frac{1}{s^2} + 7 \cdot \frac{1}{s} $$$$ = \frac{4}{s^3} + \frac{5}{s^2} + \frac{7}{s} $$ Answer: \( \boxed{\frac{4}{s^3} + \frac{5}{s^2} + \frac{7}{s}} \)
Laplace transforms are especially useful in industrial control for solving ordinary differential equations (ODEs) that describe system dynamics. By transforming time-domain equations into the \( s \)-domain, algebraic methods can be used to find solutions, which are then transformed back to the time domain.
Example 1: Solve the ODE \( \frac{dy}{dt} = 4 \), \( y(0) = 2 \), using Laplace transforms.
Given: \( \frac{dy}{dt} = 4 \), \( y(0) = 2 \)
Take Laplace transform of both sides:$$ L\{ \frac{dy}{dt} \} = L\{4\} $$ Recall: \( L\{ \frac{dy}{dt} \} = s Y(s) - y(0) \), \( L\{4\} = \frac{4}{s} \)$$ s Y(s) - 2 = \frac{4}{s} $$$$ s Y(s) = \frac{4}{s} + 2 $$$$ Y(s) = \frac{4}{s^2} + \frac{2}{s} $$ Inverse Laplace:$$ y(t) = 4t + 2 $$ Answer: \( \boxed{y(t) = 4t + 2} \)
Example 2: Solve \( \frac{d^2y}{dt^2} + y = 0 \), \( y(0) = 1 \), \( y'(0) = 0 \), using Laplace transforms.
Given: \( \frac{d^2y}{dt^2} + y = 0 \), \( y(0) = 1 \), \( y'(0) = 0 \)
Take Laplace transform:$$ L\{ \frac{d^2y}{dt^2} \} + L\{y\} = 0 $$ Recall: \( L\{ \frac{d^2y}{dt^2} \} = s^2 Y(s) - s y(0) - y'(0) \), \( L\{y\} = Y(s) \)$$ s^2 Y(s) - s \cdot 1 - 0 + Y(s) = 0 $$$$ (s^2 + 1) Y(s) = s $$$$ Y(s) = \frac{s}{s^2 + 1} $$ Inverse Laplace: \( L^{-1}\{ \frac{s}{s^2 + 1} \} = \cos t \)$$ y(t) = \cos t $$ Answer: \( \boxed{y(t) = \cos t} \)
Example 3 (Industrial Control Context): The current in a circuit satisfies \( \frac{di}{dt} + 2i = 0 \), \( i(0) = 10 \). Find \( i(t) \).
Given: \( \frac{di}{dt} + 2i = 0 \), \( i(0) = 10 \)
Take Laplace transform:$$ L\{ \frac{di}{dt} \} + 2 L\{i\} = 0 $$ Recall: \( L\{ \frac{di}{dt} \} = s I(s) - i(0) \)$$ s I(s) - 10 + 2 I(s) = 0 $$$$ (s + 2) I(s) = 10 $$$$ I(s) = \frac{10}{s + 2} $$ Inverse Laplace: \( L^{-1}\{ \frac{1}{s + a} \} = e^{-a t} \)$$ i(t) = 10 e^{-2t} $$ Answer: \( \boxed{i(t) = 10 e^{-2t}} \)
Find the Laplace transform of \( f(t) = 4t + 5 \). (2 marks)
Compute the Laplace transform of \( f(t) = 6e^{3t} \). (2 marks)
Find the Laplace transform of \( f(t) = 2t^2 - 3t + 7 \). (3 marks)
A temperature sensor in a greenhouse outputs a signal \( S(t) = 8e^{-0.5t} \). Find its Laplace transform. (2 marks)
Solve the ODE \( \frac{dy}{dt} = 7 \), \( y(0) = 2 \), using Laplace transforms. (3 marks)
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Create a free accountThis chapter provided a comprehensive overview of Laplace transforms and their critical role in solving engineering problems. It began by introducing the concept of Laplace transforms, highlighting their ability to convert complex time-domain functions into simpler algebraic expressions in the s-domain. The chapter then demonstrated how to derive Laplace transforms from first principles, reinforcing the foundational integration process involved. Key properties of the Laplace transform, such as linearity, shifting, and scaling, were outlined to illustrate how these characteristics simplify calculations and problem-solving. Methods for finding inverse Laplace transforms, including the use of partial fractions, were presented to show how to revert solutions back to the time domain. The application of Laplace transforms in solving ordinary differential equations was explored, showcasing their effectiveness in handling initial value problems. The chapter concluded by extending these techniques to systems of simultaneous differential equations, ensuring solutions satisfy given initial conditions. Overall, students gained the mathematical tools to approach a wide range of engineering challenges involving dynamic systems.
Given: \( f(t) = 3t \)
Formula:$$ \mathcal{L}\{t\} = \frac{1}{s^{2}} $$
Substitute:$$ \mathcal{L}\{3t\} = 3 \times \frac{1}{s^{2}} $$
Calculate:$$ = \frac{3}{s^{2}} $$
Final Answer: \( \frac{3}{s^{2}} \)
Given: \( f(t) = 5e^{2t} \)
Formula:$$ \mathcal{L}\{e^{at}\} = \frac{1}{s-a} $$
Substitute:$$ \mathcal{L}\{5e^{2t}\} = 5 \times \frac{1}{s-2} $$
Calculate:$$ = \frac{5}{s-2} $$
Final Answer: \( \frac{5}{s-2} \)
The emergency stop circuit in an automated bottling plant in Thika is described by the following differential equation for the relay coil voltage:$$\frac{dv}{dt} + 5v = 10$$
Given the initial voltage across the coil is zero, use Laplace transforms to determine the expression for \(v(t)\).
(4 marks)
Define the Laplace transform and state its region of convergence. (4 marks)
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