Inverse Laplace Transform Formula Steps and Solved Examples
The Laplace Transform and Inverse Laplace Transform is a powerful tool for solving non-homogeneous linear differential equations (the solution to the derivative is not zero).
The Laplace Transform finds the output Y(s) in terms of the input X(s) for a given transfer function H(s), where s = jω. The inverse Laplace Transform finds the input X(s) in terms of the output Y(s) for a given transfer function H(s), where s = jω.
What is Laplace Transform:
The Laplace Transform is a linear operator on continuous functions. It maps the function's domain onto the complex plane and transforms the function's variables from time-domain to frequency-domain. The inverse of this transformation, also taking place in the complex plane, consists of rotating counterclockwise around a point on the unit circle by 90 degrees and then scaling down by a factor of -1 in the vertical direction.
What is Inverse Laplace Transform:
The derivative of s is jω; when finding what X(s) equals when Y(s) equals zero, we find the inverse function (X(-s)) and remember that multiplying both sides by j gives complex conjugate. We can multiply both sides by j to get rid of the (-s).
The Inverse Laplace Transform takes the output Y(s) and finds what X(s) it is in terms of, for a given transfer function H(s).
Transfer Functions: The transfer function is simply s divided by jω. Since Laplace transforms are linear, the transfer function can be factored into a product of simpler functions.
Laplace Transform and Inverse Laplace Transform
The Inverse Laplace Transform can be described as the transformation into a function of time. In the Laplace inverse formula, F(s) is the Transform of F(t), while in Inverse Transform F(t) is the Inverse Laplace Transform of F(s). Therefore, we can write this Inverse Laplace transform formula as follows:
f(t) = L⁻¹{F}(t) = \[\frac{1}{2\pi i} \lim_{T\rightarrow \infty} \oint_{\gamma - iT}^{\gamma + iT} e^{st} F(s) ds\]
If the integrable functions differ on the Lebesgue measure, then the integrable functions can have the same Laplace transform. Therefore, there is an inverse transform on the very range of transform. The inverse of a complex function F(s) to generate a real-valued function f(t) is an inverse Laplace transformation of the function. If a unique function is continuous on 0 to ∞ limit and also has the property of Laplace Transform. This function is, therefore an exponentially restricted real function.
Inverse Laplace Transform Table
This inverse laplace table will help you in every way possible.
Inverse Laplace Transform Theorems
Theorem 1: When a and b are constant,
L⁻¹ {a f(s) + b g(s)} = a L⁻¹ {f(s)} + b L⁻¹{g(s)}
Theorem 2: L⁻¹ {f(s)} = \[e^{-at} L^{-1}\] {f(s - a)}
Inverse Laplace Transform Examples
Example 1 Compute the inverse Laplace transform of Y (s) = \[\frac{2}{3−5s}\].
Solution Adjust it as follows:
Y(s) = \[\frac{2}{3 - 5s} = \frac{-2}{5}. \frac{1}{s - \frac{3}{5}}\]
Thus, by linearity,
Y(t) = \[L^{-1}[\frac{-2}{5}. \frac{1}{s - \frac{3}{5}}]\]
= \[\frac{-2}{5} L^{-1}[\frac{1}{s - \frac{3}{5}}]\]
= \[\frac{-2}{5} e^{(\frac{3}{5})t}\]
Example 2 Compute the inverse Laplace transform of Y (s) = \[\frac{5s}{s^{2} + 9}\]
Solution Adjust it as follows:
Y (s) = \[\frac{5s}{s^{2} + 9} = 5. \frac{s}{s^{2} + 9}\]
Thus, by linearity,
y(t) = \[L^{-1} [5. \frac{s}{s^{2} + 9}]\]
= \[5 L^{-1} [\frac{s}{s^{2} + 9}]\]
= 5 Cos 3t
Example 3 Compute the inverse Laplace transform of Y (s) = \[\frac{2}{3s^{4}}\].
Solution Adjust it as follows:
Y (s) = \[\frac{2}{3s^{4}} = \frac{1}{9} . \frac{3!}{s^{4}}\]
Thus, by linearity,
y(t) = \[L^{-1} [ \frac{1}{9}. \frac{3!}{s^{4}}]\]
= \[\frac{1}{9} L^{-1} [\frac{3!}{s^{4}}]\]
= \[\frac{1}{9}t^{3}\]
Example 4 Compute the inverse Laplace transform of Y (s) = \[\frac{3s + 2}{s^{2} + 25}\].
Solution Adjust it as follows:
Y (s) = \[\frac{3s + 2}{s^{2} + 25}\]
= \[\frac{3s}{s^{2} + 25}\] + \[\frac{2}{s^{2} + 25}\]
= \[3. \frac{s}{s^{2} + 25} + \frac{2}{5} . \frac{5}{s^{2} + 25}\]
Thus,
\[L^{-1}[3. \frac{s}{s^{2} + 25} + \frac{2}{5} . \frac{5}{s^{2} + 25}]\]
= \[3 L^{-1} [\frac{s}{s^{2} + 25}] + \frac{2}{5} L^{-1} [\frac{5}{s^{2} + 25}]\]
= 3 Cos 5t + \[\frac{2}{5}\] Sin 5t.
Example 5 Compute the inverse Laplace transform of Y (s) = \[\frac{1}{3 - 4s} + \frac{3 - 2s}{s^{2} + 49}\]
Solution Adjust it as follows:
Y (s) = \[\frac{1}{3 - 4s} + \frac{3 - 2s}{s^{2} + 49}\]
= \[\frac{1}{-4} . \frac{1}{s - \frac{3}{4}} + \frac{3}{s^{2} + 49} - \frac{2s}{s^{2} + 49}\]
= \[\frac{1}{-4} . \frac{1}{s - \frac{3}{4}} + \frac{3}{7} . \frac{7}{s^{2} + 49} -2. \frac{s}{s^{2} + 49}\]
Thus,
y(t) = \[L^{-1} [\frac{-1}{4}. \frac{1}{s - \frac{3}{4}} + \frac{3}{7} . \frac{7}{s^{2} + 49} -2. \frac{s}{s^{2} + 49}]\]
= \[-\frac{1}{4} L^{-1} [\frac{1}{s - \frac{3}{4}}] + \frac{3}{7} L^{-1}[\frac{7}{s^{2} + 49}] -2 L^{-1} [\frac{s}{s^{2} + 49}]\]
= \[-\frac{1}{4} e^{(\frac{3}{4})t} + \frac{3}{7} sin 7t - 2 cos 7t\]
Example 6 Compute the inverse Laplace transform of Y (s) = \[\frac{5}{(s + 2)^{3}}\]
Solution The transform pair is:
\[t \Leftrightarrow \frac{2}{s^{3}}\]
According to the proposition,
\[e^{-2t}t^{2} \Leftrightarrow \frac{2}{(s + 2)^{3}}\]
Therefore,
y(t) = \[L^{-1} [\frac{5}{(s + 2)^{3}}]\]
= \[L^{-1} [\frac{5}{2} . \frac{2}{(s + 2)^{3}}]\]
= \[\frac{5}{2} L^{-1} [\frac{2}{(s + 2)^{3}}]\]
= \[\frac{5}{2} e^{-2t}t^{2}\]
Example 7 Compute the inverse Laplace transform of Y (s) = \[\frac{4(s - 1)}{(s - 1)^{2} + 4}\]
Solution The transform pair is:
\[cos 2t \Leftrightarrow \frac{s}{s^{2} + 4}\]
According to the proposition,
\[e^{t} cos 2t \Leftrightarrow \frac{s - 1}{(s - 1)^{2} + 4}\]
Hence,
y(t) = \[L^{-1} [\frac{4(s - 1)}{(s - 1)^{2} + 4}]\]
= \[4 L^{-1} [\frac{s - 1}{(s - 1)^{2} + 4}]\]
= \[4e^{t} cos 2t\]
Conclusion:
The Laplace Transform is an operator that maps the input to the output of a linear differential equation involving derivatives of functions. It transforms variables from time-domain to frequency domain. The inverse of this transformation, also taking place in the complex plane, consists of rotating counterclockwise around a point on the unit circle by 90 degrees and then scaling down by a factor of -1 in the vertical direction. The Inverse Laplace Transform takes the output and finds what X(s) it is in terms of, for a given transfer function H(s). If the integrable functions differ on the Lebesgue measure, then the integrable functions can have the same Laplace transform. Students can use this information to learn more about a topic they are studying.
FAQs on Inverse Laplace Transform Explained for Engineering Maths
1. What is the inverse Laplace transform?
The inverse Laplace transform is the process of converting a function from the Laplace domain F(s) back to the time domain f(t). If L{f(t)} = F(s), then L⁻¹{F(s)} = f(t). It is mainly used to recover the original time function after solving algebraic equations in the s-domain. This technique is widely used in differential equations, control systems, and engineering mathematics.
2. What is the formula for the inverse Laplace transform?
The formal formula for the inverse Laplace transform is given by the Bromwich integral: f(t) = (1/2πi) ∫(c−i∞ to c+i∞) e^{st} F(s) ds. In practice, this integral is rarely computed directly. Instead, we use standard inverse Laplace transform tables, partial fractions, and known transform pairs such as L⁻¹{1/s} = 1 and L⁻¹{1/(s−a)} = e^{at}.
3. How do you find the inverse Laplace transform using partial fractions?
To find the inverse Laplace transform using partial fractions, first decompose F(s) into simpler rational terms and then apply standard formulas. Steps:
- Factor the denominator of F(s).
- Express F(s) as a sum of partial fractions.
- Find constants by equating coefficients.
- Apply known inverse Laplace formulas.
Example: For F(s) = 3/(s(s+2)), write 3/(s(s+2)) = A/s + B/(s+2). Solving gives A = 3/2, B = −3/2. Thus, L⁻¹{F(s)} = (3/2) − (3/2)e^{−2t}.
4. What is L⁻¹{1/s}?
The inverse Laplace transform of 1/s is 1. This follows from the basic transform pair L{1} = 1/s. Therefore, applying the inverse operation gives L⁻¹{1/s} = 1, which represents a constant function in the time domain.
5. What is the inverse Laplace transform of 1/(s − a)?
The inverse Laplace transform of 1/(s − a) is e^{at}. This comes from the standard pair L{e^{at}} = 1/(s − a). Therefore, L⁻¹{1/(s − a)} = e^{at}, which represents an exponential function in the time domain.
6. How do you find the inverse Laplace transform of 1/(s² + a²)?
The inverse Laplace transform of 1/(s² + a²) is (1/a) sin(at). This is based on the standard formula L{sin(at)} = a/(s² + a²). Therefore, dividing by a gives L⁻¹{1/(s² + a²)} = (1/a) sin(at), which represents a sinusoidal function.
7. What is the shifting theorem in inverse Laplace transform?
The first shifting theorem states that L⁻¹{F(s − a)} = e^{at} f(t) if L{f(t)} = F(s). This means shifting s to (s − a) in the Laplace domain multiplies the time function by e^{at}. For example, if L⁻¹{1/s} = 1, then L⁻¹{1/(s − 3)} = e^{3t}.
8. How is the inverse Laplace transform used to solve differential equations?
The inverse Laplace transform is used to convert algebraic solutions in the s-domain back to time-domain solutions of differential equations. Steps:
- Take the Laplace transform of the differential equation.
- Use initial conditions to simplify.
- Solve the resulting algebraic equation for Y(s).
- Apply L⁻¹ to find y(t).
This method simplifies solving linear ordinary differential equations with constant coefficients.
9. What is the convolution theorem for inverse Laplace transform?
The convolution theorem states that L⁻¹{F(s)G(s)} = ∫₀ᵗ f(τ)g(t−τ)dτ. This means the inverse Laplace of a product in the s-domain equals the convolution of the corresponding time functions. It is especially useful when F(s) cannot be easily simplified using partial fractions.
10. What are common mistakes when finding inverse Laplace transforms?
Common mistakes in finding the inverse Laplace transform include algebra errors and misuse of standard formulas. Frequent errors:
- Incorrect partial fraction decomposition.
- Forgetting to apply the shifting theorem properly.
- Confusing formulas like 1/(s² + a²) and s/(s² + a²).
- Ignoring initial conditions in differential equations.
Careful algebra and correct use of standard Laplace transform pairs help avoid these errors.
