[Maths Class Notes] on Inverse Laplace Transform Pdf for Exam

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}{2pi i} lim_{Trightarrow 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} . fra
c{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.