Question

Find the inverse Laplace transform of the given function. $$ F(s)=\frac{2(s-1) e^{-2 s}}{s^{2}-2 s+2} $$

Short answer

Question: Determine the inverse Laplace transform of the given function, \(F(s) = \frac{2(s-1)e^{-2s}}{s^2-2s+2}\). Answer: The inverse Laplace transform of the given function is \(f(t) = -e^{(t-2)}\cos(t-2)u(t-2)\).

Step-by-step solution

Identify the exponential term in F(s)

Recall that the Laplace transform of a time-shifted function \(f(t-a)u(t-a)\) is \(e^{-as}F(s)\), where 'a' is the time-shift factor and \(u(t-a)\) is the unit step function. In the given function \(F(s) = \frac{2(s-1)e^{-2s}}{s^2-2s+2}\), the exponential term is \(e^{-2s}\), which corresponds to a time-shifted signal in the time domain.

Remove the exponential term

The given function can be rewritten as \(F(s) = e^{-2s}G(s)\), where \(G(s) = \frac{2(s-1)}{s^2-2s+2}\). We will now focus on finding the inverse Laplace transform of \(G(s)\).

Perform partial fraction decomposition on G(s)

To perform partial fraction decomposition on \(G(s) = \frac{2(s-1)}{s^2-2s+2}\), first factor the denominator: \(s^2-2s+2 = (s-1)^2 + 1\). Now, decompose G(s) into two fractions: $$ G(s) = \frac{A}{(s-1) + j} + \frac{B}{(s-1) - j} $$ where A and B are constants, and j is the imaginary unit. Multiply both sides by the denominator, \((s-1)^2 + 1\), to clear the fractions: $$ 2(s-1) = A[(s-1) - j] + B[(s-1) + j] $$ To solve for A and B, plug in values for s that eliminate one of the terms. Let \(s = 1 + j\): $$ 2(1-j) = 2(1 - j) + B(2) $$ The above equation simplifies to: $$ B = -1 $$ Now, let \(s = 1 - j\): $$ 2(1+j) = A(2) + 2(1 + j) $$ The above equation simplifies to: $$ A = -1 $$ So, \(G(s) = \frac{-1}{(s-1) + j} - \frac{1}{(s-1) - j}\).

Find the inverse Laplace transform of G(s)

The inverse Laplace transform of G(s) is: $$ g(t) = \mathcal{L}^{-1}\{G(s)\} = \mathcal{L}^{-1}\left\{-\frac{1}{(s-1) + j} - \frac{1}{(s-1) - j}\right\} $$ Using the inverse Laplace transform table, we know that \(\mathcal{L}^{-1}\left\{\frac{1}{s-a \pm jb}\right\} = e^{at}\cos(bt)\) and \(\mathcal{L}^{-1}\left\{-\frac{1}{s-a \pm jb}\right\} = -e^{at}\cos(bt)\): $$ g(t) = -e^{t}\cos(t) $$

Apply the time-shifting property to find the inverse Laplace transform of F(s)

Recall that \(F(s) = e^{-2s}G(s)\). Since \(g(t) = -e^{t}\cos(t)\), the inverse Laplace transform of F(s) is the time-shifted signal g(t-2)u(t-2): $$ f(t) = \mathcal{L}^{-1}\{F(s)\} = -e^{(t-2)}\cos(t-2)u(t-2) $$ Hence, the inverse Laplace transform of the given function is \(f(t) = -e^{(t-2)}\cos(t-2)u(t-2)\).

Key concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down complex rational expressions into simpler fractions that are easier to work with, especially when finding their inverse Laplace transforms. In this exercise, we deal with a complex fraction in the form of \[ G(s) = \frac{2(s-1)}{s^2-2s+2} \]To perform partial fraction decomposition, we first factor the denominator. Here, the denominator is expressed as \[ s^2 - 2s + 2 = (s-1)^2 + 1 \]This form tells us the roots are complex due to the "+1" term. The partial fractions then have the form:- \( \frac{A}{(s-1) + j} \)- \( \frac{B}{(s-1) - j} \)where \( j \) is the imaginary unit representing \( \sqrt{-1} \). The next step involves eliminating the fractions by multiplying through by the common denominator and solving the resulting equation to find the constants \( A \) and \( B \). By substituting appropriate values of \( s \) (here, \( s = 1+j \) and \( s = 1-j \)), we calculate \( A = -1 \) and \( B = -1 \). Thus, partial fraction decomposition simplifies the expression, facilitating its inverse Laplace transform.

Time-shifting Property

The time-shifting property is crucial in this exercise because it connects the Laplace transform of time-shifted functions with exponential terms in the s-domain. The property can be expressed as:- For a function \( f(t-a) u(t-a) \), its Laplace transform is \( e^{-as} F(s) \) - Here, \( a \) is the amount of time-shift and \( u(t-a) \) is the step function.In our exercise, the inverse Laplace transform of the function incorporates a time-shift because the function \( F(s) = \frac{2(s-1) e^{-2s}}{s^2-2s+2} \) features an exponential term \( e^{-2s} \). To find the time-shifted inverse, the calculation involves first finding the inverse Laplace transform of \( G(s) \) without the exponential component:- Here, \( G(s) = \frac{2(s-1)}{s^2-2s+2} \) is transformed to \( g(t) = -e^{t} \cos(t) \).Then, applying the time-shifting property, we replace \( t \) with \( (t-2) \) and multiply by the unit step function \( u(t-2) \) to obtain- \( f(t) = -e^{(t-2)} \cos(t-2) u(t-2) \)

Complex Numbers in Differential Equations

Complex numbers play a significant role in the solving of differential equations, especially when the roots of characteristic equations are complex. In this example, we transform and solve a differential equation incorporating complex numbers through the partial fraction decomposition method, where our denominator - \( s^2 - 2s + 2 \) leads us to the form \( (s-1)^2 + 1 \) which suggests complex roots.Complex roots have the pattern \( a \pm jb \), and the corresponding inverse Laplace transform results in:- \( e^{at} \cos(bt) \) for the real part - Here, \( a = 1 \) and \( b = 1 \), indicating an exponential decay \( e^{t} \) and oscillatory cosine function \( \cos(t) \).In simpler terms, the presence of complex numbers in equations directly influences the oscillatory behaviour and exponential changes in their solutions. This is particularly clear when observing the inverse Laplace transform of the given function.

Laplace Transform of Time-shifted Functions

The Laplace transform of time-shifted functions is centered on the unit step function \( u(t-a) \), which is used to model systems that activate at a certain time \( t = a \). In our problem, the function's inherent time-shift is due to the exponential term \( e^{-2s} \), which corresponds to a delay effect in the time domain.When computing the inverse Laplace transform of such time-shifted functions:- We first remove the exponential term to focus on the main function \( G(s) \), whose inverse is determined.- Once the inverse function \( g(t) \) is obtained, it’s time-shifted by replacing \( t \) with \( t-a \), and then it is multiplied by \( u(t-a) \) to ensure that it starts at the correct time point.In this exercise, the method ultimately allows us to express the inverse Laplace transform of \( F(s) \) as a function activated at \( t = 2 \), depicted as- \( f(t) = -e^{(t-2)} \cos(t-2) u(t-2) \)This modeling with time-shift helps analyze and design systems that start after a certain delay, common in many real-world applications.