Question

Exercises 6-8: Compute the inverse Laplace transform of the given matrix function \(\mathbf{Y}(s)\). \(\mathbf{Y}(s)=\left[\begin{array}{c}\frac{1}{s} \\ \frac{2}{s^{2}+2 s+2} \\\ \frac{1}{s^{2}+s}\end{array}\right]\)

Short answer

Question: Find the inverse Laplace transform of the following matrix function: \(\mathbf{Y}(s) = \left[\begin{array}{c} \frac{1}{s} \\ \frac{2}{s^{2}+2 s+2} \\ \frac{1}{s^{2}+s} \end{array}\right]\) Answer: The inverse Laplace transformed matrix is: \(\mathbf{y}(t) = \left[\begin{array}{c} u(t) \\ e^{-t} \sin{t} \\ u(t) - e^{-t} \end{array}\right]\)

Step-by-step solution

Find the inverse Laplace transform of the first element

To find the inverse Laplace transform of \(\frac{1}{s}\), we simply recall that the inverse Laplace transform of \(\frac{1}{s}\) is the unit step function, у(t).

Find the inverse Laplace transform of the second element

To find the inverse Laplace transform of \(\frac{2}{s^{2}+2 s+2}\), we first rewrite the denominator and set it to complete the square form. The denominator can be written as: \(s^{2} + 2s + 2 = (s+1)^{2} + 1\) Now, we have the following Laplace transform: \(\frac{2}{(s+1)^{2}+1}\) Recall the inverse Laplace transform formula: \(L^{-1} \{ \frac{k}{(s-a)^2 + k^2} \} = e^{at} \sin{kt}\) Using the formula: \(L^{-1}\{\frac{2}{(s+1)^{2}+1}\} = e^{-t} \sin{t}\)

Find the inverse Laplace transform of the third element

To find the inverse Laplace transform of \(\frac{1}{s^{2}+s}\), we perform partial fraction decomposition: \(\frac{1}{s^{2}+s} = \frac{A}{s} + \frac{B}{s+1}\) Multiplying both sides by the common denominator, we get: \(1 = A(s+1) + Bs\) Setting s = 0, we find A = 1; Setting s = -1, we find B = -1. Thus, \(\frac{1}{s^{2}+s} = \frac{1}{s} - \frac{1}{s+1}\) Now we find the inverse Laplace transform of each term: \(L^{-1}\{\frac{1}{s}\} = u(t)\) and \(L^{-1}\{-\frac{1}{s+1}\} = -e^{-t}\) By linearity property of Laplace transform, we get: \(L^{-1}\{\frac{1}{s^{2}+s}\} = u(t) - e^{-t}\)

Write down the final inverse Laplace transformed matrix

Now that we have the inverse Laplace transforms of each element in the original matrix function, we can compile them into a single matrix function that represents the inverse Laplace transform of \(\mathbf{Y}(s)\): $\mathbf{y}(t)=\left[\begin{array}{c} u(t) \\ e^{-t} \sin{t} \\ u(t) - e^{-t} \end{array}\right]$ And that's our final inverse Laplace transformed matrix.

Key concepts

Laplace Transforms

The Laplace Transform is a powerful mathematical tool used in engineering and physics to simplify complex differential equations into algebraic equations. This transformation can be particularly beneficial when analyzing linear time-invariant systems. The general form of the Laplace Transform of a function f(t), which is defined for t ≥ 0, is given by the integral formula:
\[L\{f(t)\} = \int_{0}^{\infty} e^{-st}f(t) \, dt\]
where s is a complex number. The inverse Laplace Transform, denoted as L^-1{F(s)}, is the process of converting a function in the s-domain back into the time domain. This operation essentially reverses the effect of the Laplace Transform, and is crucial for understanding dynamic behaviors in the time domain after manipulating and solving equations in the s-domain. The inverse Laplace Transform is particularly helpful in solving linear ordinary differential equations that have been transformed into algebraic ones.

Partial Fraction Decomposition

Partial fraction decomposition is an algebraic technique useful in breaking down complex rational expressions into simpler ones, which are easier to integrate, differentiate, or invert if dealing with Laplace Transforms.
Take a rational function of the form: \[ \frac{P(s)}{Q(s)} \]
where P(s) and Q(s) are polynomials, and degree of P(s) < degree of Q(s). This function can be expressed as a sum of simpler fractions with polynomials of lower degree in the denominator. When applying this method to invert a Laplace Transform, one can rewrite the original complicated expression into a sum of terms whose inverse transforms are readily known or easier to calculate. This approach is neatly illustrated in the textbook solution with the decomposition of \(\frac{1}{s^{2}+s}\), which simplifies the process of finding its inverse Laplace Transform.

Unit Step Function

The unit step function, often denoted as u(t), is a piecewise function that is zero for t < 0 and one for t ≥ 0. It is a fundamental concept in the study of system responses, notably in electrical engineering and control theory. In the context of Laplace Transforms, the unit step function represents a sudden change or a 'switch-on' event at a particular moment in time.
The inverse Laplace Transform of \( \frac{1}{s} \) is u(t), which signifies a constant input or presence starting from t = 0. It is essentially the mathematical equivalent of switching on a signal that stays on indefinitely. This simple yet important function helps in representing piecewise continuous signals within the framework of Laplace Transforms, aiding in the analysis of systems that exhibit step changes.

Complex Exponentials

Complex exponentials, featuring the base of the natural logarithm e raised to a complex power, play a central role in various areas of mathematics, including the study of Laplace Transforms. For any complex number a + bi, the complex exponential is represented as:
\[ e^{(a+bi)t} = e^{at}(\cos(bt) + i\sin(bt))\]
where i is the imaginary unit, and t is a real number. In the solutions given by the textbook, complex exponentials emerge in the form of functions like e^{-t} which are part of the inverse Laplace Transforms of terms like \( \frac{2}{(s+1)^2 + 1} \). The role of complex exponentials is particularly noteworthy when describing oscillatory or damped harmonic motion through the sine and cosine terms that appear as a result of Euler's formula. They are essential for understanding how systems respond over time, especially when these responses are oscillatory in nature, as seen in the sinusoidal components of the solutions.