Understanding the Laplace Transform of \(t \cos(at)\)

To find the Laplace transform of the function \(f(t) = t \cos(at)\), we can use a standard property of Laplace transforms related to multiplication by \(t\). The property states that if \(\mathcal{L}\{f(t)\} = F(s)\), then \(\mathcal{L}\{t f(t)\} = -\frac{d}{ds} F(s)\).

Step-by-Step Calculation of Laplace Transform

Let's break down the process:

1. First, we need to find the Laplace transform of the function \(f(t) = \cos(at)\). The Laplace transform of \(\cos(at)\) is a standard result:

   \(\mathcal{L}\{\cos(at)\} = \frac{s}{s^2 + a^2}\)

   So, in this case, \(F(s) = \frac{s}{s^2 + a^2}\).
2. Next, we apply the property for multiplication by \(t\):

   \(\mathcal{L}\{t \cos(at)\} = -\frac{d}{ds} \left(\mathcal{L}\{\cos(at)\}\right) = -\frac{d}{ds} \left(\frac{s}{s^2 + a^2}\right)\)

3. Now, we need to differentiate \(F(s) = \frac{s}{s^2 + a^2}\) with respect to \(s\). We use the quotient rule for differentiation, which states that if \(F(s) = \frac{u(s)}{v(s)}\), then \(F'(s) = \frac{u'(s)v(s) - u(s)v'(s)}{[v(s)]^2}\).

   Here, \(u(s) = s\) and \(v(s) = s^2 + a^2\).

   Let's find the derivatives of \(u(s)\) and \(v(s)\) with respect to \(s\):

   • \(u'(s) = \frac{d}{ds}(s) = 1\)
   • \(v'(s) = \frac{d}{ds}(s^2 + a^2) = 2s\)

   Now, apply the quotient rule:

   \(\frac{d}{ds} \left(\frac{s}{s^2 + a^2}\right) = \frac{(1)(s^2 + a^2) - (s)(2s)}{(s^2 + a^2)^2}\)

   Simplify the numerator:

   \(s^2 + a^2 - 2s^2 = a^2 - s^2\)

   So, the derivative is:

   \(\frac{d}{ds} \left(\frac{s}{s^2 + a^2}\right) = \frac{a^2 - s^2}{(s^2 + a^2)^2}\)

4. Finally, we apply the negative sign from the property \(\mathcal{L}\{t f(t)\} = -\frac{d}{ds} F(s)\):

   \(\mathcal{L}\{t \cos(at)\} = - \left(\frac{a^2 - s^2}{(s^2 + a^2)^2}\right)\)

   Distribute the negative sign in the numerator:

   \(\mathcal{L}\{t \cos(at)\} = \frac{-(a^2 - s^2)}{(s^2 + a^2)^2} = \frac{-a^2 + s^2}{(s^2 + a^2)^2} = \frac{s^2 - a^2}{(s^2 + a^2)^2}\)

The Laplace transform of \(t \cos(at)\) is \(\frac{s^2 - a^2}{(s^2 + a^2)^2}\).

Comparing with Options

Let's look at the given options and see which one matches our calculated result:

• Option 1: \(\frac{{{s^2} + {a^2}}}{{{{\left( {{s^2} - {a^2}} \right)}^2}}}\)
• Option 2: \(\frac{s}{{{{\left( {{s^2} - {a^2}} \right)}^2}}}\)
• Option 3: \(\frac{{{s^2} - {a^2}}}{{{{\left( {{s^2} + {a^2}} \right)}^2}}}\)
• Option 4: \(\frac{s}{{{{\left( {{s^2} + {a^2}} \right)}^2}}}\)

Our calculated result \(\frac{s^2 - a^2}{(s^2 + a^2)^2}\) matches Option 3.

Revision Table: Key Laplace Transforms and Properties

Additional Information on Laplace Transforms

The Laplace transform is a powerful mathematical tool used to convert a function of a real variable \(t\) (often time) to a function of a complex variable \(s\) (complex frequency). This transformation can simplify solving differential equations and analyzing linear time-invariant (LTI) systems.

The definition of the Laplace transform of a function \(f(t)\) for \(t \ge 0\) is given by the integral:

\(\mathcal{L}\{f(t)\} = F(s) = \int_0^\infty e^{-st} f(t) dt\)

The property \(\mathcal{L}\{t f(t)\} = -\frac{d}{ds} F(s)\) used in this problem is a specific case of the differentiation with respect to \(s\) property. This property is particularly useful for finding the Laplace transforms of functions multiplied by powers of \(t\), such as \(t^n f(t)\).

The Laplace transform exists for functions \(f(t)\) for which the integral converges. The variable \(s\) can be a real number or a complex number, typically represented as \(s = \sigma + j\omega\), where \(\sigma\) and \(\omega\) are real numbers and \(j\) is the imaginary unit.