Understanding Laplace Transforms and Function Matching

This problem requires us to match common time-domain functions \(f(t)\) with their corresponding Laplace transforms \(F(s)\). The Laplace transform is a powerful tool used in engineering and physics to transform differential equations in the time domain \(t\) into algebraic equations in the complex frequency domain \(s\). This makes solving systems much simpler.

Key Laplace Transform Formulas

To solve this matching problem, we need to know the standard Laplace transforms for the given functions. Here are the relevant formulas:

• The Laplace transform of \(e^{at}\) is \(\mathcal{L}\{e^{at}\} = \frac{1}{s-a}\).
• The Laplace transform of \(e^{-at}\) is \(\mathcal{L}\{e^{-at}\} = \frac{1}{s-(-a)} = \frac{1}{s+a}\).
• The Laplace transform of \(t^n\) is \(\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}\).
• The Laplace transform of \(\sin(\omega t)\) is \(\mathcal{L}\{\sin(\omega t)\} = \frac{\omega}{s^2 + \omega^2}\).
• The Laplace transform of \(\cos(\omega t)\) is \(\mathcal{L}\{\cos(\omega t)\} = \frac{s}{s^2 + \omega^2}\).

We also need the frequency shifting property: If \(\mathcal{L}\{f(t)\} = F(s)\), then \(\mathcal{L}\{e^{at}f(t)\} = F(s-a)\).

Step-by-Step Matching Process

Let's find the Laplace transform for each function in List I and match it with the correct expression in List II.

Matching A: \(f(t) = e^{-at}\)

Using the standard formula for \(e^{-at}\), we have:

\(\mathcal{L}\{e^{-at}\} = \frac{1}{s - (-a)} = \frac{1}{s+a}\)

This matches expression IV in List II.

So, A – IV.

Matching B: \(f(t) = te^{at}\)

This function is a product of \(t\) and \(e^{at}\). We can use the frequency shifting property. First, let's find the Laplace transform of \(f(t) = t\). Here, \(n=1\).

\(\mathcal{L}\{t^1\} = \frac{1!}{s^{1+1}} = \frac{1}{s^2}\)

Now, apply the frequency shifting property with \(a\):

\(\mathcal{L}\{e^{at} \cdot t\} = F(s-a)\)

Where \(F(s) = \frac{1}{s^2}\). Replacing \(s\) with \((s-a)\), we get:

\(\mathcal{L}\{te^{at}\} = \frac{1}{(s-a)^2}\)

This matches expression III in List II.

So, B – III.

Matching C: \(f(t) = \sin(\omega t)\)

Using the standard formula for \(\sin(\omega t)\), we have:

\(\mathcal{L}\{\sin(\omega t)\} = \frac{\omega}{s^2 + \omega^2}\)

This matches expression II in List II.

So, C – II.

Matching D: \(f(t) = \cos(\omega t)\)

Using the standard formula for \(\cos(\omega t)\), we have:

\(\mathcal{L}\{\cos(\omega t)\} = \frac{s}{s^2 + \omega^2}\)

This matches expression I in List II.

So, D – I.

Summary of Matching

Let's put our findings into a table:

Comparing with Options

The correct matching is A – IV, B – III, C – II, D – I.

Let's check the given options:

• Option 1: A – I, B – II, C – IV, D – III (Incorrect)
• Option 2: A – III, B – II, C – I, D – IV (Incorrect)
• Option 3: A – IV, B – III, C – I, D – II (Incorrect)
• Option 4: A – IV, B – III, C – II, D – I (Correct)

Option 4 correctly matches our derived Laplace transforms.

Revision Table: Common Laplace Transforms

Additional Information: Properties of Laplace Transform

Beyond the basic formulas, understanding the properties of the Laplace transform is crucial for solving more complex problems. Some important properties include:

• Linearity: \(\mathcal{L}\{c_1 f_1(t) + c_2 f_2(t)\} = c_1 F_1(s) + c_2 F_2(s)\)
• Frequency Shifting: \(\mathcal{L}\{e^{at} f(t)\} = F(s-a)\) (Used for \(te^{at}\))
• Time Shifting: \(\mathcal{L}\{f(t-a)u(t-a)\} = e^{-as} F(s)\) for \(a \ge 0\)
• Derivative Property: \(\mathcal{L}\{f'(t)\} = sF(s) - f(0)\)
• Integral Property: \(\mathcal{L}\{\int_0^t f(\tau) d\tau\} = \frac{1}{s} F(s)\)
• Multiplication by t: \(\mathcal{L}\{t^n f(t)\} = (-1)^n \frac{d^n}{ds^n} F(s)\) (Alternative method for \(te^{at}\))

These properties allow us to find Laplace transforms of more complicated functions by breaking them down or transforming known transforms.