Laplace transform of a unit parabolic function occurring at t = 0 is given by ______.

Explanation:

Laplace Transform of a Unit Parabolic Function

Definition: The Laplace transform is a widely used integral transform in engineering and physics that converts a function of time (t) into a function of complex frequency (s). The Laplace transform of a function f(t) is defined as:

\[ \mathcal{L}\{f(t)\} = F(s) = \int_{0}^{\infty} e^{-st} f(t) \, dt \]

Where:

• \( \mathcal{L} \) denotes the Laplace transform operator
• \( f(t) \) is the time-domain function
• \( F(s) \) is the Laplace-transformed function in the frequency domain
• \( s \) is a complex number frequency parameter

Unit Parabolic Function: The unit parabolic function, also known as the second-order ramp function, is given by:

\[ f(t) = \frac{t^2}{2} \]

for \( t \geq 0 \). This function represents a parabolic increase starting from \( t = 0 \).

Laplace Transform of the Unit Parabolic Function:

To find the Laplace transform of the unit parabolic function \( f(t) = \frac{t^2}{2} \), we use the definition of the Laplace transform:

\[ \mathcal{L}\left\{\frac{t^2}{2}\right\} = \int_{0}^{\infty} e^{-st} \cdot \frac{t^2}{2} \, dt \]

We can simplify this by factoring out the constant \( \frac{1}{2} \):

\[ \mathcal{L}\left\{\frac{t^2}{2}\right\} = \frac{1}{2} \int_{0}^{\infty} e^{-st} t^2 \, dt \]

To solve this integral, we use the standard Laplace transform formula for \( t^n \):

\[ \mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}} \]

For \( n = 2 \), we have:

\[ \mathcal{L}\{t^2\} = \frac{2!}{s^{2+1}} = \frac{2}{s^3} \]

Therefore, the Laplace transform of \( \frac{t^2}{2} \) is:

\[ \mathcal{L}\left\{\frac{t^2}{2}\right\} = \frac{1}{2} \cdot \frac{2}{s^3} = \frac{1}{s^3} \]

Thus, the Laplace transform of a unit parabolic function occurring at \( t = 0 \) is \( \frac{1}{s^3} \).

Correct Option Analysis:

The correct option is:

Option 1: \( \frac{1}{s^3} \)

This option correctly represents the Laplace transform of the unit parabolic function \( \frac{t^2}{2} \), as derived above.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 2: 1

This option is incorrect because the Laplace transform of the unit parabolic function is not a constant value. The correct Laplace transform has a dependency on the complex frequency parameter \( s \).

Option 3: \( \frac{1}{s^2} \)

This option is incorrect as it represents the Laplace transform of a unit ramp function \( t \), not the unit parabolic function \( \frac{t^2}{2} \). The unit ramp function's Laplace transform is indeed \( \frac{1}{s^2} \), but for the parabolic function, it is \( \frac{1}{s^3} \).

Option 4: \( \frac{1}{s} \)

This option is incorrect because it represents the Laplace transform of a unit step function, not the unit parabolic function. The unit step function (Heaviside function) has a Laplace transform of \( \frac{1}{s} \).

Conclusion:

Understanding the Laplace transform of different functions is essential for solving differential equations and analyzing systems in the frequency domain. The unit parabolic function, represented by \( \frac{t^2}{2} \), has a Laplace transform of \( \frac{1}{s^3} \). This transform is particularly useful in control systems and signal processing, where such functions are commonly encountered. By comparing the correct option with the incorrect ones, we can clearly see the importance of accurately applying the Laplace transform formulas to various functions.