Laplace transform of the function f(t) denoted by F(s) is given by
The Laplace transform is a fundamental mathematical tool widely used in engineering and physics to analyze linear time-invariant systems. It transforms a function from the time domain (\(t\)) to the complex frequency domain (\(s\)). This transformation simplifies the process of solving differential equations, making complex problems more manageable by converting them into algebraic problems.
Laplace Transform Definition
The definition of the Laplace transform for a function \(f(t)\) (where \(t \ge 0\)), denoted as \(F(s)\) or \(\mathcal{L}\{f(t)\}\), is given by a specific improper integral. This integral converts the time-domain function \(f(t)\) into a function of the complex variable \(s\).
The standard formula for the unilateral Laplace transform is:
\[ F(s) = \mathcal{L}\{f(t)\} = \int_0^{\infty} f(t) e^{-st} dt \]
In this definition:
- \(f(t)\) is the original function in the time domain, usually defined for \(t \ge 0\).
- \(s\) is a complex variable, often represented as \(s = \sigma + j\omega\), where \(\sigma\) and \(\omega\) are real numbers and \(j = \sqrt{-1}\).
- \(e^{-st}\) is the kernel of the transformation. The negative sign in the exponent is crucial for the transform's convergence and properties.
- The integral is taken from \(0\) to \(\infty\), indicating that the Laplace transform is a unilateral transform, considering only the function's behavior for non-negative time.
Formula Comparison and Analysis
Let's examine the provided options against the standard definition of the Laplace transform:
- Option 1: \( \displaystyle\int_0^{\infty} f(t) e^{st} dt \)
This option has an incorrect sign in the exponent of the exponential term (\(e^{st}\) instead of \(e^{-st}\)). This form does not represent the standard Laplace transform.
- Option 2: \( \displaystyle\int_{-\infty}^{\infty} f(t) e^{st} dt \)
This option has two fundamental errors. Firstly, the limits of integration are from \(-\infty\) to \(\infty\), which corresponds to a bilateral (two-sided) Laplace transform or resembles the Fourier transform, not the standard unilateral Laplace transform. Secondly, the exponent has an incorrect sign (\(e^{st}\) instead of \(e^{-st}\)).
- Option 3: \( \displaystyle\int_0^{\infty} f(t) e^{-st} dt \)
This option perfectly matches the standard definition of the unilateral Laplace transform, including the correct integration limits (\(0\) to \(\infty\)) and the correct negative sign in the exponent (\(e^{-st}\)).
- Option 4: \( \displaystyle\int_0^{\infty} f(t) e^{t} dt \)
This option is incorrect because the exponent of the exponential term is missing the complex variable \(s\). The integration is with respect to \(t\), and the transformed function must be a function of \(s\), not just a constant after integration.
Conclusion on Laplace Transform Formula
Based on the standard mathematical definition, the Laplace transform of the function \(f(t)\) is correctly given by the integral with limits from \(0\) to \(\infty\) and the exponential term \(e^{-st}\).
Therefore, the correct representation for the Laplace transform \(F(s)\) of the function \(f(t)\) is:
\[ F(s) = \int_0^{\infty} f(t) e^{-st} dt \]
Important Questions from Laplace Transform
Which of the following is the final value of the impulse response of the system whose transfer function is
(2s + 1)/(s 4 + 8s 3 + 16s 2 + s)
Find the Laplace transform for the following time domain.
y(t) = -2te -t + 4e -t - 4e -2t
Match List I with List II
List – I
List – II
f(t)
F(S)
A.
e -at
I.
\(\rm \frac{s}{s^2+ \omega^2}\)
B.
te at
II.
\(\rm \frac{\omega}{s^2+ \omega^2}\)
C.
sinωt
III.
\(\rm \frac{1}{(s- a)^2}\)
D.
cosωt
IV.
\(\rm \frac{1}{(s+ a)}\)
Choose the correct answer from the options given below:
The Laplace transform of sin h (at) is
The unilateral Laplace transform of f(t) is \(\frac{1}{s^2+s+1}\). The unilateral Laplace transform of t f(t) is
