Question

Let $$ H(x)= \begin{cases}0, & x<0, \\ 1, & x \geq 0 .\end{cases} $$ Find the Fourier transform of (a) \(f(x)=H(x) e^{-a x}, \quad a>0\) (b) \(f(x)=H(x) e^{-a x} \cos b t, \quad a>0, \quad b \neq 0\) (c) \(f(x)=H(x) e^{-a x} \sin b t, \quad a>0, \quad b \neq 0\)

Short answer

(a) \( F(k) = \frac{1}{a + 2\pi i k} \). (b) \( F(k) = \frac{1}{2} \left( \frac{1}{a + 2\pi i k - ib} + \frac{1}{a + 2\pi i k + ib} \right) \). (c) \( F(k) = \frac{1}{2i} \left( \frac{1}{a + 2\pi i k - ib} - \frac{1}{a + 2\pi i k + ib} \right) \).

Step-by-step solution

Understand the Heaviside Function

The Heaviside function, denoted as \( H(x) \), is defined as \( 0 \) for \( x < 0 \) and \( 1 \) for \( x \geq 0 \). It is often used to 'turn on' a function at \( x = 0 \). This function will be relevant in each part of our exercise as it defines the domain of integration for the Fourier transforms.

Set Up Fourier Transform for Part (a)

For part (a), we need to find the Fourier transform of \( f(x) = H(x) e^{-ax} \). The Fourier transform is given by the integral \[ F(k) = \int_{-\infty}^{\infty} f(x) e^{-2\pi i k x} \, dx \]. Given \( H(x) \), the integration limits change from \( 0 \) to \( \infty \). Thus, the integral becomes \[ F(k) = \int_{0}^{\infty} e^{-ax} e^{-2\pi i k x} \, dx. \]

Solve the Integral for Part (a)

Combine exponents to simplify: \( e^{-(a + 2 \pi i k)x} \). Now solve the integral: \[ F(k) = \int_{0}^{\infty} e^{-(a + 2\pi i k)x} \, dx = \frac{1}{a + 2 \pi i k}. \] The result follows from the standard form of the integral of an exponential function.

Apply the Fourier Transform to Part (b)

For part (b), the function is \( f(x) = H(x) e^{-ax} \cos(bx) \). Plug into the Fourier transform with limits from \( 0 \) to \( \infty \): \[ F(k) = \int_{0}^{\infty} e^{-ax} \cos(bx) e^{-2\pi i kx} \, dx. \] Substitute \( \cos(bx) \) using Euler's formula: \( \frac{1}{2}(e^{ibx} + e^{-ibx}) \).

Simplify and Integrate for Part (b)

Splitting the cosine into different exponential terms gives two integrals: \[ F(k) = \frac{1}{2} \left( \int_{0}^{\infty} e^{-(a + 2\pi i k - ib)x} \, dx + \int_{0}^{\infty} e^{-(a + 2\pi i k + ib)x} \, dx \right). \] Solve each using the exponential integral formula: \[ F(k) = \frac{1}{2} \left( \frac{1}{a + 2\pi i k - ib} + \frac{1}{a + 2\pi i k + ib} \right). \]

Set Up the Fourier Transform for Part (c)

For part (c), we have \( f(x) = H(x) e^{-ax} \sin(bx) \). Implement the Fourier transform: \[ F(k) = \int_{0}^{\infty} e^{-ax} \sin(bx) e^{-2\pi i kx} \, dx. \] Use \( \sin(bx) = \frac{1}{2i}(e^{ibx} - e^{-ibx}) \) to split the integral into two components.

Solve the Integral for Part (c)

Substituting the sine representation: \[ F(k) = \frac{1}{2i} \left( \int_{0}^{\infty} e^{-(a + 2\pi i k - ib)x} \, dx - \int_{0}^{\infty} e^{-(a + 2\pi i k + ib)x} \, dx \right). \] Computing each integral gives: \[ F(k) = \frac{1}{2i} \left( \frac{1}{a + 2\pi i k - ib} - \frac{1}{a + 2\pi i k + ib} \right). \]

Finalize Results and Interpret

Each transformation results in rational functions of \( k \): (a) \( F(k) = \frac{1}{a + 2\pi i k} \), (b) \( F(k) = \frac{1}{2} \left( \frac{1}{a + 2\pi i k - ib} + \frac{1}{a + 2\pi i k + ib} \right) \), and (c) \( F(k) = \frac{1}{2i} \left( \frac{1}{a + 2\pi i k - ib} - \frac{1}{a + 2\pi i k + ib} \right) \). These results indicate the frequencies each function contributes to a signal.

Key concepts

Heaviside Function

The Heaviside function, also known as the unit step function, is a simple yet powerful mathematical tool used in various areas like control theory and signal processing. Its role is to "switch on" a particular part of a function at a specific point, usually at zero. For instance, in this exercise, it helps us define when a function starts contributing to its standard form.Mathematically, it is defined as a piecewise function:

• It equals 0 for all values of \( x < 0 \), essentially "turning off" the function.
• It equals 1 for \( x \geq 0 \), "activating" the function from this point onwards.

This clear-cut function allows for controlled analysis over specific domains. It simplifies dealing with discontinuities, allowing continuity where needed. When used in Fourier transforms, like in our exercise, it adjusts the limits of integration to focus on the important range of the function. The Heaviside function aids in smooth transitions in signal analysis, breaking a complex waveform into manageable parts.

Exponential Integral

Integrating exponential functions can often seem intimidating, but there's a formula that helps demystify it. The exponential integral is central in mathematics, especially when dealing with Fourier transforms, like the ones in our exercise.In its most typical form, the integral calculates:\[ \int_{0}^{\infty} e^{-\beta x} \, dx = \frac{1}{\beta}\]where \( \beta \) is a constant. This result showcases how adding exponential decay contributes interesting behavior, pulling the evaluation towards zero, stabilizing where otherwise calculations may have been undefined.

• Exponential integrals diminish rapidly at infinity, making the whole process much more stable.
• Their simplicity is also their beauty, reducing the integral to straightforward algebraic terms.

Exponential integrals are used in a variety of applications like decay processes, financial models, and anywhere time provides a damping effect. Understanding these integrals helps better manage transformations into frequency domains, allowing precise calculations of involvement at various frequencies.

Euler's Formula

Euler's formula is a remarkable equation in mathematics that establishes a profound connection between exponential functions and trigonometric functions. It states that for any real number \( x \),\[e^{ix} = \cos x + i\sin x\]Euler's formula provides a bridge between the real and imaginary components of complex numbers, a link that is crucial in Fourier analysis.

• It transforms trigonometric expressions into exponential ones, simplifying calculations.
• This transformation helps in expressing waves and oscillations succinctly.

In our exercise, Euler's formula conveniently turned complex trigonometric functions into linear components, making integration and manipulation more approachable. Its application is essential in areas dealing with signal processing, quantum mechanics, and electrical engineering, offering smoother methods to solve otherwise intricate problems.

Frequency Analysis

Frequency analysis is a method for dissecting waves and oscillations into their constituent frequencies. This examination is particularly prevalent in fields like signal processing and acoustics. The Fourier transform is a powerful tool in frequency analysis, taking time-domain information and transforming it into the frequency domain. In simpler terms, it helps determine which frequencies are present in a particular signal and how strong they are.

• It reveals hidden frequency components in datasets, offering insights into the underlying patterns.
• By understanding each frequency's role and magnitude, we can analyze and reconstruct signals efficiently.

This process is critical for both theoretical explorations and practical applications, such as music production, seismology, and even in medical imaging. In the exercise, using the Fourier transform helped uncover how each function affects the frequency spectrum, unraveling the ongoing interactions within the signal.