Question

a) Show that \(\mathscr{F}\\{d f(t) / d t\\}=j \omega F(\omega), \quad\) where \(F(\omega)=\mathscr{F}\\{f(t)\\} .\) Hint: Use the defining integral and integrate by parts. b) What is the restriction on \(f(t)\) if the result given in (a) is valid? c) Show that \(\mathscr{F}\left\\{d^{n} f(t) / d t^{n}\right\\}=(j \omega)^{n} F(\omega),\) where \(F(\omega)=\mathscr{F}\\{f(t)\\}\).

Short answer

a) Shown using integration by parts; b) \(f(t)\) must vanish at \(t = \pm \infty\); c) Shown by iterating the derivative result.

Step-by-step solution

Understanding the Fourier Transform Definition

The Fourier Transform of a function \(f(t)\) is given by:\[ F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-j \omega t} \, dt \] This transform converts a time-domain function into a frequency-domain function.

Fourier Transform of the Derivative

To find \( \mathscr{F}\{d f(t) / dt\} \), apply the Fourier transform to the derivative:\[ \mathscr{F}\{d f(t) / dt\} = \int_{-\infty}^{\infty} \frac{df(t)}{dt} e^{-j \omega t} \, dt \] Apply integration by parts with \( u = e^{-j \omega t} \) and \( dv = \frac{df(t)}{dt} \, dt \). Then \( du = -j \omega e^{-j \omega t} dt \) and \( v = f(t) \). The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \).

Applying Integration by Parts

Using integration by parts, obtain:\[ \int_{-\infty}^{\infty} \frac{df(t)}{dt} e^{-j \omega t} \, dt = \left. f(t) e^{-j \omega t} \right|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} f(t) (-j \omega e^{-j \omega t}) \, dt \] Assuming \( f(t) \) vanishes at \( t = \pm \infty \), the boundary term is zero, and thus:\[ j \omega \int_{-\infty}^{\infty} f(t) e^{-j \omega t} \, dt = j \omega F(\omega) \] So, the Fourier transform of the derivative is \( j \omega F(\omega) \).

Restriction on \(f(t)\)

The assumption made in the integration by parts step requires that \( f(t) \) vanishes at \( t = \pm \infty \). This is the restriction needed for the transform to hold.

Generalizing for Higher Derivatives

For \( \mathscr{F}\left\{ \frac{d^n f(t)}{dt^n} \right\} \), apply the result iteratively. Each differentiation introduces a factor of \( j \omega \):\[ \mathscr{F}\left\{ \frac{d^n f(t)}{dt^n} \right\} = (j \omega)^n F(\omega) \] Thus, the Fourier transform of the \(n\)-th derivative of \( f(t) \) is \( (j \omega)^n F(\omega) \).

Key concepts

Integration by Parts

Integration by Parts is a technique commonly used to integrate the product of two functions. In the context of Fourier Transforms, it's particularly useful when dealing with derivatives. The basic formula is given by:

• \(\int u \, dv = uv - \int v \, du\)

In this expression, \(u\) and \(dv\) are chosen parts of the integrand, and \(du\) and \(v\) are their respective derivatives and integrals. To demonstrate its application, we start with the integration involved in finding the Fourier Transform of a derivative.To solve \(\mathscr{F}\{d f(t) / dt\}\), we set \(u=e^{-j \omega t}\) and \(dv=\frac{df(t)}{dt}dt\). Thus, the derivative \(du=-j \omega e^{-j \omega t} dt\) and the integral \(v=f(t)\). Applying the formula, we have:\[\int_{-\infty}^{\infty} \frac{df(t)}{dt} e^{-j \omega t} \ dt = \left[ f(t) e^{-j \omega t} \right]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} f(t) (-j \omega e^{-j \omega t}) \, dt\]This shows how integration by parts helps in transitioning work from time-domain derivative data to frequency-domain solutions.

Frequency Domain

The Frequency Domain represents signals or functions through their frequency components, typically achieved by transforming time-based functions using the Fourier Transform.Instead of dealing with values over time, the frequency domain focuses on how much of each frequency exists in a signal. This representation makes it easier to analyze, especially for filtering or signal processing.For a time-domain function \(f(t)\), its frequency domain representation is \(F(\omega)\), calculated by the integral:

• \[F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-j \omega t} \, dt\]

Here, \(\omega\) represents the angular frequency. The Fourier transform, by this integral, condenses the entire time-based information into a condensed frequency spectrum, showing the amplitude and phase for each frequency component.Understanding this concept allows for easy manipulation of signals for tasks like filtering, as you're addressing the signal's core ingredients: its frequencies.

Time-Domain Function

The Time-Domain Function describes signals or functions concerning time, showing how the function evolves at each point on the time axis.For practical purposes like audio signals or electrical waveforms, it represents how the signal changes over time. This raw structure is essential as the initial form of data before transforming into other domains, such as frequency.To illustrate, a sound wave captured by a microphone is a time-domain function where the raw waveform details the changing air pressure over time. Using the Fourier Transform, this can be translated into the frequency domain, highlighting the wave's pitch and tone.The transition from time-domain \(f(t)\) to frequency-domain \(F(\omega)\) occurs through the Fourier Transform. This function's beauty lies in its ability to unveil the intricate details of signal changes across time, setting the stage for frequency analysis, a fundamental task in areas like audio and communications engineering.

Higher Derivatives

Higher Derivatives describe the successive derivatives of a function, going beyond merely the first derivative to express rates of change of rates of change.When dealing with the Fourier Transform of derivatives, each step towards higher derivatives introduces additional factors of frequency. The relation becomes clear:

• \(\mathscr{F}\left\{\frac{d^n f(t)}{dt^n}\right\}\) = \((j \omega)^n F(\omega)\)

This equation implies that every differentiation brings a multiplication by \(j \omega\), where \(j\) is the imaginary unit.The practical result of this property is powerful for control systems and signal processing. If a system response involves high derivatives, the frequency domain expression becomes more straightforward. It reveals how increasing derivatives enhance specific frequency components' amplitude.In essence, being able to handle higher derivatives through Fourier Transforms offers a simplified, powerful tool to decode complexities in dynamic systems, where understanding various orders of change is crucial.