Question

Let \(f \in G(\mathbb{R})\). Determine the Fourier transform of (a) \(f(-x)\) (b) \(f\left(x-x_{0}\right), x_{0}\) is a real constant (c) \(f(x) e^{i \omega_{0} x}, \omega_{0}\) is a real constant (d) \(f(x) \sin \omega_{0} x\) (e) \(f(x) \cos \omega_{0} x\) (f) \(e^{i x} f(3 x)\) (g) \(f(2 x)\)

Short answer

(a) \(\mathcal{F}\{f(-x)\} = \mathcal{F}\{f(x)\}(-\xi)\). (b) \(e^{-i\xi x_0} \mathcal{F}\{f(x)\}\). (c) \(\mathcal{F}\{f(x)\}(\xi - \omega_0)\). (d) \(\frac{1}{2i}[\mathcal{F}\{f(x)\}(\xi-\omega_0) - \mathcal{F}\{f(x)\}(\xi+\omega_0)]\). (e) \(\frac{1}{2}[\mathcal{F}\{f(x)\}(\xi-\omega_0) + \mathcal{F}\{f(x)\}(\xi+\omega_0)]\). (f) \(\frac{1}{3}\mathcal{F}\{f(x)\}\left( \frac{\xi - 1}{3} \right)\). (g) \(\frac{1}{2}\mathcal{F}\{f(x)\}\left( \frac{\xi}{2} \right)\).

Step-by-step solution

Fourier Transform of Negative Argument

The Fourier transform of a function is given by \[ \mathcal{F}\{f(x)\} = \int_{-\infty}^{\infty} f(x) e^{-i\xi x} \,dx. \] For the function \( f(-x) \), the Fourier transform becomes \[ \mathcal{F}\{f(-x)\} = \int_{-\infty}^{\infty} f(-x) e^{-i\xi x} \,dx. \] By substituting \( u = -x \), which gives \( du = -dx \), we find \[ = \int_{\infty}^{-\infty} f(u) e^{i\xi u} (-du) = \int_{-\infty}^{\infty} f(u) e^{i\xi u} \,du = \mathcal{F}\{f(x)\}(-\xi), \] which shows that the Fourier transform of \( f(-x) \) is the original Fourier transform evaluated at \(-\xi\).

Fourier Transform of a Shifted Function

For the function \( f(x-x_0) \), the Fourier transform is given by \[ \mathcal{F}\{f(x-x_0)\} = \int_{-\infty}^{\infty} f(x-x_0) e^{-i\xi x} \,dx. \] By substituting \( u = x-x_0 \), which implies \( x = u+x_0 \) and \( dx = du \), we have \[ = \int_{-\infty}^{\infty} f(u) e^{-i\xi (u+x_0)} \,du \] \[ = e^{-i\xi x_0} \int_{-\infty}^{\infty} f(u) e^{-i\xi u} \,du = e^{-i\xi x_0} \mathcal{F}\{f(x)\}, \] indicating that the Fourier transform of a shifted function is the product of the original Fourier transform and a complex exponential term.

Fourier Transform of a Modulated Function

For the function \( f(x) e^{i\omega_0 x} \), the Fourier transform is \[ \mathcal{F}\{f(x) e^{i\omega_0 x}\} = \int_{-\infty}^{\infty} f(x) e^{i\omega_0 x} e^{-i\xi x} \,dx \] \[ = \int_{-\infty}^{\infty} f(x) e^{-i(\xi-\omega_0)x} \,dx = \mathcal{F}\{f(x)\}(\xi - \omega_0), \] indicating that the Fourier transform of a modulated function is the original Fourier transform shifted by \( \omega_0 \).

Fourier Transform of Function Sine Modulation

For the function \( f(x) \sin(\omega_0 x) \), write \( \sin(\omega_0 x) \) as \[ \sin(\omega_0 x) = \frac{e^{i\omega_0 x} - e^{-i\omega_0 x}}{2i}. \] The Fourier transform is\[ \mathcal{F}\{f(x) \sin(\omega_0 x)\} = \frac{1}{2i} \left[ \mathcal{F}\{f(x)e^{i\omega_0 x}\} - \mathcal{F}\{f(x)e^{-i\omega_0 x}\} \right]. \] Substitute results from the previous step: \[ = \frac{1}{2i} \left[ \mathcal{F}\{f(x)\}(\xi - \omega_0) - \mathcal{F}\{f(x)\}(\xi + \omega_0) \right]. \]

Fourier Transform of Function Cosine Modulation

For the function \( f(x) \cos(\omega_0 x) \), write \( \cos(\omega_0 x) \) as \[ \cos(\omega_0 x) = \frac{e^{i\omega_0 x} + e^{-i\omega_0 x}}{2}. \] The Fourier transform becomes\[ \mathcal{F}\{f(x) \cos(\omega_0 x)\} = \frac{1}{2} \left[ \mathcal{F}\{f(x)e^{i\omega_0 x}\} + \mathcal{F}\{f(x)e^{-i\omega_0 x}\} \right]. \] Using previous results, we find\[ = \frac{1}{2} \left[ \mathcal{F}\{f(x)\}(\xi - \omega_0) + \mathcal{F}\{f(x)\}(\xi + \omega_0) \right]. \]

Fourier Transform of a Scaled and Modulated Function

For the function \( e^{i x} f(3x) \), consider each part separately: \( e^{ix} \) is a modulation and \( f(3x) \) is a scaling. First, the scaling: \[ \mathcal{F}\{f(3x)\} = \frac{1}{3} \mathcal{F}\{f(x)\} \left( \frac{\xi}{3} \right). \] Now, combine with modulation: \[ \mathcal{F}\{e^{ix} f(3x)\} = \mathcal{F}\{f(3x)\} \left( \xi - 1 \right), \] Resulting in \[ = \frac{1}{3} \mathcal{F}\{f(x)\}\left( \frac{\xi - 1}{3} \right). \]

Fourier Transform of a Scaled Function

For the function \( f(2x) \), use the scaling property:\[ \mathcal{F}\{f(2x)\} = \frac{1}{2} \mathcal{F}\{f(x)\} \left( \frac{\xi}{2} \right). \] This shows that scaling the argument of the function results in compressing the Fourier transform by the same scaling factor, along with a scaling of the amplitude.

Key concepts

Shifted Function

In the world of Fourier Transforms, understanding the impact of shifting a function is key. When you shift a function by a constant value, say by moving it to the left or right on a graph, the Fourier Transform changes in a predictable manner.

Imagine the function is shifted by a constant \(x_0\). Mathematically, this is represented as \(f(x-x_0)\). The core idea here is that this shift results in multiplying the Fourier Transform of the original function by a complex exponential factor. The new Fourier Transform becomes \(e^{-i\xi x_0} \mathcal{F}\{f(x)\}\). This means that shifting the function in the time domain results in a phase shift in the frequency domain.

• This is an essential property when dealing with real-world signals being time-shifted, like audio signals or control systems.
• The phase shift depends directly on the amount of shift and the frequency component.

Modulated Function

Modulating a function in the context of Fourier Transforms involves multiplying it by a complex exponential or sinusoidal function. When you have a function \(f(x)\) and modulate it with \(e^{i\omega_0 x}\), the Fourier Transform of the result isn't as complicated as it might seem.

The Fourier Transform of the modulated function \(f(x) e^{i\omega_0 x}\) translates into shifting the frequency components by \(\omega_0\). It looks like \(\mathcal{F}\{f(x)\}(\xi - \omega_0)\). This means modulation in the time domain leads to a frequency shift in the frequency domain.

• This property is widely used in signal processing, such as amplitude modulation (AM) in radio broadcasting or in analytical work with different frequencies.
• Modulation affects the position of the spectrum without altering its shape.

Scaling in Fourier Transform

Scaling is another fundamental property in Fourier Transforms that can help transform signals and functions efficiently. Consider scaling the function by a factor, like \(f(ax)\). When it comes to Fourier Transforms, scaling doesn't just change one aspect of the function.

When you scale a function, it results in adjusting both the frequency components and the amplitude in a specific way. Suppose we scale the function by \(2\), resulting in \(f(2x)\). The Fourier Transform becomes \(\frac{1}{2} \mathcal{F}\{f(x)\}\left(\frac{\xi}{2}\right)\). The scaling of the input results in a compression or expansion of the frequency domain.

• The frequencies become compressed or expanded by the scaling factor.
• The amplitude is inversely scaled, balancing the overall energy of the signal.

Scaling in Fourier Transforms allows signal control and adjustments necessary in systems like communications, audio processing, and image analysis.

Sine and Cosine Modulation

Sine and cosine modulation are particularly important in Fourier analysis for decomposing and reconstructing signals. When a function \(f(x)\) is modulated with a sine or cosine wave, it involves multiplying \(f(x)\) by \(\sin(\omega_0 x)\) or \(\cos(\omega_0 x)\).

Let's start with sine modulation: Represent sine as \(\frac{e^{i\omega_0 x} - e^{-i\omega_0 x}}{2i}\). The Fourier transform of \(f(x)\sin(\omega_0 x)\) uses the properties of sine's imaginary components, ultimately leading to a combination using the shifts \(\mathcal{F}\{f(x)\}(\xi-\omega_0)\) and \(\mathcal{F}\{f(x)\}(\xi+\omega_0)\).

Now, for cosine modulation: Express cosine as \(\frac{e^{i\omega_0 x} + e^{-i\omega_0 x}}{2}\), leading to a similar resultant form. In this scenario, the resulting Fourier transform has terms added together rather than subtracted, creating a direct correspondence to shifted frequency components.

• Sine modulation results in imaginary outputs, beneficial in specific analytic applications.
• Cosine modulation leads to real outputs, widely used in signal processing.

Both play crucial roles in signal modulation, analyzing waveforms, and reconstructing signals in telecommunications and electronics.