Question

Suppose that \(f(t)=f_{1}(t) f_{2}(t),\) where $$\begin{aligned} &f_{1}(t)=\cos \omega_{0} t,\\\ &f_{2}(t)=1, \quad-\tau / 2 < t < \tau / 2;\\\ &f_{2}(t)=0, \quad \text { elsewhere }. \end{aligned}$$ a) Use convolution in the frequency domain to \(\operatorname{tin} d(\omega)\). b) What happens to \(F(\omega)\) as the width of \(f_{2}(t)\) increases so that \(f(t)\) includes more and more cycles on \(f_{1}(t) ?\)

Short answer

As \( \tau \) increases, \( F(\omega) \) becomes sharper, focusing on \( \omega_0 \) and \(-\omega_0 \).

Step-by-step solution

Express Each Function's Fourier Transform

For the given functions, start with calculating the Fourier Transform. The function \( f_1(t) = \cos(\omega_0 t) \) can be expressed in terms of complex exponentials, \( f_1(t) = \frac{1}{2}(e^{j\omega_0 t} + e^{-j\omega_0 t}) \). Its Fourier transform is two delta functions at \( \omega = \omega_0 \) and \( \omega = -\omega_0 \), which leads to \( F_1(\omega) = \frac{1}{2} \left[ \delta(\omega - \omega_0) + \delta(\omega + \omega_0) \right] \).

Fourier Transform of the Rectangular Function

\( f_2(t) \) is a rectangular function of width \( \tau \). Its Fourier Transform is a sinc function, \( F_2(\omega) = \tau \operatorname{sinc}\left(\frac{\omega \tau}{2\pi}\right) \). Here 'sinc' refers to the normalized sinc function, defined as \( \operatorname{sinc}(x) = \frac{\sin(\pi x)}{\pi x} \).

Convolution Product in Frequency Domain

In the frequency domain, the product \( f(t) = f_1(t)f_2(t) \) corresponds to the convolution of their Fourier transforms: \( F(\omega) = F_1(\omega) * F_2(\omega) \). Here, \( F_1(\omega) \) is two delta functions and \( F_2(\omega) \) is a sinc function. Thus, \( F(\omega) = \frac{\tau}{2} \left[\operatorname{sinc}\left(\frac{(\omega - \omega_0) \tau}{2\pi}\right) + \operatorname{sinc}\left(\frac{(\omega + \omega_0) \tau}{2\pi}\right)\right] \).

Consider the Effect of Increasing Width \( \tau \)

As \( \tau \) increases, the sinc function in the frequency domain narrows. Therefore, \( F(\omega) \) will show peaks that become sharper centered around \( \omega_0 \) and \(-\omega_0 \). This reflects that as \( f_2(t) \) increases in time, it allows more oscillations of \( f_1(t) \) to pass, which means a narrower frequency response in \( F(\omega) \).

Key concepts

Fourier Transform

The Fourier Transform is a powerful mathematical tool that allows us to transform a time-domain signal into its frequency-domain representation. This transformation provides insights into the different frequency components present in a signal.
By converting a function to its Fourier domain, we can analyze how different frequencies contribute to the signal. This is done by applying the Fourier Transform formula:

• For a continuous signal, the Fourier Transform is given by:\[F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-j\omega t} dt\]
• The result, \( F(\omega) \), shows us how much of each frequency \( \omega \) is present in the original function \( f(t) \).

The Fourier Transform is essential in solving problems where signal processing or frequency analysis is required.
In the exercise, the function \( f_1(t) = \cos(\omega_0 t) \) is expressed using its Fourier Transform which results in delta functions. This allows for the easier combination of \( f_1(t) \) with \( f_2(t) \) by turning convolutions in the time domain into multiplications in the frequency domain.

Sinc Function

A Sinc Function, denoted as \( \operatorname{sinc}(x) \), is a mathematical function often used in signal processing and control systems. It is expressed as:

• Normalized Sinc Function: \( \operatorname{sinc}(x) = \frac{\sin(\pi x)}{\pi x} \)

This function arises frequently in the context of the Fourier Transform, particularly when dealing with the transform of rectangular signals. It describes how the frequency distribution of a signal expands or narrows depending on the time domain characteristics.
In the given exercise, the function \( f_2(t) \) is a rectangular function. Its Fourier transform results in a Sinc Function, \( F_2(\omega) = \tau \operatorname{sinc}\left(\frac{\omega \tau}{2\pi}\right) \).
The Sinc Function's role becomes significant because its shape determines the smoothness and spread of the frequency components of \( f_2(t) \). As seen in the exercise, changing the time width \( \tau \) of \( f_2(t) \) influences the frequency domain representation by altering the Sinc Function's width, highlighting that more oscillations in time result in a narrower frequency band.

Delta Function

The Delta Function, also known as the Dirac Delta Function, is a fundamental concept in signal processing and Fourier analysis. The Delta Function \( \delta(x) \) is a "function" with the property of being zero everywhere except at \( x = 0 \), where it is infinite.
In essence, it acts like an "impulse" and is extremely useful in breaking down and representing signals in terms of their frequency components:

• The key property of the Delta Function is: \( \int_{-\infty}^{\infty} \delta(x) \, dx = 1 \)
• It selectively picks out values at specific points when integrated inside other functions.

In the Fourier Transform context, a sinusoidal function like \( \cos(\omega_0 t) \) can be represented using Delta Functions due to its precise frequencies. The Delta Function simplifies this representation by focusing only on the principal frequencies of interest.
For this problem, the Fourier Transform of \( f_1(t) \) leads to Delta Functions at \( \omega = \omega_0 \) and \( \omega = -\omega_0 \), pinpointing the exact frequencies involved without any spread, unlike the Sinc Function. This serves to highlight how specific frequencies contribute exactly to the overall signal spectrum in mixed time-frequency analyses.