91Ó°ÊÓ

Consider a square pulse \(\psi(t)\) which is zero for all \(t\) not in the interval \(t_{1}\) to \(t_{2}\). Within that interval, \(\psi(t)\) has the constant value \(1 / \Delta t\), where \(\Delta t=t_{2}-t_{1}\). Let \(t_{0}\) be the time at the center of the interval. Show that \(\psi(t)\) can be Fourier-analyzed as follows: $$ \psi(t)=\int_{0}^{\infty} A(\omega) \sin \omega\left(t-t_{0}\right) d \omega+\int_{0}^{\infty} B(\omega) \cos \omega\left(t-t_{0}\right) d \omega, $$ with the solution $$ A(\omega)=0, \quad B(\omega)=\frac{1}{\pi} \frac{\sin \frac{1}{2} \Delta t \omega}{\frac{1}{2} \Delta t \omega} $$ Sketch \(B(\omega)\) versus \(\omega .\) In the limit where \(\Delta t\) goes to zero, \(\psi(t)\) is called'a "delta function" of time, written \(\rho\left(t-t_{0}\right) .\) What is \(B(\omega)\) for this delta function of time?

Step by step solution

01

Define the Square Pulse Function

The given function \( \psi(t) \) is defined as zero for all \( t \) not in the interval \( t_1 \) to \( t_2 \) and equal to \( \frac{1}{\Delta t} \) in this interval, where \( \Delta t = t_2 - t_1 \). Thus,\[\psi(t) = \begin{cases} \frac{1}{\Delta t}, & t_1 \leq t \leq t_2 \0, & \text{otherwise}\end{cases}\]

02

Apply Fourier Sine and Cosine Transforms

The problem statement gives a Fourier representation of \( \psi(t) \) as follows:\[\psi(t)=\int_{0}^{\infty} A(\omega) \sin \omega\left(t-t_{0}\right) d \omega + \int_{0}^{\infty} B(\omega) \cos \omega\left(t-t_{0}\right) d \omega.\]Given that \( A(\omega) = 0 \), we only need to find the function \( B(\omega) \).

03

Derive the Coefficient B(ω)

To find \( B(\omega) \), compute the cosine transform of the square pulse:\[B(\omega) = \frac{2}{\pi} \int_{t_1}^{t_2} \psi(t) \cos \omega(t - t_0) dt.\]Substituting \( \psi(t) = \frac{1}{\Delta t} \) and the integration limits, we have:\[B(\omega) = \frac{2}{\pi \Delta t} \int_{t_1}^{t_2} \cos \omega(t - t_0) dt.\]

04

Calculate the Integral

Change the variable of integration to simplify the expression: let \( x = \omega(t - t_0) \), then \( dx = \omega dt \). The integral becomes:\[B(\omega) = \frac{2}{\pi \Delta t \omega} \left[ \sin \omega(t - t_0) \right]_{t_1}^{t_2}.\]Evaluating, we find:\[B(\omega) = \frac{2}{\pi \Delta t \omega} \left( \sin \omega(t_2 - t_0) - \sin \omega(t_1 - t_0) \right).\]

05

Simplify the Expression for B(ω)

Express \( \sin(\text{difference of angles}) \) using the sine subtraction identity:\[\sin c - \sin d = 2 \cos \left(\frac{c + d}{2}\right) \sin \left(\frac{c - d}{2}\right).\]This simplifies \( B(\omega) \) to:\[B(\omega) = \frac{1}{\pi} \frac{\sin \frac{1}{2} \Delta t \omega}{\frac{1}{2} \Delta t \omega}.\]

06

Analyze the Limit for Delta Function and Sketch

In the limit \( \Delta t \to 0 \), \( \psi(t) \) becomes a delta function \( \rho(t - t_0) \). For this delta function, the limit of \( B(\omega) \) is:\[B(\omega) = 1.\]On a graph, \( B(\omega) \) versus \( \omega \) is a sinc function given by the equation in Step 5, which converges to 1 as \( \Delta t \to 0 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Square Pulse Function

A square pulse function is a simplistic model often used in signal processing and Fourier analysis. In the context of a square pulse, consider a time interval between \( t_1 \) and \( t_2 \). Within this interval, a signal maintains a constant value, while outside this interval it is zero.

The expression for a square pulse \( \psi(t) \) can be written as:

• \( \psi(t) = \frac{1}{\Delta t} \) for \( t_1 \leq t \leq t_2 \)
• \( \psi(t) = 0 \) otherwise

The term \( \Delta t = t_2 - t_1 \) defines the width of the pulse, and the choice of \( \frac{1}{\Delta t} \) ensures that the integral of the pulse function over the interval is one. This normalization reflects the energy content of the signal being one unit, a typical requirement in signal processing.

Delta Function

The delta function, often referred to as Dirac's delta function, is an idealized concept in mathematics and signal processing that serves as a useful tool in various analyses, including Fourier analysis. As \( \Delta t \) approaches zero in the square pulse function, its behavior models that of a delta function.

The delta function has the property:

• \( \rho(t-t_0) = 0 \) for \( t eq t_0 \)
• The integral over the entire real line is 1: \( \int_{-\infty}^{\infty} \rho(t-t_0)dt = 1 \)

In the context of Fourier analysis, as the pulse width \( \Delta t \) compresses to zero, the spectrum \( B(\omega) \) converges to a constant value of 1, suggesting a flat frequency response. This reflects the delta function's property of containing all frequencies equally.

Cosine Transform

The cosine transform is a mathematical tool used in the analysis and computation of waveforms. It's part of the Fourier transform framework where functions or signals are expressed as a sum of cosine functions of different frequencies.

In our exercise, the square pulse is expanded using a cosine transform, recognizing that \( A(\omega) = 0 \). Therefore, we focus on:

• \( B(\omega) = \frac{1}{\pi} \frac{\sin \frac{1}{2} \Delta t \omega}{\frac{1}{2} \Delta t \omega} \)

This expression is derived by computing the cosine Fourier transform of the square pulse, and it highlights the use of the cosine basis to represent signals in terms of frequencies. This approach facilitates the decomposition of the square pulse into basic harmonic components, each characterized by a cosine function with an associated amplitude \( B(\omega) \).

Sinc Function

The sinc function is a mathematical function that arises frequently in signal processing, especially in Fourier analysis. It is defined as \( \text{sinc}(x) = \frac{\sin(\pi x)}{\pi x} \), except at \( x = 0 \), where \( \text{sinc}(0) = 1 \).

In our exercise, \( B(\omega) \) is expressed in terms of a sinc function:

• \( B(\omega) = \frac{1}{\pi} \frac{\sin \frac{1}{2} \Delta t \omega}{\frac{1}{2} \Delta t \omega} \)

This form of \( B(\omega) \) reveals that it assumes the shape of a sinc function with respect to frequency \( \omega \).
The sinc function has a pivotal role because it describes how the amplitude of each frequency component decays from the peak at zero frequency. The central peak extends to lower frequencies, with side lobes decaying steadily, exhibiting symmetry about the vertical axis. This characteristic captures the frequency response of the square pulse and explains the oscillating nature of its spectrum.