/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 Suppose \(g(t)=x(t) \cos t\) and... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose \(g(t)=x(t) \cos t\) and the Fourier transform of the \(g(t)\) is $$G(j \omega)=\left\\{\begin{array}{ll}1, & |\omega| \leq 2 \\\0, & \text { otherwise }\end{array}\right.$$ (a) Deternnine \(x(t)\) (b) Specify the Fourier transform \(X_{1}(j \omega)\) of a signal \(x_{1}(t)\) such that $$g(t)=x_{1}(t) \cos \left(\frac{2}{3} t\right)$$

Short Answer

Expert verified
(a) \( x(t) \) is a band-pass signal with nonzero spectrum from 1 to 3. (b) \( X_1(j \omega) = 1 \) for \( \frac{2}{3} \leq |\omega| \leq 2 \).

Step by step solution

01

Analyze the Given Fourier Transform

The given Fourier transform of \(g(t)\) is a rectangular function in the frequency domain: \( G(j \omega) = 1 \) for \(|\omega| \leq 2\) and zero otherwise. This Fourier transform represents a low-pass filter with a cutoff frequency at \(\omega = 2\).
02

Identify Properties of Modulation in Frequency Domain

The function \(g(t) = x(t) \cos(t)\) can be represented by modulation (multiplying \(x(t)\) by \(\cos(t)\)). In the frequency domain, modulation results in the convolution of the Fourier transforms of \(x(t)\) with the Fourier transform of \(\cos(t)\). The Fourier transform of \(\cos(t)\) is given by impulse functions at frequencies \(\pm 1\).
03

Express Convolution in Frequency Domain

The Fourier transform \(G(j \omega)\) can be expressed as the convolution of \(X(j \omega)\) with \(\frac{1}{2} (\delta(\omega - 1) + \delta(\omega + 1))\), which results in two shifted copies of \(X(j \omega)\):\[ G(j \omega) = \frac{1}{2} [X(j(\omega - 1)) + X(j(\omega + 1))] \]
04

Determine the Range of Non-Zero Frequency Components

Since \( G(j \omega) \) is nonzero from \(-2\) to \(2\), it implies that \(X(j \omega)\) has to be nonzero from \(-3\) to \(-1\) and from \(1\) to \(3\).
05

Solve for Fourier Transform \(X(j \omega)\)

To satisfy the expression \(X(j(\omega - 1)) + X(j(\omega + 1)) = 2\), using symmetry and from the given frequency window, we determine that:\[ X(j \omega) = \left\{\begin{array}{ll}1, & 1 \leq |\omega| \leq 3 \0, & \text { otherwise }\end{array}\right. \]
06

Result for \(x(t)\) from \(X(j \omega)\)

From the frequency domain representation, \(x(t)\) is the inverse Fourier transform of \(X(j\omega)\), which is a band-pass filter ranging from \(\omega = 1\) to \(\omega = 3\).
07

Formulate \(x_1(t)\) for Part (b)

For \(x_1(t)\) with \(g(t) = x_1(t) \cos(\frac{2}{3}t)\), the modulation changes the cosine frequency, requiring that the information from \(\cos(t)\) be scaled by \(\frac{2}{3}\). The Fourier spectrum will shift accordingly. We consider: \[ X_1(j \omega) = X(j \frac{3\omega}{2}) \] This shift ensures the spectrum centers on \(\pm\frac{2}{3}\) instead of \(\pm 1\).
08

Solve for Fourier Transform \(X_1(j \omega)\)

The modulation using \(\cos\left(\frac{2}{3} t\right)\) results in modifying the frequency bands seen in \(X(j \omega)\) to:\[ X_{1}(j \omega) = \left\{\begin{array}{ll} 1, & \frac{2}{3} \leq |\omega| \leq 2 \0, & \text{otherwise} \end{array}\right. \]

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Key Concepts

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

Modulation in Frequency Domain
In the world of signal processing, modulation can be easily understood as the process of varying one or more properties of a signal, known as the carrier signal, with a modulating signal that usually contains information to be transmitted. In the frequency domain, modulation serves a critical role. When a signal that's in the time domain is multiplied by another signal, like in the case of using a cosine function, this is known as amplitude modulation (AM).
In our exercise, the function we are dealing with is \( g(t) = x(t) \cos(t) \). Here, \( \cos(t) \) acts as a carrier signal. When modulating with \( \cos(t) \), the Fourier Transform of this function results in distinct frequency components. This operation translates to a convolution of the Fourier Transform of the modulating function with the Fourier Transform of the original signal.
  • The frequency content of a cosine function is concentrated entirely at two frequencies: \( \pm 1 \).
  • This means the modulation will cause shifts in the frequency components of \( x(t) \).
The result is new frequency components appearing at these shifted locations; thus, creating a spectrum of the original function centered at new frequencies.
Inverse Fourier Transform
The inverse Fourier Transform is a mathematical operation that allows us to convert data from the frequency domain back to the time domain. It's essentially undoing the Fourier Transform, which analyzes the frequencies within a signal.
When you have a Fourier Transform representation, using inverse techniques helps to reconstruct the original time-domain function from its frequency-domain representation. In the context of the exercise: if you are given \( X(j \omega) \), the inverse Fourier Transform will yield \( x(t) \), allowing you to understand its time-based behavior.
For instance, if \( X(j \omega) \) corresponds to specific non-zero values, applying the inverse Fourier Transform helps derive \( x(t) \).
  • It provides a bridge to analyze how frequency contents are synthesized into a complete signal in the time domain.
  • Allows engineers and scientists to decompose problems into time and frequency domain problems, facilitating enhanced understanding and analysis.
Through hands-on use, the inverse Fourier Transform acts as a mechanism for validating consistency between the time and frequency representations of a signal.
Frequency Shift
Imagine a signal as an orchestra with different instruments playing at different pitches. A frequency shift is like changing the pitch of these instruments. It's a fundamental concept in signal processing that involves altering the frequency content of a signal. In our case, frequency shifts occur as a result of multiplication by cosine functions.
This exercise manipulates frequencies in the signal \( g(t) = x(t) \cos(t) \), shifting the signal's spectrum. This is akin to moving the center of the frequency spectrum based on the frequency of the cosine. Each impulse component of the cosine at \( \pm 1 \) causes the original spectrum of \( x(t) \) to shift.
Here's how frequency shifting operates:
  • The original signal's frequency components are displaced by the frequency of the modulating cosine function.
  • Shifts can make it possible to move lower frequency signals into higher frequency bands, typically for transmission.
Understanding frequency shift is pivotal for techniques like radio transmission, where signals are shifted into frequency bands where they can be broadcasted.
Cosine Modulation
In regular life, cosine modulation can be thought of as adjusting the intensity of music based on volume control. Mathematically, it involves using a cosine function to "modulate" another signal. This is a fundamental concept in communication systems.
When you modulate a signal like \( x(t) \) with a cosine function, you're essentially embedding data within a carrier wave by amplitude modulation. In this exercise, cosine modulation is evident when \( g(t) = x_1(t) \cos(\frac{2}{3}t) \) comes into play.
Cosine modulation carries out several functions:
  • It allows control over which frequencies are emphasized in the signal.
  • It's particularly useful for efficient transmission and for preserving the signal over longer distances.
Mastering cosine modulation provides insight into practical applications such as transmitting audio signals over radio waves, where the information is encoded by varying the amplitude of a consistent carrier signal—a technique widely used in AM radio.

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Most popular questions from this chapter

Consider an LTI system whose response to the input $$x(t)=\left[e^{-t}+e^{-3 t}\right] u(t)$$ is $$y(t)=\left[2 e^{-t}-2 e^{-4 t}\right] u(t)$$. (a) Find the frequency response of this system. (b) Determine the system's impulse response. (c) Find the differential equation relating the input and the output of this system.

Consider the signal $$x(t)=\sum_{k=-\infty}^{\infty} \frac{\sin \left(k \frac{\pi}{4}\right)}{\left(k \frac{\pi}{4}\right)} \delta\left(t-k \frac{\pi}{4}\right)$$ (a) Determine \(g(t)\) such that $$x(t)=\left(\frac{\sin t}{\pi t}\right) g(t)$$ (b) Use the multiplication property of the Fourier transform to argue that \(X(j \omega)\) is periodic. Specify \(X( j \omega)\) over one period.

Let \(x(t)\) be a signal whose Fourier transform is $$X(j \omega)=\delta(\omega)+\delta(\omega-\pi)+\delta(\omega-5)$$ and let $$h(t)=u(t)-u(t-2)$$. (a) Is \(x(t)\) periodic? (b) Is \(x(t) * h(t)\) periodic? (c) Can the convolution of two aperiodic signals be periodic?

Let \(H(j \omega)\) be the frequency response of a continuous-time LTI system, and suppose that \(H(j \omega)\) is real, even, and positive. Also, assume that $$\max _{\omega}\\{H(j \omega)\\}=H(0)$$ (a) Show that: (i) The impulse response, \(h(t),\) is real. (ii) \(\max \\{|h(t)|\\}=h(0)\). Hint: If \(f(t, \omega)\) is a complex function of two variables, then $$\left|\int_{-\infty}^{+\infty} f(t, \omega) d \omega \leq \int_{-\infty}^{+\infty}\right| f(t, \omega) | d \omega$$ (b) One important concept in system analysis is the bandwidth of an LTI system. There are many different mathematical ways in which to define bandwidth, but they are related to the qualitative and intuitive idea that a system with frequency response \(G(j \omega)\) essentially "stops" signals of the form \(e^{j \omega t}\) for values of \(\omega\) where \(G(j \omega)\) varnishes or is small and "passes" those complex exponentials in the band of frequency where \(G(j \omega)\) is not small. The width of this band is the bandwidth. These ideas will be made much clearer in Chapter \(6,\) but for now we will consider a special definition of bandwidth for those systems with frequency responses that have the properties specified previously for \(H(j \omega) .\) Specifically, one definition of the bandwidth \(B_{w}\) of such a system is the width of the rectangle of height \(H(j 0)\) that has an area equal to the area under \(H(j \omega) .\) This is illustrated in Figure \(P 4.49(a) .\) Note that since \(H(j 0)=\max _{\omega} H(j \omega),\) the frequencies within the band indicated in the figure are those for which \(H(j \omega\\}\) is largest. The exact choice of the width in the figure is, of course, a bit arbitrary, but we have chosen one definition that allows us to compare different systems and to make precise a very important relationship between time and frequency. What is the bandwidth of the system with frequency response $$H(j \omega)=\left\\{\begin{array}{ll}\mathbf{1}, & |\omega|W\end{array}\right.$$ (c) Find an expression for the bandwidth \(B_{w}\) in terms of \(H(j \omega)\) (d) Let \(s(t)\) denote the step response of the system set out in part (a). An important measure of the speed of response of a system is the rise time, which, like the bandwidth, has a qualitative definition, leading to many possible mathematical definitions, one of which we will use. Intuitively, the rise time of a system is a measure of how fast the step response rises from zero to its final value, $$s(\alpha)=\lim _{t \rightarrow \infty} s(t)$$ Thus, the smaller the rise time, the faster is the response of the system. For the system under consideration in this problem, we will define the rise time as $$t_{r}=\frac{s\left(\infty\right)}{h(0)}$$ since $$s^{\prime}(t)=h(t)$$ and also because of the property that \(h(0)=\max _{t} h(t), t_{r}\) is the time it would take to go from zero to \(s(\infty)\) while maintaining the maximum rate of change of \(s(t) .\) This is illustrated in Figure \(\mathbf{P 4} . \mathbf{4 9}(\mathbf{b})\) Find an expression for \(t_{r}\) in terms of \(H(j \omega)\). (e) Combine the results of parts (c) and (d) to show that $$B_{w} t_{r}=2 \pi$$ Thus, we cannot independently specify both tha rise time and the bandwidth of our system. For example, eq. (P4.49-1) implies that, if we want a fast system ( \(t_{r}\) small), the system must have a large bandwidth. This is a fundamental trade-off that is of central importance in many problems of system design.

Find the impulse response of a system with the frequency response $$H(j \omega)=\frac{\left(\sin ^{2}(3 \omega)\right) \cos \omega}{\omega^{2}}$$

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