/*! 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 6 Assume that \(x(f)\) is a signal... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 6

Assume that \(x(f)\) is a signal whose Fourier transform \(X(j \omega)\) is zero for \(|\omega|>\omega_{M}\) The signal \(g(t)\) may be expressed in terms of \(x(t)\) as $$g(t)=x(t) \cos \omega_{c} t-\left\\{x(t) \cos \omega_{t} t *\left(\frac{\sin \omega_{c} t}{\pi t}\right)\right\\}$$ where \(*\) denotes convolution and \(\omega_{c}>\omega_{M} .\) Determine the value of the constant \(A\) such that $$x(t)=\left\\{g(t) \cos \omega_{c} f\right) * \frac{A \sin \omega_{M} t}{\pi t}$$

Short Answer

Expert verified
The value of A is 2\pi.

Step by step solution

01

Understand the Problem

We need to find the constant \( A \) such that performing a certain convolution with \( g(t) \) gives back the original signal \( x(t) \). To do this, we will exploit properties of Fourier transforms and convolution.
02

Use Modulation Properties

The term \( g(t) \cos( \omega_c t) \) is essentially a modulation of the signal \( g(t) \) by a cosine wave, which in the frequency domain results in shifting the spectrum of \( g(t) \). This can move a bandlimited signal sufficiently away from the origin if \( \omega_c > \omega_M \).
03

Analyze Convolution with Impulse Functions

Convolution of \( x(t) \) with \( \frac{\sin(\omega_M t)}{\pi t} \) corresponds to an ideal low-pass filter in terms of frequency, preserving components within \(-\omega_M\) to \(\omega_M\). Since \( \omega_c > \omega_M \), convolving \( g(t) \) modulated by cosine with a sinc function will filter and recover components lying within the original band-limit.
04

Derive Expression for X(\omega)

Perform Fourier transform of equation: \[G(j \omega) = X(j(\omega - \omega_c)) - H(j \omega) * X(j(\omega - \omega_t))\]where \[H(j \omega) = \frac{1}{2}\left(\delta(\omega + \omega_c) + \delta(\omega - \omega_c)\right)\]
05

Set Up the Final Equation

Given \(g(t) = x(t)\) when filtered appropriately, relate this to the inverse Fourier transform:\[X(j \omega) = \frac{1}{2\pi} \int_{-\omega_M}^{\omega_M} G(ju) A \cdot \text{sinc}\left(\frac{u}{\omega_M}\right)\, du\]The identity uses the property that convolution with a rect function in the frequency domain yields a scaled sinc function in the time domain.
06

Solve for Constant A

To solve for \( A \), realize thatconvolving with \( \frac{A \sin(\omega_M t)}{\pi t} \) should fully compensate the factor by integrating across the full \( X(j\omega) \) support without distortion. Thus, \( A \) must restore full scale of these components through a frequency gain: \( A = 2\pi \).
07

Conclusion

Hence, the value of constant \( A \) needed to ensure \( x(t) = (g(t) \cos(\omega_c t)) * \frac{A \sin(\omega_M t)}{\pi t} \) is \(2\pi\).

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Key Concepts

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

Signal Modulation
Signal modulation is a key concept in understanding how information can be embedded and transmitted using waveforms. In the given exercise, modulation is used to modify the signal \( x(t) \) with a cosine function. This involves multiplying the signal by a cosine wave with frequency \( \omega_c \). This technique shifts the frequency spectrum of \( x(t) \) in such a way that it can be effectively transmitted or processed. Especially when \( \omega_c \) is greater than \( \omega_M \), this shifting moves the spectrum away from the origin, creating a new component in the frequency domain.
  • Understanding the Shift: When you modulate a signal, you effectively translate its frequencies, which can help in reducing interference and allowing multiple signals to be transmitted simultaneously without overlapping.
  • Importance in Communication: Modulation is crucial in communication systems for enabling signals to be transmitted over long distances with minimum loss and distortion.
Overall, modulation is not just about frequency shifts; it creates avenues for more efficient signal handling, which is vital in modern technology.
Convolution Theory
Convolution is a mathematical operation that combines two functions to form a third function, providing insight into the effect of one signal on another. In this context, convolution is expressed as \( g(t) = x(t) \cos(\omega_c t) * \left(\frac{\sin(\omega_c t)}{\pi t}\right) \).
  • Function Interaction: Convolution allows us to analyze how different signals interact, particularly how a signal like a sinc function can shape or filter another signal like a modulated signal.
  • Frequency Domain Implications: In the frequency domain, convolution translates to multiplication, simplifying the process of filtering and analyzing signals.
Understanding convolution in signal processing allows engineers and scientists to alter and control signal attributes, bringing versatility to signal manipulation.
Bandlimited Signals
A bandlimited signal is a signal with a Fourier transform that is zero for frequencies above a certain threshold, \( \omega_M \). This concept is at the core of the exercise because it specifies the limits within which the signal can be accurately represented and processed.
  • Why Bandlimited?: Bandlimited signals are crucial because they allow for precise reconstruction from discrete samples, an essential property for digital signal processing.
  • Implication of Bandlimiting: If a signal is bandlimited, it suggests that all its energy is confined within a defined frequency range, and frequencies beyond are negligible or non-existent.
  • Practical Application: Understanding and utilizing bandlimited signals help in effectively designing filters that can separate desired signals from noise.
By focusing on bandlimited signals, engineers can ensure signal integrity and improve the efficiency of transmission and storage systems.

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Most popular questions from this chapter

Suppose $$x(t)=\sin 200 \pi t+2 \sin 400 \pi t$$ and $$g(t)=x(t) \sin 400 \pi t$$ If the product \(g(t)(\sin 400 \pi t)\) is passed through an jdeal lowpass filter with cutoff frequency \(400 \pi\) and passband gain of 2 , determine the signal obtained at the output of the lowpass filter.

Suppose \(x[n]\) is a real-valued discrete-time signal whose Fourier transform \(X\left(e^{j \omega}\right)\) has the property that $$X\left(e^{j \omega}\right)=0 \quad \text { for } \frac{\pi}{8} \leq \omega \leq \pi$$ We use \(x[n]\) to modulate a sinusoidal carrier \(c[n]=\sin (5 \pi / 2) n\) to produce $$y[n]=x[n] c[n]$$ Determine the values of \(\omega\) in the range \(0 \leq \omega \leq \pi\) for which \(Y\left(e^{j \omega}\right)\) is guaranteed to be zero.

Let \(x[n]\) be a discrete-time signal with spectrum \(X\left(e^{j u}\right),\) and let \(p(t)\) be a continuous come pulse function with spectrum \(P(j \omega) .\) We form the signal $$y(t)=\sum_{n=-\infty}^{+\infty} x[n] p(t-n)$$ (a) Determine the spectrum \(Y(j \omega)\) in terms of \(X\left(e^{J U}\right)\) and \(P(j \omega)\) (b) If $$p(t)=\left\\{\begin{array}{ll} \cos 8 \pi t, & 0 \leq t \leq 1 \\ 0, & \text { else where } \end{array}\right.$$ determine \(P(j \omega)\) and \(Y(j \omega)\).

In this problem, we explore an equalization method used to avoid intersymbol in interference caused in PAM systems by the channel having nonlinear phase over its bandwidth When a PAM pulse with zero-crossings at integer multiples of the symbol spacing \(T_{1}\) is passed through a channel with nonlinear phase, the received pulse may no longer have zero-crossings at times that are integer multiples of \(T_{1}\). Therefore, in order to avoid intersymbol interference, the received pulse is passed through a zero forcing equalizer, which forces the pulse to have zero-crossings at integer multiples of \(T_{1}\), This equalizer generates a new pulse \(y(t)\) by summing up weighted and shifted versions of the received pulse \(x(t)\). The pulse \(y(t)\) is given by $$y(t)=\sum_{t=-N}^{N} a_{f} x\left(t-l T_{1}\right)$$ where the \(a_{l}\) are all real and are chosen such that $$y\left(\dot{k} T_{1}\right)=\left\\{\begin{array}{ll} 1, & k=0 \\ 0, & k=\pm 1,\pm 2,\pm 3, \ldots \neq N \end{array}\right.$$ (a) Show that the equalizer is a filter and determine its impulse response. (b) To illustrate the selection of the weights \(a_{i},\) let \(u\) : consides an example. If \(x\left(0 T_{1}\right)=0.0, x\left(-T_{1}\right)=0.2, x\left(T_{1}\right)=-0.2,\) and \(x\left[\left\\{X_{1}\right)=0 \text { for }|k|>1, d e-\right.\) determine the values of \(a_{1}, a_{1},\) and \(a\), such that \(y\left(\pm T_{1}\right)=0\)

For what values of \(\omega_{0}\) in the range \(-\pi<\omega_{0} \leq \pi\) is amplitude modulation with carrier \(e^{j \omega_{1}, n}\) equivalent to amplitude modulation with carrier \(\cos \omega_{0} n ?\)
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