/*! 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 Consider the following three dis... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the following three discrete-time signals with a fundamental period of 6: \(x | n]=l-\cos \left(\frac{2 \pi}{6} n\right)\), \(y[n]=\sin \left(\frac{2 \pi}{6} n+\frac{\pi}{4}\right)\), \(z[n]=x[n] y[n]\). (a) Determine the Fourier series coefficients of \(x[n]\). (b) Determine the Fourier series coefficients of \(y[n]\). (c) Use the results of parts \((a)\) and \((b),\) along with the multiplication property of the discrete-time Fourier series, to determine the Fourier series coefficients of \(z[n]=x[n] y[n]\). (d) Determine the Fourier series coefficients of \(z[n]\) through direct evaluation, and compare your result with that of part (c).

Short Answer

Expert verified
(a) \(a_0 = 1\), \(a_1 = -\frac{1}{2}\), (b) \(b_1 = \frac{-e^{j\pi/4}}{2j}\), (c) Convolution of \(a_k\) and \(b_k\) gives \(c_k\), (d) Direct evaluation of \(c_k\) confirms results.

Step by step solution

01

Determine Fourier Coefficients of x[n]

The signal \(x[n] = 1 - \cos\left(\frac{2\pi}{6}n\right)\) can be expressed as a sum of two components: a constant term and a cosine term. The constant term corresponds to the 0th harmonic, and the cosine term corresponds to the 1st and -1st harmonics. Thus, for a period \(N=6\), the Fourier coefficients are: \(a_0 = 1\), \(a_1 = -\frac{1}{2}\), \(a_{-1} = -\frac{1}{2}\), while all other \(a_k = 0\).
02

Determine Fourier Coefficients of y[n]

The signal \(y[n] = \sin\left(\frac{2\pi}{6} n + \frac{\pi}{4}\right)\) can be decomposed into exponential form using Euler's formula: \(y[n] = \frac{1}{j2}\left(e^{j\left(\frac{2\pi}{6}n+\frac{\pi}{4}\right)} - e^{-j\left(\frac{2\pi}{6}n+\frac{\pi}{4}\right)}\right)\). The coefficients are \(b_1 = \frac{-e^{j\pi/4}}{2j}\) and \(b_{-1} = \frac{e^{-j\pi/4}}{2j}\), with the other terms being zero.
03

Use Multiplication Property for z[n] Coefficients

The multiplication property of discrete-time Fourier series states that Fourier coefficients \(c_k\) of \(z[n]=x[n]y[n]\) are the convolution of \(a_k\) and \(b_k\). This requires calculating \(c_k = \sum_{m=-\infty}^{\infty} a_m b_{k-m}\). Using the non-zero coefficients found in Steps 1 and 2, convolve these to find \(c_k\) values. The summation needs to be performed for each \(k\).
04

Direct Evaluation of Fourier Coefficients of z[n]

To directly evaluate the Fourier coefficients of \(z[n]\), compute the product \(z[n] = x[n]y[n]\) and find its periodic spectrum. Using the definition of Fourier series coefficients, perform \(c_k = \frac{1}{N}\sum_{n=0}^{N-1} z[n] e^{-j\frac{2\pi}{N}kn}\) where \(N=6\). Compute for each \(k\) to obtain the same coefficients as Step 3 by direct computation.
05

Comparison

Verify that the coefficients obtained from the direct evaluation in Step 4 match those from the convolution method in Step 3, confirming the validity of the multiplication property in discrete-time Fourier series.

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Key Concepts

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

Discrete-Time Signals
Discrete-time signals are sequences of values or functions defined at discrete intervals. These signals are essential in digital signal processing, where continuous signals from the real world are sampled to create a series of discrete values. By converting a signal into a discrete-time format, engineers and analysts can employ a variety of mathematical methods to analyze, modify, or manipulate the signal for practical applications. In this exercise, we are given three discrete-time signals with a fundamental period of 6:
  • \(x[n] = 1 - \cos\left(\frac{2\pi}{6}n\right)\)
  • \(y[n] = \sin\left(\frac{2\pi}{6} n + \frac{\pi}{4}\right)\)
  • A product signal \(z[n] = x[n] y[n]\)
Understanding these signals' characteristics and relationships is crucial for tasks such as filtering, modulation, and analysis, which are foundational in communication and audio processing.
Fourier Coefficients
Fourier coefficients are the components of a signal's Fourier series representation. These coefficients provide insight into how much each frequency component contributes to the overall signal. By determining the Fourier coefficients of a signal, we effectively transform it into the frequency domain, where examining each sinusoidal component becomes more accessible.

Fourier Coefficients of \(x[n]\)

For the given signal \(x[n]\), the coefficients are found in terms of harmonics. Since \(x[n] = 1 - \cos\left(\frac{2\pi}{6}n\right)\), it's composed of a constant and a primary cosine wave. The Fourier coefficients are:
  • \(a_0 = 1\)
  • \(a_1 = a_{-1} = -\frac{1}{2}\)
  • All other \(a_k = 0\)

Fourier Coefficients of \(y[n]\)

Similarly, deriving coefficients from \(y[n] = \sin\left(\frac{2\pi}{6} n + \frac{\pi}{4}\right)\) involves translating the sine terms using Euler's formula. This results in:
  • \(b_1 = \frac{-e^{j\pi/4}}{2j}\)
  • \(b_{-1} = \frac{e^{-j\pi/4}}{2j}\)
  • All other \(b_k = 0\)
The correct identification of these coefficients is critical for utilizing the properties of the Fourier series.
Multiplication Property
The multiplication property in the context of discrete-time Fourier series offers a powerful tool. It states that if two signals are multiplied in the time domain, their Fourier coefficients are the convolution of their individual Fourier coefficients. In practical terms, if you know the Fourier coefficients of two signals, you can easily compute the coefficients of the resulting product signal without manually re-deriving them from scratch. This significantly simplifies the process of frequency analysis for products of signals. In this exercise, we obtained the coefficients for \(x[n]\) and \(y[n]\) separately. To find the coefficients for \(z[n] = x[n] y[n]\), we convolved:
  • \(c_k = \sum_{m=-\infty}^{\infty} a_m b_{k-m}\)
This showcases the elegant nature of convolution in Fourier theory, emphasizing its efficiency and power.
Signal Processing
Signal processing encompasses the analysis, manipulation, and synthesis of signals. It's a discipline dedicated to extracting meaningful information and insights from signals, often requiring transformations between the time and frequency domains.

Application to Exercise

In this exercise, we leverage different signal processing techniques, such as:
  • Identifying discrete-time signals
  • Decomposing these signals into their frequency components using Fourier series
  • Applying the multiplication property to handle complex signal products
  • Comparing results through direct and indirect methods
The exercise underscores the beauty of signal processing by demonstrating the theoretical and practical approaches to manipulating signals. It equips students with the foundational skills necessary for tackling more advanced signal processing tasks encountered in both academic and professional environments.

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Most popular questions from this chapter

A continuous-time periodic signal \(x(t)\) is real valued and has a fundamental period \(T=8 .\) The nonzero Fourier series coefficients for \(x(t)\) are specified as \(a_{1}=a_{-1}^{*}=j, a_{5}=a_{-5}=2\). Express \(x(t)\) in the form \(x(t)=\sum_{k=0}^{\infty} A_{k} \cos \left(w_{k} t+\phi_{k}\right)\).

Consider a causal discrete-time LTI system whose input \(x[n]\) and output \(y[n]\) are related by the following difference equation: \(y[n]-\frac{1}{4} y[n-1]=x[n]\) Find the Fourier series representation of the output \(y[n]\) for each of the following inputs: (a) \(x[n]=\sin \left(\frac{3 \pi}{4} n\right)\) (b) \(x[n]=\cos \left(\frac{\pi}{4} n\right)+2 \cos \left(\frac{\pi}{2} n\right)\)

For the continuous-time periodic signal \(x(t)=2+\cos \left(\frac{2 \pi}{3} t\right)+4 \sin \left(\frac{5 \pi}{3} t\right)\), determine the fundamental frequency \(\omega_{0}\) and the Fourier series coefficients \(a_{k}\) such that \(x(t)=\sum_{k=-\infty}^{\infty} a_{k} e^{j k \omega_{0} t}\).

Suppose that a continuous-time periodic signal is the input to an LTI system. The signal has Fourier series representation \(x(t)=\sum_{k=-\infty}^{\infty} \alpha^{|k|} e^{j k(\pi / 4)}\), where \(\alpha\) is a real number between 0 and \(1,\) and the frequency response of the system is \(H(j \omega)=\left\\{\begin{array}{ll}1, & \\{\omega \leq W \\ 0, & |\omega|>W\end{array}\right.\). How large must \(W\) be in order for the output of the system to have at least \(90 \%\) of the average energy per period of \(x(f) ?\)

Suppose we are given the following information about a signal \(x(t)\): 1\. \(x(t)\) is real and odd. 2\. \(x(t)\) is periodic with period \(T=2\) and has Fourier coefficients \(a_{k}\). 3\. \(a_{k}=0\) for \(|k|>1\). 4\. \(\left.\frac{1}{2}\right|_{0} ^{2}|x(t)|^{2} d t=1\). Specify two different signals that satisfy these conditions.

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