/*! 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 58 Let \(x[n]\) and \(y[n]\) be per... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(x[n]\) and \(y[n]\) be periodic signals with common period \(N,\) and let \(z[n]=\sum_{r=} x[r] y[n-r]\) be their periodic convolution (a) Show that \(z[n]\) is also periodic with period \(N\). (b) Verify that if \(a_{k}, b_{k},\) and \(c_{k}\) are the Fourier coefficients of \(x[n], y[n],\) and \(z[n]\), respectively, then \(c_{k}=N a_{k} b_{k}\). (c) Let \(x[n]=\sin \left(\frac{3 \pi n}{4}\right)\) and \(y[n]=\left\\{\begin{array}{ll}1, & 0 \leq n \leq 3 \\ 0, & 4 \leq n \leq 7\end{array}\right.\) be two signals that are periodic with period \(8 .\) Find the Fourier series representation for the periodic convolution of these signals. (d) Repeat part (c) for the following two periodic signals that also have period 8 : \(x[n]=\left\\{\begin{array}{ll}\sin \left(\frac{3 \pi n}{4}\right) . & 0 \leq n \leq 3 \\ 0, & 4 \leq n \leq 7\end{array}\right.\), \(y[n]=\left(\frac{1}{2}\right)^{n}, 0 \leq n \leq 7\).

Short Answer

Expert verified
(a) \(z[n]\) is periodic with period \(N\). (b) \(c_k = N a_k b_k\). (c) and (d) require calculating Fourier series using given formulas.

Step by step solution

01

Verify Periodicity of the Convolution (Part a)

To show that \(z[n]\) is periodic with period \(N\), consider that both \(x[n]\) and \(y[n]\) are periodic signals with a common period \(N\). The convolution is defined as \(z[n] = \sum_{r=} x[r] y[n-r]\). Since both signals are periodic, for any \(n\), \(x[n] = x[n+N]\) and \(y[n] = y[n+N].\) Thus,\[z[n+N] = \sum_{r=} x[r] y[n+N-r] = \sum_{r=} x[r] y[(n-r)+N].\]This simplifies back to \(z[n]\), showing that \(z[n]\) is also periodic with period \(N\).
02

Verify Fourier Coefficients (Part b)

To verify the Fourier coefficients, we need to show that \(c_k = N a_k b_k\). The formula for the Fourier coefficient of \(z[n]\) is given by:\[c_k = \frac{1}{N} \sum_{n=0}^{N-1} z[n] e^{-j\frac{2\pi}{N}kn}.\]Substitute \(z[n] = \sum_{r=} x[r] y[n-r]\):\[c_k = \frac{1}{N} \sum_{n=0}^{N-1} \sum_{r=} x[r] y[n-r] e^{-j\frac{2\pi}{N}kn}.\]Interchanging the order of summation and using the periodicity property:\[c_k = \sum_{r=} x[r] \left(\frac{1}{N} \sum_{n=0}^{N-1} y[n-r] e^{-j\frac{2\pi}{N}kn}\right).\]Recognizing this as the Fourier transform relation, \(b_k\) and \(a_k\) are the Fourier coefficients so that:\[c_k = N a_k b_k.\]
03

Compute Fourier Series for Given Signals in Part (c)

Given \(x[n] = \sin\left(\frac{3\pi n}{4}\right)\) and \(y[n]\) being a rectangular pulse with period 8:1. The Fourier coefficient \(a_k\) for \(x[n]\) can be derived using the discrete Fourier transform (DFT):\[a_k = \frac{1}{8} \sum_{n=0}^7 \sin\left(\frac{3\pi n}{4}\right)e^{-j\frac{2\pi}{8}kn}\].2. Use similar computation for \(b_k\) for \(y[n].\)3. Then, the coefficient \(c_k\) is computed as \(c_k = 8 a_k b_k.\) Calculate the individual terms and combine them to get the Fourier series for \(z[n].\)
04

Compute Fourier Series for Given Signals in Part (d)

Analyze \(x[n]\) and \(y[n]\) from part (d) with \(x[n]\) and \(y[n]\) behaving differently:1. For \(x[n],\) which is \{\begin{array}{ll}\sin\left(\frac{3\pi n}{4}\right)\, & 0 \leq n \leq 3 \ 0, & 4 \leq n \leq 7 \end{array} \}, compute \(a_k\) using:\[a_k = \frac{1}{8} \sum_{n=0}^{3} \sin\left(\frac{3\pi n}{4}\right)e^{-j\frac{2\pi}{8}kn}.\]2. For \(y[n] = \left(\frac{1}{2}\right)^n,\) compute \(b_k\) based on the geometric sequence properties.3. Calculate \(c_k = 8 a_k b_k\).Expand the Fourier series using the calculated coefficients for the periodic convolution \(z[n].\)

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Key Concepts

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

Fourier Coefficients
In the realm of signal processing, Fourier coefficients are essential in breaking down a periodic signal into its constituent frequencies. Each coefficient captures the amplitude and phase of a particular frequency component of the original signal.
For our periodic convolution problem, let's denote the Fourier coefficients of the signals as follows:
  • \(a_k\) represents the Fourier coefficients for the signal \(x[n]\)
  • \(b_k\) signifies the coefficients for \(y[n]\)
  • \(c_k\) stands for the coefficients of the convolved signal \(z[n]\)
One of the fascinating results in the convolution of two periodic signals is that the Fourier coefficients of the resulting signal, \(z[n]\), are given by the product of the Fourier coefficients of the original signals. This means for each frequency component \(k\), the relationship \(c_k = N a_k b_k\) holds, where \(N\) is the common period of the signals. This relation simplifies the computation of periodic convolutions significantly, as it ties directly into the multiplication of the respective coefficients.
Periodicity
Periodicity is the mathematical property that defines a repeating pattern over time for a signal. A signal \(x[n]\) is considered periodic with period \(N\) if it satisfies the condition \(x[n] = x[n+N]\) for all \(n\).
In the case of periodic convolution, when two periodic signals \(x[n]\) and \(y[n]\) with the same period \(N\) are convolved, the resulting signal \(z[n]\) is also periodic with period \(N\). This is a crucial property that simplifies the analysis of periodic convolutions.
To understand this, consider the signal expression for convolution:
  • \(z[n] = \sum_{r=} x[r] y[n-r]\)
Given the periodicity of \(x[n]\) and \(y[n]\), the terms under the summation inherently repeat over every interval of \(N\). Consequently, \(z[n+N] = z[n]\) holds true, confirming that \(z[n]\) retains the periodicity of the original signals.
Discrete Fourier Transform (DFT)
The Discrete Fourier Transform (DFT) is a powerful mathematical tool used to analyze finite and discrete signals. It transforms a sequence of numbers into components of different frequencies, and is particularly useful in digital signal processing.
The DFT of a periodic signal \(x[n]\) with period \(N\) is calculated as:
  • \(X_k = \sum_{n=0}^{N-1} x[n] e^{-j\frac{2\pi}{N}kn}\)
This formula essentially assembles the "spectrum" of \(x[n]\), representing the distribution and amplitude of its frequency components.
When it comes to periodic convolution, the DFT plays a critical role in determining the Fourier coefficients. For each of the signals in the convolution, the DFT can be used to compute their respective Fourier coefficients. These coefficients, once obtained for \(x[n]\) and \(y[n]\), allow us to efficiently determine the coefficients \(c_k\) for \(z[n]\) via simple multiplication: \(c_k = N a_k b_k\). This succinctly captures the contribution of each signal's frequency components to the resultant convolved signal.

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Most popular questions from this chapter

When the impulse train \(x[n]=\sum_{k=-\infty}^{\infty} \delta[n-4 k]\) is the input to a particular LTI system with frequency response \(H\left(e^{j \omega}\right),\) the output of the system is found to be \(y[n]=\cos \left(\frac{5 \pi}{2} n+\frac{\pi}{4}\right)\). Determine the values of \(H\left(e^{j k \pi / 2}\right)\) for \(k=0,1,2,\) and \(3\).

Consider three continuous-time periodic signals whose Fourier series representations are as follows: $$x_{1}(t)=\sum_{k=0}^{100}\left(\frac{1}{2}\right)^{k} e^{j k \frac{2 \pi}{50}}$$, $$x_{2}(t)=\sum_{k=-100}^{100} \cos (k \pi) e^{j k \frac{2 \pi}{50} t}$$, $$x_{3}(t)=\sum_{k=-100}^{100} j \sin \left(\frac{k \pi}{2}\right) e^{j k \frac{2 \pi}{50} t}$$. Use Forurier series properties to help answer the following questions: (a) Which of the three signals is/are real valued? (b) Which of the three signals is/are even?

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 \(x(f) ?\)

As we have seen in this chapter, the concept of an eigenfunction is an extremely important tool in the study of LIT systerns. The same can be said for linear. but time-varying, systems. Specifically, consider such a system with input \(x(t)\) and output \(y(t) .\) We say that a signal \(\phi(t)\) is an eigenfunction of the system if \(\phi(t) \rightarrow \lambda \phi(t)\). That is, if \(x(t)=\phi(t),\) then \(y(t)=\lambda \phi(t),\) where the complex constant \(\lambda\) is called the eigenvalue associated with \(\phi(t)\). (a) Suppose that we can represent the input \(x(t)\) to our system as a linear combination of eigenfunctions \(\phi_{f}(r),\) each of which has a corresponding eigenvalue \(\lambda_{t}\) that is, \(x(t)=\sum_{l=-2}^{\infty} c_{k} \phi_{k}(t)\). Express the output \(y(t)\) of the systern in terms of \(\left\\{c_{k}\right\\} .\left\\{\phi_{k}(t)\right\\},\) and \(\left\\{\lambda_{k}\right\\}\). (b) Consider the system characterized by the differential equation \(y(t)=t^{2} \frac{d^{2} x(t)}{d t^{2}}+t \frac{d x(t)}{d t}\)Is this system linear? Is it time invariant? (c) Show that the functions \(\phi_{\mathrm{t}}(t)=t^{k}\) are eigenfunctions of the system in part (b). For each \(\phi_{k}(t),\) determine the corresponding eigenvatue \(\lambda_{k}\). (d) Determine the output of the system if \(x(t)=10 t^{-10}+3 t+\frac{1}{2} t^{4}+\pi\).

Consider a discrete-time LTI system \(S\) whose frequency response is \(H\left(e^{j \omega}\right)=\left\\{\begin{array}{ll}1, & |\omega| \leq \frac{\pi}{8} \\ 0, & \frac{\pi}{8}<| \omega_{i}<\pi\end{array}\right.\) Show that if the input \(x[n]\) to this system has a period \(N=3,\) the output \(y[n]\) has only one nonzero Fourier series coefficient per period.

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