Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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Please do Exercise 4 and please do part A and B and please show step by step and explain
**Analyzing Digital Signals Using Complex Roots of Unity** For the remainder of this investigation, we choose a sampling period equal to 1 and examine signals that are linear combinations of complex waves with angular frequencies \(2\pi k/N\), where \(k = 0, 2, \ldots, N - 1\) (i.e., \(f = k/N, k = 0, 2, \ldots, N - 1\)). All these waves have positive angular frequencies less than \(2\pi\) and repeat after \(N\) samples. Any signal \(s(t)\) that is a linear combination of these waves can be expressed as follows: \[ s(t) = \sum_{k=0}^{N-1} a_k e^{2\pi i kt/N}. \quad \quad (1) \] The signal's information is contained in the complex amplitudes \(\{a_k\}, k = 0, \ldots, N - 1\). We aim to recover these amplitudes from the sampled signal. The properties of roots of unity are crucial in finding a method to recover the amplitudes. To achieve this, modular arithmetic will be incorporated. Let \(\zeta = e^{2\pi i/N}\) (\(\zeta\) is the Greek letter 'zeta') and recall that \(\zeta^k\) are the \(N\)th roots of unity for \(k = 0, \ldots, N - 1\). As these are roots of unity, they are solutions to the equation \(z^N - 1 = 0\). This implies that the linear factors \(z - \zeta^k\) all divide \(z^N - 1\). Since there are \(N\) distinct linear factors, it follows that: \[ z^N - 1 = (z - \zeta^0) \cdots (z - \zeta^{N-1}). \quad \quad (2) \] **Exercise 4** (a) By computing the coefficient of \(z^{N-1}\) on the right-hand side of Equation (2), show that the sum of \(N\)th roots of unity is equal to 0. (Note that this coefficient must be equal to the coefficient of \(z^{N-1}\) on the left-hand side, which
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Transcribed Image Text:**Analyzing Digital Signals Using Complex Roots of Unity** For the remainder of this investigation, we choose a sampling period equal to 1 and examine signals that are linear combinations of complex waves with angular frequencies \(2\pi k/N\), where \(k = 0, 2, \ldots, N - 1\) (i.e., \(f = k/N, k = 0, 2, \ldots, N - 1\)). All these waves have positive angular frequencies less than \(2\pi\) and repeat after \(N\) samples. Any signal \(s(t)\) that is a linear combination of these waves can be expressed as follows: \[ s(t) = \sum_{k=0}^{N-1} a_k e^{2\pi i kt/N}. \quad \quad (1) \] The signal's information is contained in the complex amplitudes \(\{a_k\}, k = 0, \ldots, N - 1\). We aim to recover these amplitudes from the sampled signal. The properties of roots of unity are crucial in finding a method to recover the amplitudes. To achieve this, modular arithmetic will be incorporated. Let \(\zeta = e^{2\pi i/N}\) (\(\zeta\) is the Greek letter 'zeta') and recall that \(\zeta^k\) are the \(N\)th roots of unity for \(k = 0, \ldots, N - 1\). As these are roots of unity, they are solutions to the equation \(z^N - 1 = 0\). This implies that the linear factors \(z - \zeta^k\) all divide \(z^N - 1\). Since there are \(N\) distinct linear factors, it follows that: \[ z^N - 1 = (z - \zeta^0) \cdots (z - \zeta^{N-1}). \quad \quad (2) \] **Exercise 4** (a) By computing the coefficient of \(z^{N-1}\) on the right-hand side of Equation (2), show that the sum of \(N\)th roots of unity is equal to 0. (Note that this coefficient must be equal to the coefficient of \(z^{N-1}\) on the left-hand side, which
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Introduction

As per the question we are given the following equation :

zN - 1 = (z - ζ0) ... (z - ζN-1)

Where ζ0 , ... , ζN-1 is the Nth roots of unity. And using this equation we have to compute :

  1. The coefficient of zN-1 at the right side of the given equation. And show that the sum of the Nth roots of unity is zero.
  2. The coefficient of zN-2 at the right side of the given equation. And establish another property of the Nth roots of unity.
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