(a) By computing the coefficient of zN-1 on the right hand side of Equation (2), show that the sum of Nth roots of unity is equal to 0. (Note that this coefficient must be equal to the coefficient of zN-1 on the left-hand side, which is 0). (b) (This one is harder) Compute the coefficient of zN-2 on the right-hand side of Equation (2). What prop- erty of the N'th roots of unity can you conclude from this?
(a) By computing the coefficient of zN-1 on the right hand side of Equation (2), show that the sum of Nth roots of unity is equal to 0. (Note that this coefficient must be equal to the coefficient of zN-1 on the left-hand side, which is 0). (b) (This one is harder) Compute the coefficient of zN-2 on the right-hand side of Equation (2). What prop- erty of the N'th roots of unity can you conclude from this?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F892e817a-9b32-4eeb-b8fc-5dd7ffde6479%2F40749a4a-f0c8-41a6-8b3b-40ba59e61aef%2Ftjzdl9c_processed.jpeg&w=3840&q=75)
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
Expert Solution
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 :
- 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.
- 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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