Number of subdivisions in a Riemann integral (DFT)

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SUMMARY

The discussion centers on calculating the Discrete Fourier Transform (DFT) of the sequence {1, 0, 4, -1, 0, 0} using the formula G_n = ∑ g_k e^{-jnk(2π/N)}. The user successfully computed G_0 to G_5 but faced uncertainty regarding the value of N for calculating G_7, G_{11}, G_{14}, G_{22}, G_{-7}, G_{-11}, G_{-14}, and G_{-22}. The user speculated that N could be either 23 or 46, based on the range of n values provided.

PREREQUISITES
  • Understanding of Discrete Fourier Transform (DFT)
  • Familiarity with complex exponentials and Euler's formula
  • Knowledge of Riemann integrals and their subdivisions
  • Basic skills in mathematical notation and summation
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  • Research the concept of periodicity in DFT to clarify the behavior of G_n for n outside the original range.
  • Study the implications of N in DFT calculations, particularly how it affects the sequence length and frequency resolution.
  • Explore the relationship between the DFT and Fourier series, focusing on how coefficients are derived.
  • Learn about the Fast Fourier Transform (FFT) algorithm for efficient computation of DFT.
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Homework Statement


This is a combination of two questions, one being the continuation of the other

3) Calculate the DFT of the sequence of measurements
\begin{equation*}
\{ g \}_{k=0}^{5} = \{ 1,0,4,-1,0,0 \}
\end{equation*}
4a) Draw the DFT calculated in question 3 on the complex plane.
4b) What are the members of the sequence:
\begin{equation*}
G_7, G_{11}, G_{14}, G_{22},G_{-7}, G_{-11}, G_{-14}, G_{-22}
\end{equation*}

Homework Equations


The assumed values of the unknown function are the sequence
\begin{equation}
g_k = f\left(\frac{kT}{N} \right), k = 0, 1, 2, \ldots, N-1,
\end{equation}
mentioned above, where ##N## is the number of subdivisions in the Riemann integral used to approximate the coefficients ##c_n## of the Fourier series of the unknown function ##f##.

Each member, ##G_n##, of the DFT is calculated as follows:
\begin{equation}
G_n = \sum_{k=0}^{N-1} g_k e^{-jnk\frac{2\pi}{N}}, n = 0, 1, 2, \ldots, N-1
\end{equation}

The Attempt at a Solution


Questions ##3## and ##4a## were not problematic at all. The DFT was
\begin{align*}
G_0 = \sum_{k=0}^{5} g_k e^{0} &= 1 + 0 + 4 - 1 + 0 + 0 = 4,\\
%
G_1 = \sum_{k=0}^{5} g_k e^{-jk\frac{2\pi}{6}} &= 1 e^{0} + 0 + 4 e^{-j2\frac{2\pi}{6}} - 1 e^{-j3\frac{2\pi}{6}}\\
&= 1 + 4 e^{-j\frac{2\pi}{3}} + 1 \\
&= 2 + 4 e^{-j\frac{2\pi}{3}}\\
%
G_2 = \sum_{k=0}^{5} g_k e^{-j2k\frac{2\pi}{6}} &= 1 + 0 + 4 e^{-j2\frac{4\pi}{6}} - 1 e^{-j3\frac{4\pi}{6}}\\
&= 1 + 4 e^{j\frac{2\pi}{3}} - 1\\
&= 4 e^{j\frac{2\pi}{3}}\\
%
G_3 = \sum_{k=0}^{5} g_k e^{-j3k\frac{2\pi}{6}} &= 1 + 0 + 4 e^{-j2\pi} - 1 e^{-j3\pi}\\
&= 1 + 4 + 1 = 6\\
G_4 = \sum_{k=0}^{5} g_k e^{-j4k\frac{2\pi}{6}} &= 1 + 0 + 4 e^{-j2\frac{4\pi}{3}} - 1 e^{-j3\frac{4\pi}{3}}\\
&= 1 + 4 e^{-j\frac{8\pi}{3}} - 1\\
&= 4 e^{-j\frac{2\pi}{3}}\\
G_5 = \sum_{k=0}^{5} g_k e^{-j5k\frac{2\pi}{6}} &= 1 + 0 + 4 e^{-j2\frac{5\pi}{3}} - 1 e^{-j3\frac{5\pi}{3}}\\
&= 1 + 4 e^{-j\frac{10\pi}{3}} - 1 e^{-j5\pi}\\
&= 2 + 4 e^{j\frac{2\pi}{3}}
\end{align*}
Based on these, I drew the following images
H4_4Acompass.png

H4_4Aplot.png
In ##4b##, however, I ran into a bit of a problem. I don't know what ##N## is supposed to be.

##n## seems to go all the way up to ##22## and down to ##-22##, which to my mind seems to imply, that ##N=23## or ##N=46##.

I'm also assuming, that the sequence ##\{ g \}_{k=0}^{5}## doesn't change, since I'm not given any new information regarding this, but I'm not sure of this either.
So, what is ##N##, exactly?

EDIT: Added labels on the points in the pictures.
 
Last edited:
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It just occurred to me, that maybe I should be using the periodicity and symmetricity of the DFT, to find out the values of ##G_n## in 4b. Any comment on this?
 

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