Advanced Signals and Systems - Discrete Fourier Transform


15. Discrete Fourier transform.


Determine the DFT of the following sequences \(v(n)\) with length \(M\): \begin{align} \text{(a)} \;\;\;\;\;\;\;\; &v(n) = \gamma_0 (n-\kappa) & \kappa \in \{ 0,1,... M-1\} \nonumber \\ \text{(b)} \;\;\;\;\;\;\;\; &v(n) = \cos(\Omega_0n) & \Omega_0 = \frac{2\pi}{M}\mu_0, \, \mu_o \in \{ 0,1,... M-1\} \nonumber \end{align}

Amount and difficulty

  • Working time: approx. xx minutes
  • Difficulty: xx


\begin{align*} \text{(a)} \ \ V_M(\mu) &=&& \text{DFT} \left\{ v_1(n) \right\} = \sum \limits_{n=0}^{M-1} \gamma_0 (n-\kappa) e^{-j\mu \frac{ \pi}{M}n} = e^{-j\mu \frac{2\pi}{M}\kappa} \ \ \rightarrow |V_M(\mu)| = 1 \\ \text{(b)} \ \ V_M(\mu) &=&& \text{DFT} \left\{ \cos \left( \frac{2\pi}{M} \mu_0 n \right) \right\} = \cdots = \\ &&&\frac{M}{2} \left[ \sum \limits_{\lambda= -\infty}^\infty \gamma_0 \left(\mu -( \lambda M + \mu_0)\right) + \sum \limits_{ \lambda = -\infty}^\infty \gamma _0\left(\mu -(\lambda M -\mu_0)\right) \right] \end{align*}


16. Relation between DFT and Fourier transform.


The Fourier transform \(R(e^{j\Omega})\) of the following sequence \begin{equation} r(n) = \begin{cases} 1 & |n|\leq N \\ 0 & \text{else} \nonumber \end{cases}\text{ ,} \end{equation} is given by \begin{equation} R(e^{j\Omega}) = \frac{\sin \left(\frac{\Omega}{2}(2N+1) \right)}{\sin \left(\frac{\Omega}{2} \right)} \nonumber \text{ .} \end{equation} Determine the DFT \(R_M(\mu)\) of the given sequence \(r(n)\). How are these two transformations linked with each other?

Amount and difficulty

  • Working time: approx. xx minutes
  • Difficulty: xx

Additional material


Applying the known definition of the DFT for deriving \(R_M(\mu)\) we find:

\begin{align*} R_M(\mu) &= \sum \limits_{n= 0}^{M-1} r(n-N) \cdot e^{-j\frac{2\pi}{M} \mu n} = \sum \limits_{n= 0}^{2N} e^{-j\frac{2\pi}{M} \mu n} \end{align*}

The shift of \(N\) has been carried out, in order to enable the calculation of the DFT. It can easily be reversed by using the shift theorem. Two cases should be considered:

  1. \(\mu \neq 0\) \begin{align*} R_M(\mu) &= e^{-j\frac{2\pi}{M} \mu N} \cdot \frac{\sin \left( \mu \frac{2\pi}{M} \frac{1}{2} (2N+1) \right) }{\sin \left( \mu\frac{2\pi}{M} \frac{1}{2} \right) } \end{align*}
  2. \(\mu = 0\) \begin{align*} R_M(\mu) &= \cdots = 2N+1 \end{align*}

Comparing this result to the result of the Fourier transform it is obviously that DFT is given by sampling the Fourier transform \(R(e^{j\Omega})\) at \(\Omega = \mu \frac{2\pi}{M}\).

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J. Reermann, P. Durdaut, S. Salzer, T. Demming, A. Piorra, E. Quandt, N. Frey, M. Höft, and G. Schmidt: Evaluation of Magnetoelectric Sensor Systems for Cardiological Applications, Measurement (Elsevier), ISSN 0263-2241,­10.1016/­j.measurement.2017.09.047, 2017

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Prof. Dr.-Ing. Gerhard Schmidt


Christian-Albrechts-Universität zu Kiel
Faculty of Engineering
Institute for Electrical Engineering and Information Engineering
Digital Signal Processing and System Theory

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24143 Kiel, Germany

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Jugend Forscht

On November 24th, one of our DSS team members, Owe Wisch, took part in the "Jugend forscht Perspektivforum" at the CAU. Thirty young students from the "Jugend forscht" project came to Kiel and participated in three different workshops focusing on career paths in maritime climate protection. Owe Wisch from our chair lead one of the workshops and presented his research topics, beamforming ...

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