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,, 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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Jens Reermann Defended his Dissertation with Distinction

On Friday, 21st of June, Jens Reermann defended his research on signals processing for magnetoelectric sensor systems very successfully. After 90 minutes of talk and question time he finished his PhD with distinction. Congratulations, Jens, from the entire DSS team.

Jens worked for about three and a half years - as part of the collaborative research center (SFB) 1261 - on all kinds of signal ...

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