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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Christian-Albrechts-Universität zu Kiel
Faculty of Engineering
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Digital Signal Processing and System Theory

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Elke is from Amsterdam and she works in the neurology department in the university hospital in the group of Prof. Maetzler. Her research topic is movement analysis of patients with neurologic disorders. Elke cooperates with us in signal processing related aspects of her research. Elke plays ...

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